# QM: Hamiltonians and energies of a system

1. The problem statement, all variables and given known data
Hi all.

Lets I have a system, whose Hamiltonian is given by $H = p_1 + p_2$, where pi is the momentum of i, where this i can be whatever, e.g. momentum in some particular direction. When the Hamiltonian is given on this form, do I know for certain that the eigenenergies of the system are on the form:

$$E_{m,n} = f(m) + g(n),$$

where f and g are two functions that depend on respectively m and n? I.e. I am wondering if the eigenenergies of the system follow the same form as the Hamiltonian.

I hope you can help.

Best regards,
Niles.

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Tom Mattson
Staff Emeritus
Gold Member
Lets I have a system, whose Hamiltonian is given by $H = p_1 + p_2$,
Shouldn't your Hamiltonian be quadratic in the momenta? Or are you doing relativistic QM in natural units?

When the Hamiltonian is given on this form, do I know for certain that the eigenenergies of the system are on the form:

$$E_{m,n} = f(m) + g(n),$$
Yes, because Hamiltonians are linear operators. (Perhaps I should say that this can be done whenever the Hamiltonian is linear. I'm not sure if you can cook up a nonlinear one).

Yes, it should. Sorry for that.

Yes, because Hamiltonians are linear operators. (Perhaps I should say that this can be done whenever the Hamiltonian is linear ...
Hmm, can this be seen from the following?

$$H \psi = E\psi \quad \Rightarrow \quad H_1 \psi + H_2\psi = E_1\psi + E_2\psi = E\psi.$$

Thanks for taking the time to reply.

Tom Mattson
Staff Emeritus
Yes. So here $E=E_1+E_2$.
I've thought about this for the last couple of days. What I have found out is that I believe the above only works if the wavefunction satisfies $\psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2)$, so it seperates Schrödingers time-independent equation. Do you agree?