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Do expectation values vary with time?

  1. Mar 11, 2012 #1
    I'm a bit confused about the nature of probability conservation and expectation values.

    According to probability conservation,


    Does that mean that expectation values e.g. <x>, <p> and <E> depend only on the position of the particle and not on time?

  2. jcsd
  3. Mar 11, 2012 #2
    Im not sure what your [itex]P[/itex] means, but the conservation law is actually expressed
    \frac{\partial \rho}{\partial t}=-\frac{\partial j_i}{\partial x_i}
    With [itex]\rho[/itex] the probability density, and [itex]j[/itex] the probability current density. They are defined
    \rho=\Psi^{*}\Psi \quad j_i=\Psi^{*}\partial_i \Psi-\Psi\partial_i \Psi^{*}
    More on this
  4. Mar 11, 2012 #3
    P is the normalised probability of finding the particle somewhere in a given volume.

    Do expectation values vary with time with ρ and j as defined?
  5. Mar 12, 2012 #4
    Total probability is obviously conserved, but expectation values depend on the probability distribution, which can of course vary with time.

    If [itex]A[/itex] is an operator on the Hilbert space of the physical states [itex]|\psi \rangle[/itex], then

    \frac{\textbf{d}}{\textbf{d}t}\langle \psi| A |\psi\rangle = -\frac{i}{\hslash} \langle\psi|[A,H] |\psi\rangle\, .

    From this follow that the expectation value of an operator over an Hamiltonian eigenstate is constant as it should.
    Moreover the mean energy is always conserved.

  6. Mar 12, 2012 #5


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    They can, but this depends in the wave function.

    Think about a free Gaussian wave packet centered at x=0 for t=0; the expectation value is of course <x> = 0.

    Now you can "boost" this wave such that it moves along the x-axis. Its width grows with time, but of course the normalization P(-∞,+∞) i.e. the probability to find the particle somewhere in the interval (-∞,+∞) remains P = 1 = const. Due to the boost the wave packet moves along the x-axis, i.e. the expectation value for x becomes time-dependent: <x(t)> = p/m * t. The expectation value for the momentum <p(t)> = p = const. b/c there is no potential (we started with a free wave packet).

    That means that the the answer to you questioin depends on:
    - state wave function ψ you are looking at
    - the dynamics, i.e. the Hamiltonian H of your system
    - the observable A for which you want to calculate <A(t)>

    If A is conserved, i.e. [H,A]=0 the <A(t)> = const.
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