# Average electronic momentum in bound state: please see this

1. Aug 18, 2010

### ani4physics

Someone please tell me if I am thinking right:

Let's consider an unperturbed electronic state of an atom/molecule. If we denote it by [a>, then the average electronic momentum in state [a> is,

<p> = <a]p[a> = (<a]p<a])* (because p is hermitian)
= (<a]*p*[a>*)
= (<a]p*[a>) (because [a> is real.)
= - (<a]p[a>) ( because p is purely imaginary. so p* = -p)
so, <p> = 0

so, the average momentum in any unperturbed eigenstate = 0

Now, the atom/molecule is being perturbed by a time-dependent scalar potential of the form V(t) = V Cos(wt). So we can write the time-dependent wave function as a perturbation series. Since there is no magnetic field, the wave function will be real and <p> = 0.

Is it true? Please correct me if I am doing it wrong. Thanks.

2. Aug 18, 2010

### qbert

not quite
(<a|p <a|)* is nonsensical.
you could do <a| (p |a>) = <a| (<a| p^dagger)^dagger = <a| (<a| p)^dagger
or (<a|p ) |a> = (p^dagger |a>)^dagger |a> = (p |a>)^dagger |a>.

what do you mean |a> is real? |a> is a STATE, not a number.

----
here's what i think you mean.

notice p is a hermitian operator so that for any state |a> we have
(<a|p|a>)* = <a|p|a>^dagger = <a| p^dagger |a> = <a|p|a>.

suppose |a> is a state such that <x|a> = a(x) is a real valued function.
[i.e. the wave-function is real (in the x - representation)]

then <a|p|a>= $-i \hbar \int d^3x ( a^*(x) \nabla a(x) ) = -i \hbar \int d^3x ( a(x) \nabla a(x) )$
first equality is definition, and second is assumption that a(x) is real valued.

since the integral of a real valued function (a a') is real, we have
(<a|p|a>)* = - (<a|p|a>).

so that <a|p|a> = 0.
-------

this doesn't follow. What makes you think the time-dependent wave function
will be real? what does the perturbation series have to do with anything?