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Harmonic oscillator

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    particle in ground state of 1D harmonic oscillator - spring constant is doubled - what is the probability of finding the particle in the ground state of the new potential



    2. Relevant equations
    v=1/2kx^2 oscillator potential
    wavefunction ground state n=0 = (alpha/pi)^1/4*e^[(-apha*x^2)/2]
    alpha = sqroot[(k*m)/hbar^2)]


    3. The attempt at a solution

    the probability will be = integral[wavefunction old*wavefunction new]dx

    initial equation d^2psi/dx^2 + [2m(E-1/2kx^2)/hbar]psi = 0
    if you double k in the potential V(x)- the equation is d^2psi/dx^2 + [2m(E-kx^2)/hbar]psi = 0

    does this change alpha to 2k? this makes the integral very complex to solve.
    or can this be done by changing the wavefunction for the new potential to (alpha/2pi)^1/4*e^[(-apha*x^2)/2]

    the change being 2pi - doubling the range of the spring constant?
    Thanks!
     
  2. jcsd
  3. Nov 6, 2007 #2

    Avodyne

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    No. How is alpha related to k? What happens to alpha if k is doubled? Does it stay the same? Double? Or what?
     
  4. Nov 7, 2007 #3
    alpha = sqrt(k*reduced mass/hbar^2)
    so doubling k would just be alpha = sqrt(2k*reduced mass/hbar^2)

    so if i take the 2 out it is basically just multiplying alpha by sqrt(2). and substituting this into my wave function would give me the wave function for the new potential. does that make sense?
     
  5. Nov 7, 2007 #4

    Avodyne

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    Yes. So now you have the two wave functions. Earlier, you wrote

    the probability will be = integral[wavefunction old*wavefunction new]dx

    but this is not quite right. The integral gives you the amplitude; what do you need to do to get the probability?
     
  6. Nov 7, 2007 #5

    malawi_glenn

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    kl14 :

    Your original quesion was how to solve the integral. Avodyne pointed out an error you did with the sqrt(2).

    But to calculate the integral, you get something like:
    [tex] k_1 k_2 e^{-a_1 x^2 }\cdot e^{-a_2 x^2} = K \cdot e^{-(a_1 +a_2 )x^2} [/tex]
    as integrand, where a_1 is the old number you have in the old wave function and so on. There is not any harder to calc this integral then the ordinary one, just make the substitution (a_1 + a_2) = A, and integrate from x=-inf to +inf; that integral exists in standard integral-tables.
     
  7. Nov 7, 2007 #6
    yes, i realize my error in typing - the probability is the amplitude squared - the value obtained after the integration.
     
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