goropikari / solutionqcqinielsenchuang Goto Github PK

Solution for Quantum Computation and Quantum Information by Nielsen and Chuang

TeX 99.79% Dockerfile 0.21%

quantum-information quantum-computing quantum-mechanics

solutionqcqinielsenchuang's Introduction

Solution for Quantum Computation and Quantum Information

This is unofficial solution manual for "Quantum Computation and Quantum Information: 10th Anniversary Edition" (ISBN-13: 978-1107002173) by Nielsen and Chuang

There is no guarantee that these solutions are correct.

If you have some comment, please feel free to open an issue or a PR.

License

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Compile tex files with Docker

docker-compose build
docker-compose run qcqi

Solved exercises

Chapter 2

Chapter 8

Chapter 9

Chapter 11

Chapter 12

solutionqcqinielsenchuang's People

Contributors

Stargazers

Watchers

Forkers

tejasavinashshetty mathophilic chunde kun-wang c0x0c sunjinzhao xfyhero2017 shashanka300 jkginfinite esabo earlbabson haganemayumi xander23333 mastercljohnson qope diracmg3 rpplayground stilnat carollice peiyong-addwater qqamber ujjwalk reteleport yichao-liang fatcatzf dnzengou fetainerlolz techxpert99 shieldforever yukh1999 alphabstc kaihung nanaengo debrevitatevitae jacobha grand-liu zeroegg tlesaul2 kubradogmus clarkmiyamoto pjkm28 aoyuqc okuijpers sternparky dimedrolling ash-55 issy94 tanganke joelmathewc kafianan3822 jpmeng algouveia michealandy vinayananda hound77 jgec-winter-of-code yxg520 agusthya0786 lfmorale zhaoyilunnn davidsony112 jimyz13 calixphd hahamengbier joycelyx715 ssjaweid

solutionqcqinielsenchuang's Issues

The proof here for Exercise 2.32 appears to assume P^2=P implies P is a projector. I am not certain this is true (the converse is certainly true, a projector satisfies P^2=P as indicated by Exercise 2.16), but could there not be other non-projectors that also satisfy P^2=P ?

The second error in the Errata list is actually not a mistake.

The statement in the textbook can be proven correct using Exercise 9.4.

Exercise 2.81

I believe a more general treatment will consider repeated singular values. In that situation, it was useful for me to notice that unitary matrices form a group, so left multiplication by a unitary matrix on a unitary matrix is equal to right multiplication by some other unitary matrix, thus allowing us to convert U_a tensor U_b to I_a tensor U_r

9.14 Correction

In the box,

This is wrong. A positive operator makes sure that it has non-negative eigenvalues.

When performing $\sqrt A$, you need to sqrt the eigenvalues under the expression of spectral decomposition:

$$ \sqrt A = \sum_i \sqrt \lambda_i | i \rangle \langle i | $$

Exercise 2.64

The actual solution is not too hard (though I also struggled a bit).

Let $1 \le k \le m$ and denote
$H = span{\phi_1,...,\phi_m }$,
$H_k = span{\phi_1,...,\phi_{k-1}, \phi_{k+1},...,\phi_m }$.
Because of linear independence, subspace $H$ has dimension $m$ and $H_k$ has dimension $m-1$. Now denote $E_k = Proj(H) - Proj(H_k)$, where $Proj(H)$ denotes the (ortho)projection operator on the subspace $H$. Clearly, $E_k \ge 0$ and $E_k \neq 0$.
Also, denote $E_{m+1} = I - E_1-...-E_m$.
Now calculate $\langle \phi_i|E_k|\phi_i \rangle = \langle \phi_i|Proj(H)|\phi_i \rangle - \langle \phi_i|Proj(H_k)|\phi_i \rangle = 1 - \langle \phi_i|Proj(H_k)|\phi_i \rangle = 0$ if $i \neq k$ and $>0$ if $i=k$. This is what we needed.

Exercise 2.76

I'm not sure about the correctness of the formalism, but this is my main idea: extend the smaller bases with null coefficients in the expansion in order to reach a situation similar to that in the standard proof.

Exercise 2.63

We actually need to prove the existence of unitaries $U_m$, not just verify the correctness of the formula $M_m = U_m\sqrt{E_m}$.

So, the correct solution is something like this:

Suppose $M_m^\dagger M_m = E_m$.
If $E_m > 0$ then denote $U_m = M_m(\sqrt{E_m})^{-1}$. We can verify that $U_m$ is indeed unitary, because we can deduce that $U_m^\dagger U_m = I$.
If $E_m \ge 0$ then use limiting argument similar to this https://en.wikipedia.org/wiki/Cholesky_decomposition#Proof_for_positive_semi-definite_matrices
I.e. consider $E_m^\prime = E_m + \frac{1}{k}I$ and so on...

Exercise 2.72 (1) Correction

Hi, I believe in the solution given for 2.72 (1), the 1/2 factor in a and d is forgotten. So I believe it should be:
If $\rho$ is positive, all eigenvalues of $\rho$ should be non-negative.
\begin{align*}
\det (\rho - \lambda I) &= (a- \lambda) (b - \lambda) - |b|^2 = \lambda^2 - (a+d)\lambda + ad - |b^2| = 0\
\lambda &= \frac{(a+d) \pm \sqrt{(a+d)^2 - 4 (ad - |b|^2)}}{2}\
&= \frac{1/2 \pm \sqrt{1/4 - 4 \left(\frac{1 - r_3^2}{4} - \frac{r_1^2 + r_2^2}{4} \right)}}{2}\
&= \frac{1/2 \pm \sqrt{1/4 - (1 - r_1^2 - r_2^2 - r_3^2)}}{2}\
&= \frac{1/2 \pm \sqrt{|\vec{r}|^2-3/4}}{2}\
.\end{align*}
Because $|\vec{r}|^2 \le 1$, $\sqrt{|\vec{r}|^2-3/4} \le \sqrt{1/4} = 1/2$ if real, and $\lambda$ is non-negative.

Error p.408. eq (9.49)

The sum $\sum_i (p_i - q_i) = \sum_i p_i - \sum_i q_i = 1 - 1 = 0$? So the error needs to be fixed a different way?

Error in Exercise 2.71 Solution

Although the square of the density operator $\rho^2$ is positive, and thus permits a spectral decomposition, the eigenvalues associated with that spectral decomposition are not necessarily the squares of the original probabilities $p_i$ (used to construct the density operator), a assumption upon which the proof relies.

(This is because the spectral decomposition of the density operator has eigenvalues that do not necessarily coincide with the probabilities of the states used to construct the density operator, as shown in p.103 of the book.)

Chapter 1

Chapter 1 also contains 2 exercises. Although they are quite open in nature, I would have liked to see someone's attempt on answering them.