Expectation Values of x & p for Wavefunction u(x,0)

[Geocentric]

Homework Statement

A particle is represented(at t=0) by the wavefunction

u(x,0) = A(a^2 - x^2) if -a<x<a
= 0 otherwise
Determine <x> & <p>.
It is given in the book that in this case <p> $$\neq$$ m*d/dt<x>. Could someone please tell me the reason for this?

Homework Equations

The Attempt at a Solution

 

[Rick88]

because if you use <p> = m d<x>/dt, you would determine the expectation value of the momentum via that of x.
But <p> is given only by ∫u p u* dx - it's a definition.

 

[Geocentric]

because if you use <p> = m d<x>/dt, you would determine the expectation value of the momentum via that of x.
But <p> is given only by ∫u p u* dx - it's a definition.

What I don't understand is that according to Ehrenfest theorem,
<p> = m* d/dt<x>
Why is this not true in this case?

 

[Tangent87]

How does your book DEFINE p mathematically?

 

[Rick88]

The integral is just the definition of <p>.
p and <p> are not the same thing, so you can't use the formula for p to find <p>.