# Time derivative of expectation value of position

In the expression of time derivative of expectation value of position, it is mentioned in book (Introduction to Quantum Mechanics by DJ Griffith) that inside the integral, the differentiation with respect to position $$x$$ will be zero, so the time derivative only applies on probability density.

But I don't get the point, why inside the integral, time derivative of position should be zero?

Now as time goes on, $$\langle x \rangle$$ will change (because of the time dependence of $$\Psi$$), and we might be intered in knowing how fast it moves. Referring to Eequations 1.25 and 1.28, we see that $$\frac{d\langle x \rangle}{dt} = \int x \frac{\partial}{\partial t} \left \lvert \Psi \right \rvert^2 dx = \frac{i \hbar}{2m}\int x \frac{\partial}{\partial x}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \, dx \, .$$

It is necessary to distinguish between the position, operator of position, and mean value of position/average position. Here one works in Schrödinger representation, which means that all the time dependence is carried by the wave function, whereas the operators are time-independent. Moreover, in the position representation the operator of position is $$\hat{x}=x$$ - a time-independent number that should be integrated with the wave function.

In other words: the average position $$\langle x\rangle$$ is time-dependent, but its operator $$x$$ is time-independent.

You may also want to consult this answer.

• What is the difference between position and average position (Is the position due to collapse of wave function and average position is due to statistical behaviour?) @Vadim May 17, 2020 at 13:59
• Position is what you measure in an experiment. If your system is not a position eigenstate, then every measurement will produce a different value of position, and then you can average over them. QM gives a way of calculating/predicting this average via the wave function, which is found by solving the Schrödinger equation. May 17, 2020 at 14:03

An analogy might be useful.

Suppose you want to compute the time-dependence of the average weight of a population. The average weight is just \begin{align} \langle w\rangle = \int dw w N(w) \tag{1} \end{align} where $$N(w)$$ is the probability of people having weight $$w$$. Now, what changes with time is not the weight $$w$$: $$1$$kg today is the same as $$1$$kg tomorrow, but what changes in time is the probability $$N(w)$$ of having persons of a certain weight: some people will gain weight over time, some will loose weight so a better expression for the average time would be \begin{align} \langle w(t)\rangle = \int dw w N(w,t) \end{align} and of course the rate of change in this average is \begin{align} \frac{d\langle w(t)\rangle}{dt}= \int dw w \frac{N(w,t)}{dt}\tag{2} \end{align} Thus, in (2), what changes is the probability distribution. This $$N(w,t)$$ is in fact nothing but the probability distribution $$\vert \psi(x,t)\vert^2$$ in your problem.

• Thankyou sir. So with respect to my question, the $x$ inside the integral is just lenth w.r.t. some original (which is constant for some coordinate system we have chosen) and what changes is the probability density of particle being there. Is it correct? May 18, 2020 at 12:48
• yes. inside the integral the $x$ is 1m or 5cm or whatever. what changes with time is the probability of finding the particle at that point. May 18, 2020 at 13:01

One way to see this is true is that $$x$$ is used as an integration variable. You can replace it with another variable $$x\rightarrow y$$ without changing the integral $$\int x\frac\partial{\partial t}|\Psi(x,t)|^2dx=\int y\frac\partial{\partial t}|\Psi(y,t)|^2dy$$ An integration variable can't depend on time.

$$x$$ is just a position variable or operator, if you prefer. It is not the position of the particle, which instead is $$\langle x\rangle = \int dx\, x \left|\Psi\right|^2~.$$ $$\langle x\rangle$$ may depend on $$t$$, but $$x$$ does not depend on $$t$$.