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1D delta funtion potential well

  1. Oct 7, 2011 #1
    Hello forum,
    I have a question regarding the delta function potential well.
    Given the following potential:
    V(x) = -αδ(x) for -a/2 < x < a/2 (α- positive constant) and V(x) = 0 elsewhere, how would one show that the ground state is the only eigenstate with E <0. One could of course solve the problem explicitly and show that this is the case. However, I was trying to come up with a reasonable physical argument without actually solving the problem.
    So, here are my thoughts:
    1.) V(x) is even, so the eigenfunctions are either even or odd.
    2.) By the node theorem, all odd eigenfunctions have x=0 as a node. In particular, the first excited state has x=0 as a node. So, the probability of finding the particle in the first excited state at x=0 is zero (strictly speaking the probabilty density is zero). So, the particle then doesn't "feel" the delta potential. So, it must be unbounded, i.e. the nergy must be E>0.
    3.) Since any excited state has energy larger than the first excited state, it follows that all excited states have E>0.
    4.) But, every 1D potential has at least one bound state. So, it must be the ground state.

    Does this sound like a reasonable argument?

    Thank you.
     
  2. jcsd
  3. Oct 7, 2011 #2

    G01

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    I don't think it quite works. For instance, point 2 is a non sequitur. You implicitly assume a bound mode exists and describe the shape of the corresponding bound wave function. However, then you proceed to argue that the mode must be unbounded, which would mean it's shape is not as you describe.

    Also, point 4 is not correct. Not all 1-D potentials have bound states. Consider the delta "spike," [itex]V(x)=\alpha \delta (x)[/itex], as an example.

    I think the best conceptual argument is as follows: Consider a finite square well. You will notice, that as you make the well narrower and narrower, there start to be less and less bound mode solutions, until you are left with only one, even parity, bound solution. However, no matter how narrow you make the well(in other words, how close you get to the delta well limit), this one last bound mode will not go away.
     
  4. Oct 7, 2011 #3
    Hello G01,
    "Also, point 4 is not correct. Not all 1-D potentials have bound states. Consider the delta "spike," V(x)=αδ(x), as an example."
    This is correct of course. But, point 4 is still valid for the given potential. To keep a similar degree of generality it could be rephrased as: all atractive 1d potentials have at least one bound state.

    "However, then you proceed to argue that the mode must be unbounded, which would mean it's shape is not as you describe."
    I disagree on this one. In this step I merely state that an excited state cannot be bound for this potential. The node theorem is a general result ( i'm quite sure about this) and has nothing to do with the potential shape.

    "However, no matter how narrow you make the well(in other words, how close you get to the delta well limit), this one last bound mode will not go away. "
    I don't exactly see that this line of reasoning proves the point. With this statement you also implictly assume that a bound mode exists.
     
  5. Oct 8, 2011 #4

    DrDu

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    Well, but every purely attractive potential has a bound state in 1D also the delta potential.
     
  6. Oct 8, 2011 #5

    G01

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    I'm fine with that. However, I've never actually seen a full proof that all 1-D attractive potentials have at least one bound state. I'd be interested if anyone has any good proofs for that statement.


    I went over this again, and I don't think I have as much of a problem with this line of reasoning as I thought. You're trying to show by contradiction that there cannot be any excited states. It seems ok, now that I go over it again.

    No, I do not implicitly assume that a bound mode exists. The argument starts from the assumption that you have already solved the finite square well. (A problem some people have an easier time with conceptually.)

    The delta well is the limit of a finite well as the well width goes to 0 and the well depth goes to infinity while keeping their product constant. Now, if you have solved the bound well, then you can graphically find the solutions for the bound energy levels (by plotting the two sides of the transcendental equation against each other). If you look at the results you will see that no matter how narrow your well gets, there will always be one solution to the transcendental equation for a finite well. So, I'm not assuming the solution exists. The finite well problem tells me it must exist, regardless of well size! Thus, you can take your finite well, make it as narrow as you want , and be assured that the last bound mode solution will not disappear. Thus, it follows conceptually that the delta well has one bound solution as it is a limiting case of the above problem.
     
    Last edited: Oct 8, 2011
  7. Oct 8, 2011 #6

    DrDu

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    Landau Lifshetz Quantum Mechanics has a proof.
     
  8. Oct 8, 2011 #7
    "Now, if you have solved the bound well, then you can graphically find the solutions for the bound energy levels (by plotting the two sides of the transcendental equation against each other). If you look at the results you will see that no matter how narrow your well gets, there will always be one solution to the transcendental equation for a finite well. So, I'm not assuming the solution exists. The finite well problem tells me it must exist, regardless of well size! Thus, you can take your finite well, make it as narrow as you want , and be assured that the last bound mode solution will not disappear. Thus, it follows conceptually that the delta well has one bound solution as it is a limiting case of the above problem. "

    I think this is essentially the proof that every 1d attractive potential has at least one bound state.

    "You're trying to show by contradiction that there cannot be any excited states. "
    Yes, in a sense :)


    GO1 and DrDu, thank you for a meaningful discussion.

    Nemetz
     
  9. Oct 8, 2011 #8

    G01

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    I guess I'm just curious if there are any complications to the proof for wells of arbitrary shape, i.e. not square. I'll definitely check out L-L and see what they have to say.

    And yes, it was a great discussion! Thank you!
     
  10. Oct 9, 2011 #9

    DrDu

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    I had a look at Landau Lifshetz and did not find what I had in mind. I most probably confused it with Messiah Vol 1, but he also doesn't proove that an attractive 1d potential has at least one bound state.
    Thinking about it, it is not true in full generality, namely, as a counter example, you only have to consider a constant attractive potential extending from -infinity to +infinity. or a step potential So you have to restrict the class of potentials to potentials e.g. potentials which are relatively compact to H_0.
    They won't change the essential spectrum. Then argument is that an attractive potential inreases the second derivative of the wavefunction )starting from the constant solution belonging to E=0) and thus generates a node. To just remove the node, one has to lower energy and thus obtains a bound state.
    See, W. Thirring Quantum mathematical physics.
     
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