# Expectation value of non-physical observable

1. Mar 5, 2013

### Visceral

1. The problem statement, all variables and given/known data
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)?

2. Relevant equations

3. The attempt at a solution

2. Mar 6, 2013

### vela

Staff Emeritus
Nope.

3. Mar 6, 2013

### Visceral

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..?

4. Mar 6, 2013

### dextercioby

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.

5. Mar 6, 2013

### Oxvillian

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.

6. Mar 7, 2013

### Visceral

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

7. Mar 8, 2013

### Oxvillian

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