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Quantum physics - probability density,

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    consider a particle at an interval ##[-L/2, L/2]##, described by the wave function ## \psi (x,t)= \frac{1}{\sqrt{L}}e^{i(kx-wt)}##
    a) Calculate the probability density ##\rho (x,t) ## and the current density ## j(x,t)## of the particle
    b) How can you express ## j(x,t)## as a function of ##\rho (x,t) ## and the velocity ## v##?
    c) What is de probability to find the particle at the intervals ##[-L/2, L/2]##, ##[-L/2, 0]## and ##[0, L/4]##

    (note: not sure whether 'current density' is the proper translation of the term I have in Dutch for j(x,t).

    2. Relevant equations
    ##\rho (x,t) = \psi^* (x,t) \psi (x,t)##
    ## j(x,t) = - \frac{i \hbar}{2m} \big[ \psi^* (x,t) \frac{\partial}{\partial x}\psi (x,t) - \psi (x,t) \frac{\partial}{\partial x}\psi^* (x,t) \big]##
    ##P(\Omega)= \int_{\Omega} |\psi (x,t)|^2 dx##
    no clue on the relevant equation for b and c

    3. The attempt at a solution
    a) ##\rho (x,t) = \frac{1}{\sqrt{L}} e^{-i(kx-wt)} . \frac{1}{\sqrt{L}}e^{i(kx-wt)} = \frac{1}{L} e^{-i(kx-wt)+i(kx-wt)} = \frac{1}{L} e^0 = \frac{1}{L} ##
    ## j(x,t) = - \frac{i \hbar}{2m} \big[ \frac{1}{\sqrt{L}} e^{-i(kx-wt)} \frac{\partial}{\partial x}\frac{1}{\sqrt{L}}e^{i(kx-wt)} - \frac{1}{\sqrt{L}} e^{i(kx-wt)} \frac{\partial}{\partial x}\frac{1}{\sqrt{L}}e^{-i(kx-wt)} \big]##
    ## = - \frac{i \hbar}{2m} \big[ \frac{1}{L} e^{-i(kx-wt)} e^{i(kx-wt)} ik - \frac{1}{L} e^{-i(kx-wt)} e^{i(kx-wt)} (-ik) \big]##
    ## = - \frac{i \hbar}{2m} \big[ \frac{ik}{L} + \frac{ik}{L} \big]##
    ## = - \frac{i \hbar}{m} \frac{ik}{L} = \frac{\hbar k}{m L}##

    b) I was not sure how to interprete the question, express first ## j(x,t)## as a function of ##\rho (x,t) ## AND afterwards in relation to velocity, OR all at the same time?

    I have suggested for myself at least that ## j(x,t) = \frac{\hbar k}{m} \rho (x,t) ##
    I have no idea how the velocity comes into that.

    c) For the last part I have

    ## \frac{1}{L} \int_{-L/2}^{L/2} dx = \frac{x}{L}|_{-L/2}^{L/2} = 2L/2L =1##
    Since the second interval is half of the first interval the probability should be 1/2, and checking the integral that is what I got. Same logic applies for the third given interval, which is a 4th of the first, and the probability is 1/4.

    Can someone please review what I did solve, and whether I applied the correct logic or proper techniques. And can someone help me out with the velocity part?
     
  2. jcsd
  3. Jan 27, 2015 #2

    mfb

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    Staff: Mentor

    There is just one equation that combines them and has a nice expression. Your wave function describes a moving wave. What is its (phase) velocity?

    (a) and (c) look fine.
     
  4. Jan 27, 2015 #3
    I know there a continuity equation: ## \frac{\partial}{\partial t} \rho(x,t) + \frac{\partial}{\partial x} j(x,t) = 0 ##. Is that the one you mean?

    ETA: correction on the continuity equation per mfb's following comment
     
    Last edited: Jan 27, 2015
  5. Jan 27, 2015 #4

    mfb

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    Staff: Mentor

    That continuity equation does not look right (edit: the edited version is correct now), but that is not what I meant.
    What is the phase velocity v of your wave? It is the same as in the classical case.
     
    Last edited: Jan 28, 2015
  6. Jan 27, 2015 #5
    phase velocity : ##v_p = \frac{\lambda}{T} = \lambda f = \frac{\omega}{k} ##
    group velocity : ##v_g = \frac{\partial \omega}{\partial k} ##
     
  7. Jan 27, 2015 #6
    I know I could rewrite the wave in its classic notation as ## \psi (x,t) = \frac{1}{\sqrt{L}} sin (kx - wt) ##
    I know the phase velocity is ## v_p = \lambda f = \frac{\omega}{k}##
    But I don't yet see how this relates to the probability density and j(x,t)... Help very much appreciated.
     
  8. Jan 28, 2015 #7
    Found the solution:
    De Broglie: ##p = \hbar k##
    Classical mechanics: ## p=mv##

    Hence, ## j(x,t) = \frac{\hbar k}{mL} = \frac{p}{mL} = \frac{v}{L} = v \rho(x,t)##

    So, I started out correct by replacing the ## \frac{1}{L}## with the ## \rho##, but failed to recognize De Broglie relation and to apply Classical mechanics in that.
     
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