# Calculating d<p>/dt

1. Feb 2, 2014

### iScience

we were calculating $$\frac{d<p>}{dt}$$ in class and here are my class notes (sorry for the messiness):

http://i.imgur.com/gMnDDWZ.jpg

why/how does the term circled in green go to zero??..

here is a separate note when i attempted the same problem..

http://i.imgur.com/Kdcr7w9.jpg

where did i go wrong?

Last edited: Feb 2, 2014
2. Feb 2, 2014

### vanhees71

Have you heard about the abstract Hilbert-space formulation of quantum theory? Then I'd recommend to use the Heisenberg picture, where all the time dependence is at the operators that represent observables. For an observable $A$ that is not explicitly time dependent (as is the case for momentum in standard quantum mechanics), you have
$$\frac{\mathrm{d} \hat{A}}{\mathrm{d} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}].$$
The state vectors are time independent. Thus you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\frac{\mathrm{d}}{\mathrm{d} t} \langle \psi|\hat{A}|\psi \rangle = \left \langle \psi \left | \frac{\mathrm{d} \hat{A}}{\mathrm{d} t} \right | \psi \right \rangle.$$
So, finally you only need to evaluate the commutator for the time derivative of the observable's operator. The result is Ehrenfest's theorem.

Of course, all this is equivalent to the use of the Schrödinger equation and the scalar product in position representation, but it's somewhat easier to do :-).

3. Feb 2, 2014

### grzz

If iScience wants to continue in the Schrodinger picture, he can try integration by parts on the integral that he desires to equate to zero noting that ψ → 0 as x → $\pm$∞.

Last edited: Feb 2, 2014
4. Feb 2, 2014

### ChrisVer

Because the term:
$Ψ^{*} \frac{∂^{3}Ψ}{∂x^{3}} =\frac{∂}{∂x}(Ψ^{*} \frac{∂^{2}Ψ}{∂x^{2}})-\frac{∂Ψ^{*} }{∂x}\frac{∂^{2}Ψ}{∂x^{2}}=\frac{∂}{∂x}(Ψ^{*} \frac{∂^{2}Ψ}{∂x^{2}})-\frac{∂}{∂x}(\frac{∂Ψ^{*} }{∂x}\frac{∂Ψ}{∂x})+\frac{∂^{2}Ψ^{*} }{∂x^{2}}\frac{∂Ψ}{∂x}$
The last term is equal but oposite to your second term in green... the other two are total derivatives, which vanish at the integral's limits.