# Calculating entropies from purified state

I am a bit stuck with the following problem and hope you can help me.

Consider the density operator $$\rho_{AB} = \sum_{i=0}^{3} p_i |\Phi_i\rangle \langle \Phi_i|$$ where $$|\Phi_i\rangle$$ is the Bell basis.

After some intermediate steps, that are not relevant for my question, I arrive at the purification $$|\Psi\rangle_{ABE} = \sum_{=0}^{3} \sqrt{p_i} |\Psi_i\rangle_{AB} \otimes |\epsilon_i\rangle_E$$ where $$|\epsilon_i\rangle_E$$ is some orthonormal basis of the environment system $$E$$.

That can be reexpressed as $$|\Psi\rangle_{ABE}$$ can be rewritten as $$|\Psi\rangle_{ABE} = \sum_{x,y=0}^{3} |x,y\rangle \otimes |w_{x,y}\rangle$$ with

$$|w_{0,0}\rangle := \sqrt{\frac{p_0}{2}} |\epsilon_0\rangle + \sqrt{\frac{p_1}{2}} |\epsilon_1 \rangle$$

$$|w_{0,1}\rangle := \sqrt{\frac{p_2}{2}} |\epsilon_2\rangle + \sqrt{\frac{p_3}{2}} |\epsilon_3 \rangle$$

$$|w_{1,0}\rangle := \sqrt{\frac{p_2}{2}} |\epsilon_2\rangle - \sqrt{\frac{p_3}{2}} |\epsilon_3 \rangle$$

$$|w_{1,1}\rangle := \sqrt{\frac{p_0}{2}} |\epsilon_0\rangle - \sqrt{\frac{p_1}{2}} |\epsilon_1 \rangle$$

After performing some orthonormal measurement, I obtain the state:

$$\rho_{XYE} = \sum_{x,y=0}^{3} |x\rangle\langle x| \otimes |y\rangle\langle y|\otimes \sigma_E^{x,y}$$ where $$\sigma_E^{x,y} = |w_{x,y} \rangle \langle w_{x,y}|$$. I have checked my results and they should be correct up to here. I am struggling with calculating entropies related to that state.

I need to show that $$H(\rho_{XE}) = 1 + b(p_0+p_1)$$ where $$b$$ is the binary entropy function, $$b(p) = -p \log_2(p) - (1-p) \log_2(1-p)$$ Therefore, I traced out system $$Y$$, obtaining $$\rho_{XE} = \sum_{x,y} |x\rangle \langle x| \otimes \sigma_E^{x,y} = \begin{pmatrix}\sigma_E^{0,0} + \sigma_E^{0,1}&0\\0&\sigma_E^{1,0} + \sigma_E^{1,1}\end{pmatrix}$$

The (von Neumann) entropy reads $$H(\rho_{XE}) = -Tr[\rho_{XE} \log_2(\rho_{XE})] = Tr\left[\begin{pmatrix}\left(\sigma_E^{0,0} + \sigma_E^{0,1}\right) \log_2\left(\sigma_E^{0,0} + \sigma_E^{0,1}\right)&0\\0&\left(\sigma_E^{1,0} + \sigma_E^{1,1}\right) \log_2\left( \sigma_E^{1,0} + \sigma_E^{1,1}\right)\end{pmatrix}\right]$$

The eigenvalues of $$\sigma_E^{0,0} + \sigma_E^{0,1}$$ are $$\lambda = \frac{1\pm \sqrt{1-4(\sqrt{p_0p_3}+\sqrt{p_1 p_2})^2}}{2}$$ and where I used $$\sum_{i=0}^{3} p_i = 1$$ The eigenvalues of $$\sigma_E^{1,0} + \sigma_E^{1,1}$$ are the same. I do not see how I can obtain $$H(\rho_{XE}) = 1 + b(p_0+p_1)$$ with these expressions for the eigenvalues. In particular, I do not see how I can get rid of $$p_2$$ and $$p_3$$, which do not occur in the claimed result.

I have similar problems with the other entropies I want to calculate. Hence, it might help to find my error here, then I hopefully find the correct expression for all the other entropies, too.

• Hi! I have slightly edited the formatting of your equations to improve (IMO) readability. Feel free to rollback if you wish. May 12, 2021 at 10:29
• On what system are you performing the orthonormal measurement? You seem to have converted your pure state to a mixed state by simply removing all of the off-diagonal elements of the mixed state. May 12, 2021 at 14:27
• @QuantumMechanic I perform measurement on system $X$ and on system $Y$. May 13, 2021 at 12:00

Perhaps it would be worth checking the eigenvalues that you obtained. For example, we can write $$\sigma^{00}+\sigma^{01}$$ in block-diagonal form in the $$\{|\epsilon_0\rangle,|\epsilon_1\rangle,|\epsilon_2\rangle,|\epsilon_3\rangle\}$$ basis: \begin{aligned} \frac{p_0}{2}|\epsilon_0\rangle\langle\epsilon_0| +\frac{p_1}{2}|\epsilon_1\rangle\langle\epsilon_1| + \frac{\sqrt{p_0 p_1}}{2}|\epsilon_0\rangle\langle\epsilon_1| + \frac{\sqrt{p_0 p_1}}{2}|\epsilon_1\rangle\langle\epsilon_0| + \frac{p_2}{2}|\epsilon_2\rangle\langle\epsilon_2| +\frac{p_3}{2}|\epsilon_3\rangle\langle\epsilon_3| + \frac{\sqrt{p_2 p_3}}{2}|\epsilon_2\rangle\langle\epsilon_3| + \frac{\sqrt{p_2 p_3}}{2}|\epsilon_3\rangle\langle\epsilon_2| =\left( \begin{array}{cccc} \frac{p_0}{2} & \frac{\sqrt{p_0} \sqrt{p_0}}{2} & 0 & 0 \\ \frac{\sqrt{p_0} \sqrt{p_0}}{2} & \frac{p_0}{2} & 0 & 0 \\ 0 & 0 & \frac{p_0}{2} & \frac{\sqrt{p_0} \sqrt{p_0}}{2} \\ 0 & 0 & \frac{\sqrt{p_0} \sqrt{p_0}}{2} & \frac{p_0}{2} \\ \end{array} \right).\end{aligned}
Since this is block-diagonal, its eigenvalues are the same as the eigenvalues of the blocks; i.e., the noneigenvalues of $$\sigma^{00}+\sigma^{01}$$ are the same as the nonzero eigenvalues of $$\sigma^{00}$$ and of $$\sigma^{01}$$: $$\lambda\in\left(\frac{p_0+p_1}{2},\frac{p_2+p_3}{2},0,0\right)=\left(\frac{p_0+p_1}{2},\frac{p_0+p_1}{2},0,0\right).$$ These come from solutions of equations such as $$\left(\frac{p_0}{2}-\lambda\right)\left(\frac{p_1}{2}-\lambda\right)-\frac{p_0p_1}{4}=0$$.
Similarly, we can express the other element in block-diagonal form: $$\sigma^{10}+\sigma^{11}=\left( \begin{array}{cccc} \frac{p_0}{2} & -\frac{\sqrt{p_0} \sqrt{p_0}}{2} & 0 & 0 \\ -\frac{\sqrt{p_0} \sqrt{p_0}}{2} & \frac{p_0}{2} & 0 & 0 \\ 0 & 0 & \frac{p_0}{2} & -\frac{\sqrt{p_0} \sqrt{p_0}}{2} \\ 0 & 0 & -\frac{\sqrt{p_0} \sqrt{p_0}}{2} & \frac{p_0}{2} \\ \end{array} \right),$$ which has the same eigenvalues as $$\sigma^{00}+\sigma^{01}$$.