# Multi-part Hamiltonian results in special properties of eigenfunctions?

Suppose we have a Hamiltonian that is the sum of two other Hamiltonians, $\hat{H}=\hat{H}_1+\hat{H}_2$, and that they all have a discrete (at least partially) spectrum. Is it necessary for the eigenfunctions of $\hat{H}$ to be products of eigenfunctions of $\hat{H}_1$ and $\hat{H}_2$? That it is sufficient is trivial - let $\psi=\psi_1\psi_2$, where $\psi$, $\psi_1$, and $\psi_2$ are eigenfunctions of the $\hat{H}$, $\hat{H}_1$, and $\hat{H}_2$ Hamiltonians respectively.

$$\hat{H}\psi=E\psi$$

$$\hat{H}\psi=(\hat{H}_1 + \hat{H}_2)\psi_1\psi_2=(E_1+E_2)\psi$$

However, I'm not too certain about its necessity. Do you have any hints on how to show this?

EDIT: Here is the exact quote from L.L. I'm learning this material right now, so please forgive my ignorance. I'm sure I've misunderstood/misread something.

If the Hamiltonian of the system is the sum of two (or more) pasrts, $\hat{H}=\hat{H}_1+\hat{H}_2$, one of which contains only the coordinates $q_1$ and the other only the coordinates $q_2$, then the eigenfunctions of the operator $\hat{H}$ can be written down as products of the eigenfunctions of the operators $\hat{H}_1$ and $\hat{H}_2$, and the eigenvalues of the energy are equal to the sums of the eigenvalues of these operators.

Source: "Quantum Mechanics (3rd ed.)", Landau & Lifshitz, section 10 page 29

No, for a number of reasons. Look at it from a discrete sense. Hamiltonians are Hermitian matrices. There are possibly infinite ways of writing a Hermitian matrix as a sum of two others. Also there is no product operation between two vectors. You can do an inner product, an outer product or even a tensor product. You can't just multiply them together like numbers.

Answer to the edit: The key part is the quote "one of which contains only the coordinates $q_1$ and the other only the coordinates $q_2$". Then you can write the Hamiltonains as $H_1(q_1)$ and $H_2(q_2)$. And their corresponding eigenvectors as $\psi_1(q_1)$ and $\psi_2(q_2)$. $\psi_1(q_1)\psi_2(q_2)$ is the direct(tensor) product of the two. $$H_1(q_1)\psi_1(q_1)\psi_2(q_2) = E_1\psi_1(q_1)\psi_2(q_2).$$ As $H_1(q_1)$ cannot affect functions in variable $q_2$. Similarly for $H_2(q_2)$.

• I've edited my question. Feb 18, 2016 at 7:01

Sufficiency is clear. Let's do necessity.

Let's expand the eigenvector $|E\rangle$ of $\hat{H}$ in the product basis of eigenvectors of $\hat{H}_1$ and $\hat{H}_2$, i.e. $$|E\rangle = \sum_{n,m}a_{nm}|n\rangle_1|m\rangle_2$$ and act with $\hat{H}$. By assumption, $$E|E\rangle = \hat{H}|E\rangle = (\hat{H}_1+\hat{H}_2)|E\rangle.$$ On the left-hand side, we can write $$E|E\rangle= \sum_{n,m}Ea_{nm}|n\rangle_1|m\rangle_2,$$ and on the right-hand side, we can write $$(\hat{H}_1+\hat{H}_2)|E\rangle = \sum_{n,m}a_{nm}(\hat{H}_1+\hat{H}_2)|n\rangle_1|m\rangle_2 = \sum_{n,m}a_{nm}(E^{(1)}_n+E^{(2)}_m)|n\rangle_1|m\rangle_2,$$ where we have used the fact that we have expanded our eigenstate in the product basis made from eigenstates of the Hamiltonian parts.

Now, by orthonormality, we can equate coefficients which means that $a_{nm}(E^{(1)}_n+E^{(2)}_m) = a_{nm}E.$ This implies that either $a_{nm} = 0$ (in which case we can neglect that term in the expansion anyway) or that $E^{(1)}_n+E^{(2)}_m = E$, which completes the proof.

• Comment to the answer (v1) of the question (v2): The product states make sense in L&L's setting, but not necessarily in OP's setting. Feb 20, 2016 at 0:54