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Delta well + infinite barrier -> bound state

  1. Jul 13, 2010 #1
    Hi,

    I'm trying to understand the quantum mechanical solution to this potential:

    [tex]V(x) = \left\{\begin{array}{cc}\infty & \mbox{ for } x < 0,\\-\lambda\delta(x-d) & \mbox { for } x > 0\end{array}\right.[/tex]

    A particle of mass m is constrained to move on the half straight line [itex]\{x \in \mathbb{R}: x > 0\}[/itex] under this potential. I want to examine the condition for existence of bound states, and its dependence on [itex]\lambda[/itex] and [itex]d[/itex], where both [itex]\lambda[/itex] and [itex]d[/itex] are positive.

    For x < 0, [itex]\psi(x) = 0[/itex]

    For 0 < x < d, [itex]\psi(x) = A\sin(kx)[/itex] where [itex]k = \sqrt{2mE}/\hbar[/itex].

    For x > d, [itex]\psi(x) = Ce^{ikx} + De^{-ikx}[/itex].

    Enforcing continuity at x = d, [itex]A sin(kd) = Ce^{ikd} + De^{-ikd}[/itex].

    Integrating Schrodinger's equation over a small interval around [itex]x = d[/itex]

    [tex]-\frac{\hbar^2}{2m}\int_{d-\epsilon}^{d+\epsilon}\frac{d^2 \psi}{dx^2} dx - \lambda\int_{d-\epsilon}^{d+\epsilon}\delta(x-d)\psi(x) dx = E\int_{d-\epsilon}^{d+\epsilon}\psi(x) dx[/tex]

    But I don't get a condition relating [itex]\lambda[/itex] and [itex]d[/itex] for the existence of a bound state. What have I missed?

    Thanks in advance.
     
    Last edited: Jul 13, 2010
  2. jcsd
  3. Jul 13, 2010 #2
    There might not be an explicit relation between \\lambda and d, if that what's you are trying to find. But you do have the relation that the difference of the derivatives to the left and right of the peak is equal to the negative of \\lambda times 2m, divided by hbar^2. Comparing the real and imaginary part, you might get a relation between the trigonometric ratios of d and \\lambda. I think the energy levels might have the discrete band structure, as is found for the case of a solid, with the potential represented by the dirac comb.
     
  4. Jul 13, 2010 #3

    DrDu

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    For a bound state, the wavefunction has to decay exponentially for x>d.
     
  5. Jul 13, 2010 #4
    Okay, I found the source of this problem and it asks me to find a relation between d and lambda for the existence of at least one bound state.
     
  6. Jul 13, 2010 #5
    Right so that means k becomes imaginary.
     
  7. Jul 13, 2010 #6

    DrDu

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    yes, and either C or D becomes 0.
     
  8. Jul 13, 2010 #7
    Okay just to clarify matters...

    For 0 < x < d, since the wavefunction is constrained to be 0 for x <= 0, it should have the form

    [tex]\psi(x) = A\sin(kx)[/tex]

    where [itex]k = \sqrt{\frac{2mE}{\hbar^2}}[/itex] for E > 0. This makes k real.

    For x > d, the wavefunction should have the form

    [tex]\psi(x) = B e^{-kx}[/tex]

    for the same k. Now, what I want to understand is...what is the definition of a bound state here? Suppose the delta well didn't exist. Then, the "bound"
    state would just be [itex]\psi(x) = A\sin(kx)[/itex] (or in other words, no bound state, as its just an infinite step at x = 0). With the delta well in place at x = d, what really is a bound state for the composite potential? In other words, how is an E > 0 solution a bound state?
     
  9. Jul 13, 2010 #8

    DrDu

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    All solutions to the Schroedinger equation which are normalizable are bound states.
    The other solutions, like sin(kx) in the free case are not normalizable and belong to the continuum. Whether one considers non-normalizable solutions as admissible solutions at all depends on the formalism. In the orthodox view a la von Neumann, they don't belong to the Hilbert space at all and are non-admissible. In the rigged Hilbert space approach of Bogoliubov which accomodates naturally the Dirac bra-ket formalism, they are ok.
     
  10. Jul 13, 2010 #9
    Got it. So, are you saying that the solutions I listed -- the sin(kx) in (0, d) and the exponentially decaying wave for (d, \infty) are bound states for E > 0? The reason for my confusion in defining a "bound state" and distinguishing between E > 0 and E < 0 stems from the fact that we have a barrier which goes to +\infty and a well which goes to -\infty.
     
  11. Jul 13, 2010 #10
    Here's my working.

    [tex]\psi(x) = \left\{\begin{array}{cc}A\sin(kx) & 0 < x < d\\Ce^{-kx} & x > d\end{array}\right.[/tex]

    1. Continuity at x = d:

    [tex]A\sin(kd) = Ce^{-kd}[/tex]

    2. Integration of Schrodinger's equation around x = d

    [tex]\sin(kd) + \cos(kd) = \frac{2m\lambda}{\hbar^2 k}\sin(kd)[/tex]

    where [itex]k = \sqrt{\frac{2mE}{\hbar^2}}[/itex] and E > 0.

    This gives (if my algebra is correct) the "condition relating [itex]\lambda[/itex] and [itex]d[/itex] for the existence of a bound state" as

    [tex]\tan(kd) = \frac{1}{\frac{2m\lambda}{\hbar^2 k}-1}[/tex]

    Is this the desired condition, even though it involves another parameter (E)?

    PS -- The question also says that an observer can detect the existence of the potential by bouncing back waves from the wall at x = 0, of the form sin(kx) which will acquire a phase shift when the delta well is present, and look like sin(kx + phi). If this is to be believed, the exponentially decaying solution should be replaced with the complex exponential wave, don't you think?
     
    Last edited: Jul 13, 2010
  12. Jul 13, 2010 #11

    DrDu

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    I didn't check your calculations but I'd rather expect that the bound state has E<0 in that problem as the continuum of free states still starts at E=0.
     
  13. Jul 13, 2010 #12

    DrDu

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    Btw, instead of sin and cos you should also use sinh and cosh.
     
  14. Jul 13, 2010 #13
    Why? Is that because E < 0? And why is that so?

    PS -- The question also says that an observer can detect the existence of the potential by bouncing back waves from the wall at x = 0, of the form sin(kx) which will acquire a phase shift when the delta well is present, and look like sin(kx + phi). If this is to be believed, the exponentially decaying solution should be replaced with the complex exponential wave, don't you think?
     
  15. Jul 13, 2010 #14

    DrDu

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    The ansatz in post #1 will yield the correct answer for scattering states.
    For the bound state, as E<0, k will be imaginary and if you write exp(-kx) on the right, you should also write sinh (kx).
     
  16. Jul 13, 2010 #15
    Ok, here is my revised working

    [tex]k = \sqrt{-2mE/\hbar^2}[/tex]

    where E < 0

    For 0 < x < d

    [tex]\psi(x) = A\sinh(kx) + B\cosh(kx)[/tex]

    as [itex]\psi(0) = 0[/itex], B = 0.

    for x > d,

    [tex]\psi(x) = C\sinh(kx) + D\cosh(kx)[/tex]

    Integrating Schrodinger's equation,

    [tex]k(C-A)\cosh(kd) + \left(kD + \frac{2m\lambda}{\hbar^2}A\right)\sinh(kd) = 0[/tex]

    which gives (please correct me if I'm wrong)

    [tex]C = A[/tex]
    [tex]kD + \frac{2m\lambda}{\hbar^2}A = 0[/tex]

    So,

    [tex]\psi(x) = \left\{\begin{array}{cc}A\sinh(kx) & \mbox{ for } 0 < x < d\\A\sinh(kx) -\frac{2m\lambda}{\hbar^2 k}A\cosh(kx) &\mbox{ for } x > d\end{array}\right.[/tex]

    But this doesn't give me any condition connecting [itex]\lambda[/itex] and [itex]d[/itex]. Also, how can I write the wavefuntion as sin(kx + phi) when the delta barrier is present? The question suggests I should do so, to get to a transcendental equation for phi.
     
    Last edited: Jul 13, 2010
  17. Jul 13, 2010 #16

    DrDu

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    sigh, for x>d an independent combination of sinh and cosh is not allowed, as I thought I had explained before as these functions diverge when x goes to infinity. Only for D=-C, that is, for exp (-kx), a normalizable solution does result.
     
  18. Jul 13, 2010 #17
    Oops, I'm sorry...the last step I wrote here is wrong...I figured out the problem btw, and was just going to post the correction. As usual, I had messed up because of 'k'. You're right of course, I wrote sinh() and cosh() but on my notebook I still had the old sin() and cos(). Everything works out now...including the condition on lambda and d, which is

    [tex]\frac{2m\lambda d}{\hbar^2} > 1[/tex]

    (If this condition is satisfied, there is at least one bound state.)

    As for the second part of the question, what they're actually proposing is that we bounce waves back from the wall at x = 0 (when lambda = 0). That is a scattering problem, and so has nothing to do with the bound states that we just computed. I was confused by the language because the question refers to the solution of the first part.

    PS -- A bad day :-/
     
    Last edited: Jul 13, 2010
  19. Jul 15, 2010 #18
    DrDu, can you help me understand this second part properly. I am stuck somewhere.

    An observer at [itex]x = \infty[/itex] can infer the existence of [itex]V(x)[/itex] by bouncing back waves from the wall at [itex]x = 0[/itex]. For [itex]\lambda = 0[/itex], the wavefunction with energy [itex]E = \hbar^2 k^2/2m[/itex] would be [itex]\psi(x) = \sin(kx)[/itex]. But with [itex]\lambda \neq 0[/itex], the wavefunction acquires a phase [itex]\phi[/itex] so that [itex]\psi(x) = \sin(kx + \phi)[/itex] at large x.

    I have to find an equation determining [itex]\phi[/itex]. In this case, the wavefunction is [itex]\psi(x) = B\sin(kx + \phi)[/itex] (for E > 0 and real k) for large x. Can this wavefunction be used as that near the delta well? And what is the wavefunction for [itex]0 < x < d[/itex] when the delta well is present? Can that just be [itex]A\sin(kx)[/itex]?
     
  20. Jul 15, 2010 #19

    DrDu

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    You can use that ansatz for x>d and for 0<x<d, A sin kx is ok.
     
  21. Jul 15, 2010 #20
    Ok, so if I do that,

    continuity at x = d gives

    [tex]A\sin(kd + \phi) = B\sin(kd)[/tex]

    and integrating Schrodinger's equation around x = d gives

    [tex]\frac{\hbar^2 k}{2m}\cos(kd + \phi) + \left(\lambda-\frac{\hbar^2 k}{2m}\cot(kd)\right)\sin(kd + \phi) = 0[/tex]

    This equation can be solved 'exactly' for [itex]\tan(kd + \phi)[/itex] and hence for [itex]\phi[/itex]. But the questions asks for a 'transcendental equation' for [itex]\phi[/itex].
     
  22. Jul 15, 2010 #21

    DrDu

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    Hm, isn't tan(kd+phi)=y a transcendental equation for phi?
     
  23. Jul 15, 2010 #22

    DrDu

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    Btw, I have an extra problem for you.
    You found already the k for which for x>d phi is proportional to exp(-kx).
    Can you find other solutions of that equation with complex k ?
     
  24. Jul 15, 2010 #23
    According to my understanding, it would be if y is a function of phi (http://en.wikipedia.org/wiki/Transcendental_equation). I am confused here because the question says 'the transcendental equation for phi cannot be solved exactly. For small energies, assume [itex]\phi \approx a k[/itex] and find [itex]a[/itex]'. We are assuming that [itex]\psi(x) = B\sin(kx + \phi)[/itex] all the way up to [itex]x = d[/itex] (whereas the question gives it as an asymptotic form of [itex]\psi(x)[/itex]), and that [itex]\psi(x) = A\sin(kx)[/itex] for [itex]0 < x < d[/itex]. What is wrong with these assumptions?
     
  25. Jul 15, 2010 #24
    I'm not sure I follow you. Can you please restate this statement? The condition for the existence of a bound state is obtained from

    [tex]k = \frac{m\lambda}{\hbar^2}(1-e^{-2kd})[/tex]

    when [itex]k = \sqrt{-2mE/\hbar^2}[/itex] for E < 0 (bound states). I argued that this equation has a solution only if the RHS as a function of k has a slope greater than that of the LHS. That's how I got the condition for the existence of a bound state. Is this what you were referring to?
     
  26. Jul 15, 2010 #25

    DrDu

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    How shall I say? For a potential of infinite range, the statement only holds asymptotically. But for a potential of finite range d, if A exp (ikx)+B exp(ikx) =C sin (kx+phi) holds asymptotically (i.e. abs(A)=abs(B)) then it holds down to x=d.
     
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