Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In the axiomatic presentation of QM, I've seen it stated many times that the reduction postulate is not needed and/or incorrect, and could be gotten rid of.

However, without the reduction postulate, wouldn't tests of QM simply yield wrong results? If I measure a system that is in some linear superposition $\psi=\sum c_i |e_i\rangle$ of eigenstates of my observable $O$, the measurement postulate says I will get one of the eigenvalues as a result, with probability $|c_i|^2$. But now, since I no longer have the reduction postulate, I will calculate that as a result of the interaction with my measurement apparatus $\psi$ evolved linearly and unitarily into $\psi'$, and therefore still consists of (possibly a different) linear superposition $\sum c_i' |e_i\rangle$ of eigenstates of $O$. Therefore, again applying the measurement postulate, I calculate that I will now get $|e_i\rangle$'s eigenvalue with probability $|c_i'|^2$. After a sequence of $n$ such measurements, close together in time, the probability that I will get some particular $|e_j\rangle$'s eigenvalue in all the $n$ measurements is $(|c_j|^2 * |c_j'|^2 * ... * |c_j^{(n)}|^2) << 1$.

However, we know from experience that making $n$ close in time measurements of the same observable will yield the same value with a probability of $1$ (or very close to $1$).

So I don't see how the reduction postulate can just be dropped without either QM giving incorrect predictions, or without modifying some other postulates somehow?

share|improve this question
    
If you consider measurement as an interaction between the apparatus and the system, repeated measurement (with fresh subsystems for the different measurement records) will give rise to strongly correlated measurement outcomes, due to the entanglement between the measurement outcomes and the post-measurement state. –  Niel de Beaudrap Jan 7 at 7:45
    
Do I have to view the measurement as an interaction between the apparatus and the system (therefore having to consider the combined apparatus-system state)? Why can't I simply consider the system's state by itself ($\psi$), and how it will evolve under some 'external' hamiltonian that represents the measuring apparatus? In that case, it would seem that I can still make the argument above in my original question - that at each measurement the state of the system will be ∑ci|ei⟩, ∑ci'|ei⟩, ∑ci''|ei⟩, etc. –  Tim Jan 8 at 5:40
    
If you do not track the state of the aparatus, you cannot track the state of the record of the apparatus, which is to say the measurement outcome; and therefore you cannot track whether or not the outcomes of multiple measurements are consistent. –  Niel de Beaudrap Jan 8 at 5:42

3 Answers 3

You may wish to look at http://arxiv.org/abs/1107.2138 (published in Phys. Rep.; extremely long:-(, but you can look at their introduction and conclusions; or you can look at their previous much shorter article http://arxiv.org/abs/quant-ph/0702135)

They consider a specific model of measurement and show how the Born rule and the projection postulate can be derived as approximations in some cases from unitary evolution. You are right that unitary evolution, strictly speaking, does not allow definite outcomes of measurement. The reason why this is not in contradiction with experimental results is shown to be the same as the reason why reversibility of Newton's laws (or unitary evolution of quantum mechanics) is not in contradiction with practical irreversibility of thermodynamics/statistical mechanics: the recurrence times can be very large.

share|improve this answer
    
This seems to be just a rehash of Zurek's work in Einselection, i.e., they still deal with density matrices and not the states themselves. The problem with this approach is that in an ensemble of state+environment pairs, you can't tell which pairs go which way, and the density matrix formalism implicitly contains Born's rule. –  lionelbrits Feb 16 at 12:50
    
@lionelbrits: They comment on Zurek's work on page 32 of their preprint. Do you find their comments unreasonable? As for density matrices, if they implicitly contain the Born's rule, how could they find deviations from the Born's rule? And I guess density matrix formalism includes pure functions as well. Or am I mistaken? –  akhmeteli Feb 16 at 17:29
1  
Well as for your second question, the density matrix assumes born's rule for a system+environment. It's only when the environment behaves in a certain way that the system itself seems to obey Born's rule on its own (the mixed state). –  lionelbrits Feb 18 at 22:16
    
@lionelbrits: I am not sure why "the density matrix assumes born's rule for a system+environment". Again, the density matrix formalism describes pure states as well. –  akhmeteli Feb 21 at 2:36
1  
Yes but tbh if you are truly in a pure state, then "collapse" never happens and there is no notion of probability. Only by coarse graining and ignoring the environmental effects do we seem to get reduction and Born's rule. But the density matrix is specifically constructed to deal with probabilities and ensembles, so while it gives the right answer, it doesn't seem to be able to tell you why you get the right answer –  lionelbrits Feb 22 at 14:26

as I understand this: Your first measurement will yield a result "i" with probability |c′i|2 , your following measurements will give result "i" with probability 1.

share|improve this answer

Note that, even without "reduction postulate", given any normalized initial state and any normalized final state, you may find a unitary transformation which transforms the initial state into the final state.

Of course, it will be a one-shot unitary transformation, depending only on the initial state and the final state, so you would not be able to "control" this transformation , to "reproduce" it, to associate some predefined hamiltonian, and so on...

For instance, suppose that the initial state is $|i\rangle = \alpha |+\rangle + \beta |-\rangle$, with $|\alpha^2| + |\beta|^2=1$, and suppose that the final state (after measurement) is $|f\rangle = |+\rangle$

Now, it is easy to see that the unitary matrix $U(f, i) = \begin{pmatrix} \bar \alpha& \bar \beta\\ -\beta&\alpha\end{pmatrix}$ makes the job, that is $|f\rangle = U(f, i) |i\rangle$

All this is coherent with quantum mechanics, you may consider, that you have the probability $|\alpha|^2$ to have some unitary evolution wich brings you in the state $|+\rangle$, and the probability $|\beta|^2$ to have some unitary evolution wich brings you in the state $|-\rangle$

However, these unitary transformations are generally not unique, so the above example is very particular.

This can be extended also, when considering more realistic states with interactions with measurement apparatus and/or environment. For instance, the states :

$|1\rangle = (\alpha |+\rangle + \beta |-\rangle) \otimes |M_x\rangle$

$|2\rangle = \alpha |+\rangle \otimes |M_+\rangle + \beta |-\rangle \otimes |M_-\rangle$

$|3\rangle = |+\rangle \otimes |M_+\rangle$

(where $|M_x\rangle, |M_+\rangle, |M_-\rangle$ are normalized states of the measurement apparatus), are all normalized states, corresponding respectively to a state without measurement apparatus, a pre-measurement state (with measurement apparatus, but with no explicit measurement), and a post-measurement state.

So, you will be always able to find a unitary transformation, which transforms $|1\rangle$ into $|2\rangle$, and an other unitary transformation which transforms $|2\rangle$ into $|3\rangle$, and a unitary transformation which transforms $|1\rangle$ into $|3\rangle$

Of course, these unitary transformations will have the same extreme limitations already descrived above (one-shot transformation, depending only on the initial state and the final state, not "controllable", not reproducible , not associated to some predefined hamiltonian)

You may have a look to consistent histories, too, for instance, here and here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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