The wave function in the finite square well

  • #1
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Summary:

Even though the potential function is even, the scattering states of the time-independent wave function is not even or odd. (You can't use this theorem in the scattering states.) Why?

Main Question or Discussion Point

Hello! I have been recently studying Quantum mechanics alone and I've just got this question.

If the potential function V(x) is an even function, then the time-independent wave function can always be taken to be either even or odd. However, I found one case that this theorem is not applied, which is the case of scattering states in finite square well.

If V(x) is -V_0 at -a<x<a and 0 at lxl>a, V(x) is an even function(V_0 is a positive constant), so the time-independent wave function should be even or odd function but in my textbook (Quantum Mechanics by Griffiths), the author doesn't use this theorem even tho he does in case of bound states in finite square well.

I'm assuming because the wave function is not normalizable, so you can't really apply that "even-or-odd-theorem", but I want to know more precise reasons!
 

Answers and Replies

  • #3
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The bolded qualifier is key.



Are not time-independent.
That was fast and very clear. Haha Thanks!!!
 
  • #4
PeroK
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Summary:: Even though the potential function is even, the scattering states of the time-independent wave function is not even or odd. (You can't use this theorem in the scattering states.) Why?

Hello! I have been recently studying Quantum mechanics alone and I've just got this question.

If the potential function V(x) is an even function, then the time-independent wave function can always be taken to be either even or odd. However, I found one case that this theorem is not applied, which is the case of scattering states in finite square well.

If V(x) is -V_0 at -a<x<a and 0 at lxl>a, V(x) is an even function(V_0 is a positive constant), so the time-independent wave function should be even or odd function but in my textbook (Quantum Mechanics by Griffiths), the author doesn't use this theorem even tho he does in case of bound states in finite square well.

I'm assuming because the wave function is not normalizable, so you can't really apply that "even-or-odd-theorem", but I want to know more precise reasons!
In the 2nd edition there is a footnote on page 81: "We could look for even and odd functions ... but the scattering problem is inherently asymmetric."

Note that the theorem does not say all solutions must be either even or odd, only that one can construct solutions that are either even or odd. Whether you chose to do this depends on the symmetry or asymmetry of the problem.

In this case an even function would be a superposition of equal and opposite incoming waves from both directions.
 
  • #5
vanhees71
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The bolded qualifier is key.



Are not time-independent.
Of course you can define time-independent scattering states. That's what's done in the socalled "old-fashioned perturbation theory". Of course here you don't need perturbation theory. The scattering states are then defined as the energy eigenstates that are not square integrable. If you define ##V(x)## such that ##V(x) \rightarrow 0## for ##|x| \rightarrow \infty##, the that's the case for the states with energy eigenvalues ##E \geq 0##.

Now of course you can always diagonalize simultaneously the Hamilton operator and the parity operator if the potential is symmetric, ##V(x)=V(-x)## (which you can always achive for the potential well by shifting the origin of the ##x## coordinate accordingly), i.e., you can always look for parity even and parity odd energy eigenfunctions, which simplifies your task, because you need to consider only boundary conditions at one of the locations of the jumps of the potential.

But that's not what you want for a scattering state. Here you want a state which describes a particle which is incoming from one side. Of course that should not be described by an energy-eigenstate but a time-dependent wave packet, but as any solution of the time-dependent Schrödinger equation you can write it in terms of the energy eigenfunctions, and what you need to describe the situation that a particle comes in from the left and moves to the right are the energy eigensolution with positive energy eigenvalues which have only a right-moving plane wave right from the potential, i.e., ##\exp(\mathrm{i} k x)## with ##k>0## right from the potential. With this ansatz then you can solve uniquely for the complete wave function using the boundary conditions at the edges of the box potential. On the left side of the potential you necessarily have then both a right-moving and a left-moving part. This makes sense, because if you think in terms of the right moving wave packet, what happens when the wave packet gets in the region of the potential, one part is moving further to the right and another part is reflected back to the left.

This kind of "right-moving scattering solutions" are of course not a parity eigenstate, but since the potential is parity symmetric there's another linear independent solution of the same energy which is a "left-moving scattering solution", and it is given by the parity transformation applied to the right-moving one. Of course, you can build two other solutions of that energy which are parity even and parity odd, respectively.
 
  • #6
PeterDonis
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The scattering states are then defined as the energy eigenstates that are not square integrable.
These states are not normalizable, so they can't be physically realized. So yes, strictly speaking I should have said that physically realizable scattering states (or normalizable scattering states) are not time-independent.
 
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  • #7
PeterDonis
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Of course, you can build two other solutions of that energy which are parity even and parity odd, respectively.
But these solutions won't describe scattering states, correct?
 
  • #8
vanhees71
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These states are not normalizable, so they can't be physically realized. So yes, strictly speaking I should have said that physically realizable scattering states (or normalizable scattering states) are not time-independent.
That's of course an important point. These states are "normalized to a ##\delta## distribution",
$$\langle u_E|u_{E'} \rangle=\delta(E-E'),$$
and you can build wave packets with them defining the real states.
 
  • #9
vanhees71
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But these solutions won't describe scattering states, correct?
No, because with the scattering states you want to describe a different situation, namely one where a particle that can be considered free (being far away from the potential well) (asymptotic free state) moving towards it. That's for sure not a parity-symmetric situation. Of course, states do not obey the symmetry of the Hamiltonian but depend on the preparation, and in a scattering experiment we prepare scattering states, which are neither parity even nor parity odd.
 
  • #10
PeterDonis
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No, because with the scattering states you want to describe a different situation
Wouldn't that be "yes"? You appear to be agreeing with my statement that the solutions which are parity even and parity odd do not describe scattering states.
 
  • #11
vanhees71
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Yes, I agree. Perhaps, my "no" is misleading...
 

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