The 'i' in the momentum operator

[aspy591]

The momentum operator is -i*h bar * derivative wrt x. But won't this lead sometimes to complex expected values of momentum? What does this mean physically, since complex values can't be measured.

 

[Doc Al]

All operators in QM, including this one, are Hermitian. The eigenvalues of a Hermitian operator are real, not complex. (Which is a good thing, as you point out. Complex values would be a problem.)

 

[xepma]

The i is actually needed to make the momentum operator Hermitian!

 

[Doc Al]

The i is actually needed to make the momentum operator Hermitian!

Exactly!

 

[LostConjugate]

The operator d/dx is not Hermitian, so you add a -i to it and the i's cancel out to -1 and then the - cancels out to +1

 

[jtbell]

But won't this lead sometimes to complex expected values of momentum?

Others have given the reasons why the expectation value of p must always be real.

From a practical point of view, when you calculate such an expectation value, you usually get an expression that contains i, to start with. But you can always eliminate i from these expectation values by using identities such as

\sin x = \frac{1}{2i} \left( e^{ix} - e^{-ix} \right)

If you can't, then you've made a mistake in your algebra somewhere.