Does the expression:

$$\langle p\rangle_{x=a, b} =\frac{\int_a^b \Psi(x)^*\,\hat{p}\,\Psi(x)dx}{\int_a^b |\Psi(x)|^2dx} $$

have any physical meaning when $\int_a^b |\Psi(x)|^2dx\neq\int_{-\infty}^{\infty} |\Psi(x)|^2dx$?

Can it be related to an observable quantity and if so how?

I understand that from the Born interpretation that $|\Psi(x)|^2$ and $|\Psi(p)|^2$ can be interpreted as the probability density for finding a particle at x or with momentum p respectively and as such expressions such as:

$$\langle x\rangle_{x=a, b} =\frac{\int_a^b \Psi(x)^*\,x\,\Psi(x)dx}{\int_a^b |\Psi(x)|^2dx} $$

appear to have physical meaning as the expected position if x is found to be in the interval a to b.

But does this hold when the integral is not over the same variable as representation of the wavefunction?

  • $\begingroup$ @AccidentalFourierTransform I don't believe it should. If $P(x)$ is an unnormalised probability density in a region $[a,b]$ then to normalise it for that range you should divide by its integral in the region of interest. Note dividing by the integral over all space could lead to the expression giving an expected x outside of the range. It would be equivalent to the expected value in the region multiplied by the probability of it being found in that region. $\endgroup$
    – J.L.
    Commented Feb 10, 2016 at 18:20
  • $\begingroup$ @AccidentalFourierTransform Maybe I have not been clear about what I intend the physical meaning of $\langle x\rangle_{x=a,b}$ to be. It is meant to represent the expected value of x if one were to average over infinitely many measurements of x on systems in a state $\Psi(x)$ and ignore the measurements outside of the range $[a,b]$. $\endgroup$
    – J.L.
    Commented Feb 10, 2016 at 18:36


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.