# von Neumann entropy with continuous basis

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 $$\sum_{n=1}^\infty \sum_{k=0}^{n} \frac{(n-1)!}{k! (n-k)!} (-1)^k$$?

• Can't you evaluate the summations at the end? Mar 29, 2023 at 2:59
• The trace of the density matrix is always 1. Mar 29, 2023 at 3:01
• 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. Mar 29, 2023 at 3:45
• @march Yeap, you have the reason. What in fact I have proved is that von Neumann entropy is zero for pure states. Mar 29, 2023 at 4:02
• It a pretty long way of showing that $\log(1)=0$, but I like it! :)
– hft
Mar 29, 2023 at 4:17

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