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2 Qs in QM.

  1. Nov 27, 2009 #1

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    1. The problem statement, all variables and given/known data
    1.Consider a one dimensional attraction potential V(x) s.t V(x)<0 for each x.
    Using the variational principle, show that such a potential has at least one bound state.
    Hint: use a gaussian in x as a trial functio.

    2. A particle with charge e and mass m is confined to move on the circumference of a circle of radius r. Find the eigenstates and eigenvalues of the Hamiltonian.



    3. The attempt at a solution
    1. Now sure where to start, I have calculated [tex]<H(a)>=<\psi|H|\psi>/<\psi|\psi>[/tex] where [tex]\psi(x)=exp(-ax^2)[/tex], and I know that it's bigger than E0, usually in order for us to find a strict upper bound for E0, we need to differentiate <H(a)> wrt to a and find its minimum, but here V(x) is not given explicitly so I guess I need to show somehow that <H(a)> is bounded, Ive showed that it's bounded by [tex]\frac{a\hbar^2}{m}[/tex], is that enough to show that for this potential it has at least one bounded eigenstate.

    2. I have no idea, I mean the potential is: [tex]V(r)=-mw^2r^2/2[/tex], how do I proceed from here?

    Thanks in advance.
    QM rulessssssssssssssss!!! :-)
     
  2. jcsd
  3. Nov 27, 2009 #2
    For #2 you should start with Schrodinger's equation:

    [tex]
    -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)
    [/tex]

    And work from there.
     
  4. Nov 28, 2009 #3

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    I know that, is the solution different than the harmonic oscillator equation?
     
  5. Nov 28, 2009 #4

    diazona

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    For #2, why are you saying that the potential is [itex]V(r) = -m\omega^2 r^2/2[/itex]? That doesn't appear to be part of the problem. In fact, there isn't any potential at all mentioned in the problem, which means that [itex]V = 0[/itex] everywhere. Unless you've left something out of the problem statement? (actually, I guess it wouldn't even matter for a purely radial potential)
     
  6. Nov 28, 2009 #5

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    Well, if a particle is confined to move on a circular path then obviously a centriptal force acts upon it, which is m w^2 r, integrate this wrt r and multiply this by a minus sign, and you get the potential I wrote.
     
  7. Nov 28, 2009 #6

    diazona

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    Um... no. You're still thinking classically. In quantum mechanics, when a particle is confined to move along a circular path, it literally has to be confined to that path - you have to assume that the entire space in which the particle is able to exist is nothing more than that circular path. You're dealing with a one-dimensional system with a single position coordinate, let's say [itex]s[/itex], using periodic boundary conditions with a period of [itex]2\pi r[/itex] (so that [itex]s[/itex] and [itex]s + 2\pi r[/itex] represent the same point for any [itex]s[/itex]). You can, if you like, analyze the system using polar coordinates, i.e. use [itex]\theta \equiv s/r[/itex] instead of [itex]s[/itex] as the coordinate.

    If you would like to solve the Schrödinger equation using the potential you came up with, [itex]-m\omega^2 r^2/2[/itex], by all means go ahead and do it. In fact, if I may venture a guess as to what point you're at in your physics education, it's a problem you will probably wind up doing at some point in the future - the two-dimensional harmonic oscillator. But you will find that the particle is not confined to a circle; there will be some nonzero probability of finding it at any point in space.
     
  8. Nov 28, 2009 #7

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    So what is the potential in this specific question? zero?
     
  9. Nov 28, 2009 #8

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    BTW, any help on the first question is welcomed as well.
     
  10. Nov 28, 2009 #9

    diazona

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    Yeah, like I said, if the problem doesn't mention a potential, that generally means [itex]V = 0[/itex] everywhere. Keep in mind that "everywhere" for this problem is just a circle, a one-dimensional entity.

    By the way, if a problem mentions some standard system with a known potential like "harmonic oscillator" or "infinite square well", that's a backhanded way of telling you what the potential is, so be on the lookout for such non-mathematical ways to give you information. But I don't think that's the case here. (Unless you are really insistent on treating it as a two-dimensional system, then you could try [itex]V = 0[/itex] at radius [itex]r[/itex] and [itex]V = \infty[/itex] everywhere else... but that really complicates things. It is, however, how you'd construct this kind of system experimentally.)
     
  11. Nov 28, 2009 #10
    For 1, can't you break it up as this:

    [tex]
    \langle H(a)\rangle=\langle \frac{p^2}{2m}\rangle +\langle V\rangle
    [/tex]

    (you can neglect the denominator terms because the wave function should be normalized.) Now what condition makes a bound state exist?
     
  12. Nov 28, 2009 #11

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    If the Energy is negative.

    i.e I need to show that <H(a)><=0.
    Now I guess because [tex]<H(0)>=\int V(x)<=0[/tex] all I need to show is that
    d<H(a)>/da=0 for a=0, and d^2<H(0)>/da^2<0.

    Thanks.
     
  13. Nov 28, 2009 #12

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    Ok, I solved it, because <H(a)>>=E0 for any a>0, then for a=0 0>=<H(0)>>=E0, so there is at least one bound state mainly the ground state.
     
  14. Nov 28, 2009 #13
    Not quite. [itex]\langle p^2\rangle[/itex] has a value, so you need to show that

    [tex]
    \int \Psi^2V(x)dx>-\frac{1}{2m}\langle p^2\rangle
    [/tex]

    also, I know I did copy that part of it, but why are you looking at H(a)?
     
  15. Nov 29, 2009 #14

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    I am not sure how to explictly show the inequality you wrote, but I think I solved it, look at post #12 in this thread.
     
  16. Nov 29, 2009 #15
    While this is technically true, I'm not sure that this will satisfy the answer for your teacher. You are making the claim that

    [tex]
    \langle H\rangle=\frac{\langle p^2\rangle}{2m}+\langle V\rangle \sim E_0
    [/tex]

    which generally is the case, but you are not at all showing that it is bounded. The expectation value of [itex]p^2[/itex] should give you a positive number (which means, given just this there is no bound state). What you have to do next is show what I wrote in post #13:

    [tex]
    \langle H\rangle=\frac{\langle p^2\rangle}{2m}+\langle V\rangle<0
    [/tex]

    So that

    [tex]
    \langle V\rangle=\int\Psi^2V(x)\,dx>\frac{1}{2m}\langle p^2\rangle
    [/tex]

    (where in this line I've taken [itex]V[/itex] to be negative already so that the negatives cancel).
     
  17. Nov 29, 2009 #16

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    I don't see how to show this...
    <p^2> depends on a, and I don't know what this value is?
     
  18. Nov 29, 2009 #17
    I'm assuming by [itex]a[/itex] you mean the [itex]a[/itex] in [itex]\Psi=\exp[-ax^2][/itex]. If this is the case, then no, the momentum does not depend on [itex]a[/itex]--even if this were the case, with your setting [itex]a=0[/itex], you should see that exp[0]=1 and then you'd have to show that [itex]\langle V\rangle >1[/itex].

    If you use the Heisenberg uncertainty principle, you should find

    [tex]
    \langle p^2\rangle=\langle (\Delta p)^2\rangle=\frac{\hbar^2}{2\langle(\Delta x)^2\rangle}\simeq \frac{\hbar^2}{2d_0^2}
    [/tex]

    where [itex]d_0[/itex] is some characteristic length (possibly the Bohr radius, but it doesn't really matter in this problem). What you need to show now is that the expectation value of the potential is greater than this number using the equation I gave in the previous post (so that when going back to the expectation value of the Hamiltonian, you get a negative number so that there are bound states).
     
  19. Nov 29, 2009 #18

    Actually, what I should say is I think you're looking at this problem the wrong way. You should not be choosing [itex]a[/itex] to satisfy the problem by minimization via differentiation (particularly since we're told the potential depends on [itex]x[/itex], not [itex]a[/itex] and the momentum also does not depend on [itex]a[/itex]). You need to be solving the expectation values (both [itex]\langle p^2\rangle[/itex] and [itex]\langle V\rangle[/itex] should be non-zero) so that the total energy is negative. Since you know what the momentum expectation value is, you can solve the potential expectation value by using the inequality I wrote a couple posts ago.
     
  20. Nov 30, 2009 #19

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    if I write a trial function f(x)=exp(-x^2), I get that:
    <H>=<V>+hbar^2/2m sqrt(pi/2)
    I don't see why this should be negative?
     
  21. Nov 30, 2009 #20
    The condition for there to be a bound state is that the energy must be less than 0. Recall that since [itex]\langle H\rangle\simeq E_0[/itex], then by showing that

    (1) [itex]\langle V\rangle<0[/itex]

    and

    (2) [itex]\left|\langle V\rangle\right|>\left|\langle p^2\rangle\right|[/itex]

    you are showing that there is a bound state.
     
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