Expectation value of position derivative of $V(x$)

Question

In one of the proofs of Ehrenfest theorem, at the last step claims that $\int \frac{dV}{dx}$$(|\psi^2|)$dx is equal to the expectation value of $\frac{dV}{dx}$. Now according to my understanding of basic quantum physics, $|\psi^2|$ denotes the probability of a particle to exist at a particular position, which does not correlate to the probability of $\frac{dV}{dx}$ attaining a particle value, where am I going wrong?

Answer 1

Suppose $A$ is an observable of a particle on a line, with state $\psi$. The probability of finding the particle in the infinitesimal interval $dx$ is $|\psi|^2dx$. The value of the observable at point $x$ is $A(x)$.

The average value of $A$ is then the "sum" of the value $A(x)$ times the probability that the particle being in $dx$. The sum is replaced by an integral because $x$ is a continuous variable, and you get $\langle A\rangle = \int A(x) |\psi|^{2}dx$, which reproduces your quoted expression.

Answer 2

Let $\rho(x)$ be a probability distribution of some random variable $X$ defined on the real line. In that case, the probability for $X$ to attain values in $[a,b]$ is given by the integral

$$P(X\in [a,b])=\int_{a}^b\rho(x)dx.\tag{1}$$

If $f(x)$ is some function, the mean value of $f(X)$ is then defined to be $$\langle f(X)\rangle\equiv \int_{-\infty}^\infty f(x)\rho(x)dx\tag{2}.$$

In Quantum Mechanics, if $\psi(x)=\langle x|\psi\rangle$ is the position space wavefunction of a particle in one-dimension, the probability distribution for the position random variable is, by the QM postulates, $\rho(x)=|\psi(x)|^2$.

As such, given any function of the position operator, say $f(X)$, the corresponding mean value is given by $$\langle f(X)\rangle=\int_{-\infty}^\infty f(x)|\psi(x)|^2dx\tag{3},$$

in accordance to (2). Your question considers the particular case $f(x)=V'(x)$. While I have explained this in connection to standard random variables, equation (3) can also be derived just in QM. In that case, recall that if $X$ is the position operator and $|x\rangle$ its eigenvalue basis, a function $f(x)$ applied to $X$ is defined to be the operator $f(X)$ that obeys $$f(X)|x\rangle= f(x)|x\rangle\tag{4}.$$

In that case, we evaluate $\langle f(X)\rangle$ in a state $|\psi\rangle$:

$$\langle f(X)\rangle = \langle \psi|f(X)|\psi\rangle = \int_{-\infty}^\infty dx \langle \psi|f(X)|x\rangle \langle x|\psi\rangle\tag{5},$$

where we just used completeness of the basis. Now use (4) and recognize $\langle x|\psi\rangle = \psi(x)$ and $\langle \psi|x\rangle=\psi^\ast(x)$ so that

$$\langle f(X)\rangle = \int_{-\infty}^\infty dx f(x)\langle \psi|x\rangle \langle x|\psi\rangle=\int_{-\infty}^\infty f(x)|\psi(x)|^2dx\tag{6}.$$