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A QM problem on finding eigenvalues

  1. May 8, 2005 #1
    Hi. I have this problem which i am stuck at:

    Consider a one-dimensional Hamilton operator of the form

    [tex] H = \frac{P^2}{2M} - |v\rangle V \langle v| [/tex]

    where the potential strength V is a postive constant and [tex] |v \rangle\langle v| [/tex] is a normalised projector, [tex] \langle v|v \rangle = 1 [/tex]. Determine all negative eigenvalues of H if [tex] |v \rangle [/tex] has the position wave function [tex] \langle x|v \rangle = \sqrt{\kappa} e^{- \kappa |x|} [/tex] with [tex] \kappa > 0 [/tex].

    It seems to me the only step i could take is to apply Hamilton H to x bra and v ket:

    [tex] \langle x|H|v \rangle = - \frac{\hbar ^2}{2M} \left( \frac{\partial}{\partial x} \right)^2 \langle x|v \rangle -\langle x|v \rangle V [/tex]

    which gives [tex] = - \left( \frac{\hbar ^2 \kappa ^2}{2M} + V \right) \langle x|v \rangle [/tex]

    So how do I proceed (if i am right thus far)?

    Can someone give me some hints as to how to solve this problem?

  2. jcsd
  3. May 8, 2005 #2


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    Please make sure you post problems such as this in the Homework Help section.

  4. May 8, 2005 #3
    try to write the hamiltonian in the p basis.

  5. May 9, 2005 #4

    I am sorry, but i dont understand what you mean by writing the hamiltonian in the p basis.

    (and i will remember to post such problems in the Homework Help section next time, sorry about that :tongue: )
  6. May 9, 2005 #5
    I don't know how to solve this problem, but I'd like to. I'm not sure what good it does to write the hamiltonian in the p basis. Maybe seratend can elaborate on that?

    for instance,

    [tex]\langle p | v \rangle = \int_{-\infty}^{+\infty} \langle p | x \rangle \langle x | v \rangle dx
    = \sqrt{\frac{\kappa}{2 \pi \hbar}} \int_{-\infty}^{+\infty} e^{-\kappa |x| - ipx/ \hbar}dx
    = \sqrt{\frac{2}{\pi}} \frac{(\hbar \kappa)^{3/2}}{\hbar^2 \kappa^2 + p^2}[/tex]

    or something close to that at least. Then the matrix elements of H in the p basis are

    [tex]\langle p | \hat{H} | p^\prime \rangle
    = \langle p | \frac{\hat{p}^2}{2m} | p^\prime \rangle - V \langle p | v \rangle \langle v | p^\prime \rangle
    = \frac{p^2}{2m} \delta (p - p^\prime) - V \langle p | v \rangle \langle v | p^\prime \rangle[/tex]

    but I don't see how this simplifies when you substitute this in the SE

    [tex]\langle p | \hat{H} | \Psi \rangle = \int_{-\infty}^{+\infty}\langle p | \hat{H} | p^\prime \rangle \langle p^\prime | \Psi \rangle d p^\prime
    = E \langle p | \Psi \rangle[/tex]

    it looks like an ugly integral equation. Was this the procedure seratend was talking about?
  7. May 9, 2005 #6
    nobody knows how to do this problem?
  8. May 11, 2005 #7


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    The difficult part of this problem is what to do with the potential operator [tex] |v\rangle V \langle v| [/tex] . If instead, the author had asked for the same problem with just [tex] V(x) [/tex], you'd have known exactly what to do. The mixing of bra ket notation with wave function notation is confusing.

    To solve the problem, you have to look for eigenfunctions, say a wave function [tex] \psi(x) [/tex]. When you look at the problem in this way, you will realize that the potential operator when applied to the wave function will give:

    [tex]|v\rangle V \langle v| |\psi \rangle == \kappa V e^{-\kappa |x|} \int e^{-\kappa |x|} \psi(x) dx [/tex].

    In other words, you may think of [tex] |v\rangle V \langle v| [/tex] as a set of instuctions for how to operate on a wave function. In this case, the instructions say "the result of this operator on a wave function is the product of three things. There is a ket [tex]|v\rangle[/tex], a number V, and a number given by the inner product of [tex]\langle v | [/tex] with the wave function. The result of the operator is the product of these three things (one ket and two scalars)."

    I don't think that I've ruined your learning experience by pointing this out. I doubt that the author intended you to work so hard on setting it up.

    The main problem with the way that QM is taught is that it requires so much mathematical sophistication that the learners have difficulty finding time to appreciate the theory.

  9. May 11, 2005 #8
    Do you mean: [tex] \sum_{v} |v\rangle\langle v| = 1[/tex]? Or just [tex] | v \rangle \langle v|=1[/tex]?
    Last edited: May 11, 2005
  10. May 11, 2005 #9


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    Here's what i get.

    [tex] \hat{H}|\psi\rangle=E|\psi\rangle [/tex] (1)

    [tex] \hat{H}=\frac{1}{2m}\hat{P}^{2}-V|v\rangle\langle v| [/tex] (2)

    I'm gonna project (1) on a set of linear functionals [itex] \{\langle x|\} [/itex] which obey the completitude relation

    [tex] \int_{\mathbb{R}} |x\rangle\langle x| \ dx =\hat{1} [/tex] (3)


    [tex] \langle x|\hat{H}|\psi\rangle=\left\langle x\left |\frac{1}{2m}\hat{P}^{2}\right|\psi\right\rangle -V\langle x|v\rangle\langle v|\psi\rangle [/tex] (4)

    [tex] \langle x|\hat{H}|\psi\rangle=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}-V\sqrt{\kappa}e^{-\kappa \left |x\right |}\langle v|\psi\rangle [/tex] (5)

    I make use of (3)

    [tex] \langle v|\psi\rangle =\int_{\mathbb{R}} \langle v|x\rangle\langle x|\psi\rangle \ dx=\sqrt{\kappa} \int_{\mathbb{R}} \psi (x) e^{-\kappa \left |x\right |} \ dx [/tex] (6)

    So,(5) becomes

    [tex] \langle x|\hat{H}|\psi\rangle=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}-\kappa V \int_{\mathbb{R}} \psi (x) e^{-\kappa \left |x\right |} \ dx [/tex] (7)

    The equation (1) in the basis [itex] \{\langle x|\} [/itex] is

    [tex] \langle x|\hat{H}|\psi\rangle=E\langle x|\psi\rangle [/tex] (8)

    Coupling (7) & (8),we reach an integro-differential equation for the wavefunction [itex] \psi (x) [/itex].

    [tex] -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}-\kappa V \int_{\mathbb{R}} \psi (x) e^{-\kappa \left |x\right |} \ dx =E\psi (x) [/tex] (9)

    Let's solve this equation,searching for plane-wave solutions (wave packets).

    To be continued.

    Last edited: May 11, 2005
  11. May 11, 2005 #10


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    A first observation for (9) is that,for example,

    [tex] \psi (x)=-\psi (-x) [/tex] (10)

    (odd solutions),then the equation becomes

    [tex] -\frac{\hbar^{2}}{2m} \frac{d^{2}\psi (x)}{dx^{2}}=E\psi (x) [/tex] (11)

    together with the condition (10).Let's solve the equation (11),together with the energy spectrum be entirely negative,i.e.

    [tex] E=-\lambda^{2} \ ,\lambda\in\mathbb{R}-\{0\} [/tex] (12)

    Then the ODE (11) becomes

    [tex]-\frac{\hbar^{2}}{2m} \frac{d^{2}\psi (x)}{dx^{2}}=-\lambda^{2} \psi (x) [/tex] (13)

    Choosing an exponential solution [itex] \psi (x) \sim e^{\chi x} [/itex] (14),we get

    [tex] \frac{\hbar^{2}}{2m}\chi^{2}=\lambda^{2} [/tex] (15)

    with the solutions

    [tex] \chi_{1,2}=\pm\sqrt{\frac{2m\lambda^{2}}{\hbar^{2}}} =:\pm k \ ,k>0[/tex] (16)

    The wavefunction is

    [tex]\psi (x)=C_{1}e^{+ k x}+C_{2}e^{-k x} [/tex] (17)

    Coupling (17) and (10),we get

    [tex] \psi (x)=\bar{C}\sinh k x [/tex] (18)

    But the wavefunction (18) is not normalizable,therefore the negative spectrum which,by virtue of the equation (12),is the whole negative semiaxis,doesn't correspond to physical states described by ODD wavefunctions.

    Since one can show that any real function defined on all [itex] \mathbb{R} [/itex] can be written as a sum between an odd function & an even function,we conclude that the integro differential equation (9) should have EVEN wavefunctions as solutions.

    To be contnued.

  12. May 11, 2005 #11


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    So let's search for even solutions as wavepackets for the equation (9)

    [tex]\psi (x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{\mathbb{R}} \tilde{\psi}(p) \ e^{\frac{1}{i\hbar}px} \ dp [/tex] (19)

    [tex]\frac{d^{2}\psi (x)}{dx^{2}}=-\frac{1}{\hbar^{2}}\frac{1}{\sqrt{2\pi\hbar}}\int_{\mathbb{R}} p^{2} \ \tilde{\psi}(p) \ e^{\frac{1}{i\hbar}px} \ dp [/tex] (20)

    Coupling (19),(20) and (9),one gets (after a bunch of simplifications)

    [tex]\int_{\mathbb{R}} \left(\frac{p^{2}}{2m}-\kappa V e^{-\kappa |x|}-E\right) \tilde{\psi} (p) \ e^{\frac{1}{i\hbar}px} \ dp=0 [/tex] (21)

    The solution to the equation (21) is simply

    [tex] \tilde{\psi}(p)=\delta \left(\frac{p^{2}}{2m}-\kappa V e^{-\kappa |x|}-E\right) [/tex] (22)

    One can go on now and find the EVEN wavefunction,solution of (9) which are,of course,

    [tex]\psi (x)=\int_{\mathbb{R}}\delta \left(\frac{p^{2}}{2m}-\kappa V e^{-\kappa |x|}-E\right) e^{\frac{1}{i\hbar}px} \ dp \ \ \ \mbox{with} \ \ \psi (x)=\psi (-x)[/tex] (23)

    To be continued.

    Last edited: May 11, 2005
  13. May 11, 2005 #12


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    Here's the nice thing.Let's think about equation (22).What does it mean...

    [tex] \tilde{\psi}(p)=\delta\left(\frac{p^{2}}{2m}-\kappa V e^{-\kappa|x|}-E\right)=\left\{\begin{array}{ccc} 0 &\mbox{when}&\frac{p^{2}}{2m}\neq\kappa V e^{-\kappa |x|}+E \\ \infty & \mbox{when} & \frac{p^{2}}{2m}=\kappa V e^{-\kappa |x|}+E \end{array}\right \end{array}\right [/tex] (24)

    Let's concentrate on this equation:

    [tex] \frac{p^{2}}{2m}-\kappa V e^{-\kappa |x|} =E [/tex] (25)

    We want to have E<0.So it's fair to say,that

    [tex] \frac{p^{2}}{2m}<\kappa V e^{-\kappa |x|} [/tex] (26)

    In the asymptotic limit [itex] x \rightarrow \pm \infty [/itex] (27),the RHS tends to 0.The only way (26) can be then fulfilled is that [itex] p=0 \ \ \mbox{when} \ x \rightarrow \pm \infty [/itex] (28).

    So the particle is at rest at infinity.This is another requirement imposed to the wavefunction,besides the even character.The expectation value of the momentum operator computed with the wavefunction that we found must go to zero in the asymptotic limit of the particle's coordinate.

    To be continued.

  14. May 11, 2005 #13


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    And now to your dissapointment,i'll leave it to the OP (or someone else) so finish the problem.

    The integral in the equation (21) can be simply computed if you take a look at post #3 of this thread https://www.physicsforums.com/showthread.php?t=61393&highlight=delta+Dirac and aply the theory exactly on this case.

    You'll get 2 solutions.Now,due to the analysis made before,you'll see that,either the [itex] \cos [/itex] in the eventual wavefunction,or the [itex] \cosh [/itex] are nonnormalizable wavefunctions,therefore,do not describe physical states.

    So the answer to the problem should be

    "The potential

    [tex] \hat{V}=V|v\rangle \langle v| [/tex]

    does not admit physical states for a continuous energy spectrum [itex] \{E\}\subseteq\mathbb{R}^{-} [/itex]".

    Surely,one must ask whether that potential admits bound states (discrete energy spectrum contained again in the negative semiaxis).I'll invite the curious reader to solve the integro-differential eq.(9) with the requirement of even wavefunctions (i hope u see why) through other method than the one described (viz.Fourier decomposition -->wave packets--->continuous spectrum).

    Last edited: May 11, 2005
  15. May 11, 2005 #14
    Sorry, I thought it was evident. You have done almost all the job to solve it, at least formally.

    [tex]\langle p | \hat{H} | \Psi \rangle = \frac{p^2}{2m}\langle p | \Psi \rangle - V \langle p | v \rangle \int_{-\infty}^{+\infty}\ \langle v | p^\prime \rangle \Psi (p^\prime) d p^\prime=E \langle p | \Psi \rangle \ \ (a)[/tex]

    Now, you have the contant [itex] C(\Psi)= \int_{-\infty}^{+\infty}\ \langle v | p^\prime \rangle \Psi (p^\prime) d p^\prime[/itex] in the equation above (that will be used to define the energy quantification conditions).

    You have therefore the direct resolution:
    [tex] \Psi(p) = \frac{ V \langle p | v \rangle C(\psi)}{\frac{p^2}{2m}-E} \ \ (b)[/tex]

    And using the equation for the constant: [itex] C(\Psi)= \int_{-\infty}^{+\infty}\ \langle v | p^\prime \rangle \Psi (p^\prime) d p^\prime[/itex]

    you get:

    [tex] C(\Psi)= \int_{-\infty}^{+\infty}\ \langle v | p \rangle \frac{ V \langle p | v \rangle C(\psi)}{\frac{p^2}{2m}-E} dp \ \ (c)[/itex]

    Or if you prefer:

    [tex] 1= \int_{-\infty}^{+\infty}\ \frac{ V \langle v | p \rangle \langle p | v \rangle}{\frac{p^2}{2m}-E} dp \ \ (d)[/itex]

    If E <0, there is no problem. You get the result f(E)=1 and solving it you get energy values. If E>0, there is no solutions (to get [itex] \psi(p) [/itex] as a function): we need to enlarge the space to include the free solutions.


    P.S. These results assume that the solution and its derivatives are is L^2 . If not, the operator p^2 should be changed in order to take into account the discontinuities.

    P.P.S. I think dexter has done an error in the beginning of its demo in the formula:
    [tex] \langle x|\hat{H}|\psi\rangle=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}-\kappa V \int_{\mathbb{R}} \psi (x) e^{-\kappa \left |x\right |} \ dx \ \ (7) [/tex]
    He has forgotten the function exp(-k|x|) in front of the constant integral:

    [tex] \langle x|\hat{H}|\psi\rangle=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}-\kappa V e^{-\kappa \left |x\right |} \int_{\mathbb{R}} \psi (x^\prime) e^{-\kappa \left |x^\prime \right |} \ dx \ \ (7) [/tex]

    However, I have not read carefully all what dexter has written (and I may be wrong).
  16. May 11, 2005 #15


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    Yep,it should have been twice.Okay.I think the OP could solve it,the line is clear enough.

  17. May 11, 2005 #16
    What does mean OP?

  18. May 11, 2005 #17


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    Original Poster (the one who starts the thread).

  19. May 11, 2005 #18
    I thought it was the (h)orrible Person :biggrin:

  20. May 11, 2005 #19
    First of all, thanks to both seratend and dexter for all the help!

    I follow seratend's method. That was very slick. dexter, I'm still digesting your method. Some things that you find obvious I find...well...not obvious :tongue:. I know seratend had already pointed out the missing exponential factor, but in going from equation (9) to equation (21), what happened to the integral over x?

    thanks again
  21. May 11, 2005 #20


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    It disappered.Because it was an odd function integrated on a symmetrical domain wrt the origin and that is automatically 0...

    Even in the wrong form of the integro-diff.eq (9),the results in (10-21) are correct,becasue those factors do not interviene.

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