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I am computing the von Neumann entropy associated with a density operator $\hat{\rho}$ which is defined for a pure state; that is, $\hat{\rho}^2=\hat{\rho}$. Besides, we compute this entropy in terms of the continuous eigenbasis of the position operator, that is,

$$S(\hat \rho)=- \text{Tr}\left[\hat\rho \log \hat\rho \right]=-\int_{-\infty}^{+\infty} dx~\left\langle x \right| \hat\rho \log \hat\rho \left| x\right\rangle, \tag{1}$$

Where $ \left\lbrace \left| x\right\rangle\right\rbrace_{x\in \mathbb{R}}$ is the continuous eigenbasis of the position operator $\hat{X}$ (that is, $\hat X \left| x\right\rangle =x \left| x\right\rangle$ ) and we have used $\text{Tr}[\hat{A}]=\int_{-\infty}^{+\infty}dx~\left\langle x \right| \hat{A} \left| x\right\rangle$ as the trace of $\hat{A}$ in such basis. Then, we expand the logarithm in series; to get $$\hat \rho \log \hat \rho = \hat \rho \sum_{n=1}^\infty \frac{1}{n} (\mathbb{\hat{I}} - \hat \rho)^n. \tag{2}$$

Now, it must be noted that the identity operator $\hat{\mathbb{I}}$ commute with $\hat{\rho}$. Besides, $\mathbb{\hat{I}}^{k}=\mathbb{\hat{I}}$ for $k\geq 0$ and $\hat{\rho}^k=\hat{\rho}$ for $k\geq 1$ (since $\hat{\rho}^2=\hat{\rho}$). Then, using this facts, we expand $(\mathbb{\hat{I}} - \hat \rho)^n$ in a binomial series of the form:

$$(\mathbb{\hat{I}} - \hat \rho)^n= \sum_{k=0}^{n} \binom n k \mathbb{\hat{I}}^{n-k}(-\hat{\rho})^k= \mathbb{\hat{I}}+\hat{\rho}\sum_{k=1}^{n} \binom n k (-1)^k. \tag{3} $$

Substituting Eq. (3) in Eq. (2), we have

$$\hat \rho \log \hat \rho =\sum_{n=1}^\infty \frac{1}{n}\left[ \hat{\rho}+\hat{\rho}^{2}\sum_{k=1}^{n} \binom n k (-1)^k\right]. $$

$$~~~~~~~~~=\sum_{n=1}^\infty \frac{1}{n}\left[ \hat{\rho}+\hat{\rho}\sum_{k=1}^{n} \binom n k (-1)^k\right]$$

$$~~~~~~~~~=\hat{\rho}\sum_{n=1}^\infty \frac{1}{n}\left[ 1+\sum_{k=1}^{n} \binom n k (-1)^k\right]$$

$$~~~~~=\hat{\rho}\sum_{n=1}^\infty \frac{1}{n}\left[\sum_{k=0}^{n} \binom n k (-1)^k\right]. \tag{4}$$

Where in the second line we use again the fact $\hat{\rho}^2=\hat{\rho}$. Now, using the definition for binomial coeficient $\binom n k = \frac{n!}{k! (n-k)!}$, the Eq. (4) becomes

$$\hat \rho \log \hat \rho = \hat{\rho}\sum_{n=1}^\infty \sum_{k=0}^{n} \frac{(n-1)!}{k! (n-k)!} (-1)^k; \tag{5}$$

where we use $n!/n=(n-1)!$. Therefore, by substituting Eq. (5) in Eq. (1), we deduce

$$S(\hat \rho)= -\sum_{n=1}^\infty \sum_{k=0}^{n} \frac{(n-1)!}{k! (n-k)!} (-1)^k \int_{-\infty}^{+\infty} dx~\left\langle x \right| \hat\rho \left| x\right\rangle. \tag{6}$$

Then, I have one question:

(1) There exists some definition for the series $\sum_{n=1}^\infty \sum_{k=0}^{n} \frac{(n-1)!}{k! (n-k)!} (-1)^k $?

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  • $\begingroup$ Can't you evaluate the summations at the end? $\endgroup$ Mar 29, 2023 at 2:59
  • $\begingroup$ The trace of the density matrix is always 1. $\endgroup$ Mar 29, 2023 at 3:01
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    $\begingroup$ After testing some examples, it sure seems like that inner sum $\sum_{k=0}^n(\cdot)$ is equal to zero. And, well, I think the entropy of a pure state should be 0, so that tracks. $\endgroup$
    – march
    Mar 29, 2023 at 3:45
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    $\begingroup$ @march Yeap, you have the reason. What in fact I have proved is that von Neumann entropy is zero for pure states. $\endgroup$ Mar 29, 2023 at 4:02
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    $\begingroup$ It a pretty long way of showing that $\log(1)=0$, but I like it! :) $\endgroup$
    – hft
    Mar 29, 2023 at 4:17

1 Answer 1

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I am computing the von Neumann entropy...

$\hat{\rho}^2=\hat{\rho}$...

Substituting Eq. (3) in Eq. (2), we have...

$$\hat \rho \log \hat \rho =\sum_{n=1}^\infty \frac{1}{n}\left[ \hat{\rho}+\hat{\rho}^{2}\sum_{k=1}^{n} \binom n k (-1)^k\right]. $$ ...

$$~~~~~=\hat{\rho}\sum_{n=1}^\infty \frac{1}{n}\left[\sum_{k=0}^{n} \binom n k (-1)^k\right]. \tag{4}$$

...Then, I have one question:

(1) There exists some definition for the series $\sum_{n=1}^\infty \sum_{k=0}^{n} \frac{(n-1)!}{k! (n-k)!} (-1)^k $?

It's zero.

So too is the quantity in square brackets in your Eq. (4) equal to zero.

You can see this like: $$ 0 = 0^n = (1 - 1)^n = \sum_{k=0}^n\left(\begin{matrix}n\\ k\end{matrix}\right)(1)^{n-k}(-1)^k $$ $$ =\sum_{k=0}^n\left(\begin{matrix}n\\ k\end{matrix}\right)(-1)^k $$

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