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Confusion about eigenstates and wavefunctions.

  1. Jan 1, 2008 #1
    This is the way I understand it. Correct me if I'm wrong.

    A 'particle' in a given situation will be in a state [tex]|\psi>[/tex], which is determined by the Schrodinger Equation. After measurement, the particle will then go to a state [tex]|\omega>[/tex], where [tex]|\omega>[/tex] is an eigenvector of the operator corresponding to the measurement. The measurement you'll read will be the corresponding eigenvalue.


    Now this is where the confusion comes in. If I make any sort of errors here, feel free to correct me. I know there's a huge error here somewhere, but I just don't know where it is.

    Let's say we have the particle in a box scenario, with the box going from 0 to a. In this case, the (normalized) solutions of the time independent Schrodinger Equation are of the form [tex]\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})[/tex] or [tex]\frac{i}{\sqrt{2a}}(e^{\frac{-in\pi x}{a}}+e^{\frac{in\pi x}{a}})[/tex].

    Let's say we were to measure the momentum of the particle. We know that in the X basis, the momentum operator is [tex]i\hbar\frac{d}{dx}[/tex], with eigenvectors being (constant multiples of) [tex]e^{-\frac{ipx}{\hbar}}[/tex]. Since [tex]p=\pm\sqrt{2mE}=\pm\hbar k=\hbar \frac{n\pi}{a}[/tex], we can write the momentum eigenvectors as [tex]e^{\frac{-in\pi x}{a}}[/tex]. [tex]e^{\frac{in\pi x}{a}} [/tex] also happens to be a solution if you consider the negative value of momentum.

    Since the general solution of the Schrodinger equation in this well are always linear combinations of [tex]e^{ikx}[/tex] and [tex]e^{-ikx}[/tex], we know that eigenstates of the momentum operator are also solutions to the Schrodinger equation. Everything is fine so far. The problem comes when you note that in order for a linear combination of the two forms of eigenfunctions for the momentum operator isn't a solution itself (unless one of the weights are zero). In other words, [tex]C_1e^{ikx} + C_2e^{-ikx}[/tex] is only an eigenfunction of the momentum operator if [tex]C_1[/tex] or [tex]C_2[/tex] is zero. However, the boundary conditions on the Schrodinger equation requires that the state vector [tex]C_1+C_2=0[/tex] (the boundary condition at the left point). This means that the only state that is a solution to the Schrodinger equation satisfying the boundary conditions and happens to be an eigenfunction of momentum operator is the trival solution.

    What I'm wondering is what happens after you measure the momentum of a particle. What state does it go to? It would seem that it can't collapse to a momentum eigenstate because the momentum eigenstate doesn't satisfy the Schrodinger equation after applying the boundary conditions.
     
  2. jcsd
  3. Jan 1, 2008 #2

    olgranpappy

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    Well, no. Since they don't obey the boundary conditions... But anyways, your question (which I'll get to below) is a very good one, and the real issues don't come from boundary conditions. The PIAB is a funny little problem. It is a great pedagogical example in many ways, but the unphysical (infinite) potential can lead to some problems.

    So, to answer your question (as best I can). Recall that, strictly speaking, momentum eigenstates
    (and also position eigenstates) are not physical particle states. They do not belong to the set of square integrable functions! They are convienient fictions.

    Next, recall Heisenbergs uncertainty principle. Here is where the trouble with the box comes in. Because of the infinite potential the uncertainty in the position is at most 'a'. Thus, if I were to measure
    the momentum *exactly* (so that the uncertainty in the momentum is zero). We would have a violation of the uncertainty principle.

    As stated above, it's not a problem of the boundary conditions, but rather of the unphysicallity of both the problem itself (PIAB infinite potentials) and the unphysicallity of the momentum eigenstates.

    You can only measure the momentum to at best an uncertainty of about \hbar/a (which is on the order of the eigenvalues itself for the lowest few schrodinger levels).

    But say you do go measure the momentum, pretty well, not exactly, but pretty well. And say you find a value '+q' (plus or minus some relatively small amount which is nonzero due to uncertainty).

    Then the wavefunction is not exactly [itex]\frac{1}{\sqrt{a}}e^{iqx}[/itex] inside the box, but is something that looks very much like e^{iqx} except near the edges where it goes to zero (since the particle still cant be found outside the box).

    And the distribution in position space (|\psi|^2(x)) looks something very much like a constant (like |e^{iqx}|^2 would) except near the edges of the box.

    cheers.
     
  4. Jan 1, 2008 #3
    I thought momentum eigenstates did represent possible states of a particle, and that after a measurement of momentum, the particle collapses to one of these states? From the looks of it, the comment you made is in direct violation of the postulates of Quantum Mechanics (or maybe I'm just interpreting the postulates wrong).

    From Principles of Quantum Mechanics by R. Shankar, page 116, postulate III

    "If the particle is in a state [tex]|\psi>[/tex], measurement of the variable (corresponding to) [tex]\Omega[/tex] will yield one of the eigenvalues [tex]\omega[/tex] with probability [tex]P(\omega)\propto |<\omega|\psi>|^2[/tex]. The state of the system will change from [tex]|\psi>[/tex] to [tex]|\omega>[/tex] as a result of the measurement."

    In this specific case, it seems that this postulate can't be fufilled due to the fact that state vector can't collapse to an eigenstate of the measured operator. Either that, or it will collapse, but will no longer obey (the boundary conditions of) the Schrodinger equation.

    This also goes against what I understand about Quantum Mechanics. I could be wrong, but from my own understanding, the uncertainty principle doesn't place limits on a given measurement. You could measure the momentum of a particle as accurately as your equipment allows. What the uncertainty principle does is place a lower limit on the standard deviation of measurements of an ensemble of particles that are in the same state. If this is the case, an ideal measurement wouldn't violate the uncertainty principle. In fact, Shankar outlines examples of how one could make such ideal measurements (in theory).

    Also, Griffith's "Introduction to Quantum Mechanics" second edition, page 19

    "Please understand what the uncertainty principle means: Like position measurements, momentum measurements yield precise answers---the "spread" here refers to the fact that measurements on identically prepared systems do not yield identical results."

    Correct me if I'm wrong about my understanding if the uncertainty principle.

    The rest of your post makes little sense to me, but that could be because it's 12AM in my time zone, and as I get sleepy, my brain function falls to zero faster than a particle in a bound state :P (Yeah, I know my physics humor isn't all that great :tongue:)
     
    Last edited: Jan 1, 2008
  5. Jan 3, 2008 #4

    olgranpappy

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    the measurement projects on a subspace. what happens if that subspace has a dimension greater than one? I.e., if the eigenvalue you measure is degenerate?

    yes, that's so.

    For the state psi = e^{iqx} the probability density |psi|^2 is one everywhere in space. The particle is equally likely to "be anywhere". If you work out <x>, <x^2>, <p>, <p^2> for this state you can see that

    <delta x> = system size

    <delta p> = zero

    so that

    <delta x><delta p> = 0

    Whereas in general one must have

    <delta x><delta p> > hbar/2

    which is only possible for a plane wave as the system size goes to infinity.

    the mathematicians usually exclude vectors like |x> and |p> explicitly because they have infinite norm and thus, as I said, dont properly belong to the Hilbert space.... but Shankar specifically adds these vectors into his definition of a "physical" Hilbert space. He's free to do that, but it introduces subtleties that he does not address.

     
  6. Jan 3, 2008 #5

    malawi_glenn

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    woozie: have you written the SE in momentum representation when you inserted the momentum eigenfunction?

    The usual form of SE is in postition representation:
    [tex]( \dfrac{\hat{p}^2}{2m} + V(x))\psi (x) = E\psi (x) [/tex]
    (sorry for the sloppy TeX)

    In order to get the momentum wave functions, you do the (inverse) fourier transform of the position wave functions that satisy SE above.

    I dont think you have taken this into account in your original post, thats why you are confused.

    When I use SE (for finding wave functions for the deuteron for example), i first start with the SE in position representation, then i solve it. I obtain the poistion eigen functions. Then I transform them into momentum wave functions, then i calculate things like <p>, <p^2> etc.
     
    Last edited: Jan 3, 2008
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