Do expectation values vary with time?

[Darkmisc]

I'm a bit confused about the nature of probability conservation and expectation values.

According to probability conservation,

\frac{∂P(r,t)}{∂t}=0.


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


Thanks

 

[jfy4]

Im not sure what your P means, but the conservation law is actually expressed
<br /> \frac{\partial \rho}{\partial t}=-\frac{\partial j_i}{\partial x_i}<br />
With \rho the probability density, and j the probability current density. They are defined
<br /> \rho=\Psi^{*}\Psi \quad j_i=\Psi^{*}\partial_i \Psi-\Psi\partial_i \Psi^{*}<br />
More on this

 

[Darkmisc]

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?

 

[Ilmrak]

Total probability is obviously conserved, but expectation values depend on the probability distribution, which can of course vary with time.

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

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

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.

Ilm

 

[tom.stoer]

Do expectation values vary with time with ρ and j as defined?

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.