von Neumann entropy in a limiting case

Question

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 $\log{3}$.

Answer 1

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 $