# Finding a complete eigenbasis for an "entangled" Hamiltonian?

Suppose we have a tensor product Hilbert space $$\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$$ and we have a Hamiltonian defined thereon which is given by $$H = H_{1e} \otimes I+ H_1 \otimes H_2$$. Suppose also we know an eigenbasis $$|E_{2i} \rangle$$ of $$H_2$$. What can we say then about the overall eigenbasis of $$H$$? Can we "try" eigenstates of the form $$|\psi \rangle |E_{2i} \rangle$$, and so find that $$H|\psi \rangle |E_{2i} \rangle = (H_{1e}+ E_{2i}H_1\otimes I)|\psi \rangle |E_{2i} \rangle$$ and then solve this eigenvalue problem in $$\mathcal{H}_1$$ by considering the eigenstates $$|\psi \rangle$$ of $$H_{1e}+ E_{2i}H_1$$ in $$\mathcal{H_1}$$ (we see immediately that $$|\psi \rangle |E_{2i} \rangle$$ will be an $$H$$ eigenstate in this case)? My chief concern, and the gist of this question, is how do I know I am not missing any possible $$H$$ eigenstates with this "trial" method?

For example/context, in the case of a fully separable Hamiltonian, say $$H = H_{1e} \otimes I+ I \otimes H_{2e}$$, one can verify by explicit calculation that the product basis $$|E_{1j} \rangle|E_{2i} \rangle$$ are eigenstates. Then (and this is the key step which I don't know above), one can use the particular definition of a tensor product space which says that it is the vector space spanned by any product basis (a basis formed by products of bases of each $$\mathcal{H}_i$$). Thus, since the $$|E_{1j} \rangle$$ and $$|E_{2i} \rangle$$ are eigenkets of the self-adjoint operators $$H_{1e}$$ and $$H_{2e}$$, respectively, we have a complete eigenbasis of $$H$$.

In case this question is not clear, it is motivated by Problem 7.10 in Ballentine's quantum text, which reads:

Two spin 1/2 particles interact through the spin-dependent potential $$V(r)$$ = $$V_1(r)+V_2(r)σ(1)·σ(2)$$. Show that the equation determining the bound states can be split into two equations, one having the effective potential $$V_1(r)+V_2(r)$$ and the other having the effective potential $$V_1(r)-3V_2(r)$$.

Of course, the idea here is that we want to use the eigenkets of $$σ(1)·σ(2)$$ which were obtained in an earlier problem, but my concern is around using trial eigenstates of the form $$|{\psi_0}\rangle|{1/2,1/2,0,0}\rangle$$ and $$|{\psi_1}\rangle|{1/2,1/2,1,0/\pm 1}\rangle$$ ($$|j_1,j_2,J,M\rangle$$ is my notation for total angular momentum eigenstates).

P.S. I would greatly appreciate it if you can comment on why Ballentine mentions bound specifically ("that the equation determining the bound states can be split into two equations" -- i.e. why bound in particular) when this seems to be general, whether we are looking for bound or unbound states of $$H$$.

Consider the problem in the eigenbasis $$\lvert E_{2i}\rangle$$ of $$H_2$$. Then, the Hamiltonian is block-diagonal in the basis of system $$2$$, with the blocks being $$H_{1e}+E_{2i} H_1$$. By diagonalizing this operator, you can now find the eigenvectors and eigenvalues of the full Hamiltonian; in particular, the eigenvectors will indeed be of the form $$\lvert\psi\rangle\lvert E_{2i}\rangle$$.

• I am still not sure I understand, I'm afraid. You say "in the eigenbasis of $\lvert E_{2i}\rangle$", but what does that mean here when $\lvert E_{2i}\rangle$ is only an eigenstate of $\mathcal{H}_2$ but not $\mathcal{H}$? I do agree that by using the trial method I outline above, we get eigenkets $|\psi_i \rangle |E_{2i} \rangle$ where the $|\psi_i \rangle$ solve the "effective problem" which drops out once $E_{2i}$ emerges as a scalar. But I'm not sure I understand why these eigenvectors are complete. Why needn't there be any entangled eigenvectors in this basis?
– EE18
Commented Aug 27, 2023 at 15:08
• @EE18 Are you familiar how to express operators in a tensor product of two Hilbert spaces (i.e. that you get a block-matrix, where each block corresponds to a basis element of one space (here H2), and describes an operator on the other space (here H1)? Commented Aug 27, 2023 at 15:15
• I don't think so no, and perhaps that's the part of your argument which is going above my head. I mean I'm familiar with how $H_1 \otimes H_2$ acts on the tensored Hilbert space component-wise, but I don't think that's what you're referring to? Would you be able to include mention of that in your answer, or perhaps just point to a reference to consult? My textbooks have never mentioned this (I've only ever encountered dotted spin operators $\mathbf{J}_1 \cdot \mathbf{J}_2$ which are more easily solved for because of the decomposition in terms of the total angular momentum operator.
– EE18
Commented Aug 27, 2023 at 15:35
• @EE18 Maybe an easier way to see that this works is by counting the number of eigenvectors you get: It precisely matches the dimension of the Hilbert space, so at least for the finite dimensional case this shows you that you obtained all eigenstates. Commented Aug 27, 2023 at 17:06
• How would you do this counting? $\mathcal{H}$ (and $H$) here is completely general (even if we say finite-dimensional). If $\dim \mathcal{H}_2 = n_2$ we see that this trial method yields us $n_2$ equations. Let $\dim \mathcal{H}_1 = n_1$. Why should we be certain that we have $n_1$ solutions to each of these $n_2$ equations? Does one perhaps appeal to each of the "effective" Hamiltonians being Hermitian operators on $\mathcal{H}_1$, and therefore having $n_1$ eigensolutions? I suppose I can agree to that.
– EE18
Commented Aug 27, 2023 at 17:09