1
$\begingroup$

The expectation value of momentum is calculated in Quantum Mechanics (QM) as follows: $$ \langle p\rangle = \frac{\hbar}{i} \int_{-\infty}^{+\infty} \Psi^*(x,t) \frac{\partial}{\partial x} \Psi(x,t) \,dx $$

The expectation value of momentum must be real because measured values of momentum in similarly prepared states are real and the average or expectation value of real numbers is of course real.

This means that the above definite integral must have a factor of $i$ to cancel the $i$ outside to give a real expectation value for momentum.

My question: What property of the wave function can be deduced from the fact that the integral of the product of $\Psi$ and the partial space derivative of $\Psi$ over all of space is purely imaginary?

$\endgroup$

1 Answer 1

0
$\begingroup$

It tells you nothing about wavefunctions, since the momentum operator is Hermitian. The expectation value of a Hermitian operator is always real, since $$ \langle \psi \vert A \vert \psi\rangle^\ast = \langle \psi\vert A^\dagger \vert\psi\rangle \overset{A=A^\dagger}{=} \langle \psi \vert A \vert \psi\rangle.$$ If you find some wavefunction for which the quantity is naively not real, then you have run into one of the subtleties of the momentum operator on the space of wavefunctions $L^2(\mathbb{R})$, e.g. that it is only densely defined and not on wavefunctions that do not vanish at infinity, since the vanishing of a boundary term is crucial for its Hermiticity.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.