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Let's say we have a state $ \phi=\sum_i c_i \phi_i $ where the $ \phi_i $ denote energy eigen vectors with non degenerate eigen values.

Now if a measurement of the energy is done this state collapses to one of the eigen states at time $ t=t1$ say.

If the energy is measured again after a certain interval at time $t=t2$ most books say that if the second measurement is not made immediately after the first one we may not get the same result as the previous measurement (Sakurai modern quantum mechanics stern gerlach experiment).

Stationary states though do not evolve in time so then if the wave function had collapsed to an eigen state of the Hamiltonian how then can we get a different result in the second measurement of the energy (which is the same as measuring the angular momentum in case of stern gerlach experiment for a spin half particle)?

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closed as off-topic by Aaron Stevens, ZeroTheHero, Jon Custer, user191954, sammy gerbil Nov 15 '18 at 23:16

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    $\begingroup$ What have you tried? Show your work and/or efforts and explain where you are stuck. $\endgroup$ – DanielSank Nov 11 '18 at 7:21
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    $\begingroup$ This looks like an assignment question. A bit of research will turn up relevant answers on this site. $\endgroup$ – ZeroTheHero Nov 11 '18 at 7:22
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    $\begingroup$ Perhaps I am unusually trusting this morning, but this doesn't look like a homework question to me. Getting a hold of issues like this is an important step on the way to understanding QM. $\endgroup$ – John Rennie Nov 11 '18 at 7:37
  • $\begingroup$ Possible duplicate of Time evolution of quantum state after an observation $\endgroup$ – sammy gerbil Nov 15 '18 at 23:16
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The eigenstates of the hamiltonian are stationary and do not evolve with time. So if your measurement of the energy collapses the system to an eigenstate the system will not evolve with time and therefore cannot ever go back to the initial state.

However in real life the uncertainty principle guarantees no measurement will ever collapse the state to an energy eigenstate. Instead the best it can do is collapse the system to a state $\sum_i c_i \phi_i$ where one of the coefficients $c_i$ is close to unity and all the others are very small i.e. approximately an eigenstate. This state will then evolve with time as CuriousHegemon explains.

In general the post measurement state won't evolve back to the initial state. Exactly how it evolves depends on the coefficients $c_i$ post measurement, and these depend on exactly how the measurement is done.

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So I'm going to switch to bra-ket notation because we want to be manipulating vectors in Hilbert spaces and transforming to real coordinates later.

So if we have a state $|\phi\rangle = \sum_n c_n |n\rangle$, where $|n\rangle$ are the energy eigenstates of the Hamiltonian, then to see how this state evolves in time we look at the Time Evolution operator $U(t)$: $$ |\phi(t)\rangle = U(t) |\phi\rangle = e^{-(i/\hbar) H t} |\phi\rangle \\ = \sum_n c_n e^{-(i/\hbar) H t} |n\rangle = \sum_n c_n e^{-(i/\hbar) E_n t} |n\rangle \\ $$

Here, I've used the identity that if an eigenket $|a\rangle$ of an operator $A$ has the eigenvalue $a$, we get $e^{f(A)} = e^{f(a)}$. So now we can ask, what is the probability that my state stays in the original state $|\phi\rangle$ ? Let's find that probability by taking the norm: $$ P(|\phi(t)\rangle \text{ remains in } |\phi\rangle) = |\langle\phi|\phi(t)\rangle|^2 =| \Big( \sum_{n'} c_{n'}^* \langle n' | \Big) \Big( \sum_{n''} c_{n''} e^{-(i/\hbar) E_{n''} t} |n''\rangle \Big) |^2 \\ = | \sum_n |c_n|^2 e^{-(i/\hbar) E_n t} |^2 $$

If you keep going, you'll find time oscillations. Hope that helps!

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  • $\begingroup$ What if $ |\phi> $ itself is a stationary state? $\endgroup$ – pinaki nayak Nov 11 '18 at 7:33
  • $\begingroup$ Let's say $|\phi\rangle = |m\rangle$, where $|m\rangle$ is an energy eigenstate. Then all the $c_n = 0$ except $c_m = 1$. Then the probability is just $P = |e^{-(i / \hbar) H t} |^2 = 1$. So the stationary state is aptly named, since it doesn't evolve with time! $\endgroup$ – CuriousHegemon Nov 11 '18 at 10:15

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