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If I have a quantum system which I prepare in a certain state, this state then evolves unitarily via a Hamiltonian. Suppose an observer provokes a collapse of the wave function by a certain measurement, this means that it must be in an eigenstate of the measurement.

  1. What happens subsequent to that?

  2. Will it remain in the same state?

  3. Will it evolve unitarily according to the same Hamiltonian?

  4. If I make the same measurement will I get exactly the same value with certainty?

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The wave function always evolves unitarily according to the Hamiltonian relevant for the given system. It never violates this evolution - there is no discontinuous "collapse" in which it would behave differently. The moment of measurement that the laymen incorrectly describe as "collapse" just means that an answer to a question is settled - it could have been given by a probabilistic distribution only before the measurement. –  Luboš Motl Jul 30 '13 at 19:23
It means that the outcomes that were not realized may be "forgotten" (the branches of the wave function may be erased from the memory) but this is just a subjective simplification of the wave function or density matrix ready to make further predictions - we may replace the previous probability distributions etc. that depended on many variables by the conditional probabilities in which the realized outcome of the recent measurement was substituted and taken into account. But one doesn't have to do so - it's just a bookkeeping device, a psychological simplification. –  Luboš Motl Jul 30 '13 at 19:24
@LubošMotl That's no mainstream physics! $:)$ Mozibur, there is no global agreement on the measurement problem; but we can safely state that the state will evolve unitarily after the measurement, unless another measurement is made! –  Ali Jul 30 '13 at 19:31
@Ali: Ok, that presumably all good physicists accept. But, presumably Motls view, although not mainstream is accepted in certain quarters? –  Mozibur Ullah Jul 30 '13 at 20:13
What @LubošMotl says here is, absolutely, mainstream and correct physics. There's a vast amount of obfuscatory and confusing literature, including in textbooks, but that doesn't mean the correct answer isn't widely known and understood. –  Matt Reece Jul 31 '13 at 1:44

3 Answers 3

I'll answer this just to get a feedback about my own understanding of this (probably much more complicated than I think) subject.

The wave function will always evolve unitarily according to Hamiltonian. If the state of initial preparation (or a state after collapse) happens to be an eigenstate of the subsequent measurement - you will measure a per-determined eigenvalue. In other words (the story my intuition invented to settle this stuff inside my head), if you prepare (or measure) the system in a state which does not contain any undetermined information for subsequent measurement - you can predict the result of this measurement.

Once the measurement is done the wave function collapses. What does it mean? A lot of bla-bla, metaphysics, religious and cultural discussions and etc. I did not really understand this collapse completely. However, I know that this collapse brings a wave function to an eigenstate of the measured observable. This provides the following information about subsequent measurement:

  1. If the eigenstates of subsequent measurement are identical to the eigenstates of the previous measurement (I suspect that the right formulation of this is that there are "one-to-one" and "onto" mapping between these sets of eigenstates) - see the first paragraph
  2. If the eigenstates sets are not exactly identical, but there is partial correlation - you can predict some probabilities of the subsequent measurement
  3. If the eigenstates sets are "independent" - you get no information about subsequent measurement's result

In other words (my intuition is such a story-teller!), the more correlation there is between this observable and the subsequent one, the more information you can get about the subsequent measurement.

All the above feels reasonable as long as Hamiltonian does not change. If there are external factors which change Hamiltonian (as I believe the case in real measurements), there are no guarantees whatsoever. However, and this is a pure speculation, I guess that if one can predict the evolution of Hamiltonian in time - some predictions about subsequent measurements may still be made (unless the observables are completely independent).

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You prepare the system in some state. The state is described by a wavefunction which is an eigenfunction of a complete set of compatible observables( the operators for all the observables commute with each other). if you prepare an ensemble of systems identically (so they all have the same wavefunction) and measure the value of any one (or more) of these compatible observables for each member of the ensemble, you get the same value in each case. There is no collapse of the wavefunction associated with this measurement as the system is described by the same wavefunction after the measurement as before it. The wavefunction evolves in time in a manner governed by the shroedinger equation which in turn depends on the Hamiltonian for the system.

Now if you measure some observable which is incompatible with the original set which describe fully the state i.e it is represented by an operator which doesnot commute with them and for which there is consequently an uncertainty relation between this observable and those previously discussed; then up until the measurement the wavefunction evolves according to the shroedinger equation. But the measurement itself is not described by the shroedinger equation. There is a random, discontinuous jump to a new state which is one of the eigenstates of the new observable. Which new state occurs cannot be predicted. Only the probabilities of each of the possibilities can be calculated (from the inner product of the original state with the new state). Each of the ensemble of identical systems may give different measured values from others despite having been prepared identically. The wavefunction is said to have collapsed onto whichever new state we observe. Subsequently this new state evolves according to the shreodinger equation until a new measurement of an observable incompatible with the observables which characterize the new state.

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I post this answer to check my understanding.

Imagine a wavefunction in 1 dimensions with a known energy and momentum it's wavefunction will be:

$$\Psi(x, t) = e^{i(kx-\omega t)} = e^{i(px-E t)/\hbar}$$

With some calculus and algebra you can derive the momentum operator and get this:

$$-i\hbar \partial_x \Psi = p \Psi$$

There $-i\hbar \partial_x$ is the momentum operator (I used $\partial_x$ for sorthand for partial derivation). The $p$ is the momentum we measured: the eigen-value of the operator.

Since we prepared the state with a known momentum, the measurement of the momentum doesn't have any effect on the state.

Now imagine a state that is a superposition of 3 possible momenta, so it's a sum of 3 states for each momentum:

$$\Psi = \Psi_1 + \Psi_2 + \Psi_3$$

The superposition principle allows this. Applying the momentum operator on them, you'll get this:

$$-i\hbar \partial_x \left( \Psi_1 + \Psi_2 + \Psi_3 \right) = p_1\Psi_1 + p_2\Psi_2 + p_3\Psi_3 $$

That means our state have 3 different momenta at the same time, but the measurement must give one of the 3 possible eigenvalues. You can get the probability of collapse to a particular state by calculating the

$$ \langle \Psi_i| \Psi\rangle = \int_{-\infty}^\infty\Psi_i^*(x,t) \Psi(x,t) dx$$

Where the asterisk means the complex conjugate. And on the bra-side there must be one of the eigenstates of the operator (that is a pure plane wave with known momentum).

So to answer your question (partially):

  1. After the measurement the Copenhagen interpretation says that state immediately changes to one of the eigenstates. The many worlds interpretation says there is no such collapse instead all the eigenstates can coexist simultaneously in parallel worlds. If the nature have chosen $p_1$ as the measurement result, you'll know that the state is now $\Psi_1$ which is then renormalized to ensure $\langle \Psi_1 | \Psi_1 \rangle = 1$. This renormalization just a technical step for convenience since the Schrodinger-equation doesn't care if you multiply the wave function with an arbitrary constant number. You can see states as infinite dimensional vectors (you can use dimensional analogy of the finite dimensional vectors). And only the directions of these vectors matter. Not the length.

  2. An operator doesn't change the direction of an eigenstate.

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