The meaning of a good quantum number

Question

My book runs through the following argument:

Ehrenfest's theorem states that $$\frac{d\langle Q \rangle}{dt}=\frac{[Q,H]}{i\hbar}+\langle \frac{\partial Q}{\partial t} \rangle$$ and so for a time independent operator commuting with the Hamiltonian, $\langle Q \rangle=constant$. Furthermore, $Q^2$ will commute with the Hamiltonian so that $\langle Q^2 \rangle=constant$ holds too. Then the variance $\Delta Q^2=\langle Q^2 \rangle-\langle Q\rangle^2=constant$.

Now suppose at t=0 the state of a system is given by $\langle\psi\rangle=\langle q_i\rangle$, so that it is in a state of well defined $Q$. The quantum number $q_i$ is a label for this well defined state and can be used to compute the corresponding eigenvalue for the state - for simplicity, lets assume the quantum number is the eigenvalue. Then the above results imply that $\langle Q \rangle=q_i$ and $\Delta Q^2=0$ for all times. This tells us that if we begin in an eigenstate of $Q$, we stay in it at all times, and so this means that the quantum number $q_i$ is called a good quantum number.

Now I understand what is going on, but I seem to be missing the significance of all this - my brain sort of thinks it is obvious anyway for any operator regardless of whether it is time independent and whether it commutes with the Hamiltonian.

To illustrate my thoughts, consider any operator $A$ corresponding to some observable. The operator has eigenstates which we can expand our state in, and lets say that we begin such that $\langle\psi\rangle=\langle a_i\rangle$. Then if we keep measuring $A$, we always measure it to be $a_i$, and so our system is always stuck in this eigenstate. Surely $a_i$ qualifies as a good quantum number by the above logic, as we know $A$ at all times. But I haven't said $A$ is time independent nor that it commutes with the Hamiltonian. So what exactly is special about good quantum numbers?

Thanks for any help.

Answer 1

I can identify two sources of confusion here.

1. You didn't say that $A$ commutes with the Hamiltonian, but you might be assuming it. Let's say that the Hamiltonian has eigenstates $|1\rangle$, $|2\rangle$, ... $|n\rangle$. Assume for simplicity that there is no degeneracy. Then an eigenstate of $A$ can be expanded as

$$|a_i\rangle = \alpha_1 |1\rangle + \alpha_2 |2\rangle + \cdots +\alpha_n|n\rangle$$

Now if we evolve this in the Schrödinger picture,

$$|a_i (t)\rangle = \alpha_1 e^{-iE_1 t}|1\rangle + \alpha_2 e^{-iE_2 t}|2\rangle + \cdots +\alpha_n e^{-iE_n t}|n\rangle$$

Since there is no degeneracy these energies must be all different and therefore if the system started in some state $|a_i\rangle$ it will change into something else via this time evolution which changes the relative phases between the energy eigenstates. Unless, of course, $H$ and $A$ commute, in which case both operators are simultaneously diagonalizable and the above sum has exactly one term. Then time evolution merely contributes an irrelevant absolute phase.

2. You're confusing expectation values with eigenvalues. In the infinite square well, for instance, the expectation value of $x$ in the ground state is precisely the middle of the well. However, this state is very far from an eigenvector of the position operator, which would look like a delta function peaked at a position. In the ground state of the infinite square well it's true that $\langle x \rangle$ wouldn't change with time, but that's because the ground state is an energy eigenstate and not an eigenstate of the position operator.