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Free particle at time t

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    a free particle of mass m moving in one dimension is known to be in the initial state
    ψ(x,0)=sin(k_0 x)
    1. what value of p (momentum) will measurement yield at the time t,and with what probabilities will these values occur?
    2. suppose that p is measurement at t=3 s and the value (h/2pi)(k_0) is found. what is ψ(x,0)
    at t>3 s?

    2. Relevant equations quantum mechanics by Liboff chapter 6

    3. The attempt at a solutionI dont know, must I normalize it? for part 1 when I integrate the expectation value of the momentum that is infinite because we have this
    ∫_(-∞)^∞▒〖sink_0 x) cos〖k_0 x〗 dx〗
    and another question what probabilities occur?
    for parti 2. I dont know what can I do. can I write( exp i(k_0 x)-exp i(k_0 x))/2 and we know for free particle we have A exp i(k_0 x)-Bexp i(k_0 x) and I use it?
    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 12, 2008 #2
    Hi najima,

    Free particles don't exist in stationary states, so there is no definite energy of the particle. Have you learned about wave packets and Fourier transforms yet?
  4. Dec 12, 2008 #3
    yes,so you mean I use delta function .I try but I cant integrate.
  5. Dec 12, 2008 #4
    Yes, you will integrate, but "not" using the delta function. You are given the initial wave function. From here you can compute its fourier transform to construct the wave packet function, which is a function of k. Does this help?
  6. Dec 12, 2008 #5
    could you help me more?
  7. Dec 12, 2008 #6
    so you mean b(k)=∫_(-∞)^∞ sin (k_0 x)exp i(kx)/(2pi)^1/2 is it infinite?
  8. Dec 12, 2008 #7
    I think I made mistake. Let's look at this from the beginning. The initial wave function is

    [tex] \Psi\left(x,0\right) = sin\left(k_{0}x\right) [/tex]

    Are you given any boundary conditions for this wave function, or is it over the whole domain of space?

    EDIT: Actually, there is no mistake. It just leads to same result.
    Last edited: Dec 12, 2008
  9. Dec 12, 2008 #8
    No,I am not given any boundary condition. I wrote everything that are given by the question.
  10. Dec 12, 2008 #9
    Okay, good. So, what does this tell you about the localization of the particle? Think about the initial wave function when making this determination.

    EDIT: Once you determined this, think about what this means in regards to the momentum of the particle.
    Last edited: Dec 12, 2008
  11. Dec 12, 2008 #10
    I can't find out.you mean the initial state determined boundary conditions?
  12. Dec 12, 2008 #11
    Yes. The initial state is the given by wavefunction. You said it's not normalizable. Why is this? It's wavefunction does not vanish at +/- infinity. What does this say about the localization of the particle? The particle has an ill-defined position. If this still confuses you graph [tex] \Psi\left(x,0\right)^{2} [/tex], the probability density.

    Hint: Think about what this means according to the uncertainty principle. And then, how is k related to the momentum? =)
    Last edited: Dec 12, 2008
  13. Dec 12, 2008 #12


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    Homework Helper

    This is bad (imprecise) notation, and you are right to be confused. Basically, the problem is that this is not the wavefunction; it is proportional to the wavefunction (so this doesn't exactly describe the state). And, in particular, the state of a free particle is not a member of the Hilbert space, but we are sloppy in physics and we ignore this "subtle" problem. Basically, buffordboy is hinting as the resolution to the problem, but it can be handled without going there if you are willing to tolerate some physicist's slop.

    Well, strictly speaking, if they give you a wave function, then THEY must normalize it, or at least put a normalization factor out front, or they should not call it a state. (See my previous comments.).

    Yes, because you are not using the true state in your calculation. You can use the more applicable definition of the expectation value:
    This allows you to generalize the concept of "state" to any normalization.

    Something like that, but don't forget all of the i's.
  14. Dec 12, 2008 #13
    for part 1 you mean I must write ∫_(-∞)^∞〖sink_0 x) cos〖k_0 x〗 dx〗/∫_(-∞)^∞〖sink_0 x) sin〖k_0 x〗 dx〗?am I right?
  15. Dec 12, 2008 #14
    bufford means that particle can exist everywhere because of sin function? so it is infinite.yes?
  16. Dec 12, 2008 #15
    Thanks Turin. I had some confusion on this as well about the initial wave function.

    Najima have you learned about the bra/ket notation to understand Turin's remark.
  17. Dec 12, 2008 #16
    What is infinite?
  18. Dec 12, 2008 #17
    yes .I've learned. :)
    is it this∫_(-∞)^∞〖sink_0 x) cos〖k_0 x〗 dx〗/∫_(-∞)^∞〖sink_0 x) sin〖k_0 x〗 dx〗?but I can integrate it yet.
  19. Dec 12, 2008 #18
    probability of particle existence .
  20. Dec 12, 2008 #19
    najima, here was my thinking. The position of the particle is ill-defined, just like you said. The uncertainty in its position approaches infinity. But it has a well-defined wavelength, which is related to [tex] k_{0} [/tex]. And how is this related to momentum?

    I am going to grab a cup of coffee and will be right back on trying to work out the same result using Turin's suggestion.

    EDIT: Using the exponential terms for sine, Maple says that the integral is equal to zero when computing the expectation value for p. The expectation value makes sense physically, and could be deduced without any calculations. But, I don't think this is what the problem is asking for, now is it? The momentum of the particle is well-defined but is equally likely to have positive or negative direction, so this is why the expectation value is zero. You shouldn't need to any integrals to determine what the momentum of the particle is.

    Going back to what I said about using Fourier transforms, you should find that [tex] \Psi\left(x,t\right) [/tex] is constant, which may useful for the second part of the problem.
    Last edited: Dec 12, 2008
  21. Dec 12, 2008 #20


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    Homework Helper

    editting; please wait.

    Not exaclty. This is hard for me to read. This is what I see:


    Be careful about what the derivative of sine is, and what is the exact definition of P as an operator on the coordinate basis. Also, it may be easier for you to see the answer (i.e. do the integral) if you use the complex exponential decomposition that you suggested in another post. However, I suggest that you first identify why I say that this integral expression for the expectation value of P is wrong.
    Last edited: Dec 12, 2008
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