The wavefunction and degenerate states

Question

I am trying to learn QM by following MIT 804 online lectures by Prof. Adams. There are two statements mentioned in the beginning lectures which do not make sense to me at all.

1. In classical mechanics, knowing position and momentum gives complete knowledge of the system - i.e. based on this, I can calculate anything that can possibly be measured. In QM, the analog is the wavefunction which describes the system completely.
2. The eigenvectors of an operator (corresponding to measurables) form a complete orthonormal basis. Further, for a wavefunction in 1-dimension, there are no operators which have degenerate eigenvalues.

First of all, in classical mechanics, knowing the position and momentum does not tell me anything about the charge of the particle. It also does not really tell me about the mass of the particle either. So how do I exactly have complete knowledge? Regarding the second point, if indeed the wavefunction describes everything about the particle, then there must be an operator corresponding to charge, one for mass, and one for spin. The charge and mass (rest mass) are fixed quantities. The spin can be one of 2 values. Therefore the operators corresponding to these have 1 (or 2) eigenvalues. They must be degenerate if I am to form an orthonormal basis (The basis for the eigenfunction has infinitely many orthonormal vectors).

Obviously my line of reasoning is wrong. But where am I making the mistake?

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EDIT:

The answer and comments really helped resolve the issue. I would like to add some further details based on the comments for anyone else who has similar concerns:

The eigenfunction (1-dimensional) as described in the question does NOT completely specify the state of the particle. While it specifies observables such as position, momentum etc, it does not describe the spin. What the eigenfunction really is (as described in the question) are the coefficients of the basis vectors of position (delta functions). There are infinitely many basis vectors of position, and so the coefficients together can be described as a function of the eigenvalues of position - i.e. functions of x. The spin cannot be described in terms of eigenvectors of position. We need another set of eigenvectors of spin (2 such eigenvectors exist for an electron for example). The complete state for the particle can be described in terms of the combined eigenvectors of position and spin. As an example, the below state is an eigenstate of position and spin (where x=0, spin=up):

|x_0,s_u>

Now we can describe the state of the particle as a superposition of these position + spin eigenstates. The coefficients in this expansion are the complete wavefunction (of 2 variables).

The important fact that separates spin from position/momentum etc is that the spin operator commutes with operator for position. This is why we need a complete set of commuting observables (CSCO) to completely specify the state.

The same argument can be extended to mass and charge. The operators for these commute with both spin and position operators. So you could specify the eigenvectors as below. Example shown for electron below:

|x_0,s_u, m_me, q_-e>

(eigenstate for x=0, spin=up, mass=mass of electron, charge = -e)

However mass and charge commute with every other known observable (unlike position or spin - eg. position does not commute with momentum, spin along x does not commute with spin along any other direction). The fact that they commute with everything else, and they have single eigenvalues, means we might as well drop it from the eigenstate description with no loss of generality

i.e. |x_0,s_u, m_me, q_-e> and |x_0,s_u> are the same thing.

This is why we can consider mass and charge as external parameters and do not need to model them as observables.

Answer 1

Per the comment, I think we are having a semantics problem.

When doing quantum mechanics and talking about "knowing the particle from its wavefunction", it's generally accepted that we already know the particle ID.

Aside [Particle ID]: The phrase "Particle ID" is generally an experimental concern and has nothing to do with quantum foundations. Rather, you scatter an electron off a nucleus, and detect a final state particle with known momentum (from its bending radius in a magnet, assuming $|q|=e$, and measure its velocity vs a threshold (gas Cherenkov detector), which, since $p=\gamma m\beta$, it becomes a crude mass measurement, so that you are using observables to tell if it is an $e^-$ or a $\pi^-$. That's not quantum mechanics concerns--it's just something from the experimental particle physicists' toolbox.

So back to QM and $\psi(x)$: it's generally assumed that the charge and mass of the particle are known parameters, and not something to be measured.

The starting point is a Hamiltonian:

$$ H = -\hbar^2\frac{\nabla^2}{2m} + qV(\vec r) $$

so that the mass ($m$) and charge ($q$) are known.

The same is true for spin, where the Hamiltonian picks up a term:

$$ H' = -\vec\mu\cdot\vec B $$

with the magnetic moment depending on the spin

$$\vec mu = \gamma_s\vec S$$

For a particle with known spin $S$ (which means it has known and fixed intrinsic angular momentum $\hbar\sqrt{S(S+1)}$, you have to make a spin operator with $2S+1$ basis states (not just 2 as you state--that only works for $S=\frac 1 2$).

Then, with the eigenstates, you can only know the projection of $\vec S$ onto one axis.

Your point (2): no degeneracy in 1D, seems to be about systems without spin, it also doesn't include $\hat p^2$, which can be degenerate for $|\pm p\rangle$.

If you want to talk about mass, charge, spin operators formally as ways to describe your particle, that gets into quantum field theory, and is a whole 'nother thing. In quantum mechanics, they are just parameters. In relativistic QFT, it gets into charge conjugation and https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group.