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Demonstrating the oscillatory behaviour of a particle

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data

    \Psi (t) = \frac{1}{\sqrt{2}}\begin{pmatrix}

    Set b1= b2 = 1/√2 and demonstrate the oscillatory behaviour of the particle between the two minima. How long does it take the particle to completely move from one minimum to the other?

    2. Relevant equations
    Possible differentiation of a matrix (?)

    3. The attempt at a solution
    I have no reference for this question so am really unsure what it is asking, perhaps someone could point me in the right direction?

    p.s. this will likely be the last question I ask before the exams and would like to say a massive thank you to everyone on here that has helped with some of the questions I've posted it's made a big difference to my understanding of many concepts and I would have been at a loss otherwise!
    Last edited: Jan 8, 2015
  2. jcsd
  3. Jan 8, 2015 #2
    Two quick checks on the state that you provided:
    (1) Does it really involve the exponential of an exponential? I suspect its just a bracket issue, but just to be sure :p
    (2) Are both phase terms (the term in the argument of the exponential) the same?
  4. Jan 8, 2015 #3
    Indeed there was a bracket issue it is fixed now. And no, apologies, the full equation in the notes is:

    \Psi (t)= b_{1}exp(\frac{-i(E_{0}+V)t}{\hbar})\Psi _{1}+b_{2}exp(\frac{-i(E_{0}-V)t}{\hbar})\Psi _{2}= \frac{1}{\sqrt{2}}\begin{pmatrix}
  5. Jan 8, 2015 #4
    Oh okay, I think I saw one of the signs wrongly and thought the phases were the same (which would be strange).
    Were there any more details of the physical set-up? Otherwise the terms "minima" and [itex]\Psi_{1}[/itex] / [itex]\Psi_{2}[/itex] are quite hard to interpret without a context.
  6. Jan 8, 2015 #5
    The problem is taken from notes on a two state system where the following hamiltonian is applied

    H = \begin{pmatrix}
    E_{0} &V \\
    V & E_{0}

    it results in two eigen states

    \Psi _{1} = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1\\ 1

    \Psi _{1} = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1\\ -1


    "Hence, if the electron is in an eigenstate it will to populate both minima. The most general state of the system can be a linear superposition of these two syates with the appropriate time dependant factors."

    \Psi (t)= b_{1}exp(\frac{-i(E_{0}+V)t}{\hbar})\Psi _{1}+b_{2}exp(\frac{-i(E_{0}-V)t}{\hbar})\Psi _{2}= \frac{1}{\sqrt{2}}\begin{pmatrix}

    Thus, for general b1 and b2 the electron will oscillate between the two potential minima.

    Then it asks the question, Set b1 = b2 = 1/√2 and demonstrate the oscillatory behaviour..... as stated in the first post above.

    Thanks for the reply.
  7. Jan 9, 2015 #6
    Ah, okay, as I thought, it is a two coupled potential well system.
    First of all, it is helpful to eliminate the common global phase, which is irrelevant. Then, if b1 = b2, the sum of the two exponentials can be reduced to a nicer, simpler form.
    So in the end, the state can be rewritten in the form
    1\\0\end{pmatrix}f(t) +
    (Recall that [itex]
    1\\0 \end{pmatrix}[/itex] and [itex]\begin{pmatrix}
    0\\1 \end{pmatrix}[/itex] are the eigenstates of the uncoupled wells.)
  8. Jan 9, 2015 #7
    Ahh i think I see that. But how would you calculate the time it takes to go from one state to the other?

    Many thanks for the reply.

    edit and what is the "common global phase" part of the equation? what does this phrase mean? thanks again
    Last edited: Jan 9, 2015
  9. Jan 9, 2015 #8
    Well it would be quite obvious once you obtain the explicit forms of f(t) and g(t). At certain times, f(t) will be zero, at others, g(t) will be zero.

    Basically, because the wavefunction is not an observable entity, we can only define it up to a global phase factor. That is to say, for instance, adding an extra phase to the say, the state [itex]|\psi\rangle[/itex], which results in [itex]e^{i\theta}|\psi\rangle[/itex], does not change the dynamics, because the phase factor [itex]e^{i\theta}[/itex] always cancels out. Do take note however of the distinction between the global phases and phase differences - the latter is physical significant and can be observed.
  10. Jan 9, 2015 #9
    Thanks, going to need a little more on this if possible? Would you be willing to show this mathematically as I'm still a little unsure.

    That's great thanks!
  11. Jan 9, 2015 #10
    \Psi (t) = \frac{1}{2}\begin{pmatrix}
    exp(\frac{-i V t}{\hbar})+exp(\frac{i Vt}{\hbar})\\
    exp(\frac{-iVt}{\hbar})-exp(\frac{i Vt}{\hbar})
    cos \left(\frac{Vt}{\hbar}\right)\\
    - i sin \left(\frac{Vt}{\hbar}\right)
    I presume you can take it from here?
  12. Jan 10, 2015 #11
    Yes, thanks for all your help.
  13. Jan 10, 2015 #12
    Ok, I may look like an idiot here but its better to ask, look stupid and then come to understand something rather than not ask and possibly be wrong. So this is what I've got:

    therefore we have

    \Psi (t) = \frac{1}{2}cos(\frac{Vt}{\hbar})\begin{pmatrix}
    1\\ 0

    \end{pmatrix} -\frac{1}{2}isin(\frac{Vt}{\hbar})\begin{pmatrix}
    0\\ 1


    therefore at t=ħ/V π/2 the first term goes to zero and at t=ħ/V π the second term goes to zero, ie it oscillates between the two.

    is that along the right lines?
  14. Jan 10, 2015 #13
    Yup, don't forget to include the integer multiples of those times when writing your solution out as well :)
  15. Jan 10, 2015 #14
    Excellent got there in the end! thanks again!
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