Separability of Hamiltonian and Factorization of Wavefunction

Question

In Shankar's QM book Chapter 10 pg. 274, it was said that quantum mechanically, the separability of the hamiltonian $$H=H_1(x_1, p_1)+H_2(x_2,p_2)$$ leads to the factorization of the wave function: $$\psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2).$$

How can this be shown to be true?

I know that for a separable Hamiltonian, the energy eigenvalue equation $$(\hat{H_1}+\hat{H_2})\psi_E(x_1,x_2)=E\psi_E(x_1,x_2)$$ gives energy solutions which are separable: $$\psi_E(x_1,x_2)=\psi_{E_1}(x_1)\psi_{E_2}(x_2).$$

The general wavefunction will hence be a superposition of the separable energy eigenfunctions $\psi_E$:

$$\psi(x_1,x_2)=\sum_{E_1+E_2=E}\psi_{E_1}(x_1)\psi_{E_2}(x_2)$$

How do I proceed to show that the above expression can be written as

$$\sum_{E_1+E_2=E}\psi_{E_1}(x_1)\psi_{E_2}(x_2)=\psi_1(x_1)\psi_2(x_2)?$$

Answer 1

I think your interpretation of Shankar's statement is too strong and that cannot be true. Shankar's statement is that separable states make up a complete eigenbasis of the Hamiltonian $H$. Your interpretation is that every eigenstate of $H$ is a separable state. The two statements are equivalent only if the spectrum of $H$ is non-degenerate. Otherwise, you are free to add up two separable eigenstates with the same energy to make up an entangled eigenstate, which cannot be separable in any way. (Quantum entanglement is a basis-independent statement. There is no way to separate an entangled state)

Actually a simpler way to see this is to consider the trivial case $H_1=H_2=0$. Then any state, either entangled or unentangled, is an eigenstate. However, it is still possible to find a complete basis of $H=0$ which only consists of unentangled state (that is, the total Hilbert space is a tensor product of the two Hilbert spaces).