0 Nomenclature and notation

roadmap of the book

0.0 Linear algebra and quantum mechanics

  1. A positive operator A is one for which ψ|A|ψ0 for all |ψ.
  2. A positive definite operator A is one for which ψ|A|ψ>0 for all |ψ0​.
  3. The support of an operator is defined to be the vector space orthogonal to its kernel.
  4. Hermitian operator: the vector space spanned by its eigenvectors has non-zero eigenvalues.
  5. U unitary operator or matrix.
  6. For two-level quantum systems used as qubits, we shall usually identify the state |0(1,0) and |1(0,1)​​.
  7. We use the notations I,X,Y,Z to denote σ0,σ1,σ2,σ3 respectively.

0.1 Information theory and probability

  1. Probability distribution refers to a finite set of real numbers, px, such that px0 and Σxpx=1.
  2. The relative entropy of a positive operator A with respect to another positive operator B is defined by:
(1)S(AB)tr(AlogA)tr(AlogB)

0.2 Frequently used gates & symbols

common gates

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

  1. 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.
  2. 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.
  3. Strong Church–Turing thesis: Any algorithmic process can be simulated efficiently using a Turing machine.
  4. 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.
  5. 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.
  6. Hence we have a modified version of Church-Turing Thesis: Any algorithmic process can be simulated efficiently using a probabilistic Turing machine.
  7. It is comparatively difficult to come up with quantum algorithm, since 要先抑制自己已經根深蒂固的古典思維, 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.
  8. 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.
  9. 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.
  10. Shannon 的 noisy channel coding theorem 只給出在噪音通道下, 有效資訊傳輸量的上界, 但沒明確給出有哪個 error correcting code 可以達到這個上界.
  11. In 1996 two groups (= CSS codes) subsumed by the stabilizer codes. The theory of quantum error-correcting codes was developed to protect quantum states against noise.
  12. 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).
  13. 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!
  14. 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.

  1. 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 |0 & excited state |1) orbiting a single atom.

  2. Interestingly, by giving not enough light for an electron to perform |0|1, it will be moved "halfway" into |+ state.

  3. Geometric representation of a qubit: a qubit in superposition can be written as

    (2)|ψ=α|0+β|1,whereas|α|2+|β|2=1

    but since |α|2+|β|2=1, we can rewrite the equation as

    (3)|ψ=eiγ(cosθ2|0+eiφsinθ2|1)

    where θ, φ, and γ are real numbers. Since the factor eiγ have no observable effects, we can essentially write the equation

    (4)|ψ=cosθ2|0+eiφsinθ2|1

    The numbers θ and φ define a point on the unit three-dimensional sphere, which is called Bloch-sphere:

    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.

  4. Multiple qubits: say, if we express a pair of qubits in their computational basis state, we get

    (5)|ψ=α00|00+α01|01+α10|10+α11|11

    , with the normalization condition Σx{0,1}2|αx|2=1. Now if we only take measurement on the first qubit:

    (6){will measure 0 with probability|α00|2+|α01|2, leaving the post-measurement state|ψ=α00|00+α01|01|α00|2+|α01|2will measure 1 with probability|α10|2+|α11|2, leaving the post-measurement state|ψ=α10|10+α11|11|α10|2+|α11|2
  5. A particularly important two qubit state is the Bell state or EPR pair,

    (7)|00+|112

    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

  1. Quantum state written in vector notation: α|0+β|1[αβ].

  2. Quantum NOT gate X=[0110], hence X[αβ]=[βα].

  3. So quantum gates on a single qubit can be described by two by two matrices U. Are there any constraints on what matrices may be used as quantum gates? It turns out that the normalization condition |α|2+|β|2=1 yields

    (8)UU=I

    Amazingly, this unitarity constraint is the only constraint on quantum gates!

  4. Another two important gates:

    (9)Z[1001]andH12[1111]

    To get the visualization of the Hadamard gate on the Bloch sphere, the operation is just a rotation of the sphere about the y^ axis by 90°, followed by a rotation about the x^ axis by 180° (below is the illustration of an H gate acting on |+ state).

    hadamard gate visualizatioon

✏️ Decomposing single qubit operations

Later in the content we can prove that an arbitrary 2×2 unnitary matrix may be decomposed as

(10)U=eiα[eiβ/200eiβ/2][cosγ2sinγ2sinγ2cosγ2][eiδ/200eiδ/2]

where α,β,γ, and δ 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

  1. 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.

  2. CNOT can be realized as the control qubit and the target qubit are XORed and stored in the target qubit:

    Bloch Sphere

    , where is addition modulo two. Yet another way of describing the action of the CNOT is through its matrix representation:

    (11)UCNOT=[1000010000010010]

    , as for the single qubit case, the requirement that probability be conserved is expressed in the fact that UCNOT is a unitary matrix, that is, UCNOTUCNOT=I.

  3. Any multiple qubit logic gate may be composed from CNOT and single qubit gates. (proof in later content)

1.3.3 Measurements & Quantum circiuts

  1. It is possible to treat the |+ and | states as though they were the computaitonal basis states, for instance,

    (12)|ψ=α|0+β|1=α+β2|++αβ2|
  2. Generally, given ay basis states |a and |b for a qubit, it is possible to express an arbitrary state as a linear combination α|a+β|b of these states. Furthermore, provided the states are orthonormal, it is possible to perform a measurement with respect to the basis.

  3. Swap gate: Below circuit accomplishes the swap operation.

    swap gate equivalence
    |a,b|a,ab|a(ab),ab=|b,ab(13)|b,(ab)b=|b,a
  4. 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 |ψ from slot A to slot B (which starts in also a pure state |s) in a quantum machine. Thus the initial state of the copying machine is:

(14)|ψ|s

Some unitary evolution U now effects the copying procedure, ideally,

|ψ|sUU(|ψ|s)=|ψ|ψ

This works for another state, |φ, also:

(15)U(|ψ|s)=|ψ|ψ(16)U(|φ|s)=|φ|φ

Taking the inner product of these two equations gives:

(17)ψ|φ=(ψ|φ)2

From the above equation there are oly two possibilities: either |ψ=|φ, or |ψ are orthogonal to |φ​. 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

quantum teleportation scheme

Above is a demonstration of quantum teleportation:

  1. First generate an EPR pair (or Bell state): the part before the dashed line. Assume |ψ=α|0+β|1, then the state of the three qubits are:

    original state :(α|0+β|1)|00after Hadamard gate12(α|0+β|1)(|00+|10)after CNOT12(α|0+β|1)(|00+|11)
  2. We then entangle the state which we want to teleport with qubit #2:

    from previous section :12(α|000+α|011+β|100+β|111)after CNOT12(α|000+α|011+β|110+β|101)after Hadamard12(α(|0+|12)(|00+|11)+β(|0|12)(|10+|01))

    which equals to:

    (18)12(|00(α|0+β|1)+|01(α|1+β|0)+|10(α|0β|1)+|11(α|1β|0))
  3. We can then apply X and Z gate to the third qubit (depend on the measurement outcome of qubit #1 and #2) to restore |ψ 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

  1. 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:

  2. 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 H gate to a |0 state, then measure it and results in a 50/50 chance of 0 or 1.

1.4.2 Quantum parallelism

  1. quantum parallelism allows quantum computers to evaluate a function f(x) for many different values of x simultaneously. Suppose f(x):{0,1}{0,1} is a function with a one-bit domain and range. The below circuit implemented a transformation Uf:|x,y|x,yf(x) (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).

    quantum parallelism

    Now if |y=0 we will result f(x) in the second register. But if we first Hadamard transform the first qubit (or Walsh– Hadamard transform), we'll result in the state: |0,f(0)+|1,f(1)/2.

  2. Generally, the result of performing the Hadamard transform on n qubits initially in the all |0 state is

    (19)12nx|x

    , where the sum is over all possible values of x, and we write Hn to denote this action (read "" as "tensor"). Note that the Hadamard transform is super efficient since it can produce a superposition of 2n states using only n qubits.

  3. Only quantum parallelism is not enough, we also require the ability to extract information about more than merely one value of f(x) from superposition states like

    (20)12nx|x|f(x)

    , which is obtained from applying Hadamard transform and then Uf to the n+1 qubit state |0n|0.

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 |ψ0=|01 and use the same transform Uf above:

Deutsch's algorithm scheme

after two H gate, we have |ψ1=[|0+|12][|0|12], due to the unitarity of Uf, we expect the state |ψ2=(1)f(x)|x|0|12. (因為如果f(x)=0的話那Uf就等於沒做任何事, 那如果f(x)=1的話那第二個qubit出來的yf(x)就相當於反相y) So there are two possible groups of outcome |ψ2:

(21)|ψ2={±[|0+|12][|0|12], if f(0)=f(1)±[|0|12][|0|12], if f(0)f(1)

, therefore we have:

(22)|ψ3={±|0[|0|12], if f(0)=f(1)±|1[|0|12], if f(0)f(1)

, notice that if

(23){f(0)=f(1)f(0)f(1)=0f(0)f(1)f(0)f(1)=1

, hence we can technically write |ψ3 as:

(24)|ψ3=±|f(0)f(1)[|0|12]

, 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 f(x) (在此題即f(0)f(1)), using only one evaluation of f(x).

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 f(x) is balanced or constant?

  1. select a number x from 0 to 2n1.
  2. feed x into f, note that f(x) only outputs 0 or 1.

balanced: equal amount of 0​ and 1​ as outcome for all possivle value of x​,  constant: constant output disregarding different values of x​.

  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 ϵ<1/2. What is the performance of the best classical algorithm for this problem?

  solution  
(33)(2n/22)/(2n2)×2=(2n1)(2n11)/2(2n)(2n1)/2×2=2n112n1<12

A probabilistic classical computer can solve Deutsch’s problem with two evaluations with some probability of error ϵ<1/2.

1.4.5 Summarization

Generally there are 3 classes of quantum algorithms which provide an advantage over known classical algorithms:

We will now breifly describe each of these three classes of algorithms.

  1. Quantum algorithms based upon the Fourier transform The discrete Fourier transform is usually described as transforming a set x0,,xN1 of N complex numbers into a set of complex numbers y0,,yN1 defined by

    (34)yk1Nj=0N1ei2πjkNxj

    , 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 U on n qubits with the action it performs on computational basis |j:

    (35)|j12nk=02n1ei2πjk2n|k,where0j(2n1).

    , you can check the unitarity of this transformation, which makes it a valid quantum circuit; moreover if we perform same transformation on superposition states:

    (36)j=02n1xj|jj=02n1xj(12nk=02n1ei2πjk2n|k)=k=02n1yk|k

    , which is the vector version for the Fourier transform in equation (33) where N=2n.

    Time complexity:

    • Classically, the fast Fourier transform takes roughly Nlog(N)n2n steps.
    • On a quantum computer, it takes only (log(N))2n2 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 |0 or |1, preventing us from learning the transform result yk 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 f be a function from a finitely generated group G to a finite set X such that f is constant on the cosets of a subgroup K, and distinct on each coset. Given a quantum black box for performing the unitary transform U|g|h=|g|hf(g), for gG,hX, and an appropriately chosen binary operation on X, find a generating set for K.

  2. Quantum search algorithms With its basic principles were discovered by Grover, quantum search algorithm solves the following problem:

    Given a search space of size N, 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 N operations.
    • But the quantum search algorithm allows it to be solved using approximately N 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.

  3. Quantum simulation

    • In general, storing the quantum state of a system with n distinct components takes something like cn bits of memory on a classical computer, where c 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 cn 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 cn qubit simulation will collapse into a definite state, giving only cn bits of information; in other words, the total amount of cn 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

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

  1. 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.

  2. 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).

  3. What is more peculiar is that for the beamZX scheme (everytime progressing to the next arrow we only take the positive-half part of splitted beams), there are also two peaks at ±X; and for beamZXZ scheme, there are again two peaks at ±Z. (古典的想法是Z方向的磁矩已經在第一次Z磁場的時候被過濾掉了)

  4. The qubit model provides a simple explanation of this experimentally observed behavior:

    |+Z|0|Z|1|+X(|0+|1)/2|X(|0|1)/2

    , therefore we have |+Z=|+X+|X2 and |+X=|+Z+|Z2, 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

  1. 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.
  2. 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.

  3. 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.

  4. 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.

  5. 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.

  6. 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1015個溶液中的粒子, 可以看成是1015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:

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.

  1. 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 pj,j=1,2,,d. Each use of the source results in the "letter" j being emitted, chosen at random with probability pj, 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 pe>pz, 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 pj can be compressed so that on average each use of the source can be represented using H(pj) bits of information, where

      (38)H(pj)jpjlogpj

      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:

      1. Static resources X2: the bit and the information source
      2. Two-stage dynamic process X1: compress the information source and then decompress it after transmission
      3. 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:

      1. Static resources X2: the information source, and the bits being sent through the channel.
      2. Dynamic process X3: noise process + encoding process + decoding process.
      3. 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.

  2. 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 |0 and (|0+|1)/2, 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 (|ψj with probability pj): the theorem reduces to classical case saying that the source may be compressed down to but not beyond the classical limit H(pj).

      • 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 |0 with probability p and (|0+|1)/2 with probablilty 1p can be reliably compressed using fewer than H(p,1p) qubits per use of the source!)

      • compressing procedure for a certain scheme:

        1. Suppose the source emit state |0 with probability p and (|0+|1)/2 with probablilty 1p and is repeated n times, creating a total of n qubits of information.

        2. With limn we will have in average np+n(1p)2 of |0 and n(1p)2 of |1 in the above n qubits, for such configuration there are (nn(1p)/2)Stirling’s approximation2nH(1+p2,1p2)N combinations in total.

        3. We can then label these combinations from |000,|001,,|110,|111, equivalent to performing a unitary transformation:

          (39)|0n(1+p)2|1n(1p)2unitary transform|j|0(nnH(1+p2,1p2))

          , and discard the tailing |0s, leaving a compressed state of nH(1+p2,1p2) qubits.

        4. To decomopress, we just append the previously discarded |0s and do the inverse unitary transform.

        This procedure for quantum data compression and decompression results in a storage requirement of H(1+p2,1p2) qubits per use of the source, and we can deduce whenever p1/3, it will be an improvement over Shannon’s noiseless channel coding theorem scheme H(p,1p). (這種壓縮方法的核心概念是利用|0(|0+|1)/2都在|0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.

  3. Quantum distinguishability

    • As the above scheme mentioned, in general we cannot distinguish |0 from (|0+|1)/2. If we are able to distinguish {|0,|1,|+,|}, we can develop a way of communication which is faster than light! (e.g. suppose Alice and Bob share a Bell's pair |00+|112, which is also |+++|2 if you do some math, then Alice can convey messages via measuring her qubit with intended basis (she can deliberately choose Z-measurement or X-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 {|0,|1,|+,|} 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 {|0,|1,|+,|} 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 |ψ or |φ correctly identified the state, could be used to build a device which cloned the states |ψ and |φ, 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.

  1. 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 Cn, the space of all n-tuples of complex numbers (z1,,zn).

2.1.1 Bases, operators and matrices

  Exercise 2.1  

(Linear dependence: example) Show that (1,1), (1,2) and (2,1) are linearly dependent.

  solution  
(1)[11]+(1)[12]+(1)[21]=0
  Exercise 2.2  

(Matrix representations: example) Suppose V is a vector space with basis vectors |0 and |1, and A is a linear operator from V to V such that A|0=|1 and A|1=|0. Give a matrix representation for A, with respect to the input basis |0,|1, and the output basis |0,|1. Find input and output bases which give rise to a different matrix representation of A.

  solution  
(42)A=[0110]

, now if we switch from original input & output bases to {|0,|1}{|1,|0}, then A will become [1001].

  Exercise 2.3  

(Matrix representation for operator products) Suppose A is a linear operator from vector space V to vector space W , and B is a linear operator from vector space W to vector space X. Let |vi,|wj,|xk be bases for the vector spaces V, W, and X, respectively. Show that the matrix representation for the linear transformation BA is the matrix product of the matrix representations for B and A, with respect to the appropriate bases.

  solution  

From the description we know |vi=jAji|wj and |wj=kBkj|xk. Combine the two equations we get

|vi=jkBkjAji|xk=k(BA)ki|xk

, hence the linear transformation BA is the matrix product of the matrix representations for B and A.

  Exercise 2.4  

(Matrix representation for identity) Show that the identity operator on a vector space V 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  
(43)|vj=iIij|vi=|vj,jIij=δij

2.1.2 The Pauli matrices and inner products

  Exercise 2.5  

Verify that (,) just defined is an inner product on Cn

  solution  

Let |v=(v1,v2,,vn) and |w=(w1,w2,,wn), then the three conditions require satisfaction are:

(|v,iλi|wi)=jvj(iλiwij)=iλi(ivjwij)=iλi(|v,|wi)(|v,|w)=iviwi=i(wivi)=(|w,|v)(|v,|v)=ivivi=i|vi|20

Thus (,) is an inner product on Cn.

  Exercise 2.6  

Show that any inner product (,) is conjugate-linear in the first argument, (iλi|wi,|v)=iλi(|wi,|v).

  solution  
(44)(iλi|wi,|v)=(|v,iλi|wi)=(iλi(|v,|wi))=iλi(|v,|wi).
  Exercise 2.7  

Verify that |w(1,1) and |v(1,1) are orthogonal. What are the normalized forms of these vectors?

  solution  

w|v=11=0orthogonal, |w=12(1,1) and |v=12(1,1).

  Exercise 2.8  

Prove that the Gram–Schmidt procedure produces an orthonormal basis for V .

  solution  

One can proof by mathematical induction.

  Exercise 2.9  

(Pauli operators and the outer product) The Pauli matrices can be considered as operators with respect to an orthonormal basis |0,|1 for a two-dimensional Hilbert space. Express each of the Pauli operators in the outer product notation.

  solution  
X=|01|+|10|Y=i|01|+i|10|Z=|00||11|

  Exercise 2.10  

Suppose |vi is an orthonormal basis for an inner product space V. What is the matrix representation for the operator |vjvk|, with respect to the |vi basis?

  solution  

All 0 but except one 1 at jth row and kth column.

2.1.3 Eigenvectors and Hermitian operators

  Exercise 2.11  

(Eigen decomposition of the Pauli matrices) Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices X, Y, and Z.

  solution  

  Exercise 2.12  

Prove that the matrix

[1011]

is not diagonalizable.

  solution  

Characteristic function: (1λ)2=0λ=1, thus |λ1=[01]. but since c|λ1λ1|=c[0001][1011], it is not diagonalizable.

  Exercise 2.13  

If |w and |v are any two vectors, show that (|wv|)=|vw|.

  solution  

Let |wv| be M, then Mij=wivj, and (|vw|)ij=viwj, and since (viwj)=vjwi, followed by complex conjugate (vjwi)=vjwi, we have conclude that (|wv|)=|vw|.

Or, without the matrix representation, consider:

{ψ|(|wv|)ϕ=(|wv|)ψ|ϕ=ϕ|(|wv|)ψψ|(|wv|)ϕ=(ψ|wv|ϕ)=ϕ|vw|ψ

, by comparing the two expressions we have (|wv|)=|vw|.

  Exercise 2.14  

(Anti-linearity of the adjoint) Show that the adjoint operation is anti-linear,

(iaiAi)=iaiAi
  solution  
(aiAi)ψ|ϕ=ψ|(aiAi)ϕ=aiψ|(Ai)ϕ=ai(Ai)ψ|ϕ=ai(Ai)ψ|ϕ.

  Exercise 2.15  

Show that (A)=A.

  solution  

(A)ψ|ϕ=ψ|Aϕ=Aϕ|ψ=ϕ|Aψ=Aψ|ϕ, hence (A)=A.

  Exercise 2.16  

Show that any projector P satisfies the equation P2=P.

  solution  

P2=(i|ii|)(j|jj|)=i,j|ii|jj|=i,j|ij|δij=i|ii|=P

  Theorem 2.1  

(Spectral decomposition) Any normal operator M on a vector space V is diagonal with respect to some orthonormal basis for V. 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 d, the dimension of V.

  1. For d=1 case, it's trivial.

  2. For d>1 case, first let λ be one of the eigenvalues of M, and P be the projector onto the λ eigenspace, and Q the projector onto the orthogonal complement. Then M=IMI=(P+Q)M(P+Q)=PMP+QMP+PMQ+QMQ. The first term PMP=λP and the second term QMP=0 for obvious reason (can be deduced from definition), we claim the third term PMQ=0 for the following reason:

    • proof: Let |v be an element of the subspace P, then MM|v=MM|v=λM|v. This can be viewed as an element "M|v" turns out to be an eigenvector (associated with eigenvalue λ) when acted upon by the same operator M, therefore it's also an element in the subspace P. Thus MP=λPQMP=0, take its adjoint we get PMQ=0. (note that P and Q are Hermitian)

    Now we've arrived at M=PMP+QMQ, next we're gonna prove that QMQ is normal:

    • proof: QM=QMI=QM(P+Q)=QMP+QMQ=0+QMQ, and QM=QM(P+Q)=0+QMQ. Therefore, by the normality of M and property Q2=Q (one can prove it trivially), we have:

      (QMQ)(QMQ)=QMQMQ=QMMQ=QMMQ=QMQMQ=(QMQ)(QMQ)

      , thus QMQ is normal. Then, by induction (from the first step), QMQ is diagonal with respect to some orthonormal basis for the subspace Q (因為對於d=2的case, 由於P的dimension至少為1, 所以Q的dimension至多為1, 而我們已經在第一步證明過d=1的case時這定理是顯然合法的了, 所以我們根據歸納法將d成功往前從1推進到2, 再繼續下去). And on the other hand PMP is already diagonal with respect to some orthonormal basis for P, therefore it follows that M=PMP+QMQ is diagonal with respect to some orthonormal basis for the total vector space V.

This conclusion means that M can be written as M=iλi|ii|, where λi are the eigenvalues of M, |i is an orthonormal basis for V, while each individual |i being an eigenvector of M with eigenvalue λi.

We can also express M in terms of projector: M=iλiPi, where Pi are the projectors onto the λi eigenspace of M. And:

  Exercise 2.17  

Show that a normal matrix is Hermitian if and only if it has real eigenvalues.

  solution  

Therefore a normal matrix is Hermitian if and only if it has real eigenvalues.

  Exercise 2.18  

Show that all eigenvalues of a unitary matrix have modulus 1, that is, can be written in the form eiθ for some real θ.

  solution  

Suppose |λ is one of U's eigenvector, then U|λ=λ|λ. We have 1=λ|λ=λ|I|λ=λ|UU|λ=λλλ|λ=λ2=1. Therefore λ=eiθ.

  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 A and A are matrix representations of an operator A on a vector space V with respect to two different orthonormal bases, |vi and |wi. Then the elements of A and A are Aij=vi|A|vj and Aij=wi|A|wj. Characterize the relationship between A and A.

  solution  

Define Ui|wivi|, then we have:

Aij=vi|A|vj=k,lvi|wkwk|A|wlwl|vj=k,lvi|U|vkwk|A|wlvl|U|vj=k,lUikAklUlj

  Exercise 2.21  

Repeat the proof of the spectral decomposition for the case when M is Hermitian, simplifying the proof wherever possible.

  solution  

If M is Hermitian, M=M. We pick up from the part where proving QMQ is normal:

(QMQ)(QMQ)=QMQQMQ=QMQQMQ=(QMQ)(QMQ)

, therefore QMQ is normal. And keep on the induction and deduce that QMQ is diagonal, and so on...

  Exercise 2.22  

Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal.

  solution  
λi|H|λj=λjλi|λjλj|H|λi=λiλj|λiλj|H|λi=λjλj|λi=λiλj|λi (taking the adjoint of the first equation)(λjλi)λj|λi=(λjλi)λj|λi=0

, therefore if λiλj then λj|λi=0, which is the eigenvectors are necessarily orthogonal.

  Exercise 2.23  

Show that the eigenvalues of a projector P are all either 0 or 1.

  solution  

Since P2=P, the following two equations are equivalent:

{P|λ=λ|λP2|λ=λP|λ=λ2|λ

, thus λ=λ2λ={0,1}.

  Exercise 2.24  

(Hermiticity of positive operators) Show that a positive operator is necessarily Hermitian. (Hint: Show that an arbitrary operator A can be written A=B+iC where B and C are Hermitian.)

  solution  

Suppose A is an arbitrary operator, it can be expressed as

A=(A+A2)+i(AA2i)B+iC

, where B and C are both obviously Hermitian. A positive operator means that v|A|v0 and v|A|vR.

v|A|v=v|(B+iC)|v=v|B|v+iv|C|vR

, therefore C=0 and A is necessarily Hermitian.

  Exercise 2.25  

Show that for any operator A, AA is positive.

  solution  

Let ψ|AA|ψ=c with |ψ an arbitrary vector, we have c=ψ|(AA)|ψ=ψ|AA|ψ=c, hence cR. On the other hand, ψ|AA|ψ=(A|ψ,A|ψ)=A|ψ20, therefore AA is positive.

2.1.4 Tensor products

  Exercise 2.26  

Let |ψ=(|0+|1)/2. Write out |ψ2 and |ψ3 explicitly, both in terms of tensor products like |0|1, and using the Kronecker product.

  solution  
|ψ2=12(|0|0+|0|1+|1|0+|1|1)=12[1111]|ψ3=122(|0|0|0+|0|1|0+|1|0|0+|1|1|0+|0|0|1+|0|1|1+|1|01|+|1|1|1)=122[11111111]

  Exercise 2.27  

Calculate the matrix representation of the tensor products of the Pauli operators (a) X and Z; (b) I and X; (c) X and I. Is the tensor product commutative?

  solution  
XZ=[0010000110000100]IX=[0100100000010010]XI=[0010000110000100]

, the tensor product is not commutative.

  Exercise 2.28  

Show that the transpose, complex conjugation, and adjoint operations distribute over the tensor product.

(58)(AB)=AB;(AB)=AB;(AB)=AB
  solution  
(AB)=[A11BA1nBAm1BAmnB]=AB(AB)=[A11BAm1BA1nBAmnB]=AB(AB)=[A11BAm1BA1nBAmnB]=AB

  Exercise 2.29  

Show that the tensor product of two unitary operators is unitary.

  solution  

Let U1 and U2 be the two unitary operators (therefore U1U1=I,U2U2=I), then we have:

(U1U2)(U1U2)=U1U1U2U2=II=I

  Exercise 2.30  

Show that the tensor product of two Hermitian operators is Hermitian.

  solution  

Let H1 and H2 be the two Hermitian operators, then we have:

(H1H2)=H1H2=H1H2=(H1H2)

  Exercise 2.31  

Show that the tensor product of two positive operators is positive.

  solution  

Let A and B be the two positive operators, then we have:

vw|AB|vw=v|A|vw|B|w0

  Exercise 2.32  

Show that the tensor product of two projectors is a projector.

  solution  

Let P1 and P2 be the two projectors, then we have:

(P1P2)2=(P1P2)(P1P2)=P12P22=P1P2

  Exercise 2.33  

The Hadamard operator on one qubit may be written as

(59)H=12[(|0+|1)0|+(|0|1)1|]

Show explicitly that the Hadamard transform on n qubits, Hn, may be written as

(60)Hn=12nx,y(1)xy|xy|

Write out an explicit matrix representation for H2.

  solution  
Hn=12n(x1,y1(1)x1y1|x1y1|)(x2,y2(1)x2y2|x2y2|)(xn,yn(1)xnyn|xnyn|)=12nx,y(1)xy|xy|

, and the matrix representation for H2 is:

[1111111111111111]

2.1.5 Operator functions

  Exercise 2.34  

Find the square root and logarithm of the matrix

[4334]
  solution  

The spectral decomposition for the matrix is:

[4334]=1(12[11]12[11])+7(12[11]12[11])

Therefore we have:

A=12[7+171717+1]lnA=12[ln7ln7ln7ln7]

  Exercise 2.35  

(Exponential of the Pauli matrices) Let v be any real, three-dimensional unit vector and θ a real number. Prove that

(62)exp(iθvσ)=(cosθ)I+isin(θ)vσ

where vσi=13viσi.

  solution  

We first calculate the eigenvector and eigenvalue of vσ=[v3v1iv2v1+iv2v3]:

det[v3λv1iv2v1+iv2v3λ]=λ2(v12+v22+v32)=0λ=±v12+v22+v32

, but since v is stated as a unit vector, we have v12+v22+v32=1, hence λ=±1. Denote the eigenvectors with |λ1 and |λ1.

exp(iθvσ)=exp(1iθ)|λ1λ1|+exp(1iθ)|λ1λ1|=(cosθ+isinθ)|λ1λ1|+(cosθisinθ)|λ1λ1|=(cosθ)(|λ1λ1|+|λ1λ1|)+(isinθ)(|λ1λ1||λ1λ1|)=(cosθ)I+isin(θ)vσ

, since the spectral decomposition of vσ is |λ1λ1||λ1λ1|.

  Exercise 2.36  

Show that the Pauli matrices except for I have trace zero.

  solution  

tr(X)=tr(Y)=tr(Z)=0

  Exercise 2.37  

(Cyclic property of the trace) If A and B are two linear operators show that

(65)tr(AB)=tr(BA)
  solution  

Suppose Am×n and Bn×m, then we have:

tr(AB)=i=1m(j=1nAijBji)=j=1n(i=1mBjiAij)=tr(BA)

  Exercise 2.38  

(Linearity of the trace) If A and B are two linear operators, show that

(66)tr(A+B)=tr(A)+tr(B)

and if z is an arbitrary complex number show that

(67)tr(zA)=ztr(A)
  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 LV of linear operators on a Hilbert space V is obviously a vector space - the sum of two linear operators is a linear operator, zA is a linear operator if A is a linear operator and z is a complex number, and there is a zero element 0. An important additional result is that the vector space LV can be given a natural inner product structure, turning it into a Hilbert space.

(1) Show that the function (,) on LV×LV defined by

(68)(A,B)tr(AB)

is an inner product function. This inner product is known as the Hilbert–Schmidt or trace inner product. (2) If V has d dimensions show that LV has dimension d2. (3) Find an orthonormal basis of Hermitian matrices for the Hilbert space LV.

  solution  

(1) To prove this we need to assure that (,) satisfies: linearity in the second argument, (a,b)=(b,a), and (a,a)0 with equality if and only if a=0.

(A,iλiBi)=tr(AiλiBi)=tr(iλiABi)=iλitr(ABi)=iλi(A,Bi)(A,B)=(tr(AB))=tr((AB))=tr(BA)=(B,A)(A,A)=tr(AA)=i|Aii|20

, hence the function (,) on LV×LV defined by (A,B)tr(AB) is an inner product function.

(2) A vector in V has d elements, and a matrix in LV has d×d=d2 elements. Therefore dim(LV)=d2.

(3) 是不是可以只考慮上三角區域的element啊, 再用 Gram–Schmidt procedure 構造出正交歸一矩陣基底.

2.1.6 The commutator and anti-commutator

  Theorem 2.2  

(Simultaneous diagonalization theorem) Suppose A and B are Hermitian operators. Then [A,B]=0 if and only if there exists an orthonormal basis such that both A and B are diagonal with respect to that basis. We say that A and B are simultaneously diagonalizable in this case.

  proof  

The converse statement is easy to prove (if A and B are diagonal in the same orthonormal basis then [A,B]=0), so we start with the forward one.

Let |a,j be an orthonormal basis for the eigenspace Va of A with eigenvalue a, the index j 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 [A,B]=0, we have:

(71)AB|a,j=BA|a,j=aB|a,j

, therefore we can view B|a,j as an element of the eigenspace Va. Let Pa denote the projector onto the space Va and define BaPaBPa (it is defined this way so that Ba can be Hermitian on Va), therefore we can perform spectral decomposition to Ba in terms of an orthonormal set of eigenvectors which span the space Va. Suppose these eigenvectors are |a,b,k, where the indices a and b label the eigenvalues of A and Ba, and k is an extra index to allow for the possibility of a degenerate Ba.

Using the same technique of equation (69), we know that B|a,b,k is an element of Va, so B|a,b,k=PaB|a,b,k. Moreover we have Pa|a,b,k=|a,b,k, so combine the two relation we have:

(72)B|a,b,k=PaBPa|a,b,k=Ba|a,b,k=b|a,b,k

, which indicates that |a,b,k is an eigenvector of B with eigenvalue b, thus |a,b,k is an orthonormal set of eigenvectors of both A and B, spanning the entire vector space on which A and B are defined. That is, A and B are simultaneously diagonalizable.

  Exercise 2.40  

(Commutation relations for the Pauli matrices) Verify the commutation relations

(73)[X,Y]=2iZ;[Y,Z]=2iX;[Z,X]=2iY

There is an elegant way of writing this using ϵjkl, the antisymmetric tensor on three indicies, for which ϵjkl=0 except for ϵ123=ϵ231=ϵ312=1 and ϵ321=ϵ132=ϵ213=1:

(74)[σj,σk]=2il=13ϵjklσl
  solution  
[X,Y]=[i00i][i00i]=2i[1001][Y,Z]=[0ii0][0ii0]=2i[0110][Z,X]=[0110][0110]=2i[0ii0]

  Exercise 2.41  

(Anti-commutation relations for the Pauli matrices) Verify the anti-commutation relations

(75){σi,σj}=0

, where ij are both chosen from the set {1,2,3}. Also verify that for i{0,1,2,3},

(76)σi2=I
  solution  

One can proof the above relation via the commutation relations from last exercise.

  Exercise 2.42  

Verify that

(77)AB=[A,B]+{A,B}2
  solution  
[A,B]+{A,B}2=ABBA+AB+BA2=2AB2=AB

  Exercise 2.43  

Show for j,k{1,2,3},

(78)σjσk=δjkI+il=13ϵjklσl
  solution  

From the conclusion of previous exercise (2.41 and 2.42), we have:

σjσk=[σj,σk]+{σj,σk}2=[σj,σk]+02=Iδjk+il=13ϵjklσl

  Exercise 2.44  

Suppose [A,B]={A,B}=0 and A is invertible. Show that B must be 0.

  solution  

From the above assumption we have AB=BA and AB=BA, hence BA=BA. And since A is invertible, BAA1=BAA1B=B, which indicates B=0.

  Exercise 2.45  

Show that [A,B]=[B,A].

  solution  
[A,B]=(AB)(BA)=BAAB=[B,A]

  Exercise 2.46  

Show that [A,B]=[B,A].

  solution  

Proof by simply expanding the equation.

  Exercise 2.47  

Suppose A and B are Hermitian. Show that i[A,B] is also Hermitian.

  solution  

From the conclusions of the previous exercises (2.45 and 2.46), we have:

(i[A,B])=i[A,B]=i[B,A]=i[B,A]=i[A,B]

, hence i[A,B] is also Hermitian.

2.1.7 The polar and singular value decompositions

  Theorem 2.3  

(Polar decomposition) Let A be a linear operator on a vector space V. Then there exists unitary U and positive operators J and K such that

(79)A=UJ=KU

, where the unique positive operators J and K satisfying these equations are defined by JAA and K=AA. Moreover, if A is invertible then U is unique.

We call the expression A=UJ the left polar decomposition of A, and A=KU the right polar decomposition of A. 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 A, AA is positive, hence JAA is also a positive operator, so it can be given a spectral decomposition:

(80)J=iλi|ii|,(λi0)

Now we define |ψiA|i in order to construct the relation ψi|ψi=λi2, then we filter those i with λi0 and define their |ei|ψi/λi to make them be normalized. Since they are orthogonal, we have 0=ei|ej=i|AA|j/(λiλj)=i|J2|j/(λiλj) for ij.

We can complement the orthonormal set |ei using the Gram-Schmidt procedure to create the substitute basis vector for i that we omitted at the first moment due to λi=0. We then define a unitary operator Ui|eii| so that for:

{λi0,UJ|i=λi|ei=|ψi=A|iλi=0,UJ|i=0

from the above cases we see that UJ and A both performs identical actions on |i, therefore A=UJ.

Now if A is invertible, then so is J, and hence we can express U as U=AJ1. Here the right polar decomposition naturally comes in because A=UJ=UJUU=KU, where KUJU is a positive operator due to AA=KUUK=KUUK=K2,K=AA.

The following singular value decomposition combines the polar decomposition + spectral theorem:

  Corollary 2.4  

(Singular value decomposition) Let A be a square matrix. Then there exist unitary matrices U and V , and a diagonal matrix D with non-negative entries such that

(81)A=UDV

The diagonal elements of D are called the singular values of A.

  proof  

By polar decomposition we have A=SJ with unitary S and positive J (left-polar decomposition). And by spectral theorem, J=TDT with unitary T and diagonal D (notice that D has no negative entries since J is positive). Combining the two relatinos we get:

(82)A=STDT=UDV

, where UST and VT.

  Exercise 2.48  

What is the polar decomposition of a positive matrix P ? Of a unitary matrix U ? Of a Hermitian matrix, H ?

  solution  

  Exercise 2.49  

Express the polar decomposition of a normal matrix in the outer product representation.

  solution  

Normal matrix A is diagonizable: A=iλi|ii|. We utilize this to calculate J:

(83)J=AA=ijλjλi|jj|ii|=i|λi|2|ii|=i|λi||ii|

, therefore A=Ui|λi||ii|.

  Exercise 2.50  

Find the left and right polar decompositions of the matrix A=[1011].

  solution  

A=[1101],AA=[2111],AA=[1112]

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.

As for how a state |ψ 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 t1|ψ is related to the state of the system at t2|ψ by a unitary operator U which depends only on the times t1 and t2:

(85)|ψ=U|ψ

  Exercise 2.51  

Verify that the Hadamard gate H is unitary.

  solution  
HH=HH=12[1111][1111]=12[2002]=I

  Exercise 2.52  

Verify that H2=I.

  solution  
H2=12[1111][1111]=12[2002]=I

  Exercise 2.53  

What are the eigenvalues and eigenvectors of H?

  solution  

Characteristic equation: λ21=0, thus λ=±1.

{λ=1,12[1111][ab]=[ab],|λ1=14+22[112]λ=+1,12[1111][ab]=+[ab],|λ+1=1422[11+2]

 

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,

(86)id|ψdt=H|ψ

, where is the reduced Planck's constant and H is the Hamiltonian (not Hadamard matrix) of the closed system.

  Exercise 2.54  

Suppose A and B are commuting Hermitian operators. Prove that eAeB=eA+B.

  solution  

We utilize Theorem 2.2: [A,B]=0 if and only if there exists an orthonormal basis such that both A and B are diagonal with respect to that basis. Therefore we can write A=iai|ii|, B=jbj|jj|, and A+B=k(ak+bk)|kk|.

eAeB=ijeaiebj|ii|jj|=ijeai+bjδij|ij|=ieai+bi|ii|=keak+bk|kk|=eA+B

  Exercise 2.55  

Prove that U(t1,t2) defined here is unitary.

  solution  

Note that H=EE|EE|, therefore we have:

U(t1,t2)U(t1,t2)=exp[iH(t2t1)]exp[iH(t2t1)]=(Eexp[iE(t2t1)]|EE|)(Eexp[iE(t2t1)]|EE|)=EEexp[i(EE)(t2t1)]|EδEEE|=Eexp(0)|EE|=I

, similarly we have U(t1,t2)U(t1,t2)=I, hence the operator is unitary.

  Exercise 2.56  

Use the spectral decomposition to show that Kilog(U) is Hermitian for any unitary U, and thus U=exp(iK) for some Hermitian K.

  solution  

Since U 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 eiθ for some real θ.

Kilog(U)=iklog(k)|kk|=ikiθk|kk|=kθk|kk|

, since θkR for all k, we have θk=θk and thus K is Hermitian.

2.2.2 Quantum measurement

  Postulate 3  

Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is |ψ immediately before the measurement then the probability that result m occurs is given by

(90)p(m)=ψ|MmMm|ψ

, and the state of the system after the measurement is

(91)Mm|ψψ|MmMm|ψ

. The measurement operators satisfy the completeness equation,

(92)mMmMm=I

. The completeness equation expresses the fact that probabilities sum to one: 1=mp(m)=mψ|MmMm|ψ.

  Exercise 2.57  

(Cascaded measurements are single measurements) Suppose {Ll} and {Mm} are two sets of measurement operators. Show that a measurement defined by the measurement operators {Ll} followed by a measurement defined by the measurement operators {Mm} is physically equivalent to a single measurement defined by measurement operators {Nlm} with the representation NlmMmLl.

  solution  

It is essential to prove the post-measurement states for two ways are equivalent.

2.2.3 Distinguishing quantum states

✏️ Proof that non-orthogonal states can’t be reliably distinguished

A proof by contradiction shows that no measurement distinguishing the non-orthogonal states |ψ1 and |ψ2 is possible. Suppose such a measurement is possible, define Eij:f(j)=iMjMj (意思即測量到 j1 或是 j2 都代表量子態是 i), then these observations may be written as:

(93)ψ1|E1|ψ1=1;ψ2|E2|ψ2=1

Clearly we have iEi=I (因為E就是M的組合), therefore iψ1|Ei|ψ1=1, but since this 1 comes entirely from the contribution of E1, we have ψ1|E2|ψ1=0E2|ψ1=0.

Now we decompose |ψ2 into components parallel and perpendicular to |ψ1: |ψ2=α|ψ1+β|φ, therefore E2|ψ2=βE2|φ, and since |ψ1 and |ψ2 are non-orthogonal we have |β|<1, thus implies:

(94)1=ψ2|E2|ψ2=|β|2φ|E2|φ|β|21

, which leads to contradiction (the second last "" is due to φ|E2|φiφ|Ei|φ=φ|I|φ=1).

2.2.4 Projective measurements

  Exercise 2.58  

Suppose we prepare a quantum system in an eigenstate |ψ of some observable M, with corresponding eigenvalue m. What is the average observed value of M, and the standard deviation?

  solution  
(ΔM)2=M2M2=ψ|(m,mmmPmPm)|ψψ|M|ψ2=mψ|(mmPm)|ψm2=m2m2=0

, and obviously M=m.

✏️ The Heisenberg uncertainty principle

Suppose A and B are two Hermitian operators, from simple derivation we have:

(100)|ψ|[A,B]|ψ|2+|ψ|{A,B}|ψ|2=4|ψ|AB|ψ|2|ψ|[A,B]|ψ|24|ψ|AB|ψ|2

(because if we assume ψ|AB|ψ=x+iy where x,yR, then ψ|[A,B]|ψ=2iy and ψ|{A,B}|ψ=2x, therefore it follows the relation.)

Now we can then apply Cauchy–Schwarz inequality to the last term 4|ψ|AB|ψ|2 in the above equation:

(101)|ψ|AB|ψ|2ψ|A2|ψψ|B2|ψ

Hence we have:

(102)|ψ|[A,B]|ψ|24ψ|A2|ψψ|B2|ψ

Now we substitute A and B with A=CC and B=DD where C and D are two observables, which makes:

ψ|[A,B]|ψ=ψ|CDCDCD+CDDC+DC+DCDC|ψ=ψ|[C,D]|ψψ|A2|ψ=ψ|C2|ψψ|C|ψ2=(ΔC)2ψ|B2|ψ=ψ|D2|ψψ|D|ψ2=(ΔD)2

, and will ultimate leads to the most used form of Heisenberg uncertainty principle:

(103)ΔCΔD|ψ|[C,D]|ψ|2

(有人可能會用錯誤的方法詮釋不確定性原理:「為了測量 C 到一定的準度, 過程中勢必會影響(disturb)到 D 的量值」, 這是不對的. 正確的理解方法應為: 今天準備一百萬個一模一樣的量子態 |ψ, 然後前五十萬個我們都測它的 C, 後五十萬個都測量它的 D, 那麼兩者的標準差將符合上式規範.)

Example: If we use observables X and Y to measure the quantum state |0, the uncertainty principle tells us that,

ΔXΔY|0|2iZ|0|2=1

One elementary consequence of this is that ΔX and ΔY must both be strictly greater than 0, as can be verified by direct calculation.

  Exercise 2.59  

Suppose we have qubit in the state |0, and we measure the observable X. What's the average value & standard deviation of X?

  solution  
X=0|X|0=[10][0110][10]=0Δ(X)2=0|X2|002=[10][1001][10]=1Δ(X)=1

  Exercise 2.60  

Show that vσ has eigenvalues ±1, and that the projectors onto the corresponding eigenspaces are given by P±=I±vσ2.

  solution  

We've already shown in Exercise 2.35 the eigenvalues are ±1. Now we calculate their eigenvectors:

. Hence we have P±=I±vσ2.

  Exercise 2.61  

Calculate the probability of obtaining the result +1 for a measurement of vσ, given that the state prior to measurement is |0. What is the state of the system after the measurement if +1 is obtained?

  solution  
p(+1)=0|P+|0=0|I+vσ2|0=12+v32=1+v32

, and the post-measurement state is

P+|0p(+1)=11+v3212[1+v3v1+iv2]=12(1+v3)[1+v3v1+iv2]=1+v32[1v1+iv21+v3]=1+v321v1iv2[v1iv21v3]=1+v3211v32[v1iv21v3] (這步驟其實不太數學)=12(1v3)[v1iv21v3]=|λ+

2.2.5 POVM measurements

✏️ 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 PiPj=δijPi 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 Pm 測量一次過後, 再用一次 Pm, 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 Mm, writing down the statement mathematically we have

Em=MmMm=Mm

, from Exercise 2.25 we deduce that Mm is positive, hence Hermitian, therefore we have MmMm=Mm2=Mm. The last equal sign indicates that Mm are projective operators.

  Exercise 2.63  

Suppose a measurement is described by measurement operators Mm. Show that there exist unitary operators Um such that Mm=UmEm, where Em is the POVM associated to the measurement.

  solution  

 

From polar decomposition (Theorem 2.3), A=UJ,JAA, we have Mm=UmMmMm=UmEm.

  Exercise 2.64  

Suppose Bob is given a quantum state chosen from a set |ψ1,,|ψm of linearly independent states. Construct a POVM {E1,E2,,Em+1} such that if outcome Eioccurs (1im), then Bob knows with certainty that he was given the state |ψi. (The POVM must be such that ψi|Ei|ψi>0 for each i.)

  solution  

這就代表給定 i, 量到其他 i 的機率就是 0, 那構成這個 Ei 的元素就要剔除任何平行其他 i 的分量, 即:

(108)Ei=A|ψiψi|

, where

(109)|ψi=|ψij=1,jimψi|ψj|ψj|ψj2

, and A is chosen in order to maintain the positivity of Em+1=Ii=1mEi.

2.2.6 Phase & Composite systems

  Exercise 2.65  

Express the states |0+|12 and |0|12 in a basis in which they are not the same up to a relative phase shift.

  solution  

Apply them to new basis |+|0+|12 and ||0|12 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 1 through n, and system number i is prepared in the state |ψi, then the joint state of the total system is |ψ1|ψ2|ψn.

  Exercise 2.66  

Show that the average value of the observable X1Z2 for a two qubit system measured in the state |00+|112 is zero.

  solution  
(110)00|+11|2(X1Z2)|00+|112=00|+11|2|10|012=0

  Exercise 2.67  

Suppose V is a Hilbert space with a subspace W. Suppose U:WV is a linear operator which preserves inner products, that is, for any |w1 and |w2 in W,

w1|UU|w2=w1|w2

Prove that there exists a unitary operator U:VV which extends U. That is, U|w=U|w for all |w in W, but U is defined on the entire space V. Usually we omit the prime symbol and just write U to denote the extension.

  solution  

Since W is a subspace for V, we can find an orthogonal complement W of it, that is, V=WW. Let |wi, |wj, and |uj be othonormal bases for W, W, and (U act on W), respectively.

Now we define U:VV as UidimW|uiwi|+jdimW|ujwj|, where |uiU|wi. Now we can verify the unitarity of U by direct calculation:

(U)U=(i|wiui|+j|wjuj|)(i|uiwi|+j|ujwj|)=i|wiwi|+j|wjwj|=I

, and similarly we have:

U(U)=(i|uiwi|+j|ujwj|)(i|wiui|+j|wjuj|)=i|uiui|+j|ujuj|=I

. Thus U is unitary. Moreover we can check for all |wW,

U|w=(i|uiwi|+j|ujwj|)|w=i|uiwi|w+j|ujwj|w=i|uiwi|w+0(since|wj|w) =iU|wiwi|w=U|w

, implying that U is an extension of U.

  Exercise 2.68  

Prove that |ψ|a|b for all single qubit states |a and |b.

  solution  

Suppose |ψ=|00+|112=|a|b=(a0|0+a1|1)(b0|0+b1|1), therefore we have:

{a0b0=1/2a0b1=0a1b0=0a1b1=1/2

, 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:

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

 

  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:

|Φ+=12[1001]|Φ=12[1001]|Ψ+=12[0110]|Ψ=12[0±110]

Check their linear independence:

a|Φ++b|Φ+c|Ψ++d|Ψ=0a+d=ad=b+c=bc=0a=b=c=d=0

, hence they are linear independent (also with norm 1), and form a set of orthonormal basis (for two qubit space).

  Exercise 2.70  

Suppose E is any positive operator acting on Alice's qubit. Show that ψ|EI|ψ takes the same value when |ψ 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 00,01,10,11 Alice is trying to send? If so, how, or if not, why not?

  solution  

You may check by simple matrix calculation that ψi|EI|ψi=E00+E11=0|E|0+1|E|1 where |ψiBell states. Now if Eve "intercepts" Alice's qubit by performing measurement operators Mm on it, the probability that Eve get outcome m is p(m)=ψi|MmMmI|ψi, and since MmMm is positive, we have immediate conclusion that Eve won't be able to distinguish which Bell pair Alice is transiting toward Bob since all p(m) are the same.

2.4 The Density Operator

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

  1. The density operator language provides a convenient means for describing quantum systems whose state is not completely known.

  2. To be more precise, suppose we only know a quantum system having probabilities pi for corresponding states |ψi, then we call {pi,|ψi} an ensemble of pure states. The density operator for the system is defined:

    (113)ρipi|ψiψi|

    , which is often known as the density matrix. (the two terms can be used interchangeably)

  3. 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:

    (114)ρ=ipi|ψiψi|UipiU|ψiψi|U=UρU
  4. Measurements are also easily described in the language of density operators. Suppose the initial state is |ψi, then the probability of getting result m is

    (115)p(m|i)=ψi|MmMm|ψi=tr(MmMm|ψiψi|)

    (if you can't get the last equality, see here for reference). By the law of total probability, the total probability of obtaining m is:

    p(m)=ip(m|i)pi=ipitr(MmMm|ψiψi|)=tr(MmMmρ)
  5. What is the density operator of the system after obtaining the measurement result m? If the initial state was |ψi then the state after obtaining the result m is

    (116)|ψim=Mm|ψiψi|MmMm|ψi

    , which implies after a measurement which yields the result m we have an ensemble of states |ψim with respective probabilities p(i|m). The corresponding density operator ρm is therefore:

    (117)ρm=ip(i|m)|ψimψim|=ip(i|m)Mm|ψiψi|Mmψi|MmMm|ψi

    , here we do some math manipulation for calculation convenience: (以下轉換你可以自己驗證)

    (118)p(i|m)=p(i,m)p(m)=p(m|i)pip(m)

    , therefore the equation becomes:

    ρm=ip(i|m)Mm|ψiψi|Mmψi|MmMm|ψi=ip(m|i)pip(m)Mm|ψiψi|Mmψi|MmMm|ψi=iψi|MmMm|ψipitr(MmMmρ)Mm|ψiψi|Mmψi|MmMm|ψi=ipitr(MmMmρ)Mm|ψiψi|Mm(119)=MmρMmtr(MmMmρ)
  6. A quantum system whose state is known exactly (just one |ψ, prob.=1) is called pure state. In that way the density operator is simply ρ=|ψψ|; otherwise, ρ is in a mixed state (it is a mixture of the different pure states in the ensemble for ρ). In the exercise later we will prove:

    (120){a pure statetr(ρ2)=1a mixed statetr(ρ2)<1
  7. 如果一個系統是由多個 ρi 構成, 每個對應到機率 pi, then the system may be described as ipiρi. proof: 假設 ρi 是來自純態集合 {pij,|ψij} 的混態, 那這個系統處於 ρi 的總機率即 pipij, 根據此關係我們可以求出此系統的 density matrix:

    (121)ρ=ijpipij|ψijψij|=ipiρi

    , where ρi=jpij|ψijψij|. We say that ρ is a mixture of the states ρi with probabilities pi.

  8. 這種 mixture 的概念之後會很常用到. For example, if we have a quantum system in the state ρm with prob. p(m), but know nothing about the actual value of m, then the state of such a system can be described by density operator:

    (122)ρ=mp(m)ρm=mtr(MmMmρ)MmρMmtr(MmMmρ)=mMmρMm

    , 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 ρ is the density operator associated to some ensemble {pi,|ψi} if and only if it satisfies the conditions:

(1) Trace condition: ρ has trace equal to one. (2) Positivity condition: ρ is a positive operator.

  proof  

Suppose ρ=ipi|ψiψi| is a density operator. Then

(123)tr(ρ)=ipitr(|ψiψi|)=ipi=1

, hence thte trace condition is satisfied. Suppose |φ is an arbitrary vector in state space. Then

(124)φ|ρ|φ=ipiφ|ψiψi|φ=ipi|φ|ψi|20

, hence the positivity condition is satisfied.

Conversely, suppose ρ is any operator satisfying the trace and positivity conditions. Since ρ is positive, it must have a spectral decomposition

(125)ρ=jλj|jj|

, where the vectors |j are orthogonal, and λj are real, non-negative eigenvalues of ρ. And from the trace condition we see that jλj=1. herefore, a system in state |j with probability λj will have density operator ρ. That is, the ensemble {λj,|j} is an ensemble of states giving rise to the density operator ρ.

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 ρ 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 ρ with trace one, acting on the state space of the system. If a quantum system is in the state ρi with probability pi, then the density operator for the system is ipiρi.

Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. That is, the state ρ of the system at time t1 is related to the state ρ of the system at time t2 by a unitary operator U which depends only on the times t1 and t2,

(126)ρ=UρU

Postulate 3: Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is ρ immediately before the measurement then the probability that result m occurs is given by

(127)p(m)=tr(MmMmρ)

, and the state of the system after the measurement is

(128)MmρMmtr(MmMmρ)

The measurement operators satify the completeness equation,

(129)mMmMm=I

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 1 through n, and system number i is prepared in the state ρi, then the joint state of the total system is ρ1ρ2ρn.

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:

  Exercise 2.71  

(Criterion to decide if a state is mixed or pure) Let ρ be a density operator. Show that tr(ρ2)1, with equality holds if and only if ρ is a pure state.

  solution  
tr(ρ2)=tr(i,jpipj|ii|jj|)=tr(i,jpipj|iδijj|)=tr(ipi2|ii|)=ipi2

Since ρ is positve (which implies 0pi1 hence pi2pi for all i) and the completeness relation ipi=1, we have ipi21. The equality stands if and only if ρ is a pure state (which is ρ=|ψψ|).

注意!! 很多人會自作聰明, 把一個 ρ 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

ρ=34|00|+14|11|

must be in state |0 with probability 34 and in the state |1 with probability 14. 現在就來舉個反例, now if we define:

|a34|0+14|1|b34|014|1

, then this ensemble will also give rise to ρ=12|aa|+12|bb|=34|00|+14|11|. 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, |ψ~i and |φ~i generate the same operator ρ? (notation: |ψ~i=pi|ψi so that ρ=i|ψ~iψ~i|, 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 |ψ~i and |φ~j generate the same density matrix if and only if

(130)|ψ~i=juij|φ~j

, where uij is a unitary matrix of complex numbers, with indicies i and j, and we "pad" whichever set of vectors |ψ~i or |φ~j is smaller with additional vectors 0 so that the two sets have the same number of elements. 如果用 normalized states |ψi,|φj 來形容, we have pi|ψi=juijqj|φj for some unitary matrix uij.

  proof  

Suppose |ψi~=juij|φj~ for some unitary uij. Then

i|ψi~ψi|~=ijkuij|φj~uikφk|~=jk(iukiuij)|φj~φk|~=jkIkj|φj~φk|~=j|φj~φj|~

, which shows that |ψi~ and |φj~ generate the same operator.

Conversely, suppose

(131)A=i|ψi~ψi|~=j|φj~φj|~.

Let A=kλk|kk| be a decomposition for A such that |k are orthonormal, and λk are strictly positive. 我們的證明策略是 to relate the states |ψi~ to the states |k~λk|k, and similarly relate the states |φj~ to the states |k~, and finally combine the two relations, done.

  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

(133)ρ=I+rσ2

, where r is a real three-dimensional cevtor such that r1. This vector is known as the Bloch vector for the state ρ. (2) What is the Bloch vector representation for the state ρ=I/2 ? (3) Show that a state ρ is pure if and only if r=1. (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 ρ 是 hermitian 的所以只有三個自由度 a,b,c.)

(134)I+rσ2=12[1+r3r1ir2r1+ir21r3]must beρ=[abbc]

Another property of density matrix is that it's positive, i.e. real non-negative eigenvalues:

det(ρλI)=14((1λ)2(r12+r22+r32))=0λ=2±44(1r2)2=1±r0(positive matrix)

, hence r must be 1. (2) ρ=I/2 implies r=0, which corresponds to the origin of Bloch sphere. (3) From here we know that ρ is pure if and only if tr(ρ2)=1, hence (可參考 Exercise 2.40)

(135)tr(14(I+2rσ+(rσ)2))=tr(14(I+2rσ+r2I))=14(2+2r2)=1

, it naturally follows that r=1 in order to fulfill the relation. (4) Suppose |ψ=α|0+β|1, and with ρ=|ψψ| and the constrain tr(ρ)=0, we have α2+β2=1 (which cioncides with here from section 1.2), it follows that we can rewrite them in the form of:

(136)|ψ=eiγ(cosθ2|0+eiφsinθ2|1)

  Exercise 2.73  

Let ρ be a density operator. A minimal ensemble for ρ is an ensemble {pi,|ψi} containing a number of elements equal to the rank of ρ. Let |ψ be any state in the support of ρ. (The support of a Hermitian operator A is the vector space spanned by the eigenvectors of A with non-zero eigenvalues.) Show that there is a minimal ensemble for ρ that contains |ψ, and moreover that in any such ensemble, |ψ must appear with probability

(137)pi=1ψi|ρ1|ψi

, where ρ1 is defined to be the inverse of ρ, when ρ is considered as an operator acting only on the support of ρ. (This definition removes the problem that ρ may not have an inverse.)

  solution  

Referencing from the last two pages in this paper and from the solution of this paper.

 

 

 

to be continued...

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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