# von Neumann entropy in a limiting case

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 $$\log{3}$$.

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 $$n \to \infty$$