A positive definite operator is one for which for all .
The support of an operator is defined to be the vector space orthogonal to its kernel.
Hermitian operator: the vector space spanned by its eigenvectors has non-zero eigenvalues.
unitary operator or matrix.
For two-level quantum systems used as qubits, we shall usually identify the state and .
We use the notations to denote respectively.
0.1 Information theory and probability
Probability distribution refers to a finite set of real numbers, , such that and .
The relative entropy of a positive operator with respect to another positive operator is defined by:
0.2 Frequently used gates & symbols
1 Introduction and overview
Each remaining section in the chapter gives a brief introduction to one or more fundamental concepts from the field: quantum bits, quantum computers, quantum gates and quantum circuits, quantum algorithms, experimental quantum information processing, quantum information and quantum communication.
1.1 Global perspectives
Whether it is possible to signal faster than light whether it is possible to clone an unknown quantum state:no-cloning theorem (early 1980s), is one of the earliest results of quantum computation and quantum information.
Church–Turing thesis: if an algorithm can be performed on any piece of hardware (say, a modern personal computer), then there is an equivalent algorithm for a Universal Turing Machine which performs exactly the same task as the algorithm running on the personal computer.
Strong Church–Turing thesis:Any algorithmic process can be simulated efficiently using a Turing machine.
The first major challenge to the strong Church–Turing thesis arose in the mid 1970s, which is that it's possible to test whether an integer is prime or composite using a randomized algorithm. That is, the Solovay–Strassen test for primality used randomness as an essential part of the algorithm.
Untill today there is no known deterministic test for primality computers with access to a random number generator would be able to efficiently perform computational tasks with no efficient solution on a conventional deterministic Turing machine.
Hence we have a modified version of Church-Turing Thesis:Any algorithmic process can be simulated efficiently using a probabilistic Turing machine.
It is comparatively difficult to come up with quantum algorithm, since ①要先抑制自己已經根深蒂固的古典思維, ②發明出新的演算法必須比當前已知所有解同樣問題的古典演算法還要快.
At the same time computer science was exploding in the 1940s, another revolution was taking place in our understanding of communication. The key step taken by Shannon was to mathematically define the concept of information.
Shannon’s noiseless channel coding theorem, quantifies the physical resources required to store the output from an information source. Shannon’s second fundamental theorem, the noisy channel coding theorem, quantifies how much information it is possible to reliably transmit through a noisy communications channel.
In 1996 two groups (= CSS codes) the stabilizer codes. The theory of quantum error-correcting codes was developed to protect quantum states against noise.
As for transmitting ordinary classical information using a quantum channel, we have superdense coding (transmit two classical bits of information, while only transmitting one quantum bit from sender to receiver).
No unifying theory of networked information theory exists for quantum channels. But it is of much consequence! One example may suffice its value: 兩個 zero-capacity 的古典通道併聯, 還是 zero-capacity; 但如果是兩個反向的量子通道併聯, 就竟然可以通訊!
Another field is quantum cryptography: for private key cryptosystem, there is QKD (quantum key distribution), which makes use of the property that eavesdropping changes the quantum state of the qubit; for public key cryptosystem, Shor's algorithm can break RSA, and Shor's quantum algorithm for solving the discrete logarithm problem can break other public key systems.
1.2 Quantum bits
In this section we introduce the properties of single and multiple qubits, comparing and contrasting their properties to those of classical bits.
How a qubit can be realized: as the two different polarizations of a photon, as the alignment of a nuclear spin in a uniform magnetic field, as two states of an electron (ground state & excited state ) orbiting a single atom.
Interestingly, by giving not enough light for an electron to perform , it will be moved "halfway" into state.
Geometric representation of a qubit: a qubit in superposition can be written as
but since , we can rewrite the equation as
where , , and are real numbers. Since the factor have no observable effects, we can essentially write the equation
The numbers and define a point on the unit three-dimensional sphere, which is called Bloch-sphere:
However, it must be kept in mind that this intuition is limited because there is no simple generalization of the Bloch sphere known for multiple qubits.
Multiple qubits: say, if we express a pair of qubits in their computational basis state, we get
, with the normalization condition . Now if we only take measurement on the first qubit:
A particularly important two qubit state is the Bell state or EPR pair,
is the key ingredient in quantum teleportation and superdense coding. We can know from the measurement on the first qubit which state the second qubit would have if we then measures it. The two qubits are strongly correlated.
1.3 Quantum computation
Changes occurring to a quantum state can be described using the language of quantum computation. Analogous to the way a classical computer is built from an electrical circuit containing wires and logic gates, a quantum computer is built from a quantum circuit containing wires and elementary quantum gates to carry around and manipulate the quantum information.
1.3.1 Single qubit gates
Quantum state written in vector notation: .
Quantum NOT gate , hence .
So quantum gates on a single qubit can be described by two by two matrices . Are there any constraints on what matrices may be used as quantum gates? It turns out that the normalization condition yields
Amazingly, this unitarity constraint is the only constraint on quantum gates!
Another two important gates:
To get the visualization of the Hadamard gate on the Bloch sphere, the operation is just a rotation of the sphere about the axis by 90°, followed by a rotation about the axis by 180° (below is the illustration of an gate acting on state).
✏️ Decomposing single qubit operations
Later in the content we can prove that an arbitrary unnitary matrix may be decomposed as
where , and are real-valued. Notice that the second matrix is just an ordinary rotation. It turns out that the first and last matrices can also be understood as rotations in a different plane. This decomposition can be used to give an exact prescription for performing an arbitrary single qubit quantum logic gate.
1.3.2 Multiple qubit gates
常見的 classical multiple gates 有 AND, OR, XOR, NAND, 和NOR. An important theoretical result is that any function on bits can be computed from the composition of NAND gates alone, which is thus known as a universal gate.
CNOT can be realized as the control qubit and the target qubit are XORed and stored in the target qubit:
, where is addition modulo two. Yet another way of describing the action of the CNOT is through its matrix representation:
, as for the single qubit case, the requirement that probability be conserved is expressed in the fact that is a unitary matrix, that is, .
Any multiple qubit logic gate may be composed from CNOT and single qubit gates. (proof in later content)
1.3.3 Measurements & Quantum circiuts
It is possible to treat the and states as though they were the computaitonal basis states, for instance,
Generally, given ay basis states and for a qubit, it is possible to express an arbitrary state as a linear combination of these states. Furthermore, provided the states are orthonormal, it is possible to perform a measurement with respect to the basis.
Swap gate: Below circuit accomplishes the swap operation.
There are a few features allowed in classical circuits that are not usually present in quantum circuits:
Acyclic: there are no loops in quantum circuits (feedback from one part of the quantum circuit to another).
No FANIN: wires can't be joined together, since this is a bitwise-OR operation and irreversible, xhence not unitary.
No FANOUT: quantum mechanics forbids the copying of a qubit, which is the no-cloning theorem.
✏️ The no-cloning theorem
Suppose we are copying the pure quantum state from slot to slot (which starts in also a pure state ) in a quantum machine. Thus the initial state of the copying machine is:
Some unitary evolution now effects the copying procedure, ideally,
This works for another state, , also:
Taking the inner product of these two equations gives:
From the above equation there are oly two possibilities: either , or are orthogonal to . Thus a cloning device can only clone states which are orthogonal to one another, and therefore a general quantum cloning device is impossible.
Even if one allows non-unitary cloning devices, the cloning of non-orthogonal pure states remains impossible unless one is willing to tolerate a finite loss of fidelity in the copied states.
1.3.4 Bell states & Quantum teleportation
Above is a demonstration of quantum teleportation:
First generate an EPR pair (or Bell state): the part before the dashed line. Assume , then the state of the three qubits are:
We then entangle the state which we want to teleport with qubit #2:
which equals to:
We can then apply and gate to the third qubit (depend on the measurement outcome of qubit #1 and #2) to restore on qubit #3.
Quantum teleportation emphasizes the interchangeability of different resources in quantum mechanics, showing that one shared EPR pair together with two classical bits of communication is a resource at least the equal of one qubit of communication.In particular, in Chapter 10待補 we explain how teleportation can be used to build quantum gates which are resistant to the effects of noise, and in Chapter 12待補 we show that teleportation is intimately connected with the properties of quantum error-correcting codes.
1.4 Quantum algorithms
1.4.1 Simulating Classical Computer
Though quantum circuits can't be used to directly simulate classical circuits (because unitary quantum logic gates are inherently reversible, whereas many classical logic gates such as the NAND gate are inherently irreversible), but we can make use of the Toffoli gate:
Now if a quanutm computer wants to simulate a classical computer which is non-deterministic (e.g. generate random number), one can achieve that by apply gate to a state, then measure it and results in a 50/50 chance of 0 or 1.
1.4.2 Quantum parallelism
quantum parallelism allows quantum computers to evaluate a function for many different values of simultaneously. Suppose is a function with a one-bit domain and range. The below circuit implemented a transformation (引進是為了要讓這個transformation是unitary的, 只要apply兩次就回到原來的state).
Now if we will result in the second register. But if we first Hadamard transform the first qubit (or Walsh– Hadamard transform), we'll result in the state: .
Generally, the result of performing the Hadamard transform on qubits initially in the all state is
, where the sum is over all possible values of , and we write to denote this action (read "" as "tensor"). Note that the Hadamard transform is super efficient since it can produce a superposition of states using only qubits.
Only quantum parallelism is not enough, we also require the ability to extract information about more than merely one value of from superposition states like
, which is obtained from applying Hadamard transform and then to the qubit state .
1.4.3 Deutsch's Algorithm
Deutsch’s algorithm combines quantum parallelism with a property of quantum mechanics known as interference, as illustrated below. We set the input state and use the same transform above:
after two gate, we have , due to the unitarity of , we expect the state . (因為如果的話那就等於沒做任何事, 那如果的話那第二個qubit出來的就相當於反相) So there are two possible groups of outcome :
, therefore we have:
, notice that if
, hence we can technically write as:
, which is very interesting indeed! (compare to the classical scheme) Because the quantum circuit has given us the ability to determine a global property of (在此題即), using only one evaluation of .
The essence of the design of many quantum algorithms is that a clever choice of function and final transformation allows efficient determination of useful global information about the function (information which cannot be attained quickly on a classical computer).
1.4.4 The Deutsch–Jozsa algorithm
First we need to deal with the Deutsch’s problem:
How many times do we need to repeat the following procedure to decide whether the function is balanced or constant?
select a number from to .
feed into , note that only outputs or .
balanced: equal amount of and as outcome for all possivle value of , constant: constant output disregarding different values of .
Classical case: The best deterministic classical algorithm require queries (for the worst case)
Quantum case:Analogous to the Duetsch's algorithm scheme, but now prepare query-qubits instead of , we result in
, after the Hadamard transform on the query registers and applying gate on the answer register we get
. Next, the classical function is evaluated using , giving
we now has a set of qubits in which the result of function evaluation is stored in the amplitude of the qubit superposition state.
Now we want to derive the expression for the Hadamard transform on superpositioned qubits, we start from case:
thus,
, and this can be summarized in a more terse term:
where is the bitwise inner product of and , modulo .
We can then evaluate :
consider the amplitude of , we found it solely depends on the function type of .
Caveats:
Deutsch’s problem is not an especially important problem; it has no known applications.
The method for evaluating the function is quite different in the two cases, render it uncomparable between classical and quantum scenario.
If one adopts a probabilistic instead of deterministic classical computer, one can quickly decide the function type with high confidence.
Exercise 1.1
(Probabilistic classical algorithm) Suppose that the problem is not to distinguish between the constant and balanced functions with certainty, but rather, with some probability of error . What is the performance of the best classical algorithm for this problem?
solution
A probabilistic classical computer can solve Deutsch’s problem with two evaluations with some probability of error .
1.4.5 Summarization
Generally there are 3 classes of quantum algorithms which provide an advantage over known classical algorithms:
algorithms based upon quantum versions of the Fourier transform (e.g. Deutsch–Jozsa algorithm, Shor's algorithm for factoring and discrete algorithm)
quantum search algorithms
quantum simulation
We will now breifly describe each of these three classes of algorithms.
Quantum algorithms based upon the Fourier transformThe discrete Fourier transform is usually described as transforming a set of complex numbers into a set of complex numbers defined by
, the Fourier transformed version of a problem is often easier than the original problem, enabling a solution.
The Fourier transform has proved so useful that a beautiful generalized theory of Fourier transforms has been developed which goes beyond the definition of equation (33). What is important is that the Hadamard transform used in the Deutsch–Jozsa algorithm is an example of this generalized class of Fourier transforms. Moreover, many of the other important quantum algorithms also involve some type of Fourier transform.
Shor’s fast algorithms for factoring and discrete logarithm, are two examples of algorithms based upon the Fourier transform. We start by imagining that we define a linear transformation on qubits with the action it performs on computational basis :
, you can check the unitarity of this transformation, which makes it a valid quantum circuit; moreover if we perform same transformation on superposition states:
, which is the vector version for the Fourier transform in equation (33) where .
Time complexity:
Classically, the fast Fourier transform takes roughly steps.
On a quantum computer, it takes only steps. (The quantum circuit to do this is explained in [Chapter 5][#5]待補)
Though we can compute QFT faster on quantum computer, Fourier transform is being performed on the information hidden in the amplitudes of the quantum state. This information is not directly accessible to measurement. (if the output state is measured, it will collapse each qubit into the state or , preventing us from learning the transform result directly.
Problems like Deutsch’s problem, and Shor’s algorithms for discrete logarithm and factoring, are all cases for Kitaev's discovery of a method to solve the Abelian stabilizer problem, and the generalization to the hidden subgroup problem:
Let be a function from a finitely generated group to a finite set such that is constant on the cosets of a subgroup , and distinct on each coset. Given a quantum black box for performing the unitary transform , for , and an appropriately chosen binary operation on , find a generating set for .
Quantum search algorithmsWith its basic principles were discovered by Grover, quantum search algorithm solves the following problem:
Given a search space of size , and no prior knowledge about the structure of the information in it, we want to find an element of that search space satisfying a known property. How long does it take to find an element satisfying that property?
Time complexity:
Classically, this problem requires approximately operations.
But the quantum search algorithm allows it to be solved using approximately operations.
Though it offers only a quadratic speedup (whereas QFT offers an exponential speedup), it has richer application and can adapt to a wide range of problems, in [Chapter 6][#6]待補 which will it be explained in detail.
Quantum simulation
In general, storing the quantum state of a system with distinct components takes something like bits of memory on a classical computer, where is a constant which depends upon details of the system being simulated, and the desired accuracy of the simulation.
Whereas on a quantum computer, it only needs qubits to perform simulation, but this does not mean that the fast simulation will allow the desired information about the quantum system to be obtained. (e.g. when we take measurement on the quantum circuit, the qubit simulation will collapse into a definite state, giving only bits of information; in other words, the total amount of bits information is not entirely accessible)
Thus, a crucial step in making quantum simulations useful is development of systematic means by which desired answers can be efficiently extracted.
The power of quantum computation
First we need to understand some of Computational complexity theory:
Two of the most important complexity classes are and . Roughly speaking, is the class of computational problems that can be solved quickly on a classical computer. is the class of problems which have solutions which can be quickly checked on a classical computer.
For instance, factoring (finding the prime factors of a given number) is a problem not in , but a problem in .
It is clear that is a subset of , since the ability to solve a problem implies the ability to check potential solutions.
But it's not so clear is whether or not there are problems in that are not in , which is:
if , then there are no problem which can be efficiently solved on a classical computer. (可以把NP-complete問題想成是NP問題裡面比較具代表性的問題)
Besides and , there're also other complexity classes:
: consists of those problems which can be solved using resources which are few in spatial size (that is, the computer is small), but not necessarily in time (long computations are fine). (有點無腦暴力法的感覺), is believed to be strictly larger than both and although, again, this has never been proved.
: problems that can be solved using randomized algorithms in polynomial time, if a bounded probability of error (say ) is allowed in the solution to the problem. (他被認為甚至比更能在classical computer上是efficiently solvable的)
No matter what do we not know, what is already clear is that the theory of quantum computation poses interesting and significant challenges to the traditional notions of computation. What makes this an important challenge is that the theoretical model of quantum computation is believed to be experimentally realizable, because – to the best of our knowledge – this theory is consistent with the way Nature works. If this were not so then quantum computation would be just another mathematical curiosity.
1.5 Experimental quantum information processing
In the next two sections , we begin with a review of the famous Stern–Gerlach experiment, which provides evidence for the existence of qubits in Nature. We then widen our scope, addressing the broader problem of how to build practical quantum information processing systems.
1.5.1 The Stern–Gerlach experiment
Qubit is the term we used to describe the fundamental element of QCQI, but is there any quantum system that behaves like a qubit in real world? It turns out that we may now understand qubits in terms of two level quantum systems.
1927 Stern–Gerlach experiment: 用烤箱加熱氫原子(本來在1922時是用銀原子), then form a beam of atoms, which subsequently fly through a magnetic field (會偏折原子束的方向), and finally project on to a screen. Theoretically, there would only be one peak since the magnetic dipole moment is 0 for hydrogen atom, but surprisingly there appears two peaks, which then suggests there may be some hidden variables (即後來提出的電子自旋 electron spin).
What is more peculiar is that for the scheme (everytime progressing to the next arrow we only take the positive-half part of splitted beams), there are also two peaks at ; and for scheme, there are again two peaks at . (古典的想法是方向的磁矩已經在第一次磁場的時候被過濾掉了)
The qubit model provides a simple explanation of this experimentally observed behavior:
, therefore we have and , and this explains why everytime there will be two beams split upon meeting new magnetic fields. This example demonstrates how qubits could be a believable way of modeling systems in Nature.
1.5.2 Prospects for practical quantum information processing
There are 2 possible points which will prohibit us from doing one or more forms of quantum information processing:
Noise: there is a threshold theorem for quantum computation, which states that if the level of noise in a quantum computer can be reduced below a certain constant threshold value, quantum error-correcting codes can be used to push it down even further, essentially ad-infinitum!The theorem indicates that the effects of noise can be made essentially negligible for quantum information processing. In Chapters 8, 10 and 12 待補 we will discuss quantum noise, quantum error-correction and the threshold theorem in detail.
QM is incorrect: If this occurs, it will be a momentous discovery in the history of science, and can be expected to have considerable consequences in other areas of science and technology, as did the discovery of quantum mechanics.
Quantum state tomography and quantum process tomography are two elementary processes whose perfection is of great importance to quantum computation and quantum information, as well as being of independent interest in their own right.
Quantum state tomography is a method for determining the quantum state of a system: by performing repeated preparations of the same quantum state, which is then measured in different ways in order to build up a complete description of the quantum state.
Quantum process tomography is a more ambitious (but closely related) procedure to completely characterize the dynamics of a quantum system. (e.g. can be used to characterize the performance of an alleged quantum gate or quantum communications channel, or to determine the types and magnitudes of different noise processes in a system.)
Quantum state tomography and quantum process tomography are described in more detail in Chapter 8 待補.
Various small-scale communications primitives like quantum cryptography and quantum teleportation are also of great interest. Chapter 12 待補 will show that teleportation may be an extremely useful primitive for transmitting quantum states between distant nodes in a network, in the presence of noise. 量子瞬移(我真會翻譯)其實更注重EPR pair的純度, since the EPR pairs may be corrupted during communication, but special entanglement distillation protocols can then be used to clean up the EPR pairs, enabling them to be used to teleport quantum states from one location to another.
For the medium-scale, a promising of quantum information processing is to the simulation of quantum systems. 因為60個qubit的系統就不足以讓超級電腦負荷過來了, 更不用說要跑模擬.
For large-scale applications, there are the factoring of large numbers, taking discrete logarithms, and quantum searching. 前兩者在長遠方向上不重要, 因為他們只威脅到現今我們所使用的加密技術的可靠度; 但quantum searching就應用範圍很廣了, we will discuss some possible applications in Chapter 6 待補.
Physical realization of QC:
Optical system turns out to be suitable for small-scale quantum computer, since photons are highly stable carriers for quantum information, which in other words, hard to directly interact with another, so the quantum gates (a form of interaction) must be mediated by something else, like an atom (光子一號先跟一顆原子作用, 該原子再跟光子二號作用, 來達成兩顆光子的間接作用)
Ion traps and neutral atom traps are alternative schemes for storing qubits. Photons are also used in this scheme, but instead of carring quantum information itself, they manipulate the quantum information stored inside the atom in trap. (e.g. Single qubit quantum gates can be performed by applying appropriate pulses of electromagnetic radiation to individual atoms; quantum measurement can be accomplished in these systems using the long established quantum jumps technique, which implements with superb accuracy the measurements in the computational basis used for quantum computation.)
Nuclear Magnetic Resonance, or NMR schemes, store quantum information in the nuclear spin of atoms in a molecule, and manipulate that information using electromagnetic radiation. (但他不太能操控單一顆原子, 通常是一次操控個溶液中的粒子, 可以看成是台電腦平行運算).
To conclude, note that it is important not to assess quantum information processing as though it were just another technology for information processing.
1.6 Quantum Information
We can identify a few fundamental goals uniting work on quantum information theory:
Identify elementary classes of static resources in quantum mechanics. (e.g. qubit, classical bit, or a Bell state shared between two distant parties 都可以作為存儲/傳遞訊息的單位)
Identify elementary classes of dynamical processes in quantum mechanics. (e.g. memory⇒the ability to store a quantum state over some period af time; quantum information transmission; or the process of protecting quantum information processing against the effects of noise)
Quantify resource tradeoffs incurred performing elementary dynamical processes. (e.g. what are the minimal resources required to reliably transfer quantum information between two parties using a noisy communications channel?)
The remainder of this section describes some examples of questions studied by quantum information theory, in each case emphasizing the fundamental static and dynamic elements under consideration, and the resource tradeoffs being considered.
Classical information through quantum channels
The fundamental results of classical information theory are Shannon’s noiseless channel coding theorem (quantifies how many bits are required to store information being emitted by a source of information), and Shannon’s noisy channel coding theorem (quantifies how much information can be reliably transmitted through a noisy communications channel)
Definition of an information source (這邊先暫時定義): a classical information source is described by a set of probabilities . Each use of the source results in the "letter" being emitted, chosen at random with probability , independently for each use of the source. (比如說如果information source是English的話, 因為一個段落中字母"e"出現的機率比"z"還高, 故 , 那我們就可以改用比較少bit的資訊量來表示字母"e", 使整體文章的bit使用量減少, 而Shannon's noiseless channel coding theorem可以告訴我們how well we can compress such a source)
The noiseless channel coding theorem tells us that a classical source described by probabilities can be compressed so that on average each use of the source can be represented using bits of information, where
is a function of the source probability distribution known as the Shannon entropy. Moreover, the noiseless channel coding theorem tells us that to attempt to represent the source using fewer bits than this will result in a high probability of error when the information is decompressed. (See chapter 12 待補 for more detail)
Verify that Shannon’s noiseless coding theorem satisfies the previous three goals:
Static resources X2: the bit and the information source
Two-stage dynamic process X1: compress the information source and then decompress it after transmission
Finally a quantitative criterion for determining the resources consumed (goal 3) by an optimal data compression scheme is found.
Shannon’s second major result, the noisy channel coding theorem, quantifies the amount of information that can be reliably transmitted through a noisy channel. 訊息傳輸可能是兩個空間點之間的傳輸, 或是兩個時間點之間的傳輸(即儲存資訊, 且是在有噪音的情況下). Both ways to overcome noises is to introduce error-correcting codes, the idea is to add an amount of redundancy into the information so that even some bits corrupt after the channel, one is still possible to recover the original message. (比如說一個channel的錯誤率是50%, 那它的 capacity 就是 half a bit) Shannon’s noisy channel coding theorem provides a general procedure for calculating the capacity of an arbitrary noisy channel.
Verify that Shannon’s noisy channel coding theorem satisfies the previous three goals:
Static resources X2: the information source, and the bits being sent through the channel.
Dynamic process X3: noise process + encoding process + decoding process.
For a fixed noise model, Shannon’s theorem tells us how much redundancy must be introduced by an optimal error-correction scheme if reliable information transmission is to be achieved (goal 3).
One might wonder if the medium used to carry information changed from a bit to a qubit, are the Shannon's two theorems still valid? (e.g. if using qubits allows a better compression rate than is possible classically?) Unfortunately, we will eventually prove that qubits do not allow any significant saving in the amount of communication required to transmit information over a noiseless channel.
For the problem of transmitting classical information through a noisy quantum channel, in Chapter 12 待補 we’ll prove the HSW (Holevo–Schumacher–Westmoreland) theorem, which provides a lower bound on the capacity of such a channel. Will the encoding scheme using entangled states raise the capacity beyond the lower bound provided by the HSW theorem? All evidence to date suggests that this doesn’t help raise the capacity, but it is still a fascinating open problem of quantum information theory.
Quantum information through quantum channels
For quantum information, we can again compress them in another way before sending them into channels, but since in general we cannot distinguish between and , we aren't able to apply Shannon entropy to quantum informations. The way we use to compress the quantum information is not error-free, instead we use a measure: fidelity, to quantify the correctness of the information after decompressing from the compressed form, in the limit of large block lengths, it should tend towards the no error limit of 1.
Schumacher’s noiseless channel coding theorem quantifies the resources required to do quantum data compression, with the restriction that it be possible to recover the source with fidelity close to 1:
source only produces orthogonal quantum states ( with probability ): the theorem reduces to classical case saying that the source may be compressed down to but not beyond the classical limit .
for a more general case (source that produces non-orthogonal states): how much a quantum source may be compressed is not Shannon entropy ever, but von Neumann entropy (他在正交態的時候會化簡成Shannon entropy, 其餘比較廣義的情況會 strictly smaller than Shannon entropy). (e.g. a source producing the state with probability and with probablilty can be reliably compressed using fewer than qubits per use of the source!)
compressing procedure for a certain scheme:
Suppose the source emit state with probability and with probablilty and is repeated times, creating a total of qubits of information.
With we will have in average of and of in the above qubits, for such configuration there are combinations in total.
We can then label these combinations from , equivalent to performing a unitary transformation:
, and discard the tailing s, leaving a compressed state of qubits.
To decomopress, we just append the previously discarded s and do the inverse unitary transform.
This procedure for quantum data compression and decompression results in a storage requirement of qubits per use of the source, and we can deduce whenever , it will be an improvement over Shannon’s noiseless channel coding theorem scheme . (這種壓縮方法的核心概念是利用和都在方向上有分量的這個redundancy, 來進行壓縮的)
Can we find an analogue of Shannon’s noisy channel coding theorem? Considerable progress on this important question has been made, using the theory of quantum error-correcting codes; however, a fully satisfactory analogue has not yet been found.
Quantum distinguishability
As the above scheme mentioned, in general we cannot distinguish from . If we are able to distinguish , we can develop a way of communication which is faster than light! (e.g. suppose Alice and Bob share a Bell's pair , which is also if you do some math, then Alice can convey messages via measuring her qubit with intended basis (she can deliberately choose -measurement or -measurement), and Bob's qubit will collapse into corresponding state as soon as Alice made her measurement, now Bob then use his device which is capable of distinguishing and he can instantaneously know what measurement does Alice perform).
但有時候這種quantum indistinguishability反而是一種advantage. (e.g. it can be used to built quantum-money: only the bank knows the exact sequence of on the note it publishes, it can attribute a unique classical serial number to each note, and then verify it by telling the merchant what will the result be after measuring with a sequence of basis the bank provides, after the merchants provide them the serial number on individual note. The counterfeiter cannot duplicate same note due to the quantum indistinguishability.)
Exercise 1.2
Explain how a device which, upon input of one of two non-orthogonal quantum states or correctly identified the state, could be used to build a device which cloned the states and , in violation of the no-cloning theorem. Conversely, explain how a device for cloning could be used to distinguish non-orthogonal quantum states.
solution
If states are distinguishable, you can determine which state you send and then, employing an appropreate designer of Hamiltonian, build a second system in the same state.
Conversely, if you can prepare many identical copies of a qubit, then it is possible to measure the mean value of noncommuting observables.
Creation and transformation of entanglement
Creating entanglement is a simple dynamical process of interest in quantum information theory. How many qubits must two parties exchange if they are to create a particular entangled state shared between them, given that they share no prior entanglement?
Transforming entanglement from one form into another. (e.g. Alice and Bob share between them a Bell state, and wish to transform it into some other type of entangled state. What resources do they need to accomplish this task?)
Answering these and more complex questions about the creation and transformation of entanglement forms a fascinating area of study in its own right, and also promises to give insight into tasks such as quantum computation. (e.g. a distributed quantum computation may be viewed as simply a method for generating entanglement between two or more parties.)
Problem 1.1
(Feynman-Gates conversation) Construct a friendly imaginary discussion of about 2000 words between Bill Gates and Richard Feynman, set in the present, on the future of computation. (Comment: You might like to try waiting until you’ve read the rest of the book before attempting this question. See the 'History and further reading' below for pointers to one possible answer for this question.)
solution
From Bill Gates:
Thirty years ago I went on vacation and fell for Richard Feynman.
A friend and I were planning a trip together and wanted to mix a little learning in with our relaxation. We looked at a local university’s film collection, saw that they had one of his lectures on physics, and checked it out. We loved it so much that we ended up watching it twice. Feynman had this amazing knack for making physics clear and fun at the same time. I immediately went looking for more of his talks, and I’ve been a big fan ever since. Years later I bought the rights to those lectures and worked with Microsoft to get them posted online for free.
Problem 1.2
What is the most significant discovery yet made in quantum computation and quantum information? Write an essay of about 2000 words for an educated lay audience about the discovery. (Comment: As for the previous problem, you might like to try waiting until you’ve read the rest of the book before attempting this question.)
solution
Maybe try waiting until I've read the rest of the book before attempting this question.
2 Introduction to quantum mechanics
In this chapter we will acquire familiarity with elementary linear algebra while introducing the notation used by physicists to describe quantum mechanics (which is different to that used in most introductions to linear algebra). We then review the basic postulates of quantum mechanics. Later on we will introduce superdense coding, a surprising and illuminating example of quantum information processing which combines many of the postulates of quantum mechanics in a simple setting. Next there are powerful tools like density operator, purifications, and the Schmidt decomposition, which are especially useful in the study of quantum computation and quantum information.
2.1 Linear algebra
The basic objects of linear algebra are vector spaces, the vector space of most interest to us is , the space of all -tuples of complex numbers .
2.1.1 Bases, operators and matrices
A spanning set () for a vector space is a set of vectors such that any vector in the vector space can be written as a combination . For example, the two vectors and span the vector space . Generally, a vector space may have many different spanning sets.
A set of non-zero vectors are linearly dependent if there exists a set of complex numbers with for at least one , such that .
It can be shown that any two sets of linearly independent vectors which span a vector space contain the same number of elements. The number of elements in the basis is defined to be the dimension of .
Exercise 2.1
(Linear dependence: example) Show that , and are linearly dependent.
solution
A linear operator is linear to its inputs:
, when we say that a linear operator is defined on a vector space , we mean that is a linear operator from to .
The most convenient way to understand linear operators is in terms of their matrix representations. Suppose is a linear operator. is a basis for V and is a basis for . Then for each in range , there exist complex numbers such that:
Exercise 2.2
(Matrix representations: example) Suppose is a vector space with basis vectors and , and is a linear operator from to such that and . Give a matrix representation for , with respect to the input basis , and the output basis . Find input and output bases which give rise to a different matrix representation of .
solution
, now if we switch from original input & output bases to , then will become .
Exercise 2.3
(Matrix representation for operator products) Suppose is a linear operator from vector space to vector space , and is a linear operator from vector space to vector space . Let be bases for the vector spaces , , and , respectively. Show that the matrix representation for the linear transformation is the matrix product of the matrix representations for and , with respect to the appropriate bases.
solution
From the description we know and . Combine the two equations we get
, hence the linear transformation is the matrix product of the matrix representations for and .
Exercise 2.4
(Matrix representation for identity) Show that the identity operator on a vector space has a matrix representation which is one along the diagonal and zero everywhere else, if the matrix representation is taken with respect to the same input and output bases. This matrix is known as the identity matrix.
solution
2.1.2 The Pauli matrices and inner products
Four extremely useful matrices (which we shall often have occasion to use) are the Pauli matrices:
An inner product is a function which take input as two vectors and from a vector space and produce a complex number as output. The standard quantum mechanical notation for the inner product is . Note that .
Exercise 2.5
Verify that just defined is an inner product on
solution
Let and , then the three conditions require satisfaction are:
Thus is an inner product on .
Exercise 2.6
Show that any inner product is conjugate-linear in the first argument, .
solution
Discussions of quantum mechanics often refer to Hilbert space. In the finite dimensional complex vector spaces that come up in quantum computation and quantum information, a Hilbert space is exactly the same thing as an inner product space.
Exercise 2.7
Verify that and are orthogonal. What are the normalized forms of these vectors?
solution
, and .
Suppose is a basis set for some vector space with an inner product. There is a useful method, the Gram–Schmidt procedure, which can be used to produce an orthonormal basis set for the vector space :
define
for define inductively by:
, we can then easily verify the created vectors form an orthonormal set which is also a basis for .
Exercise 2.8
Prove that the Gram–Schmidt procedure produces an orthonormal basis for V .
solution
One can proof by mathematical induction.
completeness relation: .
One application of the completeness relation is to give a means for representing any operator in the outer product notation: suppose is a linear operator, is an orthonormal basis for , and an orthonormal basis for . By applying the completeness relation twice we obtain:
, which is the outer product representation for . We also see from this equation that has matrix element in the column and row, with respect to the input basis and output basis .
A second application illustrating the usefulness of the completeness relation is the Cauchy–Schwarz inequality:
✏️ Proof of the Cauchy–Schwarz inequality
(書本上是用completeness relation證, 我這邊用另一種更嚴謹的方法證)Suppose and , then we have:
Therefore we have:
Exercise 2.9
(Pauli operators and the outer product) The Pauli matrices can be considered as operators with respect to an orthonormal basis for a two-dimensional Hilbert space. Express each of the Pauli operators in the outer product notation.
solution
Exercise 2.10
Suppose is an orthonormal basis for an inner product space . What is the matrix representation for the operator , with respect to the basis?
solution
All but except one at row and column.
2.1.3 Eigenvectors and Hermitian operators
An eigenvector of a linear operator on a vector space is a non-zero vector such that , where is a complex number known as the eigenvalue of corresponding to .
The characteristic function is defined to be . (it can be shown that the characteristic function depends only upon the operator , and not on the specific matrix representation used for ) The solutions of the characteristic equation are the eigenvalues of the operator.
The eigenspace corresponding to an eigenvalue is the set of vectors which have eigenvalue . It is a vector subspace of the vector space on which acts. When an eigenspace is more than one dimensional we say that it is degenerate. For example, the matrix defined by:
has two eigenvectors and corresponding to the eigenvalue 2. The two eigenvectors are said to be degenerate because they are linearly independent eigenvectors of with the same eigenvalue.
A diagonal representation for an operator on a vector space is a representation , where the vectors form an orthonormal set of eigenvectors for , with corresponding eigenvalues . An operator is said to be diagonalizable if it has a diagonal representation. e.g. the Pauli Z matrix can be written:
, where the matrix representation is with respect to orthonormal vectors and , respectively. Diagonal representations are sometimes also known as orthonormal decompositions.
Exercise 2.11
(Eigen decomposition of the Pauli matrices) Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices , , and .
solution
for matrix, with and , diagonal representations .
for matrix, with and , diagonal representations .
for matrix, with and , diagonal representations .
Exercise 2.12
Prove that the matrix
is not diagonalizable.
solution
Characteristic function: , thus . but since , it is not diagonalizable.
Exercise 2.13
If and are any two vectors, show that .
solution
Let be , then , and , and since , followed by complex conjugate , we have conclude that .
Or, without the matrix representation, consider:
, by comparing the two expressions we have .
Exercise 2.14
(Anti-linearity of the adjoint) Show that the adjoint operation is anti-linear,
solution
Exercise 2.15
Show that .
solution
, hence .
An operator whose adjoint is is known as a Hermitian or self-adjoint operator.
An important class of Hermitian operators is the projectors: Suppose is a -dimensional vector subspace of the -dimensional vector space . Using the Gram–Schmidt procedure it is possible to construct an orthonormal basis for such that is an orthonormal basis for . By definition,
is the projector onto the subspace .
From the definition it can be shown that is Hermitian for any vector , so is also Hermitian: .
The orthogonal complement of is the operator , which is a projector onto the vector space spanned by
Exercise 2.16
Show that any projector satisfies the equation .
solution
An operator is said to be normal if . Obviously, an Hermitian operator is also normal. There is a remarkable representation theorem for normal operators known as the spectral decomposition, which states that an operator is a normal operator if and only if it is diagonalizable.
Theorem 2.1
(Spectral decomposition) Any normal operator on a vector space is diagonal with respect to some orthonormal basis for . Conversely, any diagonalizable operator is normal.
proof
The converse statement is easy to prove, hence we start with the forward one, proving the implication by the method of induction on , the dimension of .
For case, it's trivial.
For case, first let be one of the eigenvalues of , and be the projector onto the eigenspace, and the projector onto the orthogonal complement. Then . The first term and the second term for obvious reason (can be deduced from definition), we claim the third term for the following reason:
proof: Let be an element of the subspace , then . This can be viewed as an element "" turns out to be an eigenvector (associated with eigenvalue ) when acted upon by the same operator , therefore it's also an element in the subspace . Thus , take its adjoint we get . (note that and are Hermitian)
Now we've arrived at , next we're gonna prove that is normal:
proof: , and . Therefore, by the normality of and property (one can prove it trivially), we have:
, thus is normal. Then, by induction (from the first step), is diagonal with respect to some orthonormal basis for the subspace (因為對於的case, 由於的dimension至少為, 所以的dimension至多為, 而我們已經在第一步證明過的case時這定理是顯然合法的了, 所以我們根據歸納法將成功往前從推進到, 再繼續下去). And on the other hand is already diagonal with respect to some orthonormal basis for , therefore it follows that is diagonal with respect to some orthonormal basis for the total vector space .
This conclusion means that can be written as , where are the eigenvalues of , is an orthonormal basis for , while each individual being an eigenvector of with eigenvalue .
We can also express in terms of projector: , where are the projectors onto the eigenspace of . And:
These projectors satisfy the completeness relation ,
and the orthonormality relation .
Exercise 2.17
Show that a normal matrix is Hermitian if and only if it has real eigenvalues.
solution
forward proof:If is Hermitian then . Let be eigenvectors of with eigenvalues , respectively. We have . We can reverse it: , it has real eigenvalues.
backward proof:Since is normal, it can be spectral decomposed: , taking its adjoint: , obviously the two equations are the same because , that is, .
Therefore a normal matrix is Hermitian if and only if it has real eigenvalues.
A matrix is said to be unitary if , same to the operators: an operator is said to be unitary if . A unitary operator also satisfies , and therefore is normal and has a spectral decomposition.
Geometrically, unitary operators are important because they preserve inner products between vectors:The inner product of and is the same as the inner product of and .
Therefore we can imagine the outer product representation of any unitary operator . It starts from letting be any orthonormal basis set, then define , we can conclude that is also a orthonormal basis set since it preserves the inner products. Clearly, .
Exercise 2.18
Show that all eigenvalues of a unitary matrix have modulus , that is, can be written in the form for some real .
solution
Suppose is one of 's eigenvector, then . We have . Therefore .
Exercise 2.19
(Pauli matrices: Hermitian and unitary) Show that the Pauli matrices are Hermitian and unitary.
solution
Show by simple calculation.
Exercise 2.20
(Basis changes) Suppose and are matrix representations of an operator on a vector space with respect to two different orthonormal bases, and . Then the elements of and are and . Characterize the relationship between and .
solution
Define , then we have:
A special subclass of Hermitian operators is the positive operators (extremely important!):
A positive operator is defined to be an operator such that for any vector , is a real non-negative number.
is positive definite if is strictly greater than zero for all . (In Exercise 2.24 one will show that any positive operator is automatically Hermitian, and therefore by the spectral decomposition has diagonal representation , with non-negative eigenvalues )
Exercise 2.21
Repeat the proof of the spectral decomposition for the case when is Hermitian, simplifying the proof wherever possible.
solution
If is Hermitian, . We pick up from the part where proving is normal:
, therefore is normal. And keep on the induction and deduce that is diagonal, and so on...
Exercise 2.22
Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal.
solution
, therefore if then , which is the eigenvectors are necessarily orthogonal.
Exercise 2.23
Show that the eigenvalues of a projector are all either or .
solution
Since , the following two equations are equivalent:
, thus .
Exercise 2.24
(Hermiticity of positive operators) Show that a positive operator is necessarily Hermitian. (Hint: Show that an arbitrary operator can be written where and are Hermitian.)
solution
Suppose is an arbitrary operator, it can be expressed as
, where and are both obviously Hermitian. A positive operator means that and .
, therefore and is necessarily Hermitian.
Exercise 2.25
Show that for any operator , is positive.
solution
Let with an arbitrary vector, we have , hence . On the other hand, , therefore is positive.
2.1.4 Tensor products
The tensor product is a way of putting vector spaces together to form larger vector spaces.
Suppose and are vector spaces of dimension and respectively, and are both Hilbert spaces. Then (read "V tensor W") is an dimensional vector space.
The elements of are linear combinations of tensor products .
In particular, if and are orthonormal bases for the spaces and , then is a basis for .
We often use the abbreviated notations , , or for the tensor product .
By definition the tensor product satisfies the following properties:
For an arbitrary scalar and elements of and os ,
For arbitrary and in and in ,
For arbitrary and in and in ,
Now about the linear operator acting on the space, if and are the linear operators on and , respectively, then we can define a linear operator on by the equation:
, we can then extend this definition to all elements of in the natural way to ensure the linearity of :
The inner product on the space is naturally defined as,
From this inner product, the inner product space inherits the other structure we are familiar with, such as notions of an adjoint, unitarity, normality, and Hermiticity.
The above discussion and definition can be made more concrete by switching to a convenient matrix representation known as the Kronecker product. Suppose is a matrix, and is a matrix, then we have the matrix representation:
. For example, the tensor product of the vectors and is the vector:
, and the tensor product of the Pauli matrices and is:
There is a useful notation , which means tensored with itself times. For example, . An analogous notation is also used for operators on tensor product spaces.
Exercise 2.26
Let . Write out and explicitly, both in terms of tensor products like , and using the Kronecker product.
solution
Exercise 2.27
Calculate the matrix representation of the tensor products of the Pauli operators (a) and ; (b) and ; (c) and . Is the tensor product commutative?
solution
, the tensor product is not commutative.
Exercise 2.28
Show that the transpose, complex conjugation, and adjoint operations distribute over the tensor product.
solution
Exercise 2.29
Show that the tensor product of two unitary operators is unitary.
solution
Let and be the two unitary operators (therefore ), then we have:
Exercise 2.30
Show that the tensor product of two Hermitian operators is Hermitian.
solution
Let and be the two Hermitian operators, then we have:
Exercise 2.31
Show that the tensor product of two positive operators is positive.
solution
Let and be the two positive operators, then we have:
Exercise 2.32
Show that the tensor product of two projectors is a projector.
solution
Let and be the two projectors, then we have:
Exercise 2.33
The Hadamard operator on one qubit may be written as
Show explicitly that the Hadamard transform on qubits, , may be written as
Write out an explicit matrix representation for .
solution
, and the matrix representation for is:
2.1.5 Operator functions
Generically, given a function , it is possible to define a corresponding matrix function on normal matrices.
Let be a spectral decomposition for a normal operator , we then define:
, one can show that is uniquely defined.
This procedure can be used to define the square root of a positive operator, the logarithm of a positive-definite operator, or the exponential of a normal operator. As an example,
Exercise 2.34
Find the square root and logarithm of the matrix
solution
The spectral decomposition for the matrix is:
Therefore we have:
Exercise 2.35
(Exponential of the Pauli matrices) Let be any real, three-dimensional unit vector and a real number. Prove that
where .
solution
We first calculate the eigenvector and eigenvalue of :
, but since is stated as a unit vector, we have , hence . Denote the eigenvectors with and .
, since the spectral decomposition of is .
The trace of (another matrix function) is defined to be the sum of its diagonal elements:
The trace is easily seen to be cyclic, that is, ; and linear also, , and , where and are arbitrary matrices, and a complex number.
From the cyclic property it follows that the trace of a matrix is invariant under the unitary similarity transformation: , (可以參考前面練習過的 Basis Change) that is, , hence it makes sense to define the trace of an operator to be the trace of any matrix representation of .
Useful result: suppose is a unit vector and is an arbitrary operator. To evaluate one can use the Gram–Schmidt procedureto create an orthonormal basis with being the first element. In that way we have:
, which is extremely useful in evaluating the trace of an operator.
Exercise 2.36
Show that the Pauli matrices except for have trace zero.
solution
Exercise 2.37
(Cyclic property of the trace) If and are two linear operators show that
solution
Suppose and , then we have:
Exercise 2.38
(Linearity of the trace) If A and B are two linear operators, show that
and if is an arbitrary complex number show that
solution
Expand the matrices in terms of element, then one can proof by simple calculation.
Exercise 2.39
(The Hilbert–Schmidt inner product on operators) The set of linear operators on a Hilbert space is obviously a vector space - the sum of two linear operators is a linear operator, is a linear operator if is a linear operator and is a complex number, and there is a zero element . An important additional result is that the vector space can be given a natural inner product structure, turning it into a Hilbert space.
(1) Show that the function on defined by
is an inner product function. This inner product is known as the Hilbert–Schmidt or trace inner product.(2) If has dimensions show that has dimension .(3) Find an orthonormal basis of Hermitian matrices for the Hilbert space .
solution
(1) To prove this we need to assure that satisfies: linearity in the second argument, , and with equality if and only if .
, hence the function on defined by is an inner product function.
(2) A vector in has elements, and a matrix in has elements. Therefore .
The commutator between two operators A and B is defined to be:
If , we say commutes with .
Similarly, the anti-commutator of two operators and is defined by:
We say anti-commutes with if .
Theorem 2.2
(Simultaneous diagonalization theorem) Suppose and are Hermitian operators. Then if and only if there exists an orthonormal basis such that both and are diagonal with respect to that basis. We say that and are simultaneously diagonalizable in this case.
proof
The converse statement is easy to prove (if and are diagonal in the same orthonormal basis then ), so we start with the forward one.
Let be an orthonormal basis for the eigenspace of with eigenvalue , the index is used to label possible degeneracies (因應一個本徵值對到多個本徵向量, 要區分開來這些向量). Since , we have:
, therefore we can view as an element of the eigenspace . Let denote the projector onto the space and define (it is defined this way so that can be Hermitian on ), therefore we can perform spectral decomposition to in terms of an orthonormal set of eigenvectors which span the space . Suppose these eigenvectors are , where the indices and label the eigenvalues of and , and is an extra index to allow for the possibility of a degenerate .
Using the same technique of equation (69), we know that is an element of , so . Moreover we have , so combine the two relation we have:
, which indicates that is an eigenvector of with eigenvalue , thus is an orthonormal set of eigenvectors of both and , spanning the entire vector space on which and are defined. That is, and are simultaneously diagonalizable.
Exercise 2.40
(Commutation relations for the Pauli matrices) Verify the commutation relations
There is an elegant way of writing this using , the antisymmetric tensor on three indicies, for which except for and :
solution
Exercise 2.41
(Anti-commutation relations for the Pauli matrices) Verify the anti-commutation relations
, where are both chosen from the set . Also verify that for ,
solution
One can proof the above relation via the commutation relations from last exercise.
Exercise 2.42
Verify that
solution
Exercise 2.43
Show for ,
solution
From the conclusion of previous exercise (2.41 and 2.42), we have:
Exercise 2.44
Suppose and is invertible. Show that must be .
solution
From the above assumption we have and , hence . And since is invertible, , which indicates .
Exercise 2.45
Show that .
solution
Exercise 2.46
Show that .
solution
Proof by simply expanding the equation.
Exercise 2.47
Suppose and are Hermitian. Show that is also Hermitian.
solution
From the conclusions of the previous exercises (2.45 and 2.46), we have:
, hence is also Hermitian.
2.1.7 The polar and singular value decompositions
The polar and singular value decompositions are useful ways of breaking linear operators up into simpler parts.
In particular, these decompositions allow us to break general linear operators up into products of unitary operators and positive operators.
Theorem 2.3
(Polar decomposition) Let be a linear operator on a vector space . Then there exists unitary and positive operators and such that
, where the unique positive operators and satisfying these equations are defined by and . Moreover, if is invertible then is unique.
We call the expression the left polar decomposition of , and the right polar decomposition of . Most often, we’ll omit the 'right' or 'left' nomenclature, and use the term 'polar decomposition' for both expressions, with context indicating which is meant.
proof
From exercise we know that for any operator , is positive, hence is also a positive operator, so it can be given a spectral decomposition:
Now we define in order to construct the relation , then we filter those with and define their to make them be normalized. Since they are orthogonal, we have for .
We can complement the orthonormal set using the Gram-Schmidt procedure to create the substitute basis vector for that we omitted at the first moment due to . We then define a unitary operator so that for:
from the above cases we see that and both performs identical actions on , therefore .
Now if is invertible, then so is , and hence we can express as . Here the right polar decomposition naturally comes in because , where is a positive operator due to .
The following singular value decomposition combines the polar decomposition + spectral theorem:
Corollary 2.4
(Singular value decomposition) Let be a square matrix. Then there exist unitary matrices and , and a diagonal matrix with non-negative entries such that
The diagonal elements of are called the singular values of .
proof
By polar decomposition we have with unitary and positive (left-polar decomposition). And by spectral theorem, with unitary and diagonal (notice that has no negative entries since is positive). Combining the two relatinos we get:
, where and .
Exercise 2.48
What is the polar decomposition of a positive matrix ? Of a unitary matrix ? Of a Hermitian matrix, ?
solution
positive matrix :Since is positive, it has spectral decomposition: with all , and we have
, thus the polar decomposition is for all , hence obviously.
unitary matrix :, thus for all .
hermitian matrix :, thus for all .note that generically since but .
Exercise 2.49
Express the polar decomposition of a normal matrix in the outer product representation.
solution
Normal matrix is diagonizable: . We utilize this to calculate :
, therefore .
Exercise 2.50
Find the left and right polar decompositions of the matrix .
solution
For , its eigenvalues and associated eigenvectors are:
, then we can calculate:
, and notice that:
, therefore we may calculate :
and then :
, thus we can calculate
. Hence the left polar decomposition of is:
For the case, or the right polar decomposition, perform similar process we get:
Tedious work, but still worth a try!
2.2 The postulates of quantum mechanics
2.2.1 State space & Evolution
The first postulate of quantum mechanics sets up the arena in which quantum mechanics takes place.
Postulate 1
Associated to any isolated physical system is a complex vector space with inner product (Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system’s state space.
A qubit has two-dimensional state space. Suppose and form an orthonormal basis for that state space, then an arbitrary state vector in that space can be written
, where and are complex numbers.
is a unit vector, which is , is equivalent to , often called normalization condition for state vectors.
As for how a state of a quantum mechanical system change with time, we have the second postulate:
Postulate 2
The evolution of a closed quantum system is described by a unitary transformation. The state of the system at is related to the state of the system at by a unitary operator which depends only on the times and :
Some common and important unitary operators on a single qubit:
Pauli matrix: quantum NOT gate, bit flip matrix.
Pauli matrix: phase flip matrix.
Hadamard gate.
Exercise 2.51
Verify that the Hadamard gate is unitary.
solution
Exercise 2.52
Verify that .
solution
Exercise 2.53
What are the eigenvalues and eigenvectors of ?
solution
Characteristic equation: , thus .
Actually we have a revised version of postulate 2 for continuous time case:
Postulate 2 (refined)
The time evolution of the state of a closed quantum system is described by the Schrödinger equation,
, where is the reduced Planck's constant and is the Hamiltonian (not Hadamard matrix) of the closed system.
The above refined postulate implies that uf we know the Hamiltonian of a system, then we understand its dynamics completely. But the question is, figuring out the Hamiltonian of the system is difficult even for today's physicists.
The spectral decomposition of Hamiltonian (因為它是Hermitian所以是normal的)
with the states be the energy eigenstates, or sometimes stationary states, and the energy of that state. The lowest energy is known as the ground state energy for the system, and the corresponding energy eigenstate (or eigenspace) is known as the ground state.
The reason why eigen states of are called stationary states is that the effect of time evolution on them only acquires a numerical factor:
驗證 postulate 2&2' 是等價的: First write down the solution of the Schrödinger equation,
, and then go to exercise 2.55. And furthermore, any unitary operator can be realized in the form for some Hermitian operator .
Unitary operators to one-to-one correspondence between the discrete-time description of dynamics.Hamiltonians continuous time description.
Exercise 2.54
Suppose and are commuting Hermitian operators. Prove that .
solution
We utilize Theorem 2.2: if and only if there exists an orthonormal basis such that both and are diagonal with respect to that basis. Therefore we can write , , and .
, similarly we have , hence the operator is unitary.
Exercise 2.56
Use the spectral decomposition to show that is Hermitian for any unitary , and thus for some Hermitian .
solution
Since is unitary, it is also normal, hence has a spectral decomposition. Also from Exercise 2.18, we know that all eigenvalues of a unitary matrix can be written in the form for some real .
, since for all , we have and thus is Hermitian.
Sometimes we may describe a quantum system which is not closed using a time-varying Hamiltonian, and it indeed serve as a good approximation to a closed system, that way we can apply unitary operators on the system withoout feeling too guilty.
2.2.2 Quantum measurement
When we make a measurement to a closed system, the interaction with the system is sufficient to make it not-closed, hence cannot be strictly described by a unitary evolution, therefore how do we compensate to this?
Postulate 3
Quantum measurements are described by a collection of measurement operators. These are operators acting on the state space of the system being measured. The index refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is immediately before the measurement then the probability that result occurs is given by
, and the state of the system after the measurement is
. The measurement operators satisfy the completeness equation,
. The completeness equation expresses the fact that probabilities sum to one: .
A simple but important example of a measurement is the measurement of a qubit in the computational basis. In this case the measurement operators are defined by , . Notice that the completeness equation is obeyed.
Suppose the state right before the measurement is . Then the probability of obtaining outcome after measurement is , similarly the probabolity of obtaining is . The state after measurement in the two cases is therefore and .
We will see in later sections that the factors with modulus , like and can effectively be ignored, so the two post-measurement states are effectively and .
Exercise 2.57
(Cascaded measurements are single measurements) Suppose and are two sets of measurement operators. Show that a measurement defined by the measurement operators followed by a measurement defined by the measurement operators is physically equivalent to a single measurement defined by measurement operators with the representation .
solution
It is essential to prove the post-measurement states for two ways are equivalent.
Suppose we do , followed by , the middle state and the final state are:
Now if we perform a single measurement instead, we have:
, which is identical to the first scenario.
2.2.3 Distinguishing quantum states
In chapter 1 we've discussed that non-orthogonal quantum states cannot be reliably distinguished. With postulate 3 as a firm foundation we can now give a much more convincing demonstration of this fact.
We demonstrate "distinguishability" using Alice and Bob plot. Suppose they have been previously informed a fixed set of states , then Alice pick one state from the basket and send it to Bob, how will he arrage the measurement operators to identify the index of the state?
If the states are orthonormal, then Bob can define measurement operators in this way: , where is all possible index, all with an additional in order to satisfy the completeness relation: . By this way Bob can measure one specific state with probability .
If the states are not orthonormal, if Bob can reliably distinguish different states, he may work this way: if he measures , his rule in mind tell him that this corresponds to the state , but in reality this is not pragmatic since if another state have component parallel to , chance are that it also gives measurement outcome , which ultimately leads to wrong decision according to Bob's rule. Now let's see a more rigorous proof below.
✏️ Proof that non-orthogonal states can’t be reliably distinguished
A proof by contradiction shows that no measurement distinguishing the non-orthogonal states and is possible. Suppose such a measurement is possible, define (意思即測量到 或是 都代表量子態是 ), then these observations may be written as:
Clearly we have (因為就是的組合), therefore , but since this comes entirely from the contribution of , we have .
Now we decompose into components parallel and perpendicular to : , therefore , and since and are non-orthogonal we have , thus implies:
, which leads to contradiction (the second last "" is due to ).
2.2.4 Projective measurements
A projective measurement is described by an observable, , a Hermitian operator on the state space of the system being observed. The obersvable has a spectral decomposition,
, where is the projector ontoo the eigenspace of with eigenvalue . (測量後可能得到的所有值就是 observable 的所有 eigenvalues 們, 即 )
Upon measuring the state , the probability of getting result is
, and if was really measured, the state immediately after the measurement is
與 postulate 3 的異同: The measurement operators in postulate 3 need merely satisfying the completeness relation . Now if we add extra restrictions like they () must be orthogonalprojectors, that is,
are Hermitian, and
the postulate 3 will reduce to equivalent with projective measurement.
Nice properties of projective measurements:
Easy to calculate average values:
Able to calculate standard deviation:
These formulation of measurement and standard deviations in terms of observables gives rise in an elegant way to results such as the Heisenberg uncertainty principle.
Exercise 2.58
Suppose we prepare a quantum system in an eigenstate of some observable , with corresponding eigenvalue . What is the average observed value of , and the standard deviation?
solution
, and obviously .
✏️ The Heisenberg uncertainty principle
Suppose and are two Hermitian operators, from simple derivation we have:
(because if we assume where , then and , therefore it follows the relation.)
Now we can then apply Cauchy–Schwarz inequality to the last term in the above equation:
Hence we have:
Now we substitute and with and where and are two observables, which makes:
, and will ultimate leads to the most used form of Heisenberg uncertainty principle:
Example:If we use observables and to measure the quantum state , the uncertainty principle tells us that,
One elementary consequence of this is that and must both be strictly greater than , as can be verified by direct calculation.
Two widely used nomenclatures for measurements deserve emphasis:
Rather than giving an observable to describe a projective measurement, often people simply list a complete set of orthogonal projectors satisfying the relations and . The corresponding observable implicit in this usage is .
Another widely used phrase, to "measure in a basis ", where form an orthonormal basis, simply means to perform the projective measurement with projectors .
Example: If we make a measurement of observable to the state , we will get either or since has eigenvalues . To be more detailed, we will measure with probability
, and similarly the result with probability.
More generally, suppose is any real three-dimensional unit vector. Then we can define an observable:
. Measurement of this observable is sometimes referred to as a measurement of spin along the axis.
Exercise 2.59
Suppose we have qubit in the state , and we measure the observable . What's the average value & standard deviation of ?
solution
Exercise 2.60
Show that has eigenvalues , and that the projectors onto the corresponding eigenspaces are given by .
solution
We've already shown in Exercise 2.35 the eigenvalues are . Now we calculate their eigenvectors:
For :
, then we can calculate projector:
For :
, then we can calculate projector:
. Hence we have .
Exercise 2.61
Calculate the probability of obtaining the result for a measurement of , given that the state prior to measurement is . What is the state of the system after the measurement if is obtained?
solution
, and the post-measurement state is
這步驟其實不太數學
2.2.5 POVM measurements
Postulate 3 gives us two core values:
a rule describing the measurement statistics (即給出測量到各種 outcome 的機率是多少).
a rule describing the post-measurement state of the system.
But sometimes, we just don't care about the post-measurement state of the system. All we want to know is what we've got for the measurement exactly. In such instances there is a mathematical tool known as the POVM formalism which is especially well adapted to the analysis of the measurements.
POVM stands for Positive Operator-Valued Measure. (先不管這名字是怎麼來的)
Suppose a measurement described by measurement operators is performed upon a quantum system in state . Then the probability getting is .
Now, we define .
Then according to elementary linear algebra and postulate 3, is a positive operator such that and .
The operators is POVM elements and the complete set is POVM.
Example: consider a projective measurement described by measurement operators , where are projectors such that and . In this instance (and only this instance) all the POVM elements are the same as the measurement operators themselves, since .
✏️ General measurements, projective measurements, and POVMs
為什麼先學 general measurements, 再學 projective measurements 或 POVM? 一般物理學家都直接從 projective measurement 開始, 但 QCQI 講求對量子系統的精確控制, 因此相較於一般只能粗略測量的系統, 從 general measurement 開始介紹會比較合適.
General measurement 又有幾個優點:
Mathematically easier than projective measurements, they have fewer restrictions, 例如它就沒有像是 這種規定.
There are important problems in QCQI (such as the optimal way to distinguish a set of quantum states) the answer to which involves a general measurement, rather than a projective measurement.
一般我們生活中見到的測量方法都是「不可重現性」的 (e.g. if we use a silvered screen to measure the position of a photon we destroy the photon in the process. This certainly makes it impossible to repeat the measurement of the photon’s position!) 但 projective measurement 都是可重現性的, 因為用 測量一次過後, 再用一次 , 並不會改變系統的 state. 所以這時候勢必要引入 general measurement formalism 來描述測量過程.
Where do POVMs fit in this picture? POVMs are best viewed as a special case of the general measurement formalism, providing the simplest means by which one can study general measurement statistics, without the necessity for knowing the post-measurement state. They are a mathematical convenience that sometimes gives extra insight into quantum measurements.
Exercise 2.62
Show that any measurement where the measurement operators and the POVM elements coincide is a projective measurement.
solution
Assume the measurement operators we're considering are , writing down the statement mathematically we have
, from Exercise 2.25 we deduce that is positive, hence Hermitian, therefore we have . The last equal sign indicates that are projective operators.
Suppose now that is some arbitrary set of positive operators such that . We will show below that there exists a set of measurement operators defining a measurement described by the POVM .
Define in order to make .
For this reason it is convenient to define a POVM to be any set of operators such that:
each operator is positive.
the completeness relation is obeyed (probabilities sum to one).
To complete the description of POVMs, we note again that given a POVM , the probability of getting outcome is given by .
到現在為止還看不出 POVM 能發揮什麼鳥用, 所以來舉個例子. Suppose Alice gives Bob a qubit prepared in one of two states:
, though it is impossible for Bob to determine whether he has been given which with reliability, it is possible for him to perform a measurement which distinguishes the states some of the time, but never makes an error of mis-identification:
Consider a POVM containing three elements:
, you can verify they satisfy positive operators and completeness relation, therefore form a legitimate POVM.
If Bob received , he then performs the measurement described by the POVM :
There is zero probability that he will observe the result . (這表示如果Bob測量到的話, 就代表他一定收到的是)
If Bob received , he then performs the measurement described by the POVM :
There is zero probability that he will observe the result . ( 的設計很巧妙吧!!)
Therefore he can perform the measurement with the rule in mind:
, 這種永遠不會出錯的代價就是 Bob 有時候做的測量會給出0資訊.
Exercise 2.63
Suppose a measurement is described by measurement operators . Show that there exist unitary operators such that , where is the POVM associated to the measurement.
solution
From polar decomposition (Theorem 2.3), , we have .
Exercise 2.64
Suppose Bob is given a quantum state chosen from a set of linearly independent states. Construct a POVM such that if outcome occurs (), then Bob knows with certainty that he was given the state . (The POVM must be such that for each .)
solution
這就代表給定 , 量到其他 的機率就是 , 那構成這個 的元素就要剔除任何平行其他 的分量, 即:
, where
, and is chosen in order to maintain the positivity of .
2.2.6 Phase & Composite systems
State is equal to state up to the global phase factor.
The statistics of measurement predicted for these two states are the same.proof: , therefore from an observational point of view these two states are identical.
Relative phase: consider state and , they are the same up to a relative phase shift because the amplitudes of are identical, and the amplitudes of differ only by a relative phase factor of .
the relative phase is a basis-dependent concept unlike global phase.
this give rise to physically observable differences in measurement statistics.
Exercise 2.65
Express the states and in a basis in which they are not the same up to a relative phase shift.
solution
Apply them to new basis and to see the difference.
How should we describe states of the composite system? Therefore a postulate 4 is needed:
Postulate 4
The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered through , and system number is prepared in the state , then the joint state of the total system is .
Notation convention: for example, in a system containing three qubits, is a Pauli operator acting on the second qubit.
Exercise 2.66
Show that the average value of the observable for a two qubit system measured in the state is zero.
solution
Projective measurements together with unitary dynamics are sufficient to implement a general measurement. The proof of this statement makes use of composite quantum systems, and is a nice illustration of Postulate 4 in action:
Suppose we have a quantum system with state space , and we want to perform a measurement described by measurement operators on the system .
To do this, we first introduce an ancilla system with state space , having an orthonormal basis in one-to-one correspondence with the possible outcomes of the measurement we wish to implement. (這個擴充系統可被視為數學上虛構的東西或是真實的物理系統.)
Let be any fixed state of , define an operator on products of states from with the state by
Due to the completeness relation and the orthonormality of , we can see that preserves inner products between states of the form :
And by the results of Exercise 2.67 it follows that can be extended to a unitary operator on the space , which can be also denoted bu .
After letting act on , suppose we perform a projective measurement on the two systems described by projectors . The outcome occurs with probability
, just as given in Postulate 3. The joint state of the system after measurement, conditional on result occurring, is given by:
, notice that the state of system is just as exactly the same as given in Postulate 3. Thus unitary dynamics, projective measurements, and the ability to introduce ancillary systems, together allow any measurement of the form described in Postulate 3 to be realized.
Exercise 2.67
Suppose is a Hilbert space with a subspace . Suppose is a linear operator which preserves inner products, that is, for any and in ,
Prove that there exists a unitary operator which extends. That is, for all in , but is defined on the entire space . Usually we omit the prime symbol and just write to denote the extension.
solution
Since is a subspace for , we can find an orthogonal complement of it, that is, . Let , , and be othonormal bases for , , and , respectively.
Now we define as , where . Now we can verify the unitarity of by direct calculation:
, and similarly we have:
. Thus is unitary. Moreover we can check for all ,
, implying that is an extension of .
Postulate 4 also enable us to define entanglement (interesting!!), consider a two qubit state , no one can write it in form of . Hence we have the following exercise:
Exercise 2.68
Prove that for all single qubit states and .
solution
Suppose , therefore we have:
, which is algebraically impossible.
We say that a state of a composite system having this property (that it can’t be written as a product of states of its component systems) is an entangled state. For reasons which nobody fully understands, entangled states play a crucial role in quantum computation and quantum information, and arise repeatedly through the remainder of this book.
A global view of quantum mechanics:
Postulate 1 sets the arena for quantum mechanics, by specifying how the state of an isolated quantum system is to be described.
Postulate 2 tells us that the dynamics of closed quantum systems are described by the Schrödinger equation, and thus by unitary evolution.
Postulate 3 tells us how to extract information from our quantum systems by giving a prescription for the description of measurement.
Postulate 4 tells us how the state spaces of different quantum systems may be combined to give a description of the composite system.
Might it be possible to reformulate quantum mechanics in a mathematically equivalent way so that it had a structure more like classical physics? It turns out by proving Bell's Inequality we show that quantum mechanics can's excape from its counter-intuitive nature.
2.3 Superdense coding
Alice can send 2 classical bits of information via transmission of only a single qubit to Bob.
First Alice and Bob share a Bell pair , then if Alice wishes to:
These four states are known as the Bell basis, Bell states, or EPR pairs. Since they form an orthonormal basis, they can be distinguished by an appropriate quantum measurement.
Now after Alice done performing actions to her own qubit, she then send it to Bob.
Bob then do a measurement in the Bell basis, and he can determine which of the four possible bit strings Alice sent.
Exercise 2.69
Verify that the Bell basis forms an orthonormal basis for the two qubit state space.
solution
The bell states defined here are labeled as convention:
Check their linear independence:
, hence they are linear independent (also with norm ), and form a set of orthonormal basis (for two qubit space).
Exercise 2.70
Suppose is any positive operator acting on Alice's qubit. Show that takes the same value when is any of the four Bell states. Suppose some malevolent third party ('Eve') intercepts Alice's qubit on the way to Bob in the superdense coding protocol. Can Eve infer anything about which of the four possible bit strings Alice is trying to send? If so, how, or if not, why not?
solution
You may check by simple matrix calculation that where . Now if Eve "intercepts" Alice's qubit by performing measurement operators on it, the probability that Eve get outcome is , and since is positive, we have immediate conclusion that Eve won't be able to distinguish which Bell pair Alice is transiting toward Bob since all are the same.
2.4 The Density Operator
An alternative formulation of quantum mechanics other than using the language of state vectors is the tool of density operator or density matrix.
They are mathematically equivalent, but it serves as a more convenient language for thinking about some commonly encountered scenarios in quantum mechanics.
In the following 3 sections, we first intorduce the density operator using the concept of an ensemble of quantum states; next we derive some general properties of the density operator; last we describe an application, as the density operator being a tool for the description of individual subsystems of a composite quantum system.
2.4.1 Ensembles of Quantum States
The density operator language provides a convenient means for describing quantum systems whose state is not completely known.
To be more precise, suppose we only know a quantum system having probabilities for corresponding states , then we call an ensemble of pure states. The density operator for the system is defined:
, which is often known as the density matrix. (the two terms can be used interchangeably)
All postulates regarding quantum mechanics from section 2.3 can be reformulated into the language for density operator. For example, the evolution of the density operator is described by the equation:
Measurements are also easily described in the language of density operators. Suppose the initial state is , then the probability of getting result is
(if you can't get the last equality, see here for reference). By the law of total probability, the total probability of obtaining is:
What is the density operator of the system after obtaining the measurement result ? If the initial state was then the state after obtaining the result is
, which implies after a measurement which yields the result we have an ensemble of states with respective probabilities . The corresponding density operator is therefore:
, here we do some math manipulation for calculation convenience: (以下轉換你可以自己驗證)
, therefore the equation becomes:
A quantum system whose state is known exactly (just one , prob.) is called pure state. In that way the density operator is simply ; otherwise, is in a mixed state (it is a mixture of the different pure states in the ensemble for ). In the exercise later we will prove:
如果一個系統是由多個 構成, 每個對應到機率 , then the system may be described as . proof: 假設 是來自純態集合 的混態, 那這個系統處於 的總機率即 , 根據此關係我們可以求出此系統的 density matrix:
, where . We say that is a mixture of the states with probabilities .
這種 mixture 的概念之後會很常用到. For example, if we have a quantum system in the state with prob. , but know nothing about the actual value of , then the state of such a system can be described by density operator:
, which is a nice compact formula which may be used as the starting point for analysis of further operations on the system.
2.4.2 General Properties of the Density Operator
In this section we move to an intrinsic characterization of density operators that does not rely on an ensemble interpretation.
Theorem 2.5
(Characterization of density operators) An operator is the density operator associated to some ensemble if and only if it satisfies the conditions:
(1) Trace condition: has trace equal to one.(2) Positivity condition: is a positive operator.
proof
Suppose is a density operator. Then
, hence thte trace condition is satisfied. Suppose is an arbitrary vector in state space. Then
, hence the positivity condition is satisfied.
Conversely, suppose is any operator satisfying the trace and positivity conditions. Since is positive, it must have a spectral decomposition
, where the vectors are orthogonal, and are real, non-negative eigenvalues of . And from the trace condition we see that . herefore, a system in state with probability will have density operator . That is, the ensemble is an ensemble of states giving rise to the density operator .
這個 theorem 給我們純純根據 density operator 的內秉性質定義它的方式, that is, we can define a density operator to be a positive operator which has trace equal to one. We can now reformulate the postulates of quantum mechanics in the density operator picture:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. (到這邊都跟前面一樣) The system is completely described by its density operator, which is a positive operator with trace one, acting on the state space of the system. If a quantum system is in the state with probability , then the density operator for the system is .
Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. That is, the state of the system at time is related to the state of the system at time by a unitary operator which depends only on the times and ,
Postulate 3: Quantum measurements are described by a collection of measurement operators. These are operators acting on the state space of the system being measured. The index refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is immediately before the measurement then the probability that result occurs is given by
, and the state of the system after the measurement is
The measurement operators satify the completeness equation,
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems. Moreover, if we have systems numbered through , and system number is prepared in the state , then the joint state of the total system is .
以上的四條量子力學假設重建構是與 state vector-based 的詮釋方式 mathematically equivalent 的. 但換成這種思考方式的好處聽說是可以比較好的描述:
the quantum systems whhose state is not known
the subsystems of a composite quanutm system
Exercise 2.71
(Criterion to decide if a state is mixed or pure) Let be a density operator. Show that , with equality holds if and only if is a pure state.
solution
Since is positve (which implies hence for all ) and the completeness relation , we have . The equality stands if and only if is a pure state (which is ).
注意!! 很多人會自作聰明, 把一個 的 eigenvalue & eigenvector 當成是構成它的 quantum states 們, 或是認為兩者之間有什麼神秘關係, 我現在就來讓你們美夢破碎! For example, one might suppose a quantum system with density matrix
must be in state with probability and in the state with probability . 現在就來舉個反例, now if we define:
, then this ensemble will also give rise to . That is, these two different ensembles of quantum states give rise to the same density matrix!
取而代之的是, 我們好奇到底哪種類 (which class of) 的 ensemble does give rise to same density matrix? That is, when do two sets of vectors, and generate the same operator ? (notation: so that , 就是把 probability 吃進去 statevector 裡面, may become not normalized in length) The answer to this question has many applications in QCQI:
Theorem 2.6
(Unitary freedom in the ensemble for density matrices) The sets and generate the same density matrix if and only if
, where is a unitary matrix of complex numbers, with indicies and , and we "pad" whichever set of vectors or is smaller with additional vectors so that the two sets have the same number of elements. 如果用 normalized states 來形容, we have for some unitary matrix .
proof
Suppose for some unitary . Then
, which shows that and generate the same operator.
Conversely, suppose
Let be a decomposition for such that are orthonormal, and are strictly positive. 我們的證明策略是 to relate the states to the states , and similarly relate the states to the states , and finally combine the two relations, done.
Let be any vector orthonormal to the space spanned by the hence for all , thus .
This implies for all and all orthonormal to the space spanned by .
It follows that each can be expressed as a linear combination of the (因為它跟「垂直於它的空間」垂直), that is, .
Since , we see that
, the operators are easily seen to be linearly independent (因為前面說過 的 spectral decomposition 裡面的 們是 orthonormal 的)
Therefore we have , which ensures that we may append extra columns (可以自己從矩陣角度想想看為什麼是columns) to to obtain a unitary matrix such that , and similarly find another unitary matrix such that .
Thus we have , where . (別忘了 unitary 的性質)
Exercise 2.72
(Bloch sphere for mixed states) The Bloch sphere picture for pure states of a single qubit was introduced in Section 1.2. This description has an important generalization to mixed states as follows.
(1) Show that an arbitrary density matrix for a mixed state qubit may be written as
, where is a real three-dimensional cevtor such that . This vector is known as the Bloch vector for the state .(2) What is the Bloch vector representation for the state ?(3) Show that a state is pure if and only if .(4) Show that for pure states the description of the Bloch vector we have given coincides with that in Section 1.2
solution
(1) From Exercise 2.35 we have: (注意 是 hermitian 的所以只有三個自由度 .)
Another property of density matrix is that it's positive, i.e. real non-negative eigenvalues:
, hence must be .(2) implies , which corresponds to the origin of Bloch sphere.(3) From here we know that is pure if and only if , hence (可參考 Exercise 2.40)
, it naturally follows that in order to fulfill the relation.(4) Suppose , and with and the constrain , we have (which cioncides with here from section 1.2), it follows that we can rewrite them in the form of:
Exercise 2.73
Let be a density operator. A minimal ensemble for is an ensemble containing a number of elements equal to the rank of . Let be any state in the support of . (The support of a Hermitian operator is the vector space spanned by the eigenvectors of with non-zero eigenvalues.) Show that there is a minimal ensemble for that contains , and moreover that in any such ensemble, must appear with probability
, where is defined to be the inverse of , when is considered as an operator acting only on the support of . (This definition removes the problem that may not have an inverse.)
solution
Referencing from the last two pages in this paper and from the solution of this paper.