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Particle oscillating between two wells

  1. Nov 24, 2008 #1

    Suppose we a pair of symmetric wells of finite potential and the particle is given to be in the initial state

    [tex]|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|\psi_{s}\rangle + |\psi_{a}\rangle)[/tex]

    (a = antisymmetric state, s = symmetric state)

    For t > 0, we have

    [tex]|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_{S}t/\hbar}(|\psi_{s}\rangle + e^{-it/\tau}|\psi_{a}\rangle)[/tex]

    where [itex]\tau = \hbar\pi/(E_{a}-E_{s})[/itex]

    We see that the particle oscillates between the two wells, but the expectation value of the energy


    is constant and equals [itex](E_{s}+E_{a})/2[/itex].

    I have two questions:

    1. What is the physical significance of this?

    2. Is this due to the specific initial state given?

    Thanks in advance.
  2. jcsd
  3. Nov 24, 2008 #2


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    Science Advisor

    1. Energy is conserved!

    2. No. Take any time-independent hamiltonian, and express the initial state as a superposition of energy eigenstates |n>, with coefficients c_n. The probability that the system has energy E_n is then |c_n|^2. Time evolution changes the phase, but not the magnitude, of each c_n. So the probability |c_n|^2 is constant in time. The expectation value of the energy is just the sum of |c_n|^2 E_n, so this is constant as well.
  4. Nov 24, 2008 #3
    Yes, of course ... I forgot :rolleyes:

    Thanks Avodyne.
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