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Probability of finding particle in shifted ground state

  1. Nov 24, 2014 #1
    1. The problem statement, all variables and given/known data
    At t<0 a particle is in the ground state of the potential V(x)= [tex] \frac{1}{2} mw^2x^2 [/tex]. At t=0 the potential is suddenly displaced by an amount x0 to V(x)= [tex] \frac{1}{2} mw^2(x-x_0)^2 [/tex] .

    a) What is the probability of the particle being in the ground state; the first excited state?
    b) At t=[tex] \frac{2pi}{w} [/tex] write the wave function

    2. Relevant equations


    3. The attempt at a solution
    I think that the wave function should be [tex] Ψ_0(x,0)= (\frac{mw}{πh})^\frac{1}{4} e^ \frac{-mw(x-x_0)^2}{2h} [/tex]
    I'm not sure what I should do though in order to find the probability in the ground state.
    For the first excited energy I know I need to use the raising operator in order to get the wave function, which would give me [tex] (\frac{mw}{h})^\frac{1}{2} [/tex] times [tex] Ψ_0 [/tex] but again i'm unsure how to find the probability.
     
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  3. Nov 24, 2014 #2

    Simon Bridge

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    Express the original state in terms of the new ones.
     
  4. Nov 24, 2014 #3
    I'm not exactly sure what you mean in order to go about doing that.
     
  5. Nov 24, 2014 #4

    Simon Bridge

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    The particle still has the same wavefunction after the potential has been changed.
    However, that no longer corresponds to an energy eigenstate.
    Expand the original wavefunction as a linear sum of the new energy eigenstates.
    What do the coefficients of the expansion represent?
     
  6. Nov 24, 2014 #5
    So I should expand Ψ =∑CnΨn where |Cn|^2 is the probability of the new energy eigenstate?
     
  7. Nov 24, 2014 #6
    Does the change in time from t<0 to t=0 not affect it in any way?
     
  8. Nov 24, 2014 #7

    Simon Bridge

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    The effect is in the possible outcome of future measurements.
     
  9. Nov 24, 2014 #8

    vela

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    This is the ground state of the system after the potential is displaced. It's not the state of the particle at ##t=0##, which is still in the state
    $$\Psi(x,0) = \left(\frac{m\omega}{πh}\right)^\frac{1}{4} e^{-\frac{m\omega x^2}{2h}}.$$ By the way, the frequency is denoted by the Greek letter ##\omega##, not ##w##.

    Look up the definition of the probability amplitude. This is a basic definition you need to know.
     
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