0
$\begingroup$

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 $H$.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

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 $\lvert\psi\rangle\lvert E_{2i}\rangle$.

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$
    – EE18
    Aug 27, 2023 at 15:08
  • $\begingroup$ @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)? $\endgroup$ Aug 27, 2023 at 15:15
  • $\begingroup$ 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. $\endgroup$
    – EE18
    Aug 27, 2023 at 15:35
  • $\begingroup$ @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. $\endgroup$ Aug 27, 2023 at 17:06
  • $\begingroup$ 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. $\endgroup$
    – EE18
    Aug 27, 2023 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.