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In some cases, such as finite and infinite square wells, the Hamiltonian has energy eigenstates which correspond to physical wavefunctions.

In other cases, such as a one dimensional universe with constant potential, it doesn't. It has the plane wave, but that's not normalizable. (Also, for all Hamiltonians, the constant zero wavefunction is an eigenstate, but that's equally non normalizable.)

Is there any physical meaning to whether the Hamiltonian has energy eigenstates? Under what sort of situations does it have energy eigenstates?

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what's not to like with plane waves? –  Emilio Pisanty Apr 23 '13 at 12:30
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Hi Buck Shlegeris: Are you essentially asking For which potentials an 1D quantum mechanical system has normalizable bound energy states? –  Qmechanic Apr 23 '13 at 12:31
    
@EmilioPisanty guessing he's refering to the fact that plane waves are non-normalisable. Of course if you go to a lattice or a box or content yourself with rigged Hilbert space everything is fine and the distinction makes no practical difference. –  Michael Brown Apr 23 '13 at 12:34
    
Yeah, my issue with plane waves is that they're non-normalizable. –  Buck Shlegeris Apr 23 '13 at 12:44
    
And most of the examples I gave were 1D, but I'm just as interested in the question for more dimensions. –  Buck Shlegeris Apr 23 '13 at 12:47
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3 Answers 3

The time-independent Schrodinger equation is mainly useful for describing standing waves. It has serious shortcomings when used to describing traveling waves. If you have an example like a constant potential, then there are only traveling-wave solutions, and the time-independent Schrodinger equation may be the wrong tool for the job.

Physically, the constant-potential example has realistic solutions that are wave packets. These packets have to be constructed from a superposition of different energies. The time-independent equation restricts you to describing a single energy eigenstate, so the physically realistic packets aren't solutions. The packets spread out over time, and this is physically necessary behavior (caused ultimately by the Heisenberg uncertainty principle) that the time-independent Schrodinger equation can't describe.

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In my opinion,if a potential function is authentic, the solution of the time independent Schrodinger equation will have physics meaning. You say a one dimensional universe with constant potential,in fact ,this kind of potential is not existed. I think the plane wave cannot be normalized is a reflection of this.

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Picking up on your comment that plane waves are not renormalisable:

Only infinite plane waves are not renormalisable, and an infinite plane wave is not physically realistic simply because we can't make, or indeed even observe) infinite objects. Any plane wave we can observe will be finite and therefore normalisable.

An infinite plane wave represents an object with zero uncertainty in the momentum, and therefore infinite uncertainty in position. The converse of this is that it's also physically unrealistic to have a solution representing a particle with a perfectly known position, i.e. a delta function, because then it has infinitely uncertain momentum.

These limitations aren't really to do with special properties of the potential. It's perfectly possible to take a superposition of plane wave solutions and construct perfectly realistic wavepackets.

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"It's perfectly possible to take a superposition of plane wave solutions and construct perfectly realistic wavepackets." That only works with the time-dependent Schrodinger equation. In the time-independent equation, you're limited to describing energy eigenstates. –  Ben Crowell Apr 23 '13 at 15:41
    
Ah, OK. Off to the chat now, I'll return to this question at six. –  John Rennie Apr 23 '13 at 16:07
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