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Expectation value of momentum for bound states

  1. Jan 30, 2015 #1

    blue_leaf77

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    1. The problem statement, all variables and given/known data
    I'm curious in proving that expectation value of momentum for any bound state is zero. So the problem is how to prove this.


    2. Relevant equations
    $$ \langle \mathbf{p_n} \rangle \propto \int \psi^*(\mathbf{r_1}, \dots ,\mathbf{r_N}) \nabla_n \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) d^3\mathbf{r_1} \dots d^3\mathbf{r_N} $$
    where ## n = 1,2, \dots N ##.

    3. The attempt at a solution
    If I assume that ## \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) ## is either even or odd then its first derivative will be its counterpart, so ## \nabla_n \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) ## has opposite parity in ## \mathbf{r_n} ##. Thus the integral will result zero.
    My question is that is this true that all bound states can only be even or odd? If yes why?
     
    Last edited: Jan 30, 2015
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  3. Jan 30, 2015 #2

    jfizzix

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    All bound states of symmetric potentials have a definite parity.

    This can be shown because in that case the entire Hamiltonian will have even parity, and will be invariant under a coordinate swap.

    Whether this is true for bound states of asymmetric potentials (say three randomly placed square wells), I couldn't say. The wavefunctions would no longer have definite parity, but it may yet be the case that the expectation of the momentum is still zero.
     
    Last edited: Jan 30, 2015
  4. Jan 30, 2015 #3

    jfizzix

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    Okay, the eigenfunctions would definitely not have a definite parity if the potential is asymmetric. As an extreme example, consider the potential, which is an infinite square well for [itex]x<0[/itex], and is a quadratic potential (a quantum spring) for [itex]x\geq 0[/itex]. Here the eigenfunctions are definitely zero for [itex]x\leq 0[/itex], and nonzero for [itex]x> 0[/itex].
     
  5. Jan 30, 2015 #4

    blue_leaf77

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    So is there a more formal way to prove that expectation value of momentum for bound states is zero?
     
  6. Jan 30, 2015 #5

    jfizzix

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    Yes there is.
    Ehrenfest's theorem.
     
  7. Jan 31, 2015 #6

    blue_leaf77

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    Ok thank you very much, now it's pretty clear why it should be zero for bound states. But first my question is does this theorem also applicable to free particle? If yes then the momentum of free particle is zero??
     
  8. Jan 31, 2015 #7

    jfizzix

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    For the free particle, the expectation of the position need not be a constant value like it would need to be for a bound state. Because of this, the expectation for the momentum need not be zero.

    Whenever the expectation of the position is constant, the expectation of the momentum will be zero.
     
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