How to prove the following bosonic entanglement expression?

Based on the article given by J. L. Ball, I. Fuentes-Schuller, and F. P. Schuller, Phys. Lett. A 359, 550 (2006) had used the following expression of von-Neumann entropy

$$$$S = - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right)$$$$ where, $$$$| \gamma | ^ { 2 } \equiv \left| \beta _ { k } / \alpha _ { k } \right| ^ { 2 }$$$$ provided with, $$$$\varrho = | \overline { 0 } _ { - k } \overline { 0 } _ { k } \rangle \left\langle \overline { 0 } _ { k } \overline { 0 } _ { - k } |\right.$$$$ $$$${\label{eq34}} | \overline { 0 } \rangle = \sum _ { n = 0 } ^ { \infty } c _ { n } | n _ { k } n _ { - k } \rangle$$$$ $$$$c _ { 0 } = \sqrt { 1 - \left| \frac { \beta _ { k } } { \alpha _ { k } } \right| ^ { 2 } }$$$$ $$$$c _ { n } = \left( \frac { \beta _ { k } ^ { * } } { \alpha _ { k } ^ { * } } \right) ^ { n } c _ { 0 }$$$$

I tried to substitute by these expressions but without success. I appreciate your answers.

• Would you be able to type up your working so that it's possible to figure out what's gone wrong? Commented Jun 18, 2019 at 18:54
• Since the calculation takes long time to retype on Stackexchange, the last step I arrived is, I have to show that, $\sum_{n=0}^{\infty} x^n \log (x^n) = \log\left(x^{x/(1-x)^2}\right)$ Commented Jun 19, 2019 at 21:33
• And that is guaranteed if I didn't miss this, $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ Commented Jun 19, 2019 at 21:35

We have, \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right) = \log _ { 2 } \left( \frac { \gamma ^ { \gamma / \gamma -1 } } { 1 - \gamma } \right) \\&= - \left(\frac{\gamma}{1-\gamma} \log _ { 2 }\gamma + \log_{2}(1-\gamma) \right) \end{aligned} \label{a} \tag{1} where we shortened our expressions by, $$\gamma = \gamma_B^2$$.
On the other hand we believe that, \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = - \sum_{i=0}^{\infty} c_n^2 \log(c_n^2) = \sum_{i=0}^{\infty} \gamma^n(1-\gamma) \log_2\left(\gamma^n(1-\gamma)\right) \\&= \sum_{i=0}^{\infty} \gamma^n(1-\gamma) \left( \log_2\ (\gamma^n)+\log_2(1-\gamma )\right) \\&= (1-\gamma ) \sum_{i=0}^{\infty} \gamma^n \log_2\ (\gamma^n)+ (1-\gamma ) \log_2 (1-\gamma ) \sum_{i=0}^{\infty} \gamma^n \end{aligned} \label{b} \tag{2} Now, those sequences have the following closed form providing that, $$|\gamma|<1$$, \begin{aligned} \sum_{n=0}^\infty \gamma^n = \frac1{1-\gamma} \\ \text{and},\\ \sum_{n=0}^\infty \gamma^n\log \gamma^n &= \sum_{n=0}^\infty \gamma^n n \log \gamma=\sum_{n=0}^\infty \gamma \gamma^{n-1} n \log \gamma =\log \gamma\left(\gamma\sum_{n=0}^\infty \left(\gamma^n\right)'\right) \\&= \log \gamma\cdot \frac \gamma{(1-\gamma)^2}=\log \gamma^{\gamma/(1-\gamma)^2} \end{aligned} It follows that we can subsitute these results in eq.(\ref{b}) to obtain eq.(\ref{a})