# Entanglement entropy and reduced density matrix

I'm learning about the definition of von Neumann entanglement entropy

$$S(\rho_1)=-\text{Tr}[\rho_1\ln\rho_1]$$ where $$\rho_1$$ is the reduced density matrix $$\rho_1=\text{Tr}_2(\rho)$$.

I was confused that the entanglement entropy of a bipartite pure state is different if it is defined using the von Neumann entropy of the reduced density matrix of subsystem 2? i.e., is the following equation correct? $$S(\rho_1)=-\text{Tr}[\rho_1\ln\rho_1]=S(\rho_2)=-\text{Tr}[\rho_2\ln\rho_2]$$ If it's true, how to prove this?

• Hint: Make sure to understand that (and how) the von Neumann entropy only depends on the spectrum of the corresponding density matrix. Then, for a pure state, you can use results of the Schmidt decomposition (both reduced density matrices have the same non-zero eigenvalues) to show that the last equality holds true. Commented Oct 9, 2023 at 6:04
• Just to be very clear here: The equality in general is false for mixed states $\rho$ and one can construct easy counter examples! Commented Oct 9, 2023 at 9:49

## 1 Answer

Using Schmidt decomposition, we can generally write a bipartite pure state as $$|\psi\rangle=\sum_i a_i |\varphi_i\rangle\otimes|\phi_i\rangle$$ and the density matrix of the total system is $$\rho=\sum_{ij}a_ia^*_j|\varphi_i\rangle\otimes|\phi_i\rangle\langle\varphi_j|\otimes\langle\phi_j|$$ thus the reduced density matrices are $$\rho_1=\text{Tr}_2(\rho)=\sum_i|a_i|^2|\varphi_i\rangle\langle\varphi_i|$$ $$\rho_2=\text{Tr}_1(\rho)=\sum_i|a_i|^2|\phi_i\rangle\langle\phi_i|$$ since $$\langle\varphi_i|\varphi_j\rangle=\langle\phi_i|\phi_j\rangle=\delta_{ij}$$.

Thus the entanglement entropy is $$S(\rho_1)=S(\rho_2)=-\sum_i|a_i|^2\ln|a_i|^2$$