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I The symmetry argument and expectation value

  1. Jun 17, 2017 #1
    In 1D QM:
    I understand that if a given potential well, U(x), is symmetric about x = L, then the expectation value for operator [x] would be <x> = L. (I am not even entirely sure why this is, guessing that the region where x<L and x>L are equally probable)

    Is it possible to draw conclusion about expectation values of other operators?
    Heres one example:

    Given ψ(x) = sin(nπx/2L) for a infinite potential well with barriers at x=0 and x=2L,
    Using symmetry arguments or otherwise, explain why or show that the expectation value of the particle momentum in this infinite well is <p> = 0

    I got <p> = 0 by applying the momentum operator [p]ψ and evaluated it at the point of symmetry x=L, which gave [p]ψ = 0 (at x=L). => <p> = ∫ψ*[p]ψ = 0

    Is that correct? Could someone explain to me what really is the "symmetry argument"?

    Cheers
     
  2. jcsd
  3. Jun 17, 2017 #2

    blue_leaf77

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    Only if the expectation value is evaluated with respect to a wavefunction with definite parity.
    It's easiest to assume a symmetric potential centered around x=0. The action of symmetry (or parity) operator ##\Pi## on the operator x is defined by ##\Pi^\dagger x \Pi = -x##. If the potential is symmetric then the Hamiltonian commutes with ##\Pi## and thus the eigenfunctions of the Hamiltonian must also have a definite parity (since the momentum operator also obeys ##\Pi^\dagger p \Pi = -p##), that is it must be either symmetric or antisymmetric. With this information you can get two results when trying to calculate the quantity ##\langle u_n |\Pi^\dagger x \Pi | u_n\rangle##, where ##|u_n\rangle## is an eigenfunction of the Hamiltonian. One way using the definition of the action of parity operator on x you get
    $$\langle u_n |\Pi^\dagger x \Pi | u_n\rangle = -\langle u_n | x | u_n\rangle$$
    The other way using the fact that ##\Pi |u_n\rangle = \pm |u_n\rangle## you get
    $$ \langle u_n |\Pi^\dagger x \Pi | u_n\rangle = \langle u_n | x | u_n\rangle $$
    So, in the end you find that ##\langle u_n | x | u_n\rangle = -\langle u_n | x | u_n\rangle## and there is only one solution namely ##\langle u_n | x | u_n\rangle =0##.
     
  4. Jun 17, 2017 #3
    It's a purely classical situation. A symmetric physical situation implies a symmetric distribution of ##x##. Given a distribution ##f(x)##, the expectation is ##<x> = \int^\infty_{-\infty}x f(x)dx##. If ##f(x)## is symmetric about ##x=L##, put ##y = x-L## and split the integral about ##y=0##. Then ##f(x) = g(y) = g(-y)## and ##<y> = \int_0^\infty y g(y)dy - \int_0^\infty (-y)g(-y)d(-y) = 0## so ##<x> = <y> + L = L##.

    (The QM identification of ##f(x)= |\psi(x)|^2## adds nothing of significance.)
     
  5. Jun 17, 2017 #4

    vanhees71

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    As I understood it, it's about the expectation value of the position for an energy eigenstate. It's clear that you can in the described case you can use a basis with states of good parity, and then ##|u_n(x)|^2=u_n(-x)|^2##. So indeed, the assumption in the OP is only true, if you work with parity eigenstates (see #2).
     
  6. Jun 18, 2017 #5
    Thanks very much for your reply. I've not learnt in great details parity and eigenstates, but I get the main idea.
     
  7. Jun 18, 2017 #6
    Thanks!
     
  8. Jun 18, 2017 #7
    Thanks!!
     
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