# 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?

-

## 2 Answers

The confusion seems to stem from a) not understanding what kind of objects you are dealing with and b) usual custom of not writing (all) arguments of functions when they are understood.

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9549856185913086, "perplexity_flag": "head"}