# Why is $\frac{dx}{dt}=0$ in this average momentum calculation?

In the following excerpt from S. Gasiorowicz's Quantum Physics, he derives an expression for the average momentum of a free particle. $\psi(x,t)$ is the wave function of a free particle, $\psi^*$ denotes its complex conjugate.

We try the following: since classically,

$$p = mv = m\frac{dx}{dt}$$

we shall write

$$<p> = m\frac{d}{dt}<x> = m\frac{d}{dt}\int{dx \psi^*(x,t) x \psi(x,t)}$$

This yields

$$<p> = m\int_{-\infty}^\infty{dx\left( \frac{\partial\psi^*}{\partial t} x \psi + \psi^* x \frac{\partial\psi}{\partial t} \right)}$$

Note that there is no $dx/dt$ under the integral sign. The only quantity that varies with time is $\psi(x,t)$, and it is this variation that gives rise to a change in $x$ with time.

I seem to have trouble understanding the difference between the position $x$ and the average position $<x>$. Why can it be assumed that $\frac{dx}{dt}=0$? What is x?

-

To clarify a) note that the position operator $\hat x$ does not depend on time, and so also its kernel $\left< x \right | \hat{x} \left | x' \right> = x\delta(x-x')$ with respect to "position vectors" $\left | x \right >$ also doesn't depend on time. This $x$ is the one that is present in your integral. So in particular ${{\rm d} x \over {\rm d} t} = 0$.
On the other hand, the average of the operator $\hat A$ in the state $\psi$ (which depends on time) obviously depends on the state $\psi$: $\left< \hat {A} \right> := \left< \psi \right | \hat {A} \left | \psi \right>$ and so if you perform averages on a family of vectors $\psi(t)$ so also the average will depend on time. In your case, this should be written $\left< \hat{x} \right > (t)$ to make it obvious that one is dealing with a function of time. But this dependence is usually understood and omitted.
+1, Marek. Interestingly, the question contains the first formula for $\langle p \rangle$ which shows perfectly what the objects depend upon. It says that $\psi$ and $\psi^*$ depend on $x,t$ while $x$ doesn't depend on anything, especially not on $t$. So despite this detailed notation, the dependence was ignored by the OP. Moreover, to claim that $x$ depends on $t$ in those formulae would be totally preposterous because $x$ is being integrated over in the formula - it takes all real values simultaneously and "all real values" (the real axis) clearly can't depend on $t$. – Luboš Motl May 11 '11 at 9:08
This $x$ is a position in the reference frame's coordinate system, which is just static by design. You can imagine it as a ruler, with a probability cloud in a foreground; the ruler stays on its place while the cloud moves and deforms changing its mean position.