A Momentum operator -- Why do we use the plane wave solution?

[zb23]

Why in order to derive the QM momentum operator we use the plane wave solution. Why later on in field theory and particle physics, the plane wave ansatz is so physically important?

 

[simonjech]

Because we can write any field in qft as a superposition of plane waves.

 

[zb23]

Yes, I agree, but still, is there some deeper meaning behind plane waves? Is it "just" most physically appropriate?

 

[simonjech]

I think that it is quite nice tool when we can decompose any field into a set of harmonic oscillators. One of the reasons is that we can solve the harmonic oscillator problem in QM completely analytically. No perturbation theory or variations or any numerical method.

 

[zb23]

Yes, I understand that but aren't the things you said just consequences? It is like it bothers me in some fundamental QM way. I mean I can't be satisfied with the argument that it gives nice solutions. I mean, it is obviously correct considering the experiments in QM.

 

[PeroK]

I mean I can't be satisfied with the argument that it gives nice solutions.

That's the main basis for mathematical physics. Why represent force as a vector? Why is time a parameter in non-relativistic physics? Why have a spacetime manifold in relativity?

 

[zb23]

Of course, I understand, I mean I am currently doing field theory in particle physics and some arguments for plane wave solutions, for me, lack some structure, therefore I asked...I just wanted to get some more insight into some basics...

 

[PeterDonis]

some arguments for plane wave solutions, for me, lack some structure

Can you give some specfic references to arguments that you find lacking?

 

[strangerep]

Why in order to derive the QM momentum operator we use the plane wave solution.

You don't have to do it in that old-fashioned way.

A more modern approach is to examine the Lie algebra of Galilean transformations as an abstract group. Then, representing that algebra in coordinate representation, implies that the operator of spatial translation, ($P_i = -i \hbar \partial_i$), must satisfy $P = MV$ (for a free particle), which is the usual expression for momentum.

For a detailed exposition of this, see Ballentine chapters 3 and 4.