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Strange QM question.

  1. Sep 26, 2005 #1

    quasar987

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    I got this question here that says there are 1000 neutrons in a one-dimensional infinite potential barrier box of lenght L and at t=0, each has a wave function given by

    [tex]\Psi(x,0) = Ax(x-L)[/tex]

    But isn't this a contradiction? We have derived earlier in class that if a particle is to be a in a box at V = 0, then its wave function will have the form of a sine, not this polynomial.

    The fact that there are many neutrons in the box shouldn't change the form of the wave function, since there is not potential energy associated with the presence of other neutrons around.

    There is also the possibility that it is not really possible that the wave function be like that and this problem is purely artificial. I think it is never a good idea to do this without making mention of it as it confuses the hell out of us. :grumpy:
     
    Last edited: Sep 26, 2005
  2. jcsd
  3. Sep 26, 2005 #2

    Galileo

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    The stationary states (eigenfunctions of H) are the sines you are looking for. The general solution is a linear combination of those sines. Actually, ANY wavefunction which satisfies the boundary conditions can be an initial wavefunction. Your Psi is clearly 0 at x=0 and x=L. To determine its evolution though, you'd have to figure out how to write it in terms of the stationary states.

    So remember. The sines are the eigenfunctions of the Hamiltonian. The wavefunctions for which [itex]H\psi = E \psi[/itex], so they are solutions to the time-independent SE. The real SE though is [itex]i\hbar \frac{\partial}{\partial t}\Psi(x,t)= H \Psi(x,t)[/itex].
     
    Last edited: Sep 26, 2005
  4. Sep 26, 2005 #3

    quasar987

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    :bugeye: This makes sense... in a Galileo type of way (which is good :smile: ).
     
  5. Sep 26, 2005 #4

    quasar987

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    Ok, this detail aside, there is a question in the problem that I don't understand. It asks "How many particles have an energy [itex]E_5[/itex] at t=0 ?"

    I'm guessing he means "What is the probable amount of particle with energy E_5 at t=0?" So I'm thinking that the idea is to find the momentum of the particles with such energy: E_5 = p²/2m <==> [itex]p_5=5\hbar\pi/L[/itex]. Now the question boils down to "What is the probability of finding a particle with momentum p_5 at t=0? Multiply that probability by the number of particles and that is the anwer." But the thing is, the probability of finding a particle with a certain definite precise momentum is zero because it would imply calculating the integral

    [tex]\phi(p) = \int_{p_5}^{p_5} \psi(x,0)e^{-ipx/\hbar}dx[/tex]

    which is 0 because of the integration bounds. :grumpy:
     
  6. Sep 26, 2005 #5

    Galileo

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    Yah, I think the exact meaning of the question is: "If you measure the energy of each particle, how many times will you find an energy of E_5?" Ofcourse this is probabilistic, but for a large # of particles (1000 is not that big, but it will do) the fraction of particles that you'll find with E_5 will be equal to the probability of measuring that energy on a single particle (with neglegible uncertainty). So now you ask: What is the probability of measuring E_5 if a particle is in the given state?

    Why are you fussing with momentum? Apply the general psotulate: If the wavefunction is [itex]|\psi >[/itex] and [itex]|q_n>[/itex] is the eigenvector corresponding to the eigenvalue [itex]q_n[/itex] of some observable Q, what's the probability of finding [itex]q_n[/itex] when you measure Q?
     
  7. Sep 26, 2005 #6

    quasar987

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    We haven't seen that yet! Observable is not defined, and so is the notation [itex]|>[/itex].

    I would really like to know what is wrong with what I said. Isn't energy directly associated with momentum via E = p²/2m for a particle in a box?

    and isn't the probability of finding the particle with momentum between p_1 and p_2 the integral

    [tex]\phi(p) = \int_{p_1}^{p_2} \psi(x,0)e^{-ipx/\hbar}dx[/tex]


    ??
     
  8. Sep 26, 2005 #7

    quasar987

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    Ugh, wait a second, this doesn't make sense; I'm integrating over x and have momenta bounds.

    Right, the equation is

    [tex]P_{[p_1,p_2]} = \int_{p_1}^{p_2} \phi \phi^* dp[/tex]
     
    Last edited: Sep 26, 2005
  9. Sep 26, 2005 #8

    quasar987

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    Is this general or just for the case of the free particle?

    For the case of the free particle, I can see why it is true in the light of Fourier's theorem. Any function satisfying the boundary conditions f(0)=f(L)=0 can be described as a serie of sin wave with carefully chosen coefficient. Since in this case, the eigenfunctions ARE those sine waves, the Fourier serie is just a linear combination of the eigenfunctions, so by linearity of the SE, it is a solution.
     
  10. Sep 26, 2005 #9
    Is this question correct?
    Neutrons are fermions, they obey Pauli Exclusion Principle.
    Can all of them occupy a single state at t=0?
     
  11. Sep 27, 2005 #10

    CarlB

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    It must be 1000 boxes, each with 1 neutron with that wave function.

    Carl
     
  12. Sep 27, 2005 #11

    Galileo

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    It'll be something that'll defniately get covered in the course. It is general. The eigenstates of any observable form a complete set, you can make an orthonormal basis out of them and expand any state in the system as a linear comb. of those basis-states. The Fourier series you get here is just a special case (you're expanding in eigenstates of the energy). If you take the Fourier transform, you're expanding in eigenstates of the momentum.

    I don't really have time now to check if the momentum approach is correct, but suspect so. I do know it's more cumbersome. Haven't you learned how to find the probability of measuring a certain energie E_n? (After, you know how to find the probability of getting a certain momentum).

    That's true. I agree with CarlB that they should be considered 1000 different systems.
     
  13. Sep 27, 2005 #12

    quasar987

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    No but after you mentioned that, I searched the book and saw that it was the very next section of the book that hadn't been covered yet. It should have been covered in class this morning but I didn't go cuz I twisted my ankle badly and can't move. Still that's only 2 days before the handout date of homework. :grumpy:

    I'll get to readin' now. cya
     
  14. Sep 27, 2005 #13

    quasar987

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    Anybody got a commentary regarding the "momentum method", namely the "paradox" that the probability of finding the particle with momentum between p_1 and p_2 the integral

    [tex]P_{[p_1,p_2]} = \int_{p_1}^{p_2} \phi \phi^* dp[/tex]

    but this implies that the probability of finding the particle with momentum exactly p_5 is 0?
     
  15. Sep 28, 2005 #14
    The way I see it is that you're approaching the problem in the wrong way. You're integrating over an interval of zero length and naturally that gives zero as the result. If the particle is in a superposition of momentum states then

    [tex]\phi=\Sigma c_n \phi_n [/tex]

    To find the probability that the particle is in a momentum state p5 you use Fourier's trick to extract c5^2 which gives you the probability of finding the particle in that state.
     
  16. Sep 28, 2005 #15

    quasar987

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    Could it be that simple?

    [tex]P_{[p_1,p_2]} = \int_{p_1}^{p_2} \phi \phi^* dp[/tex]

    gives the probability of finding the paticle with momentum btw p_1 and p_2. So [itex]\phi(p) \phi^*(p)[/itex] gives the prbability of finding the particle with momentum p. No? And phi is just the fourier transform of psi. Piece of cake.
     
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