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Time-dependent Schrodinger equation problem

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data

    For a harmonic oscillator in a state such that a measurement of energy would give either [itex]1/2\hbar\omega[/itex] or [itex]3/2\hbar\omega[/itex] with equal probabily. Write the wavefunction solution to the time-dependent Schrodinger equation.

    2. The attempt at a solution

    Given those energies I know [itex]\psi_0[/itex] and [itex]\psi_1[/itex] are the two states but I'm asked to chose the coefficients in front of them in the most general way, i.e. don't assume they are real. How do I do this?
     
  2. jcsd
  3. May 27, 2009 #2

    Hao

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    Re: Ho

    Separation of variables of the time-dependent Schrodinger's equation gives a spatial component and a time-dependent component.

    Unless you have already accounted for it, [tex]\psi_0[/tex] and [tex]\psi_1[/tex] only represent the spatial component. The key is to find the time-dependent component to tack on.

    Once you find that, you will be able to see how the coefficients are pretty arbitrary.
     
  4. May 27, 2009 #3
    Re: Ho

    Won't the time dependent part simply be [itex]\exp(-iE_nt/\hbar)[/itex]? And this will vanish for [itex]\Psi^*\Psi[/itex] where,

    [tex]\Psi = \alpha \psi_0 \exp(-iE_0t/\hbar) + \beta \psi_1 \exp(-iE_1t/\hbar)[/tex]

    So can I then normalize each term to 1/2?

    I don't see how I'd know when to use [itex]i/\sqrt{2}[/itex] instead of [itex]1/\sqrt{2}[/itex] as a coefficient for a sate that makes up 1/2 the wavefunction.
     
  5. May 27, 2009 #4

    Hao

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    Re: Ho

    It is true that the time dependent factor will vanish for [itex]
    \Psi^*\Psi
    [/itex] when [itex]
    \Psi
    [/itex] is a pure state, ie. only one basis component.

    However, if you were to work it out explicitly for [itex]
    \Psi
    [/itex] with more than one component, such as for [tex]
    \Psi = \alpha \psi_0 \exp(-iE_0t/\hbar) + \beta \psi_1 \exp(-iE_1t/\hbar)
    [/tex], you will find out that the exponential factor does not cancel out.

    As the exponential factors are not in phase, as time varies, the coefficients can flip signs and phases.

    Once you complete the above, it helps to consider the time average of [itex]
    \Psi^*\Psi
    [/itex].
     
  6. May 27, 2009 #5
    Re: Ho

    Is it correct to say that this [itex]\Psi^*\Psi[/itex] reduces back to [itex]|\alpha|^2 + |\beta|^2 = 1[/itex] by integrating over all space because the pure states are already normalized and they are orthonormal?

    [tex]\Psi^*\Psi = |\alpha|^2|\psi_0|^2 + \alpha^* \psi_0^* \beta \psi_1 \exp(-i(E_0-E_1)t/\hbar) + \beta^* \psi_1^* \alpha \psi_0 \exp(-i(E_1-E_0)t/\hbar) + |\beta|^2|\psi_1|^2[/tex]

    [tex]\int \Psi^*\Psi dx = 1[/tex]

    But then can't alpha be [itex]-i/\sqrt{2}p[/itex] OR [itex]i/\sqrt{2}[/itex] without changing anything?
     
    Last edited: May 27, 2009
  7. May 27, 2009 #6

    Pengwuino

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    Gold Member

    Re: Ho

    Yes, there is a phase of [tex]e^{i\theta }[/tex] that typically goes with the coefficients. The coefficient can be [tex]\frac{i}{{\sqrt 2 }},\frac{{ - i}}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},\frac{{ - 1}}{{\sqrt 2 }}[/tex]. They all result in the same solution in this case, however, so it's no big deal.
     
  8. May 27, 2009 #7
    Re: Ho

    Thanks... it's just the question was worded in a way that made me think I could calculate one of those explicitly.
     
  9. May 27, 2009 #8

    Hao

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    Re: Ho

    That is correct. The orthogonality of the states mean that the space integral of the cross terms is zero, while orthonormality means that the integral of the pure states component gives [tex]|\alpha|^2[/tex] and [tex]|b|^2[/tex].
     
  10. May 27, 2009 #9
    Re: Ho

    does finding the expectation value of momentum of [itex]\Psi[/itex] in the form,

    [tex]<\Psi|P|\Psi> = a \exp(i(E_0-E_1)t/\hbar) + b \exp(i(E_1-E_0)t/\hbar)[/tex] ;

    make sense? Where `a' and `b' are just some constants.
     
    Last edited: May 27, 2009
  11. May 27, 2009 #10

    Pengwuino

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    Gold Member

    Re: Ho

    If you mean does the part of the integral that goes something like [tex]\int {\psi _0 P\psi _1 } [/tex] disregarding the propagator term (since it's just a constant and can be pulled out of the integral), yes, that kind of integral is perfectly valid and makes perfect sense. When you get your answer, think about exactly what a oscillator is and what an expectation value is and it'll make sense!
     
  12. May 27, 2009 #11
    Re: Ho

    I got that result for [tex]\int \Psi^* P \Psi[/tex] where [tex]\Psi[/tex] is the mixed state that I found above.

    That expression looks like it'll give me some sines and cosines, and it makes sense that the momentum would oscillate.
     
  13. May 27, 2009 #12

    Pengwuino

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    Gold Member

    Re: Ho

    The momentum would oscillate but remember, that's not what an expectation value tells you. The expectation value is the actual value of on average, what the momentum would be.
     
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