# Separability of Hamiltonian and Factorization of Wavefunction

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

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

• Does this answer your question? Separable Hamiltonian systems in quantum mechanics Jun 23 at 9:55
• The answers under that post do not show that the eigenstates are factorizable, only that they are common eigenstates of $H_1(x_1,p_1)$ and $H_2(x_2,p_2)$ Jun 23 at 9:59
• I think Shankar was being vague and what he really meant was that the energy eigenfunctions are separable. In a few lines later, he wrote that symmetrical states like $|a\rangle \otimes |b\rangle + |b\rangle \otimes |a\rangle$ in general fails to be factorized. Jun 23 at 11:01

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).