Complex Conjugate applied to operators?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 4K views
Master J
Messages
219
Reaction score
0
I have a rather fundamental question which I guess I've never noticed before:

Firstly, in QM, why do we define the expectation values of operators as integral of that operator sandwiched between the complex conjugate and normal wavefunction. Why must it be "sandwiched" like this?


From this comes my problem. In deriving an equation for current density, I multiplied the electron velocity, which is the momentum over mass, times the density, which is the wavefunction times its complex conjugate.

Yet I have noticed that in a text, the momentum operator is ALSO conjugated. That is to say, since the momentum involves a derivative, and I have a product of wavefunctions, I use the product rule, but the second term has ih d/dx instead of the usual -ih d/dx.


Can someone shed some light on this??
 
Physics news on Phys.org
The Wikipedia article on the expectation value is clear. http://en.wikipedia.org/wiki/Expectation_value_(quantum_mechanics)#Formalism_in_quantum_mechanics

For the second point, I guess you're trying to turn the time derivative of a wave function to the form involving the momentum operator by the Schroedinger equation. To get the time derivative of [tex]\psi^*[/tex], you need to use the complex conjugate of the Schroedinger equation. The mathematical reason that you need to conjugate the operator as well is that [tex]\psi^*[/tex] is a dual vector, and so is [tex]\frac{d \psi^{*}}{dt}[/tex]. For any operator A applying to a dual vector [tex]\langle x|[/tex], [tex]\langle x|A=\langle A^{\dagger}x|[/tex] where [tex]A^{\dagger}[/tex] is the adjoint of A.