Solution of An Introduction to Quantum Computing

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• Solution of An Introduction to Quantum Computing
  • Contribute
  • Book Information
  • Exercise 1.5.1
  • Exercise 2.3.1
  • Exercise 2.3.2
  • Exercise 3.2.1
  • Exercise 3.3.1
  • Exercise 3.4.1
  • Exercise 3.4.2
  • Exercise 3.4.3
  • Exercise 3.4.4
  • Exercise 3.4.5
  • Exercise 3.5.1
  • Exercise 3.5.2
  • Exercise 3.5.3
  • Appendix
    • eigenvalues of projectors
    • Hermitian operator's eigenvalues are real
    • Hermitian operator's expected value is real
    • Trace is sum of eigenvalues

Contribute

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Book Information

• Authors: Phillip Kaye, Raymond Laflamme, Michele Mosca
• Edition: 1st
• Publisher: Oxford University Press
• Publish Date: 2007/01/18 (yyyy/mm/dd)
• ISBN-10: 019857049X
• ISBN-13: 978-0198570493

Exercise 1.5.1

Question

A sequence of

$n$ CNOT gates with the target bits all initialized to 0 is the simplest way to copy an

$n$-bit string

$y$ stored in the control bits. However, more sophisticated copy operations are also possible, such as a circuit that treats a string

$y$ as the binary representation of the integer

$y_1+2y_2+4y_3+\cdots 2^{n-1}{y}_n$ and adds

$y$ modulo

$2^n$ to the copy register (modular arithmetic is defined in Section 7.3.2).

Describe a reversible 4-bit circuit that adds modulo 4 the integer

$y\in \{0,1,2,3\}$ represented in binary in the first two bits to the integer

$z$ represented in binary in the last two bits.

Answer

Not yet

Exercise 2.3.1

Question

Prove that the trace has the cyclic property

$\mathrm{Tr}\left(ABC\right)=\mathrm{Tr}\left(BCA\right)$.

Answer

Exercise 2.3.2

Question

Using the result of the previous exercise, together with the fact that a change of orthonormal basis can be written as a unitary operator, show that

$\mathrm{Tr}\left(A\right)$ is independent of the basis in which it is expressed. Notice that in the matrix representation,

$\mathrm{Tr}\left(A\right)$ equals the sum of the diagonal entries of the square matrix
representing

$A$.

Answer

If the unitary operator

$U$'s columns are composed by the new basis representing in the original basis, the transformed

$A$, which is represented as

$A^{\prime }$ is

$$A^{\prime }=U^{-1}AU$$

Calculate trace of the both sides.

$$\begin{array}{rl}\mathrm{Tr}\left(A^{\prime }\right)& =\mathrm{Tr}\left(U^{-1}AU\right)\\ & =\mathrm{Tr}\left(U{U}^{-1}A\right)(\mathrm{Tr}\left(ABC\right)=\mathrm{Tr}\left(BCA\right))\\ & =\mathrm{Tr}\left(A\right)\end{array}$$

The trace doesn't change after changing the basis. This result holds for all linear independent basis.

Q.E.D.

Exercise 3.2.1

Question

Show that (3.2.12) is a solution of the time-independent Schrödinger equation.

$$\begin{array}{}\text{(3.2.12)}& |\psi (t_2)⟩=e^{\frac{-i}{\hslash }H(t_2-t_1)}|\psi (t_1)⟩\end{array}$$

Answer

From Schrödinger equation

$$i\hslash \frac{\partial }{\partial t}|\psi (t)⟩=H|\psi (t)⟩,$$

we can get the general solution of

$|\psi (t)⟩$ is

$$|\psi (t)⟩=e^{-\frac{i}{\hslash }Ht}|C⟩,$$

where

$|C⟩$ is a time-independent state. Substituting

$t$ by

$t_1$ and

$t_2$, we get

$$\begin{array}{rl}|C⟩=e^{\frac{i}{\hslash }H{t}_2}|\psi (t_2)⟩& =e^{\frac{i}{\hslash }H{t}_1}|\psi (t_1)⟩\\ |\psi (t_2)⟩& =e^{\frac{-i}{\hslash }H(t_2-t_1)}|\psi (t_1)⟩\end{array}$$

Q.E.D.

Exercise 3.3.1

Question

Consider the 2-qubit state

$$|\psi ⟩=\frac{1}{\sqrt{2}}|0⟩|0⟩+\frac{1}{\sqrt{2}}|1⟩|1⟩$$

Show that this state is entangled by proving that there are no possible values

${\alpha }_0$,

${\alpha }_1$,

${\beta }_0$,

${\beta }_1$ such that

$$|\psi ⟩=({\alpha }_0|0⟩+{\alpha }_1|1⟩)({\beta }_0|0⟩+{\beta }_1|1⟩).$$

(Note, the state

$|\psi ⟩$ above is called an EPR pair, named for Einstein, Podolsky, and Rosen, who considered such states in their investigations of quantum mechanics.)

Answer

Exercise 3.4.1

Question

(a) Prove that if the operators

$P_i$ satisfy

$P_i^{†}=P_i$ and

$P_i^2=P_i$, then

$P_i{P}_j=0$ for all

$i\ne j$.

(b) Prove that any pure state

$|\psi ⟩$ can be decomposed as

$|\psi ⟩=\sum _i{\alpha }_i|{\psi }_i⟩$ where

${\alpha }_i=\sqrt{p(i)}$,

$p(i)=⟨\psi |P_i|\psi ⟩$, and

$|{\psi }_i⟩=\frac{P_i|\psi ⟩}{\sqrt{p(i)}}$.
Also prove that

$⟨{\psi }_i|{\psi }_j⟩={\delta }_{i,j}$.

(c) Prove that any decomposition

$I=\sum _i{P}_i$ of the identity operator on a Hilbert space of dimension

$N$ into a sum of nonzero projectors

$P_i$ can have at most

$N$ terms in the sum.

Answer

(a) This statement is not true. Here is a counter-example.

$$\begin{array}{rl}& P_i=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],P_j=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\\ & \Rightarrow P_i{P}_j=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\ne 0\end{array}$$

To make the statement true, we have to add one more condition.

$$\begin{array}{}\text{(1)}& I=P_1+P_2+\cdots +P_i+\cdots +P_j+\cdots +P_N\end{array}$$

Each

$P_k$, where

$1\le k\le N$, is a

$N\times N$ matrix and statifies

$P_k^2=P_k$.

$$R_k\equiv \{P_{k}v:v\in N\times 1\text{ vectors}\equiv S\}$$

In fact,

$R_k$ is the image of

$P_k$. Because, for every

$v\in S$, the summation of projected

$v$ equals

$v$ itself,

$$\begin{array}{rl}{P}_{1}v+\cdots +P_{N}v& =(P_1+\cdots +P_N)v\\ & =Iv& & (\text{by (1)})\\ & =v\end{array}$$

the sum of subspaces

$R_k$ is

$S$.

$$\begin{array}{rl}\sum _k{R}_k& =\sum _k\{P_{k}v:v\in S\}\\ & =\{\sum _k{P}_{k}v:v\in S\}\\ & =\{\left(\sum _k{P}_k\right)v:v\in S\}\\ & =\{Iv:v\in S\}& & (\text{by (1)})\\ & =\{v:v\in S\}\\ \text{(2)}& & =S\end{array}$$

In the next step, it shows the sum of dimensions of

$R_k$ is

$N$.

$a_k$ is the number of zero-valued eigenvalues of

$P_k$ and

${\lambda }_{km}$ is the

$m$th eigenvalue of

$P_k$

$$\text{Misplaced \&}$$

*One eigenspace corresponding to eigenvalue 0 contributes one dimention of the kernel.

As a result, if two different

$R_k$ contribute the same dimension, it disobeys (2). This implies the direct sum of

$R_k$ is

$S$.

$$\underset{k}{⨁}{R}_k=S$$

Because the direct sum of the image and the kernel of a projector is the original vector space,

$$P_{j}v\in R_j\subset \mathrm{Ker}(P_i)$$

for every

$v\in S$. By the definition of kernel, we get

$$\begin{array}{r}{P}_i{P}_{j}v=0,\mathrm{\forall }v\in S\Rightarrow P_i{P}_j=0\end{array}$$

Q.E.D.

(b)

The statement of the question is only true if

${\alpha }_i$ is a real number. Here is the counter-example that

${\alpha }_i$ is not a real number. Let

$$\begin{array}{r}{P}_i=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],|\psi ⟩=\left[\begin{array}{c}i\\ 0\end{array}\right].\end{array}$$

$P_i$'s normalized eigenvector corresponding to the non-zero eigenvalue is

$$|{\psi }_i⟩=\left[\begin{array}{c}1\\ 0\end{array}\right].$$

We get

$$⟨\psi |P_i|\psi ⟩=p(i)=1$$

and

$$\begin{array}{rl}\phantom{\Rightarrow }{P}_i|\psi ⟩={\alpha }_i|{\psi }_i⟩& \Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}i\\ 0\end{array}\right]=i\left[\begin{array}{c}1\\ 0\end{array}\right]\\ & \Rightarrow {\alpha }_i=i\ne \sqrt{p(i)}=1\end{array}$$

We change the statement of the question. Let

$|{\alpha }_i{|}^2=p(i)$ and

$|\psi ⟩$ be normalized to make the statement true. Below is the proof.

$$\begin{array}{r}\phantom{\Rightarrow }I=\sum _i{P}_i\Rightarrow I|\psi ⟩=\sum _i{P}_i|\psi ⟩\Rightarrow |\psi ⟩=\sum _i{\alpha }_i|{\psi }_i⟩\end{array}$$

$|{\psi }_i⟩$ is the result of projecting

$|\psi ⟩$ by

$P_i$. We will show

$|\psi ⟩$ should be normalized under the requirement from the question.

$$\begin{array}{rl}& \phantom{\Rightarrow }{P}_i|\psi ⟩={\alpha }_i|{\psi }_i⟩\\ & \Rightarrow ⟨\psi |P_i^{†}{P}_i|\psi ⟩=⟨{\psi }_i|{\alpha }_i^{\ast }{\alpha }_i|{\psi }_i⟩\\ & \Rightarrow ⟨\psi |P_i{P}_i|\psi ⟩=⟨{\psi }_i|{\alpha }_i^{\ast }{\alpha }_i|{\psi }_i⟩& & (P_i^{†}=P_i)\\ & \Rightarrow ⟨\psi |P_i|\psi ⟩=⟨{\psi }_i|{\alpha }_i^{\ast }{\alpha }_i|{\psi }_i⟩& & (P_i^2=P_i)\\ & \Rightarrow ⟨\psi |P_i|\psi ⟩=|{\alpha }_i{|}^2⟨{\psi }_i|{\psi }_i⟩& & ({\alpha }_i\text{ is a number})\\ & \Rightarrow ⟨\psi |P_i|\psi ⟩=|{\alpha }_i{|}^2& & (|{\psi }_i⟩\text{ is normalized})\\ & \Rightarrow p(i)=|{\alpha }_i{|}^2& & (\text{Hermitian's expected value is real*})\end{array}$$

*Refer to Hermitian's expected value is real

Below prove

$⟨{\psi }_i|{\psi }_j⟩={\delta }_{i,j}$.

$$\begin{array}{rlrl}⟨{\psi }_i|{\psi }_j⟩& =⟨{\psi }_i|P_i^{†}{P}_j|{\psi }_j⟩& & (\text{definition of }|{\psi }_i⟩)\\ & =⟨{\psi }_i|P_i{P}_j|{\psi }_j⟩& & (P_i^{†}=P_i)\\ & =⟨{\psi }_i|0|{\psi }_j⟩& & (\text{if }i\ne j,P_i{P}_j=0,\text{proved at (a)})\\ & =0\end{array}$$

When

$i=j$,

$⟨{\psi }_i|{\psi }_i⟩=1$ because

$|{\psi }_i⟩$ is normalized. Combined these two cases, we get

$⟨{\psi }_i|{\psi }_j⟩={\delta }_{i,j}$

Q.E.D.

(c)

Denote the

$N$-dimension Hilbert space as

$S$.

$$\begin{array}{rlrl}\phantom{\Rightarrow }I=\sum _{1\le i\le N}{P}_i& \Rightarrow Iv=\sum _i{P}_{i}v& & (v\in S)\\ & \Rightarrow v=\sum _i{P}_{i}v\end{array}$$

$v$ and each

$P_{i}v$ are in

$S$. There are at most

$N$ linear independent vector in a

$N$-dimension Hilbert space. If there is

$N+1$ term in the sum,

$P_{N+1}v$ can be represent as the linear combination of the other

$N$ terms.

$$P_{N+1}v=a_1{P}_{1}v+\cdots +a_N{P}_{N}v(a_i\in \mathbb{C})$$

And because

$P_{N+1}v$ is not zero in our assumption, there must be an

$a_i\ne 0$. Assume

$a_i\ne 0$.

$$\begin{array}{rl}& \phantom{\Rightarrow }{P}_{N+1}v=a_1{P}_{1}v+\cdots +a_N{P}_{N}v\\ & \Rightarrow P_i{P}_{N+1}v=a_1{P}_i{P}_{1}v+\cdots +a_i{P}_i{P}_{i}v+\cdots +a_N{P}_i{P}_{N}v\\ & \Rightarrow 0=a_i{P}_i{P}_{i}v& & (P_i{P}_j=0)\\ & \Rightarrow 0=a_i{P}_{i}v\ne 0& & (P_i^2=P_i)\end{array}$$

We get a contradictory result, so it has at most N terms in the sum.

Q.E.D

Exercise 3.4.2

Question

Show that measuring the observable

$|1⟩⟨1|$ is equivalent to measuring the observable

$Z$ up to a relabelling of the measurement outcomes.

Answer

Exercise 3.4.3

Question

Verify that a measurement of the Pauli observable

$X$ is equivalent to a complete measurement with respect to the basis

$\{\frac{1}{\sqrt{2}}|0⟩+|1⟩,\frac{1}{\sqrt{2}}|0⟩-|1⟩\}$ basis.

Answer

Exercise 3.4.4

Question

(a) Prove that performing a projective measurement with respect to

$P_0$ and

$P_1$ (defined above) on an

$n$-qubit state is equivalent to measuring the observable

$Z^{\otimes n}$.

(b) Explain why performing a complete Von Neumann measurement with respect to the computation basis, and then outputting the parity of the resulting string is not equivalent to performing a projective measurement of the parity.

Answer

Exercise 3.4.5

Question

The observable formalism gives a convenient way to describe the expected value of a quantum measurement, where the eigenvalues

$m_i$ correspond to some relevant physical quantity. Such expectation values are particularly relevant for quantum experiments (particularly the early ones) where one does not measure individual quantum systems, but rather an ensemble of many independent and identically prepared systems, and where the measurement apparatus provides only cumulative results (e.g. the net magnetic field induced by an ensemble of nuclear spins). Consider a projective measurement described by projectors

$P_i$, and suppose we measure the state

$|\psi ⟩$. Show that the expected value of

$m_i$ is

$$E(m_i)=\mathrm{Tr}(M|\psi ⟩⟨\psi |)$$

Answer

Exercise 3.5.1

Question

Find the density matrices of the following states:
(a)

$\{(|0⟩,\frac{1}{2}),(|1⟩,\frac{1}{2})\}$
(b)

$\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩$
(c)

$\{(\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩,\frac{1}{2}),(\frac{1}{\sqrt{2}}|0⟩-\frac{1}{\sqrt{2}}|1⟩,\frac{1}{2})\}$

Answer

Density operator of a mixed state

$\{(|{\psi }_1⟩,p_1),(|{\psi }_2⟩,p_2),\cdots \}$ is

$p_1|{\psi }_1⟩⟨{\psi }_1|+p_2|{\psi }_2⟩⟨{\psi }_2|+\cdots$.

(a)

$$\rho =\frac{1}{2}|0⟩⟨0|+\frac{1}{2}|1⟩⟨1|=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]$$

(b)

$$\begin{array}{rl}\rho & =(\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩)(\frac{1}{\sqrt{2}}⟨0|+\frac{1}{\sqrt{2}}⟨1|)\\ & =\frac{1}{2}|0⟩⟨0|+\frac{1}{2}|0⟩⟨1|+\frac{1}{2}|1⟩⟨0|+\frac{1}{2}|1⟩⟨1|& & (\text{expansion})\\ & =\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]\end{array}$$

(c)

$$\begin{array}{rl}\rho & =\frac{1}{2}(\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩)(\frac{1}{\sqrt{2}}⟨0|+\frac{1}{\sqrt{2}}⟨1|)+\frac{1}{2}(\frac{1}{\sqrt{2}}|0⟩-\frac{1}{\sqrt{2}}|1⟩)(\frac{1}{\sqrt{2}}⟨0|-\frac{1}{\sqrt{2}}⟨1|)\\ & =\frac{1}{2}(\frac{1}{2}|0⟩⟨0|+\frac{1}{2}|0⟩⟨1|+\frac{1}{2}|1⟩⟨0|+\frac{1}{2}|1⟩⟨1|)+\frac{1}{2}(\frac{1}{2}|0⟩⟨0|-\frac{1}{2}|0⟩⟨1|-\frac{1}{2}|1⟩⟨0|+\frac{1}{2}|1⟩⟨1|)& & (\text{expansion})\\ & =\frac{1}{2}|0⟩⟨0|+\frac{1}{2}|1⟩⟨1|\\ & =\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]\end{array}$$

Exercise 3.5.2

Question

(a)
Prove that the density operator

$\rho$ for an ensemble of pure states satisfies the following conditions:
(i)

$\mathrm{Tr}\left(\rho \right)=1$.
(ii)

$\rho$ is a positive operator (i.e. for any

$|v⟩$,

$⟨v|\rho |v⟩$ is real and non-negative; equivalently, the eigenvalues of

$\rho$ are non-negative).

(b)
Show that for any matrix

$\rho$ satisfying conditions 1 and 2, there exists a finite list of probabilities

$p_i$ and pure states

$|{\psi }_i⟩$ such that

$\rho$ is the density matrix of the mixed state

$$\{(|{\psi }_1⟩,p_1),(|{\psi }_2⟩,p_2),\dots ,(|{\psi }_k⟩,p_k)\}.$$

Answer

(a)
(i)

Trace is the sum of eigenvalues.

$$\mathrm{Tr}\left(\rho \right)=\sum _i^N{\lambda }_i$$

$N$ is the dimension of the state,

$\{{\lambda }_i\}$ are the eigenvalues of

$\rho$.

The definition of the density operator

$\rho \equiv |\psi ⟩⟨\psi |$, where

$|\psi ⟩$ is the state with unit length. Let's prove the eigenvectors of

$\rho$ are

$|\psi ⟩$ and other

$N-1$ vectors which are orthogonal to

$|\psi ⟩$ and each other. Denote the

$N-1$ orthogonal vectors as

$\{|{\psi }_i^{\perp }⟩|1\le i\le N-1\}$.

First, prove

$|\psi ⟩$ is one of the eigenvector of

$\rho$.

$$\begin{array}{rlrl}\rho |\psi ⟩& =|\psi ⟩⟨\psi |\psi ⟩& & (\text{definition of density operator})\\ & =|\psi ⟩1& & (\text{state is an unit vector})\\ & =1|\psi ⟩\end{array}$$

Second, prove

$|{\psi }_i^{\perp }⟩$ is the eigenvector of

$\rho$.

$$\begin{array}{rlrl}\rho |{\psi }_i^{\perp }⟩& =|\psi ⟩⟨\psi |{\psi }_i^{\perp }⟩& & (\text{definition of density operator})\\ & =|\psi ⟩0& & (\text{inner product of orthogonal vectors is zero})\\ & =0\\ & =0|{\psi }_i^{\perp }⟩\end{array}$$

Denote the corresponding eigenvalue of

$|{\psi }_i^{\perp }⟩$ is

${\lambda }_i$ and the eigenvalues to

$|\psi ⟩$ is

${\lambda }_N$. From the proof above, we get

${\lambda }_i=0$ for

$1\le i\le N-1$ and

${\lambda }_N=1$.

$$\begin{array}{rl}\mathrm{Tr}\left(\rho \right)& =\sum _i^N{\lambda }_i\\ & ={\lambda }_N\\ & =1\end{array}$$

Q.E.D.

(ii)

From (i), we get the eigenvectors of

$\rho$ form an orthonormal basis

$\{|{\psi }_j⟩|1\le j\le N\}$. We can express any vector

$|v⟩$ as

$$|v⟩=\sum _j^N{a}_j|{\psi }_j⟩,a_j\in \mathbb{C}$$

$$\begin{array}{rl}⟨v|\rho |v⟩& =⟨v|\psi ⟩⟨\psi |v⟩\\ & =(\sum _j^N{a}_j^{\ast }⟨{\psi }_j|\psi ⟩)(\sum _j^N{a}_j⟨\psi |{\psi }_j⟩)\\ & =(\sum _j^N{a}_j^{\ast }⟨{\psi }_j|{\psi }_N⟩)(\sum _j^N{a}_j⟨{\psi }_N|{\psi }_j⟩)& & (|\psi ⟩\text{ is denoted as }|{\psi }_N⟩\text{, see (i)})\\ & =a_N^{\ast }⟨{\psi }_N|{\psi }_N⟩a_N⟨{\psi }_N|{\psi }_N⟩& & (\{|{\psi }_j⟩\}\text{ are orthogonal to each other})\\ & =a_N^{\ast }{a}_N& & (\{|{\psi }_j⟩\}\text{are nomalized})\\ & =|a_N{|}^2\in \mathbb{R}\end{array}$$

And in (i), we have proved the eigenvalues of

$\rho$ are

$0$s and

$1$ , which are non-negative.

Q.E.D.

(b)

Because positive operators on

${\mathcal{H}}_{\mathbb{C}}$ are normal operators. See

• functional analysis - Show that a positive operator on a complex Hilbert space is self-adjoint - Mathematics Stack Exchange
• Positive operator (Hilbert space) - Wikipedia

$\rho$ can be written as a spectral decomposition by an unitary operator.

$$\rho =U\mathrm{\Lambda }{U}^{-1}=U\mathrm{\Lambda }{U}^{†}$$

The columns of

$U$ are the

$\rho$'s eigenvectors

$\{|{\psi }_i⟩\}$ and

$\mathrm{\Lambda }$ is a diagonal matrix composed by

$\{{\lambda }_i|\rho |{\psi }_i⟩={\lambda }_i|{\psi }_i⟩\}$ the eigenvalues of

$\rho$. In the equations below, we use

$⟨e_i|$ to denote the elements of the Cartesian basis.

$$\begin{array}{rl}⟨e_i|& =\left[\begin{array}{ccc}\underset{i-1\text{ times}}{\underbrace{0\dots 0}}& 1& \underset{N-i\text{ times}}{\underbrace{0\dots 0}}\end{array}\right]\phantom{\underset{N}{\underbrace{1}}}\\ |e_i⟩& ={\left[\begin{array}{ccc}\underset{i-1\text{ times}}{\underbrace{0\dots 0}}& 1& \underset{N-i\text{ times}}{\underbrace{0\dots 0}}\end{array}\right]}^T\phantom{\underset{N}{\underbrace{1}}}\end{array}$$

$$\begin{array}{rl}\rho & =U\mathrm{\Lambda }{U}^{†}\\ & =U[\sum _i^N{\lambda }_i|e_i⟩⟨e_i|]U^{†}\\ & =\sum _i^{N}U{\lambda }_i|e_i⟩⟨e_i|U^{†}\\ & =\sum _i^N\left[{\lambda }_i(U|e_i⟩)(⟨e_i|U^{†})\right]\\ & =\sum _i^N\left[{\lambda }_i(U|e_i⟩){(U|e_i⟩)}^{†}\right]& & ((AB{)}^{†}=B^{†}{A}^{†})\\ & =\sum _i^N{\lambda }_i|{\psi }_i⟩{(|{\psi }_i⟩)}^{†}\\ & =\sum _i^N{\lambda }_i|{\psi }_i⟩⟨{\psi }_i|\end{array}$$

Because

$\mathrm{Tr}\left(\rho \right)=1$,

$\sum _i^N{\lambda }_i=1$. The sum above is the finite list we need.

$$\{(|{\psi }_1⟩,{\lambda }_1),(|{\psi }_2⟩,{\lambda }_2),\dots ,(|{\psi }_N⟩,{\lambda }_N)\}.$$

Q.E.D.

Exercise 3.5.3

Question

Consider any linear transformation

$T$ on a Hilbert space

$\mathcal{H}$ of dimension

$N$. This linear transformation

$T$ induces a transformation

$\rho \mapsto T\rho T^{†}$ on the set of linear operators on the Hilbert space

$\mathcal{H}$. Prove that the above transformation is also linear.

Answer

To prove

$\rho \mapsto T\rho T^{†}$ is linear, we have to prove

1. Additivity:
   $T({\rho }_1+{\rho }_2)T^{†}=T{\rho }_1{T}^{†}+T{\rho }_2{T}^{†}$
2. Homogenity:
   $T(a\rho )T^{†}=a(T\rho T^{†})$

Additivity:

$$\begin{array}{rlrl}T({\rho }_1+{\rho }_2)T^{†}& =[T({\rho }_1+{\rho }_2)]T^{†}& & (\text{associative})\\ & =(T{\rho }_1+T{\rho }_2)T^{†}& & (T\text{ is linear transformation})\\ & ={[T(T{\rho }_1+T{\rho }_2{)}^{†}]}^{†}& & ((AB{)}^{†}=B^{†}{A}^{†})\\ & ={\left\{T[(T{\rho }_1{)}^{†}+(T{\rho }_2{)}^{†}]\right\}}^{†}& & ((A+B{)}^{†}=A^{†}+B^{†})\\ & ={[T(T{\rho }_1{)}^{†}+T(T{\rho }_2{)}^{†}]}^{†}& & (T\text{ is linear transformation})\\ & ={[T(T{\rho }_1{)}^{†}]}^{†}+{[T(T{\rho }_2{)}^{†}]}^{†}& & ((A+B{)}^{†}=A^{†}+B^{†})\\ & =(T{\rho }_1)T^{†}+(T{\rho }_2)T^{†}& & ((AB{)}^{†}=B^{†}{A}^{†})\\ & =T{\rho }_1{T}^{†}+T{\rho }_2{T}^{†}& & (\text{associative})\end{array}$$

Q.E.D.

Homogenity:

$$\begin{array}{rlrl}T(a\rho )T^{†}& =[T(a\rho )]T^{†}& & (\text{associative})\\ & =(aT\rho )T^{†}& & (T\text{ is linear transformation})\\ & =a(T\rho T^{†})& & (\text{associative})\end{array}$$

Q.E.D.

Appendix

eigenvalues of projectors

Hermitian operator's eigenvalues are real

Prove if

$$T=T^{†}$$

, then

$$T|\psi ⟩=\lambda |\psi ⟩,\lambda \in \mathbb{R}$$

proof:

$$\begin{array}{}\text{(1)}& \begin{array}{rlrl}{(T|\psi ⟩)}^T|{\psi }^{\ast }⟩& ={\left(T^{\ast }|\psi {⟩}^{\ast }\right)}^{†}|{\psi }^{\ast }⟩& & (A^{\ast }{B}^{\ast }=(AB{)}^{\ast })\\ & =⟨\psi {|}^{\ast }{T}^T|{\psi }^{\ast }⟩& & ({(T|\psi ⟩)}^{†}=⟨\psi |T^{†})\\ & =⟨\psi {|}^{\ast }{T}^{\ast }|{\psi }^{\ast }⟩& & (T=T^{†})\\ & =⟨{\psi }^{\ast }|(T^{\ast }|{\psi }^{\ast }⟩)\\ & =⟨{\psi }^{\ast }|{(T|\psi ⟩)}^{\ast }& & (A^{\ast }{B}^{\ast }=(AB{)}^{\ast })\\ & =⟨{\psi }^{\ast }|{(\lambda |\psi ⟩)}^{\ast }& & (T|\psi ⟩=\lambda |\psi ⟩)\\ & =⟨{\psi }^{\ast }|{\lambda }^{\ast }|{\psi }^{\ast }⟩& & (A^{\ast }{B}^{\ast }=(AB{)}^{\ast })\\ & ={\lambda }^{\ast }⟨{\psi }^{\ast }|{\psi }^{\ast }⟩\end{array}\end{array}$$

$$\begin{array}{}\text{(2)}& \begin{array}{rlrl}{(T|\psi ⟩)}^T|{\psi }^{\ast }⟩& ={(\lambda |\psi ⟩)}^T|{\psi }^{\ast }⟩& & (T|\psi ⟩=\lambda |\psi ⟩)\\ & =\lambda |\psi {⟩}^T|{\psi }^{\ast }⟩& & \left(\lambda \text{ is a number}\right)\\ & =\lambda ⟨{\psi }^{\ast }|{\psi }^{\ast }⟩& & (|\psi {⟩}^T=|{\psi }^{\ast }{⟩}^{†}=⟨{\psi }^{\ast }|)\end{array}\end{array}$$

Because

$|\psi ⟩\ne 0$, by comparing (1) and (2), we get

$\lambda ={\lambda }^{\ast }$, hence

$\lambda \in \mathbb{R}$.

Hermitian operator's expected value is real

$$\begin{array}{rlrl}⟨\psi |T|\psi ⟩& =⟨\psi |T|\psi {⟩}^T& & (⟨\psi |T|\psi ⟩\text{ is a number})\\ & ={\left(\left[\begin{array}{cc}{v}_1^{\ast }& v_2^{\ast }\end{array}\right]T\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\right)}^T& & (\text{represent }|\psi ⟩as\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right])\\ & =\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]T^T\left[\begin{array}{c}{v}_1^{\ast }\\ v_2^{\ast }\end{array}\right]& & ((ABC{)}^T=C^T{B}^T{A}^T)\\ & =\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]T^{\ast }\left[\begin{array}{c}{v}_1^{\ast }\\ v_2^{\ast }\end{array}\right]& & (T^{†}=T)\\ & =⟨{\psi }^{\ast }|T^{\ast }|{\psi }^{\ast }⟩\\ & =⟨\psi |T|\psi {⟩}^{\ast }& & (A^{\ast }{B}^{\ast }=(AB{)}^{\ast })\end{array}$$

$\because ⟨\psi |T|\psi ⟩=⟨\psi |T|\psi {⟩}^{\ast }\therefore ⟨\psi |T|\psi ⟩\in \mathbb{R}$

Q.E.D.

Trace is sum of eigenvalues

Refer to Shayan Oveis Gharan's lecture note: Lecture 10: Linear Algebra Background on P.10-3.

tags: Quantum Computing Solution