# Shouldn't the operator be applied to both the wf anD its modulus?

1. Mar 29, 2013

### jshrager

Why is it:

$$\langle A \rangle = \int dV ~ \psi^* (\hat{A} \psi)$$

As opposed to:

$$\langle A \rangle = \int dV ~ (\hat{A} \psi^*) (\hat{A} \psi)$$

The op can substantially change the wf, so it would seem to make more mathematical sense (at least from a linear algebra pov) to operate on both the wf and its mod, otherwise, as we do it now (first expr) you're getting a very odd product. Is this is a property of the way that wfs are formulated? Are they a special case where this isn't required? Or maybe it's a magical property of hermetian operators? Or maybe it just works and I shouldn't think too hard about it.

2. Mar 29, 2013

### Staff: Mentor

1) it works
2) the second formula is not linear
3) There is probably some deeper reason why this works

3. Mar 29, 2013

### kith

If you start with the standard definition of the expectation value of a random variable <A> = Ʃi piai -where the ais are the possible outcomes and pi is the probability to obtain the i-th- your first formula follows from the QM postulates about these quantities. I don't know if this is satifying to you. ;-)

Last edited: Mar 29, 2013