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Expectation Value of Momentum

  1. Nov 15, 2012 #1
    Good Evening Fellows,
    I have the following question,
    So far I have learned that the expectation value of momentum is equal the time derivative of the expectation value of position. If the potential only depends upon position and not on time. Then, if we use the time independent schrodinger equation the wavefuntion will be separable into a purely function of x and a function of t. Therefore, is it correct to assert that the expectation value of momentum will always be zero for this case, since the expectation value of position will be a constant?
     
  2. jcsd
  3. Nov 15, 2012 #2

    jtbell

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    Staff: Mentor

    For the stationary states (energy eigenstates) that you get by solving the time-independent SE, this is true.

    However, it is not true for states that are superpositions of energy eigenstates. Consider for example a superposition of the first two energy eigenstates of the "particle in a box:"

    $$\Psi(x,t) = a_1 \psi_1(x)e^{-iE_1 t / \hbar} + a_2 \psi_2(x)e^{-iE_2 t / \hbar}$$

    For this wavefunction, the expectation values of position and momentum are not constant.
     
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