2
$\begingroup$

I am stuck with a question from the book Quantum theory by Asher Peres.

Excercise (9.11):

Three different preparation procedures of a spin 1/2 particle are represented by the vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\frac{1}{2} \begin{pmatrix} -1\\ \pm \sqrt{3} \end{pmatrix} $. If they are equally likely, the Shannon entropy is $\log{3}$, and the von Neumann entropy is $\log{2}$. Show that if there are $n$ such particles, all prepared in the same way, the von Neumann entropy asymptotically tends to $\log{3}$ when $n \to \infty$.

Hint: Consider three real unit vectors making equal angles: $\langle u_i,u_j \rangle = c $ if $ i \neq j$ . Show that the eigenvalues of $\sum u_i u_i^\dagger$ are 1-c, 1-c and 1+2c."

The Shannon entropy can be easily calculated to be $\log{3}$. The density matrix $ \hat\rho$ comes out to be $$\begin{pmatrix} \frac{1}{2} & 0\\ 0 & \frac{1}{2} \end{pmatrix}. $$ Therefore, the von Neumann entropy also comes out to be $\log{2}$. However, in the second part, I am not able to get von Neumann entropy equal to $\log{3}$.

$\endgroup$
3
$\begingroup$

Let us first prove the hint.

Consider three $d$- dimensional unit vectors $u_i$ and define $ A = \sum_{i=1}^{3} u_i u_i^{\dagger} $.

A simple calculation shows that \begin{align*} A \big(u_1 + u_2 + u_3\big) & = (u_1 + c\cdot u_2 + c\cdot u_3) + (c\cdot u_1 + u_2 + c\cdot u_3) + ( c\cdot u_1 + c\cdot u_2 + u_3) \\ &= (1 + 2c) \cdot \big(u_1 + u_2 + u_3\big) \end{align*} meaning $ \xi = u_1 + u_2 + u_3 $ is an eigenvector of $ A $ with $ 1 + 2c $ eigenvalue.

A similar calculation shows that $$ A \big(u_1 - u_2\big) = (1 - c) \cdot \big(u_1 - u_2\big), \hspace{1.5em} A \big(u_1 - u_3\big) = (1 - c) \cdot \big(u_1 - u_3\big) $$ Thus we have found 3-linear independent eigenvectors with eigenvalues $ 1-c, 1-c, 1+2c $. The other $ d - 3 $ eigenvalues are, of course, zero with eigenvectors orthogonal to $ V = \text{span}\{u_1, u_2, u_3\} $.

This means that the Von-Neumann entropy of the density matrix $ \rho = \frac{1}{3} \sum_{i=1}^{3} u_i u_i^{\dagger} $ is \begin{align*} S_{\rho} = &- 2 \cdot \frac{1 - c}{3} \cdot \text{log}\big( \frac{1 - c}{3} \big) - \frac{1 + 2c}{3} \cdot \text{log}\big( \frac{1 + 2c}{3} \big) \\= &- 2 \cdot \frac{1 - c}{3} \cdot \text{log}\big(1 - c\big) - \frac{1 + 2c}{3} \cdot \text{log}\big(1 + 2c\big) + \text{log}(3) \end{align*} and so $ S_{\rho} \to \text{log}(3) $ if $ c \to 0 $.

Why is this enough?

Because for n particles the states are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}^{\otimes n} $, $\frac{1}{2^n} \begin{pmatrix} -1\\ \pm \sqrt{3} \end{pmatrix}^{\otimes n} $ with dot product $ v_i^{\dagger} v_j = \big(-\frac{1}{2}\big)^n $ for $ i \neq j $, so $ c \to 0 $ as $ n \to \infty $

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.