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Infinite square well potential

  1. Aug 2, 2005 #1
    In the infinite square well potential, the obtained wavefunction is,

    [tex] \psi = \sqrt\frac{2}{a} sin \frac{n\pi x}{a} [/tex]

    and we know that the Hamiltonian commutes with the momentum operator, which implies that the eigenfunctions for the Hamiltonian is exactly the same for the momentum operator. However, when I operate the momentum operator on the wavefunction above, I failed to get a momentum eigenvalue, together with the eigenfunction (wavefunction) again.

    Is there something wrong with the wavefunctions, in that they are actually not the Hamiltonian's eigenfunction, or?
     
    Last edited: Aug 2, 2005
  2. jcsd
  3. Aug 2, 2005 #2
    I think it's because the hamiltonian and the momentum operator don't commute:

    [tex]\left[H,p\right] \psi(x) = \left[V(x),i \hbar \partial_x\right] \psi(x) = -(\partial_x V(x))\psi(x)[/tex]

    So H and p commute only if the potential V is constant. And for the infinite square well this certainly is not the case.
     
  4. Aug 2, 2005 #3
    For an infinite square well potential, the V is not dependent on x. The V is expressed as:

    [tex] V = 0 for \mid x \mid = a [/tex]
    [tex] and V = \infty for others [/tex]
     
    Last edited: Aug 2, 2005
  5. Aug 2, 2005 #4

    Gokul43201

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    What on earth does [itex]|V| = a[/itex] mean ?
     
  6. Aug 2, 2005 #5
    Sorry. I have changed the expression. Thanks for pointing out.
     
  7. Aug 2, 2005 #6

    Gokul43201

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    Touqra, that is not how an infinite square well is defined.
     
  8. Aug 2, 2005 #7
  9. Aug 2, 2005 #8
    Thanks for the site.
    I have read it and from my point of view, the wavefunction expressed in the site are the eigenfunctions of the Hamiltonian, and the site gave also the eigenvalues, E.
    But, are the wavefunctions given in the site, also the eigenfunctions of the momentum operator?
     
  10. Aug 2, 2005 #9

    Gokul43201

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    I was on my way to writing up what values V(x) takes for different x, but marlon did better.

    Nevertheless, the OP still has a relevant question, which I believe, is as yet unanswered. For |x| < a/2, V(x) = 0 and H = p^2/2m

    So, H and p should commute here and hence there should exist a simultaneous eigenbasis for H and p.
     
  11. Aug 2, 2005 #10
    To be sure of that, the Hamiltonian must commute with the momentum operator. Like Gokul explained, this is the case...

    regards
    marlon
     
  12. Aug 2, 2005 #11
    Mathematical theorem will tell us that if both of them commute, they share exactly the same set of eigenfunctions. But, when I operate the momentum operator on the wavefunction,

    [tex] p\Psi = -i\hbar \frac {d \Psi} {dx}[/tex]

    and I will get an imaginary eigenvalue with a different eigenfunction (wavefunction). To note, the sine function, upon differentiate once, will change to cosine; and the cosine function will change to sine.
    All in all, the final eigenfunction is not the same as the initial eigenfunction.

    There is no problem with the Hamiltonian, since you differentiate it twice.
     
    Last edited: Aug 2, 2005
  13. Aug 2, 2005 #12
    If we wouldn't look at the boundary conditions at x=0 and x=a we would find eigenstates of the hamiltonian of the form

    [tex]\psi(x) = e^{- i k x} (k\in \mathbf{Z})[/tex]

    with eigenvalues [tex]k^2 \hbar^2/(2 m)[/tex].

    These are indeed at the same time the eigenstates of the momentum operator at eigenvalue [tex]\hbar k[/tex]. But now we insist that the wavefunctions vanishes at x=0 and x=a. To achieve this we add two eigenstates of the hamiltonian which are eachother complex conjugates:

    [tex]\psi(x) = e^{i k x} - e^{-i k x} = (2i)\sin(kx)[/tex]

    Now this is still an eigenstate of the hamiltonian (because both have eigenvalue [tex]k^2 \hbar^2/(2 m)[/tex]). But we're adding two eigenstates of the momentum operator with different eigenvalues ([tex]\hbar k[/tex],[tex]-\hbar k[/tex]), so this surely isn't an eigenvalue of the momentum operator anymore.

    The point is: the theory doesn't tell you that two operators that commute have exactly the same eigenstates, rather it tells you that there exist some common basis of eigenstates (in this case [tex]e^{i k x}[/tex]). The reason why these two statements are not equivalent in our case, is that the eigenstates of the momentum operator are not degenerate while the eigenstates of the hamiltonian are: both [tex]\psi(x) = e^{- i k x}[/tex] and [tex]\psi(x) = e^{i k x}[/tex] have the same eigenvalue when operated on by the hamiltonian.
     
    Last edited by a moderator: Aug 2, 2005
  14. Aug 2, 2005 #13

    Gokul43201

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    Notice that

    [tex] p\psi = -i\hbar \frac {d \psi} {dx}~~~---(1)[/tex]

    and


    [tex] 2mE \psi = - \hbar ^2 \frac {d^2 \psi} {dx^2}~~~---(2)[/tex]

    are exactly the same equation. You can see this by differentiating (1) and then substituting for [itex]\psi'(x) [/itex] in terms of [itex]\psi(x) [/itex] from (1), and you get....equation (2) !

    So...the eigenfunctions of H must be the eigenfunctions of p, and in the general case, they are both of the form [itex]exp(ipx/ \hbar ) [/itex].

    When you apply the boundary conditions, you end up choosing a convenient linear combination of the energy/momentum eigenfunctions. Since the Schrodinger equation is a linear differential equation, any linear combination of its solutions is also a solution. So, in this case, the chosen energy eigenfunctions are not "the same" as the momentum eigenfunctions, but are merely made by a linear combination of them.

    Edit : Timbuqtu beat me to it...and is essentially saying the same thing.
     
    Last edited: Aug 2, 2005
  15. Aug 2, 2005 #14
    Please if A and B are observables and [A,B]=0 they share the same eigenvectors.

    Now, the quantum well potential is an excellent exercise to understand what is the quantum momentum operator and why it does not commute with the Hamiltonian H=p^2/2m+V(x).
    the QM momentum p is the generator of space translations, an unbounded operator (p=-i.hbar d/dx).
    (it operates on the L^2(R,dx) functions and not on L^2([0,a],dx) functions)

    If [H,p]=0 => the Hamiltonian should be translation invariant (trivial result). This is not the case for the Hamiltonian H of the quantum well: V(x)=/=V(x+a) => V is not translation invariant.
    Any Hamiltonian of the form H=p^2/2m + V(x) where V(x)=/=V(x+a) has this property (hopefully otherwise the predictions of QM would be very trivial).

    Seratend.
     
  16. Aug 2, 2005 #15
    V and p do not commute

    which reflects the fact, already pointed out, that V is not invariant by translation.

    The fact that V and p do not commute can be checked by direct calculation.
    It suffices to carefully note that the derivative (p) of a step function is a Dirac distribution.

    To avoid the subtilities of the distribution theory, it is better to avoid such examples and replace them by potentials with steep -but finite- edges !
    For any steep-edged potential, V and p obviously do not commute. The limit case makes no exception.
     
  17. Aug 2, 2005 #16

    Gokul43201

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    Okay, I may be completely off as I haven't thought about this enough. Isn't V(x) locally translationally invariant [ie : inside (0,a)] ? Or is it not meaningful to talk about H(p,x) for x in (0,a) and say that the Hamiltonian in this region must commute with p ?
     
  18. Aug 2, 2005 #17

    reilly

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    Note that for unbounded space, sin(kx-wt) is a valid solution for a free particle hamiltonian. A superpostion of various Lz or Sz states can be a eigenstate of total orbital or spin (or total) angular momenum. Unbroken symmetries lead to degeneracies -- the free hamiltonian is invariant under parity; momentum is not. So they cannot share identical eigenstates. Under a parity xform, the eigenstate |E,p> goes into |E,-p> where E=p**2/2m, unchanged by parity.

    regards,
    Reilly Atkinson
     
  19. Aug 3, 2005 #18
    local invariance, local commute, local share of eigenfunctions

    Gokul,

    You can indeed talk about local invariance as long as the "wavelength" of the solution is short compared to the width of the potential well. In this approximation, it will be no surprise that the eigenfunctions will be approximatively common to p and H. This will immediately be reflected in the solution itself: it will be approximately harmonic modes, as long as you don't go too close to the edges. You can check that easily.
     
  20. Aug 3, 2005 #19
    Yes. Quantum well problem is simple as long as we do not care on the mathematical rigor. It may be viewed as a good introduction to the differences and similarities between L2 and l^2 spaces (and why we may see the l^2 space as the space of periodic functions).

    Yes because V(x)= cte inside (0,a). But what does means locally translationally invariant? (what properties do we have/deduce with such a definition?). The genuine momentum p is defined on the real line (note the translation group is define on |R and not on a finite interval).

    The difficulty of the quantum well problem comes from the fuzziness on the space of wave functions (always skipped in most of QM courses). Do we consider {L^2(R, dx), f(x)=0 is x not in (0,a)} functions or L^2[(0,a),dx] functions (equivalently periodic functions)?

    It is somewhat like considering a periodic function on the real line and the same function on a given interval period. The first one may be viewed as a l^2 vector and the second one as a L^2(R,dx) vector if we say it is null outside the given interval period. All these functions give the same result inside e.g. (0,a) but they are different functions.


    This is not meaningful if you consider p as the space (|R) translation generator.
    Now you may define another "translation" operator that commutes with H. For example Tn(psi(x))=psi(x+na) for all n in Z when we consider the hilbert space of periodic functions vanishing at x=0 and x=a or the “translation” operator TxTy=T_moda(x+y) on (0,a] with may be some problems on the boundaries.

    Seratend.
     
  21. Aug 3, 2005 #20

    reilly

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    Another way to examine seratend's correct approach is simply to note that the Hamiltonian in question is T + V*THETA{(X-A)*(B-X)} with T=KE and as V-> infinity -- a mathematical nightmare, but just another day's work for a physicist. Let Y be outside the interval, say Y= B+z, with z>0. But if z goes negative,then the particle can be either in the potential well, or outside. So a shift in coordinate can change the form of the wave function: bound state <-> free state and free state <-> free state ( the scattering solutions).

    The system is not translationly invariant in the relative distance between particle and potential-- things would look pretty different to a batter if the pitcher were only 25 feet away rather than 60. Momentum is clearly not
    conserved -- there are forces, highly localized ones to be sure, but forces nonetheless.
    On the other hand, the entire system, particle + potential, is translationally invariant. The latter makes sense if we think of a center of Momentum system, with the coordinates of the CM factored out.

    However, as I pointed out above, this problem holds for a free, unconstrained particle with H=T. The problem exists whether or not momentum is conserved. So what's going on?

    The solution to all this was given by Arnold Sommerfeld roughly 100 years ago, and is termed the Sommerfeld Radiation Condition. Differential equations need boundary conditions, and that's Sommerfelds solution. He looks at incoming and outgoing waves, appropriate for scattering problems, and requires wavelike solutions of any DE to approach , generically, exp (ikx), as x->infinity. This BC fixes the appropriate Hilbert Space of functions wih Fourier transfoms -- including distributions.

    The "in" and "out" states of QFT are exactly the asymptotic states of Sommereld.

    Note: Parity and momentum do not commute with each other, but they both commute with H = T. That means parity helps define one basis for H's eigenstates, and p another -- sinkx vs exp(ikx)

    Regards,
    Reilly Arkinson
     
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