# Expectation value of non-physical observable

## Homework Statement

This may be incredibly obvious, but I just need to check. Of course we all know that physical observables must yield real expectation values. What if you tried to calculate, say, <xV(d/dx)>, where x is the position, d/dx is a first derivative, and V is the potential? This isn't an observable...so do we automatically know the answer is zero(or non-real)?

## The Attempt at a Solution

vela
Staff Emeritus
Homework Helper
Nope.

Nope.

So after searching around I agree with your answer, but I am not sure why this is so. For example, <xp> is not zero but has a complex component. xp isn't really considered a physical observable though...is it? Is the only restriction on for a real expectation value that the quantity be a "physically meaningful" observable such as momentum, energy, etc..?

dextercioby
Homework Helper
The basic requirement is for the linear operators which describe quantum observables to be self-adjoint. xp isn't. Self-adjoint operators have a spectral decomposition with a purely real spectrum, so that every expectation value on a state in their domain is real.

However, the combination

$$D = \frac{XP+PX}{2},$$

which looks the same classically, is self adjoint. One can often construct valid observables by getting the operator ordering right.

However, the combination

$$D = \frac{XP+PX}{2},$$

which looks the same classically, is self adjoint. One can often construct valid observables by getting the operator ordering right.

So since this is self adjoint...then how to we interpret this physically?

Simply as the product of position and momentum, I would think. We could in principle design a clever experiment to measure D.