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On page 96 of his book, Griffiths explains that determinate states of some observable $Q$ are eigenfunctions of that operator. So if a particle starts out in that state it will continue to be in that state as long as a measurement of an observable is not being made. This is all well and good and seemed to make sense as I went through the book but then I encountered an example (not in Griffiths) of a spin $1$ particle which starts out in the spin state $(1,1)$ which is an eigenstate of $S_z$ and evolved out of that state (with the Hamiltonian being $H = kS_x$ where $k$ is a constant) when $t > 0$. But how can that be? we know that if particle is in a determinate state it should remain in that state for all time unless a measurement is made.

More generally I considered the following scenario: Suppose you have a definite state of Angular momentum, call it $\psi(0)$ [we will consider it the initial condition which we need to evolve in time]. Then (suppose for simpicity) you can expand this definite state in terms of two eigenstates of the Hamiltonian so:

$\psi(0) = aE_1 + bE_2$

But then to obtain the state at $t>0$ we just tack on the wiggle factor corresponding to each energy eigenstate and we can easily see that the state will evolve out of our initial angular momentum eigenstate! So again: How come that be given that it was a determinate state of an observable.

The conclusion I came to was that given any observable $Q$, it will have determinate states only if they are also energy eigenstates, i.e. only if $Q$ and $H$ are compatible observables in which case they will have a common set of eigenfunctions. But there's not even a hint of that in Griffiths who explicitly defines determinate states as eigenfunctions of observables regardless of whether or not they commute with $H$. So given an observable $Q$, get the eigenfunction and you're done: you got the determinate states. But that contradicts what I've stated above so I must be missing something.

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  • $\begingroup$ if a particle starts out in that state it will continue to be in that state as long as a measurement of an observable is not being made AND as long as the time is short compared to the inverse of commutator of observable and Hamiltonian. It remains indefinitely in that eigenstate only when the commutator vanishes $\endgroup$
    – lurscher
    Apr 2, 2020 at 18:39
  • $\begingroup$ You can use the blockquotes feature for quoting things in your text by prefacing the text with “>” without the quotes. This helps us differentiate between your words and the quoted $\endgroup$ Apr 3, 2020 at 6:00

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It is a different application of the word determinate. Griffiths simply means that the result of a measurement is determinate if the state is an eigenfunction. That is not the same as determinism in time evolution.

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  • $\begingroup$ OK. So just to be clear. It's determinate in the sense that at t = 0 we are certain to get that particular angular momentum eigenvalue, but at t > 0 not necessarily. Am i correct? If that's the case then Griffiths comments seem very misleading from what I can tell... $\endgroup$
    – Leonid
    Apr 2, 2020 at 17:54
  • $\begingroup$ You are correct. I am not in a position to comment in more detail about Griffiths as I don't have the book, but for myself I have always found any treatment which is not done with complete mathematical rigour misleading, and that includes most physics text books. You have to work out what he means by working out what is true, not simply reading what he appears to say, and I think you have done that. $\endgroup$ Apr 2, 2020 at 18:05

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