6
$\begingroup$

Let $H_1$ and $H_2$ be Hamiltonians on Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$. My question in general concerns how one would form a Hamiltonian on the tensor product Hilbert space $$\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2.$$ The most obvious way, to me, is to take the tensor product of Hamiltonians: $$H_t = H_1 \otimes H_2.$$ However, I find that, in most physical applications, like when one studies open quantum systems or many body physics, usually people take the Kronecker sum as the Hamiltonian on the tensor product space: $$H_k = H_1\otimes I_2 + I_1\otimes H_2,$$ where $I_{1,2}$ are identity operators on $\mathcal{H}_{1,2}$.

I am aware of the mathematical differences between these two constructions, but my question is: can someone provide physical intuition for how these Hamiltonians differ, and, under what circumstances would I use one over the other?

$\endgroup$
4
  • $\begingroup$ The two represent different things. The first represents an interaction term between two quantum systems, and the second represents the free Hamiltonians of each system separately. (So, e.g. the second might be the kinetic and potential energies of two individual oscillators, and then the first might represent the interaction Hamiltonian if, for instance, the two oscillators are connected by a spring). $\endgroup$
    – march
    Commented Oct 6, 2021 at 17:32
  • $\begingroup$ This is part of what confuses me. If I fix a non-entangled state $|\Psi\rangle = |\psi_1\rangle|\psi_2\rangle$, then $H_t|\Psi\rangle$ is non-entangled, but $H_k|\Psi\rangle$ might be entangled. So $H_t$ describes an interacting system, but evolution by $H_t$ cannot cause entanglement, and $H_k$ describes a non-interacting system, but evolution by $H_k$ can cause entanglement. It seems by this argument that $H_t$ should describe a non-interacting system and $H_k$ should describe an interacting system? $\endgroup$ Commented Oct 6, 2021 at 17:46
  • $\begingroup$ I see! I think I can sort that out for you in a simple way. Give me a couple minutes, and I'll write an answer. $\endgroup$
    – march
    Commented Oct 6, 2021 at 17:51
  • $\begingroup$ @J.Murray Ah yes I made a mistake there. I've edited that out. Thanks for the clarification. $\endgroup$ Commented Oct 6, 2021 at 19:11

1 Answer 1

7
$\begingroup$

The two different Hamiltonian forms represent different physical things. The first, given by $$ \hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1}, $$ represents the free Hamiltonians of each system separately, and the second, given by $$ \hat{H}_1\otimes\hat{H}_2\tag{2}, $$ represents an interaction term between two quantum systems. (So, e.g. the first might be the kinetic and potential energies of two individual oscillators, and then the second might represent the interaction Hamiltonian if, for instance, the two oscillators are connected by a spring).

Importantly, Hamiltonians of the form (1) evolve unentangled states into unentangled ones, but Hamiltonians of the form (2) can evolve unentangled states into entangled ones. To see this, consider the following calculations.


Consider two quantum systems described by Hamiltonians $H_1$ and $H_2$, whose eigensystems are separately described by $$ \hat{H}_1|\psi_n\rangle = \epsilon_n|\psi_n\rangle,~~~~~\hat{H}_2|\phi_n\rangle = \mu_n|\phi\rangle. $$ Let's consider case 1 first, in which the combined Hamiltonian of the combined system happens to be \begin{equation} \hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1} \end{equation} Then, the eigenstates of this operator are the the product of the eigenstates of the individual Hamiltonians, and the eigenvalues are sums of the individual eigenvalues, which we can show by computing $$ \left(\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right) =(\epsilon_n+\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right). $$ (The details involve just using linearity.) Now, given an arbitrary unentangled initial state $|\Psi(0)\rangle$, it can be written as the product of vectors expanded in the individual energy eigenbases as $$ |\Psi(0)\rangle = \left(\sum_{n}a_{n}|\psi_n\rangle\right)\otimes\left( \sum_{m}b_n|\phi_m\rangle\right) = \sum_{nm}a_nb_m \left(|\psi_n\rangle\otimes|\phi_m\rangle\right). $$ We can then get the full time-dependence by attaching the exponential factors to the eigenvectors in the usual way, yielding $$ |\Psi(t)\rangle = \sum_{nm}e^{-i(\epsilon_n+\mu_m)t/\hbar}a_nb_m \left(|\psi_n\rangle\otimes|\phi_m\rangle\right). $$ Crucially, this state factors as $$ |\Psi(t)\rangle=\left(\sum_{nm} e^{-i\epsilon_nt/\hbar} a_n|\psi_n\rangle\right)\otimes\left( \sum_{m} e^{-i\mu_mt/\hbar} b_m|\phi_m\rangle\right), $$ and so if the system started unentangled, it remains unentangled. This is purely a consequence of the fact that the Hamiltonian is a sum of single-system Hamiltonians because this is what leads to the eigenenergies being sums of the individual ones, which allows us to factor the exponential.

Now, to see that that doesn't work in the case of a Hamiltonian of the second form, given by $$ \hat{H}_1\otimes\hat{H}_2\tag{2}, $$ we first note that that the product of eigenvectors is still an eigenvector, but the eigenvalues are now products of the individual eigenvalues, i.e., $$ \left(\hat{H}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right) =(\epsilon_n\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right). $$ If we again start with the initially unentangled state shown above, then the full time-dependent state is given by $$ |\Psi(t)\rangle = \sum_{nm}e^{-i(\epsilon_n\mu_m)t/\hbar}a_nb_m \left(|\psi_n\rangle\otimes|\phi_m\rangle\right), $$ which can no longer be factored in general!


This can also be seen by exponentiating the Hamiltonians directly to get the unitary time-evolution operators. For Hamiltonians of the form (1), the time-evolution operator is $$ U_1(t) = \exp\left( -\frac{it}{\hbar} (\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2) \right) = \exp\left( -\frac{it}{\hbar} \hat{H}_1\otimes\hat{I}_2 \right) \exp\left( -\frac{it}{\hbar} \hat{I}_1\otimes\hat{H}_2 \right), $$ which is allowed because the two operators commute with each other. Furthermore, it is relatively straight-forward to show that this can be written as $$ U_1(t) = \left( \exp\left( -\frac{it}{\hbar} \hat{H}_1 \right) \otimes\hat{I}_2\right) \left(\hat{I}_1\otimes \exp\left( -\frac{it}{\hbar} \hat{H}_2 \right)\right) = \exp\left( -\frac{it}{\hbar} \hat{H}_1 \right) \otimes \exp\left( -\frac{it}{\hbar} \hat{H}_2 \right). $$ Thus, this time-evolution operator can then be seen to "conserve unentangled-ness". The other one does not factor in that same way.

$\endgroup$
7
  • $\begingroup$ Thanks for the great answer! So it seems that the unitaries show opposite behaviour w.r.t. entanglement than the Hamiltonians, since, for $|\Psi(0)\rangle$ unentangled, we have: $H_t|\Psi(0)\rangle$ is unentangled while $U_t|\Psi(0)\rangle$ is entangled, whereas $H_k|\Psi(0)\rangle$ is entangled while $U_k|\Psi(0)\rangle$ is unentangled. But I guess it makes more sense to look at the unitaries when studying how the Hamiltonian entangles states. $\endgroup$ Commented Oct 6, 2021 at 19:10
  • 1
    $\begingroup$ That seems right to me! Hamiltonians on their own can't entangle states, because the process of entanglement clearly requires some physical process that takes some amount of time. So studying the time-evolution unitaries is necessary. In any case, the action of an arbitrary operator on a quantum state doesn't correspond to a physical process unless the operator is a (unitary) time-evolution operator or the operator (which actually generally acts on density matrices) is a member of a POVM (in which case, the operator corresponds to a measurement process.) $\endgroup$
    – march
    Commented Oct 6, 2021 at 19:18
  • $\begingroup$ The tensor product does a product of eigenvalues, whereas tensor sum does a sum (note : the operator (matrix) sum does not yields sum of eigenvalues). Does it mean that If one sums kinetic and potential energies eigenvalues it reads : $(-\frac{\hbar^2\partial^2}{2m\partial x^2}+V(y))u(x,y)=Eu(x,y)$ for a single particle ? $\endgroup$ Commented Jan 6, 2023 at 15:32
  • $\begingroup$ @Cretin2 I do not understand your question, exactly. $\endgroup$
    – march
    Commented Jan 6, 2023 at 16:58
  • 1
    $\begingroup$ @QuantumPotatoïd I still don't really understand your question, although I'm perhaps close to understanding it. I think you should just ask a real question here, and expand on your thinking. The comments are not a good place to follow up on an already-answered question. $\endgroup$
    – march
    Commented Nov 24, 2023 at 23:32

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.