Dependence of operators on the wave function

[avz2611]

in the time-dependent Schrödinger equation , our sir told us about energy and momentum operators . He just defined them , the equation was of the form Aexp(i(kx−ωt)) .if we take the equation of the form Aexp(i(kx+ωt)) will those operators change . if so generally for a wave how do we determine the operators ?

 

[atyy]

The wave function is the state of the system and can be anything at t=0. After t=0, the wave function of the system is governed by the Schroedinger equation.

In the following, I leave out many mathematical details like "self-adjoint", "commuting" etc, and just sketch the rough idea. Operators corresponding to observables like energy, momentum or position do not depend on the wave function of the system. However, to each operator there is a special set of functions called eigenfunctions, which are a subset of possible wave functions (this is actually wrong, but someone can fix it later, the rough idea is ok). In other words, the eigenfunctions of the operator belong to the operator, and different operators generally have different eigenfunctions. When an operator acts on an arbitrary wave function, the result will usually be complicated. But when an operator acts on a function that is an eigenfunction, its action is simple multiplication by a constant. This constant depends on the operator and the eigenfunction, and is called an eigenvalue. Thus an operator has many eigenfunctions, each with its corresponding eigenvalue.

 

[avz2611]

so \hat{p} = -i\hbar \frac{\partial }{\partial x} , will remain the same ?

 

[atyy]

so $\hat{p} = -i\hbar \frac{\partial }{\partial x}$ , will remain the same ?

I edited the quote by putting "##" before and after the latex.

Yes, provided you are in the position basis, ie. you write the wave function and eigenfunctions as a functions of position.

 

[naima]

A first thing to be said is that given a wave function this does not always give a definite value for the momentum.