# Do expectation values vary with time?

1. Mar 11, 2012

### 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

2. Mar 11, 2012

### jfy4

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

3. Mar 11, 2012

### 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?

4. Mar 12, 2012

### 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

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

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.

Ilm

5. Mar 12, 2012

### tom.stoer

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.