QM: Hamiltonians and energies of a system

[Niles]

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 p_i 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.Niles.

 

[quantumdude]

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).

 

[Niles]

Shouldn't your Hamiltonian be quadratic in the momenta?.

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.

 

[quantumdude]

Yes. So here $E=E_1+E_2$.

 

[Niles]

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 separates Schrödingers time-independent equation. Do you agree?