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.


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.

  1. 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.
  • $\begingroup$ I thought the point of eigenstates was that they were the special states of a system such that when your system is in an eigenstate and you keep measuring the relevant observable, the system keeps giving you the same reading. E.g $|\psi\rangle=|x\rangle$, a position eigenstate and then every position measurement you make, you always get $x$, the eigenvalue corresponding to $|x\rangle$. Position won't necessarily commute with the Hamiltonian so I suppose this is a good example. So what you're saying is that actually because you can expand $|x\rangle$ in energy eigenstates that evolve in time... $\endgroup$ – Watw Mar 19 '15 at 18:35
  • $\begingroup$ then the state $|x\rangle$ evolves to something else in time? This doesn't really seem to make sense - if I measure position $x$ and then a short time later measure position, surely I should get $x$ again - isn't that the whole 'collapse of the wavefunction' postulate? This just seems to be contradicting my general understanding of how we expand states in terms of eigenkets and what the eigenkets actually mean. $\endgroup$ – Watw Mar 19 '15 at 18:36
  • $\begingroup$ The exponential factors in front of each term in the eigenbasis will change at difference rates. This is a RELATIVE phase between each term which then changes the probability of each term as a function of time. The superposition of eigenstates of the Hamiltonian is not a stationary state. In terms of collapse of the wavefunction if you let $t \rightarrow 0$ then you will repeatedly get the same measurement for $\lvert x\rangle$ as you suggest. $\endgroup$ – Chris2807 Mar 19 '15 at 18:47
  • $\begingroup$ @Chris2807 So the whole collapse of the wavefunction/repeat measurement giving the same result idea applies at a single instant in time, and here we are concerned about the full time evolution? That would clear things up I believe... $\endgroup$ – Watw Mar 19 '15 at 18:54
  • $\begingroup$ @Watw I think your problem is the difference between smooth unitary time evolution which is what the Schrodinger equation does for us, if we are in some state at time $t=0$ what will it look like at some later time. It says nothing about the stomach-turning collapse of the wavefunction which instantaneously happens when we make a measurement. $\endgroup$ – Chris2807 Mar 19 '15 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.