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QM: Hamiltonians and energies of a system

  • Thread starter Niles
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  • #1
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1. The problem statement, all variables and given known data
Hi all.

Lets I have a system, whose Hamiltonian is given by [itex]H = p_1 + p_2[/itex], 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:

[tex]
E_{m,n} = f(m) + g(n),
[/tex]

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.
 

Answers and Replies

  • #2
Tom Mattson
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Lets I have a system, whose Hamiltonian is given by [itex]H = p_1 + p_2[/itex],
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:

[tex]
E_{m,n} = f(m) + g(n),
[/tex]
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).
 
  • #3
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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?

[tex]
H \psi = E\psi \quad \Rightarrow \quad H_1 \psi + H_2\psi = E_1\psi + E_2\psi = E\psi.
[/tex]

Thanks for taking the time to reply.
 
  • #4
Tom Mattson
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Yes. So here [itex]E=E_1+E_2[/itex].
 
  • #5
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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 [itex]
\psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2) [/itex], so it seperates Schrödingers time-independent equation. Do you agree?
 

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