So I understand, for example, the free particle propagator for a physical ket state in Hilbert space is evaluated by seeing how the unitary representation of time acts on the position kets etc... Now, suppose I have two particles, where I am considering a physical ket state in the tensor product Hilbert space, what is the way to go about finding the two-particle state propagator given also they strongly interact with one another (so they are not free)?


1 Answer 1


Assuming that you don't make any simplifying assumptions, you'll just have to solve the Schrodinger equation. You presumably know the two particle Hamiltonian, so you can solve for the propagator $U$ by writing the operator equation $$i\hbar\frac{\mathrm d}{\mathrm d t}\hat U = \hat H\hat U$$ and then going from there, just like you would for a single particle (except that the operators $\hat H$ and $\hat U$ now act on two-particle states).

Example: Two particles of mass $m$ in an external potential $\hat V_{\mathrm{ext}}$ with an interaction term $\hat V_{\mathrm{int}}$ would have a Hamiltonian of the form

$$\hat H = \frac{\hat P_1^2}{2m}+\frac{\hat P_2^2}{2m}+\hat V_\mathrm{ext}(\hat X_1) + \hat V_{\mathrm{ext}}(\hat X_2) + \hat V_{\mathrm{int}}(\hat X_1,\hat X_2)$$

and a propagator formally given by $$\hat U(t-t_0) = \exp\left[\frac{-i(t-t_0)}{\hbar}\hat H\right]$$

We could define $\hat H_i\equiv \frac{\hat P_i^2}{2m} + \hat V_{\mathrm{ext}}(\hat X_i)$, in which case the Hamiltonian would be

$$\hat H = \hat H_1 + \hat H_2 + \hat V_{\mathrm{int}}(\hat X_1,\hat X_2)$$

If $\hat V_\mathrm{int}$ were small, then we could treat it as a perturbation; in this case, we would start by solving the single-particle Schrodinger equations and then calculating successively smaller corrections. If not, then there's no general recipe for solving such a problem. You can search for eigenvalues and eigenstates like you would with any other operator, it's just that your differential equations will involve two position variables rather than one.

  • $\begingroup$ Where does that equality come from? Also, could you give me an example of two particle state with a kinetic term, a common potential and an interacting potential? I don’t understand how to formulate all this when particles are interacting. $\endgroup$
    – TheDawg
    Jan 28, 2021 at 15:33
  • $\begingroup$ @TheDawg That's the Schrodinger equation for the propagator, obtained by letting $|\psi(t)\rangle = U(t-t_0)|\psi_0\rangle$. I've updated my answer with an example "template" Hamiltonian for two interacting particles in an external potential. $\endgroup$
    – J. Murray
    Jan 28, 2021 at 15:53
  • $\begingroup$ So formally speaking, the individual hamiltonians act on their corresponding hilbert space through the tensor product of the two particle state, how does the interacting term interact in terms of its mapping from tensor Hilbert space to tensor Hilbert space? $\endgroup$
    – TheDawg
    Jan 28, 2021 at 16:36
  • $\begingroup$ @TheDawg Formally, we should write $\hat H = \hat H_1 \otimes \mathbb 1 + \mathbb 1 \otimes \hat H_2 + \hat V$. The first two operators act on one factor of the tensor product state while leaving the other alone, but the third operator cannot be separated in that way. Other than that I'm not sure what you're asking; it's an operator which eats a tensor product state and spits out another tensor product state. If you want to be more specific, you need a concrete example. $\endgroup$
    – J. Murray
    Jan 28, 2021 at 16:39
  • $\begingroup$ I agree with the expression, so for example, the coulomb interaction between two particles, how does that act on the tensor space? $\endgroup$
    – TheDawg
    Jan 28, 2021 at 16:44

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