Hello Aziza,
you have interesting questions.
I know that the average momentum <p> is defined as mddt<x>. But why is this also equal to :
∫ψ*h(2π)i∂ψ∂xdx ?
One possible way to see it is this. We write down the expression for average value <x>:
<br />
\langle x\rangle =\int \psi^* x \psi dx,<br />
which is due to Born (probabilistic understanding of \psi). With this assumption, we calculate average momentum
<br />
m \frac{d}{dt} \langle x \rangle = m \int \partial_t \psi^*x\psi + \psi^* x \partial_t \psi ~dx~~(*)<br />
Now we assume that time change of \psi is described by Schroediger's time dependent equation:
<br />
\partial_t \psi = \frac{1}{i\hbar} \hat{H} \psi<br />
with the simple Hamiltonian (more complicated ones can be used)
<br />
\hat{H} = \frac{\hat{p}^2}{2m} + U(x),<br />
where
<br />
\hat{p} = -i\hbar \frac{\partial}{\partial x}<br />
is assumed.
Plugging these two equations to express the time derivatives in the equation (*) and using Hermitian property of H, we obtain
<br />
m \frac{d}{dt} \langle x \rangle = \int \psi^* \frac{im}{\hbar}[\hat{H}x - x\hat{H}] \psi ~dx<br />
It is easily checked that
<br />
\frac{im}{\hbar}[\hat{H}x - x\hat{H}] = \hat{p}<br />
Also, why is it in general that for any average value <q> there is an operator [q] such that
<q> = ∫ψ*[q] ψ dx ?
Well, it works for x due to Born's rule and for p due to above derivation, so it is possible that it will work also also for other more complicated expressions based on them. But there could be some problems with operators containing higher derivatives than second and I think it is always good to check for every new operator whether <q> = ∫ψ*[q] ψ dx works.
Also, how come for the potential energy operator [U(x)]:
[U(x)] = U([x]) = U(x) ??
There are two aspects.
First, the question is, why the operator U in Schroedigner's equation is the multiplication by the same function of coordinates as given by classical theory and not something more complicated?
Simple answer is, this is what Schroedinger tried and it worked (mostly for explaining hydrogen and other spectra). There is presently no deduction of general Schroedinger's equation or the rule "take the classical expression for U to from H" from something more simple or more clear.
Second, the question is, why average of potential energy is calculated by the formula
<br />
\langle U\rangle = \int \psi^* U(x) \psi ~dx<br />
where U(x) is just the standard potential energy function of coordinate of particle, and not with U(x) more general?
This is probably what you refer to in
I understand that [x] = x, so U([x]) must equal U(x), but I don't understand the jump from [U(x)] to U([x])..
This can be explained as consistency with or consequence of the classical theory.
In probability theory, assuming d(x) is the probability density that the particle is at x and assuming that the potential energy for particle at x is given correctly by classical theory as U(x), the average value is calculated by means of probability density as
<br />
\langle U \rangle = \int d(x) U(x) dx.<br />
In the theory using \psi, Born introduced the prescription that d(x) should be calculated from the \psi function as
<br />
d(x) = \psi^*(x)\psi(x),<br />
which gives us
<br />
\langle U\rangle = \int \psi^* U(x) \psi dx<br />
with simple function U(x), so nothing more complicated than this seems to be necessary.
