- #1

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When I solve the eigenvalue equation for H, and get zero or negative numbers or zero for E, is that physical? If it is not physical, do the negative numbers or zero energy mean anything?

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- Thread starter chill_factor
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- #1

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When I solve the eigenvalue equation for H, and get zero or negative numbers or zero for E, is that physical? If it is not physical, do the negative numbers or zero energy mean anything?

- #2

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The only physical requirement is that a ground state has to exist. That is, the set of energy eigenvalues has to be bounded from below.

- #3

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The only physical requirement is that a ground state has to exist. That is, the set of energy eigenvalues has to be bounded from below.

Thank you.

If there were just a limited amount of states, say, 3, then if 1 was negative and the others positive or zero, it doesn't matter that its negative, it'll just be the ground state.

- #4

tom.stoer

Science Advisor

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- #5

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However, we solve that problem for a position basis. Is this true for a general basis?

I ask this because I was given a problem in matrix mechanics with a general basis which asked "what states are possible" and I got negative energies.

- #6

tom.stoer

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Of course it is true for a general basis.

Suppose you are able to construct the energy eigenbasis exactly, like for the qm harmonic oscillator.

The eigenfunctions U_{E}(a) in a specific basis 'a' are nothing else but a representation, a projection on the a-basis. If you have a complete spectrum with energy values {E} and the corresponding basis vector {|E>} consisting of a discrete bound-state spectrum and a continuous scattering-state problem such a state in a-basis can be written as

[tex]u_E(a) = \langle a|E\rangle[/tex]

If you chose 'a' = 'x' you get the position space basis for the energy eigenvectors; if you chose 'a' = 'p' you get the momentum space basis for the energy eigenvectors. So for the hydrogen atom you will find

[tex]u_E(x) = \langle x|E\rangle \to u_{nlm}(r,\theta,\phi)[/tex]

Chosing the momentum space the wave functions will look different, but the underlying spectrum of the operator is not affected.

Suppose you are able to construct the energy eigenbasis exactly, like for the qm harmonic oscillator.

The eigenfunctions U

[tex]u_E(a) = \langle a|E\rangle[/tex]

If you chose 'a' = 'x' you get the position space basis for the energy eigenvectors; if you chose 'a' = 'p' you get the momentum space basis for the energy eigenvectors. So for the hydrogen atom you will find

[tex]u_E(x) = \langle x|E\rangle \to u_{nlm}(r,\theta,\phi)[/tex]

Chosing the momentum space the wave functions will look different, but the underlying spectrum of the operator is not affected.

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- #7

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Thanks alot! I think I understand now.

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