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I Particle in the box eigenfunctions

  1. Jan 30, 2017 #1
    Hello Everyone,

    For the particle in the 1D box of width ##L##, the time invariant Schrodinger equation is cast in the form of the Hamiltonian operator and automatically leads to the energy eigenfunctions

    $$\Psi(x) = \sqrt{\frac 2L} sin(n\pi x/L)$$
    I know that these energy eigenfunctions ##\Psi(x)## have each a well defined energy and are NOT simultaneous eigenfunctions of the linear momentum operator ##\hat p##. The energy eigenfunctions ##\Psi(x)## are simultaneous eigenfunctions of the magnitude squared of the momentum operator ##\hat p##.

    To find the eigenfunctions ##\Phi## of linear momentum operator, should we simply apply the operator ##-i \hbar \frac \partial {\partial x}## to the energy eigenfunctions? The result would be a new function that represents the linear momentum eigenfunction ##\Phi(x)##...

    What other eigenfunctions can we find for the particle in the box? What other operators can we apply?
    I know that the particle in the box does not have any "rotational attributes" so the angular momentum operator is not applicable...But what if we applied the angular momentum operator (in position space) to the functions ##\Psi(x)##? What would we obtain?

    Thanks!
    Fog37
     
  2. jcsd
  3. Jan 30, 2017 #2

    PeterDonis

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    No. An eigenfunction of momentum would be a function ##\Phi## that satisfies ##\hat{p} \Phi = p \Phi## (notice no hat on the RHS, ##p## there is the eigenvalue, i.e., the value of the linear momentum). It should be easy to show that the energy eigenfunctions ##\Psi(x)## do not satisfy this equation.

    Have you tried it?
     
  4. Jan 30, 2017 #3
    Given
    $$\Psi(x) = \sqrt{\frac 2L} sin(n\pi x/L)$$

    $$-i \hbar \frac \partial {\partial x} \Psi(x) = -i \hbar \sqrt{\frac 2L} (n\pi/L) cos(n\pi x/L)$$

    So the linear momentum eigenfunction will be ##-i \hbar \sqrt{\frac 2L} (n\pi/L) cos(n\pi x/L)## which is clearly different from the energy eigenfunction ##\Psi(x)##

    Fog37
     
  5. Jan 30, 2017 #4

    PeterDonis

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    No, it won't. Did you read my previous post?
     
  6. Jan 30, 2017 #5

    PeterDonis

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    To clarify further, it is also easy to show that the functions you get by applying ##\hat{p}## to the energy eigenfunctions don't satisfy the momentum eigenfunction equation either.
     
  7. Jan 30, 2017 #6
    Sorry. I have read it completely just now and I am now clear on the fact that the energy eigenfunctions are NOT also eigenfunctions of the linear momentum operator.

    Now, given that I know the expression for the linear momentum operator, which is ##-i \hbar \frac \partial {\partial x}##, and that its eigenfunctions must satisfy the eigenvalue equation ##-i \hbar \frac \partial {\partial x} \Phi(x) = p \Phi(x)##, which is how do we go about finding these functions ##\Phi(x)##?
     
  8. Jan 30, 2017 #7
    Ok, I have made some progress:

    We need to hypothesize that the linear momentum eigenfunction have a certain form, like ##e^{ikx}##. This function has a specific momentum eigenvalue ##k \hbar##.

    The energy eigenfunction ##sin(n\pi x/L)## will be a linear combination of two momentum eigenfunctions: ##e^{ikx}## and ##e^{-ikx}##. When we measure the momentum we would obtain either ##- k \hbar## or ##k \hbar##

    Is that correct?
     
  9. Jan 30, 2017 #8

    PeterDonis

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    You guess. :wink: There isn't any cookie cutter procedure to do it.

    Knowledge of general properties of functions can help. For example, the eigenvalue equation is basically telling you that if you differentiate the function ##\Phi(x)## with respect to ##x##, you get the same function (multiplied by some constant). What kind of function has that property, that it stays the same when differentiated?

    Knowledge of basic QM can also help. For example, if we express a momentum eigenfunction as a function ##\Phi(p)## of momentum ##p## instead of position ##x##, what should it be? That should be an easier question to answer than the question of what ##\Phi(x)## should be. Then once you have ##\Phi(p)##, you can just Fourier transform to get ##\Phi(x)##.
     
  10. Jan 30, 2017 #9

    PeterDonis

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    Yes, that's a good guess. :wink: For extra credit, you could try the other method I suggested, of guessing a function ##\Phi(p)## that expresses a momentum eigenfunction as a function of momentum (instead of position), and then Fourier transforming it.

    Yes, which is why it isn't itself an eigenfunction of momentum.

    Yes. Can you see what this is telling you physically?
     
  11. Jan 30, 2017 #10
    Ok, I can see that if the momentum eigenfunction ##\Phi(p)= \delta (p-p_0)##, we would obtain, by Fourier transform, the momentum eigenfunction ##e ^{ikx}## in position space.

    As far as your question "Can I see what this is telling me physically?" I guess it means that....we cannot know the energy and the momentum exactly simultaneously. There is an uncertainty in momentum when the system is in a state of exact energy.

    I guess the whole discussion would also apply to position eigenfunctions of the the particle in the box. By the uncertainty principle, momentum and position are uncertain. But now, what are the position eigenfunctions ##\Theta(x)## of the position operator ##\hat x##, i.e. ##\hat x \Theta(x)= x \Theta(x)## ?
     
  12. Jan 30, 2017 #11

    PeterDonis

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    Exactly.

    That is true, but notice that the two possible momentum values, ##k \hbar## and ##- k \hbar##, have the same magnitude but opposite directions. (In 3 dimensions, the possible values would be 3-vectors with the same magnitude pointing in all possible directions.) So the magnitude of the momentum (more precisely, its square) is determined by the energy; what is not determined is its direction.

    Good question. Can you make a guess? (Hint: try a guess similar to the one you made for momentum eigenfunctions as functions of momentum.)
     
  13. Jan 30, 2017 #12
    well, I know they are ##\delta(x-x_0)## but I am not sure these are the position eigenfunctions specific to the particle in the box.
    In general, aren't the eigenfunctions of a specific operator specific to the particular physical situation and it all depends on the type of ##V(x)##? For instance, the position eigenfunctions for a free particle and the position eigenfunctions for the particle in the box may be different.

    Assuming the ##\delta(x-x_0)## are indeed position eigenfunctions for the particle in the box, an energy eigenfunction like ##sin(kx)## would be a superposition of many position eigenfuctions....
     
  14. Jan 30, 2017 #13

    PeterDonis

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    The position operator is the same as for any other case, so the eigenfunctions will be the same as well.

    No. The specific operator itself, the one that is relevant to whatever physics you are trying to capture, is what depends on the particular physical situation. But given a particular operator, its eigenfunctions are what they are; that's a matter of math, not physics.

    The potential ##V(x)## will affect the energy operator (the Hamiltonian); a different ##V(x)## means a different Hamiltonian. So there will be different energy eigenfunctions, because the Hamiltonian operator is different.
     
  15. Jan 31, 2017 #14
    Thanks PeterDonis.

    Also, once we find the energy eigenfunctions of the particle in the box problem with infinite potential, we need to multiply each function by the time dependent part
    $$e^{-iEt/ \hbar}$$

    By linear superposition, i.e. by a weighted summation of stationary states each multiplied by their time-dependent function, we can generate a general solution to the time-dependent Schrodinger equation. But Isn't the particle in a box a time-invariant problem that is described only by the time independent Schrodinger equation? I guess not, since, our starting point was to assume solutions of the form ##\Psi(x) \Phi(t) ## so the particle in a box still looks for solutions to the time dependent Schrodinger equation which are just separable and are obtained through the time independent equation. The point is that all wavefunctions will be functions of the type ##\Psi(x,t)## but the probability density function may or may not depend time depending on the problem...
     
  16. Jan 31, 2017 #15

    PeterDonis

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    No. The particle's state does not have to be time-independent.
     
  17. Jan 31, 2017 #16
    You are right. I thought about it deeply today.

    We start with the full time dependent Schrodinger equation, we assume we can find separable solutions like ##\Psi(x,t)= \Phi(x) \Theta(t)## and insert that into the TDSE. From there we obtain two ODEs. One of them is the TISE since it only applies to the function ##\Phi(x)## and there is not time parameter in it. By solving the TISE we find the functions ##\Phi## which we later multiply by ##\Theta(t) = e^{i\omega t}##.
    The full and final solutions ##\Psi(x,t)= \Phi(x) \Theta(t)## are the stationary, bound solutions because their associated probability function is time invariant.

    In essence, the particle in a 1D box problem is eventually about finding solutions to the time-dependent Schrodinger equation like any other problem in QM. The TISE is just an intermediate step...
     
  18. Jan 31, 2017 #17

    PeterDonis

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    This is not just an arbitrary assumption. If you want to find a solution whose probability function is time independent, it must be separable in this way (plus, the function ##\Theta(t)## must have the specific form you give later on). Can you see why?
     
  19. Jan 31, 2017 #18
    Ok, the fact that the solution is separable leads to to an ODE for the time part that has ##\Theta(t)## solutions that are time-harmonic, i.e. of the form ##e^{i\omega t}##.
    The product ##\Psi(x,t)^* \Psi(x,t)## is the probability density function which does not depend on time ##t## if the function ##\Theta(t)## is a complex exponential ##e^{i\omega t}##. Is that what you meant?
     
  20. Jan 31, 2017 #19

    PeterDonis

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    That's part of it; but it assumes that ##\Psi(x, t) = \Phi(x) \Theta(t)##. But that, i.e., separability, also can be proved given the requirement that the probability density function does not depend on ##t##; you don't have to assume it.
     
  21. Jan 31, 2017 #20
    Ok. Thanks.

    Why all eigenfunctions of various operators do not depend of time and can have a position space representation like ##\Psi(x,y,z)## that does not involve time? Is it because the operators don't depend on time themselves?
     
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