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I used to believe that the wavefunction collapse came from the interaction of the system we want to measure {S} with the measurement apparatus {M} : {S} undergoing a non unitary transformation, but {S+M} undergoing a perfectly unitary transformation, given by the Schrödinger equation but with too many degrees of freedom to be calculated.

In that picture there is no unitarity problem for the measurement since the whole system undergoes a unitary transformation.

But that picture must be wrong in some way: let's say we want to measure a spin along $z$, and that the initial state of the measurement apparatus is $ \langle I |_{M} $ : we must find a unitary transformation $U$ for {S+M} such that :

-Spin eigenstates remain unchanged under $U$ (up to a phase): $$ \langle \uparrow |_{S} \langle I |_{M} \xrightarrow{U} \langle \uparrow |_{S} \langle F_{1} |_{M}$$ $$ \langle \downarrow |_{S} \langle I |_{M} \xrightarrow{U} \langle \downarrow |_{S} \langle F_{2} |_{M}$$

($\langle F_{i}|$ being any measurement apparatus final state)

-A superposition can be projected, on $\langle \uparrow |$ or $\langle \downarrow |$ depending on $\langle I |$ and/or $U$: $$ \frac{1}{\sqrt2}(\langle \uparrow |_{S}+\langle \downarrow |_{S}) \langle I |_{M} \xrightarrow{U} \langle \uparrow |_{S} \langle F_{3} |_{M}$$

But this is not possible : $$ (\langle \uparrow |+\langle \downarrow |) \langle I | = \langle \uparrow |\langle I |+\langle \downarrow | \langle I | \xrightarrow{U} \langle \uparrow | \langle F_{1} | + \langle \downarrow | \langle F_{2} |$$

$$ \langle \uparrow | \langle F_{1} | + \langle \downarrow | \langle F_{2} | = \langle \uparrow | \langle F_{3} |$$

except if $\langle \downarrow | \langle F_{2} |=0$, which can't be since $U$ is unitary.

Does it mean that we expect the measurement apparatus initial state to be correlated to the measured system initial state, or even {S+M} to be in an initial non factorizable state ? In any measurement experiment this is true since we prepare the system in some KNOWN state. I cannot find any other explanation (maybe many world interpretation).

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    $\begingroup$ Why do you write everything as a bra instead of the more standard ket? Also, it's really confusing that you use $S$ to refer to the Measurement apparatus and $M$ to refer to the System under study. This seems backwards. Finally, you should define $|F\rangle$ in the text. $\endgroup$ – DanielSank Nov 2 '14 at 4:56
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Your assumption that the final state is factorizable in that way is the problem. You have let wave function collapse sneak into the back door. Why not let the final state be in a superposition also, as quantum mechanics requires? The alive cat is not aware of the dead cat, because the Schrödinger equation is linear.

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  • $\begingroup$ "Why not let the final state be in a superposition also, as quantum mechanics requires?" What do you mean, QM says nothing about what happen to the measurement apparatus, and F can be a superposition here. But for the measured system it must not be a superposition after a measurement according to the collapse postulate. $\endgroup$ – agemO Dec 22 '14 at 20:41
  • $\begingroup$ Where can I buy an apparatus that is not made out of quantum mechanical atoms? $\endgroup$ – lionelbrits Dec 22 '14 at 20:48
  • $\begingroup$ I agree, what do you exactly want to say ? $\endgroup$ – agemO Dec 22 '14 at 20:50
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    $\begingroup$ Not everyone agrees that there has to be a definite state after measurement, I.e. collapse. If your measurement apparatus is quantum, then the whole system must be a solution to the S.E. and obey unitarity. Hence, the apparatus becomes entangled with the system, and does not collapse it. Decoherence prevents you from detecting this... $\endgroup$ – lionelbrits Dec 22 '14 at 20:54
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    $\begingroup$ Ok, So I think the answer to my question is "Collapse can't happen" ? $\endgroup$ – agemO Dec 22 '14 at 22:05
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You are wrong in a couple of things:

1) As long as your particle (the system {S}) evolves unperturbed, it evolves unitarily - i.e. its evolution can be described as a unitary transformation. (There is an exception from this rule, but if the internal structure of the particle doesn't change during the particle evolution, the exception is not relevant). By unperturbed, I mean that the particle does not interact in any way with another particle. (It may pass though through classical fields, but let's assume that neither this happens).

2) The measurement apparatus is a macroscopic body. That means a conglomerate with an infinity of particles, and the number of particles cannot even be considered constant. So, to speak of a "state" of the apparatus doesn't make sense. We build indeed this apparatus to INDICATE the result F_1 when a particle of spin up impinges, and F_2 when a particle of spin down impinges. But we CANNOT SAY more than this. The interaction of the particle {S} with the apparatus is NON-unitary.

3) To convince you, let me tell you that a unitary transformation can be reversed (undone). The measurement by a macroscopic apparatus as I described above CANNOT be undone. Assuming that the measurement is non-destructive, and the particle exist the apparatus, say in the state spin up, if you send it back to the apparatus you won't restore the initial polarization, (1/sqrt(2)) (|up> + |down>).

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  • $\begingroup$ So you are saying that QM is not consistent with itself, that is to say, if I use Shrodinger to find the evolution of the whole system (or even the entire universe), I will not find the same result as if I separate it in two subsystem ? In one case I will have a unitary transformation, in the other case no $\endgroup$ – agemO Nov 6 '14 at 3:50
  • $\begingroup$ "As long as your particle (the system {S}) evolves unperturbed, it evolves unitarily - i.e. its evolution can be described as a unitary transformation" I totally agree, that's also why the whole universe should follow a unitary evolution $\endgroup$ – agemO Nov 6 '14 at 15:41
  • $\begingroup$ What evidence do you have for the assertion that macroscopic systems can't have wave functions? $\endgroup$ – lionelbrits Dec 22 '14 at 16:12
  • $\begingroup$ The laws of classical physics are invariant under time reversal, yet we speak of irreversible processes. In the same way, we can have unitary evolution that cannot be undone due to coarse gaining $\endgroup$ – lionelbrits Dec 22 '14 at 17:00
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    $\begingroup$ Was directed at Sofia. $\endgroup$ – lionelbrits Dec 22 '14 at 21:01

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