If we have some system $\rho_{AB}$ we can find the entanglement entropy as


But why are the entropies of the subsystems $\rho_A$ and $\rho_B$ equal?

  • $\begingroup$ What do you know? $\endgroup$ – Norbert Schuch Jan 12 at 16:30
  • $\begingroup$ @NorbertSchuch the trace of all are 1, they are all hermitian. Also, the entanglement entropy should give "the amount of bits" that are entangled, and so it makes sense that it is the same on both subsystems. But why is this the case? $\endgroup$ – James Jan 12 at 17:13
  • $\begingroup$ No, I mean: What do you know about bipartite pure quantum states? Or quantum information in general? Have you for instance heard about the Schmidt decomposition? I.e.: Someone who answers, where should they start with their explanation? $\endgroup$ – Norbert Schuch Jan 12 at 17:14
  • 1
    $\begingroup$ Good that you added the details to the argument! You should have also caught the mistake I made ;) $\endgroup$ – Norbert Schuch Jan 13 at 20:37

For a state $|\psi\rangle=\sum_{ij}C_{ij}|i\rangle|j\rangle$, the density matrix is

$$\rho=|\psi\rangle \langle \psi|=\sum_{ijkl}C_{ij}{C_{kl}}^*|ij\rangle\langle kl|$$

We find the reduced density matrix of $A$ by tracing out the $B$ subsystem $$\rho_A=\text{Tr}_B\left(\rho\right)=\sum_x {}_{B}\langle x|\sum_{ijkl}C_{ij}{C_{kl}}^*|ij\rangle\langle kl|x\rangle _B=\sum_{ikx}C_{ix}{C_{kx}}^*|i\rangle\langle k|=\sum_{ikx}C_{ix}{C^\dagger_{xk}}|i\rangle\langle k|\ .$$ Equivalently, for B we find $$\rho_B=\text{Tr}_A\left(\rho\right)=\sum_x {}_{A}\langle x|\sum_{ijkl}C_{ij}{C_{kl}}^*|ij\rangle\langle kl|x\rangle _A=\sum_{xjl}C_{xj}{C_{xl}}^*|j\rangle\langle l|=\sum_{xjl}C_{jx}^T{C_{xl}}^*|j\rangle\langle l|\ .$$

The reduced states for $A$ and $B$, respectively, are $\rho_A=CC^\dagger$ and $\rho_B=C^T C^*=(C^\dagger C)^*$. Now $CC^\dagger$ and $C^\dagger C$ have the same non-zero eigenvalues, and since they are real, also $C^TC^*$ has the same eigenvalues; and thus also the same entropy (which is a function of their eigenvalues).


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