Calculating entropies from purified state

Question

I am a bit stuck with the following problem and hope you can help me.
Consider the density operator $$rho_{AB} = sum_{i=0}^{3} p_i |Phi_irangle langle Phi_i|$$ where $|Phi_irangle$ is the Bell basis.
After some intermediate steps, that are not relevant for my question, I arrive at the purification $$|Psirangle_{ABE} = sum_{=0}^{3} sqrt{p_i} |Psi_irangle_{AB} otimes |epsilon_irangle_E$$ where $|epsilon_irangle_E$ is some orthonormal basis of the environment system $E$.
That can be reexpressed as $|Psirangle_{ABE}$ can be rewritten as $$|Psirangle_{ABE} = sum_{x,y=0}^{3} |x,yrangle otimes |w_{x,y}rangle$$ with
$$|w_{0,0}rangle := sqrt{frac{p_0}{2}} |epsilon_0rangle + sqrt{frac{p_1}{2}} |epsilon_1 rangle$$
$$|w_{0,1}rangle := sqrt{frac{p_2}{2}} |epsilon_2rangle + sqrt{frac{p_3}{2}} |epsilon_3 rangle$$
$$|w_{1,0}rangle := sqrt{frac{p_2}{2}} |epsilon_2rangle - sqrt{frac{p_3}{2}} |epsilon_3 rangle$$
$$|w_{1,1}rangle := sqrt{frac{p_0}{2}} |epsilon_0rangle - sqrt{frac{p_1}{2}} |epsilon_1 rangle$$
After performing some orthonormal measurement, I obtain the state:
$$rho_{XYE} = sum_{x,y=0}^{3} |xranglelangle x| otimes |yranglelangle y|otimes sigma_E^{x,y}$$ 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 $$rho_{XE} = sum_{x,y} |xrangle langle x| otimes sigma_E^{x,y} = begin{pmatrix}sigma_E^{0,0} + sigma_E^{0,1}&0&sigma_E^{1,0} + sigma_E^{1,1}end{pmatrix}$$
The (von Neumann) entropy reads $$H(rho_{XE}) = -Tr[rho_{XE} log_2(rho_{XE})] = Trleft[begin{pmatrix}left(sigma_E^{0,0} + sigma_E^{0,1}right) log_2left(sigma_E^{0,0} + sigma_E^{0,1}right)&0&left(sigma_E^{1,0} + sigma_E^{1,1}right) log_2left( sigma_E^{1,0} + sigma_E^{1,1}right)end{pmatrix}right]$$
The eigenvalues of $sigma_E^{0,0} + sigma_E^{0,1}$ are $$lambda = frac{1pm sqrt{1-4(sqrt{p_0p_3}+sqrt{p_1 p_2})^2}}{2}$$ 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.
Thank you for your help!

Quantum Mechanic · Answer

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 ${|epsilon_0rangle,|epsilon_1rangle,|epsilon_2rangle,|epsilon_3rangle}$ basis:
begin{aligned} frac{p_0}{2}|epsilon_0ranglelangleepsilon_0| +frac{p_1}{2}|epsilon_1ranglelangleepsilon_1| + frac{sqrt{p_0 p_1}}{2}|epsilon_0ranglelangleepsilon_1| + frac{sqrt{p_0 p_1}}{2}|epsilon_1ranglelangleepsilon_0|
+
frac{p_2}{2}|epsilon_2ranglelangleepsilon_2| +frac{p_3}{2}|epsilon_3ranglelangleepsilon_3| + frac{sqrt{p_2 p_3}}{2}|epsilon_2ranglelangleepsilon_3| + frac{sqrt{p_2 p_3}}{2}|epsilon_3ranglelangleepsilon_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 $sigma^{01}$:
$$lambdainleft(frac{p_0+p_1}{2},frac{p_2+p_3}{2},0,0right)=left(frac{p_0+p_1}{2},frac{p_0+p_1}{2},0,0right).$$ These come from solutions of equations such as $left(frac{p_0}{2}-lambdaright)left(frac{p_1}{2}-lambdaright)-frac{p_0p_1}{4}=0$.
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 $sigma^{00}+sigma^{01}$.
The rest of your formulas look correct, so using these corrected eigenvalues should fix your calculation.

Calculating entropies from purified state

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