Collapse and Peres' Coarse Graining

[rubi]

I think vanhees is right. The projection postulate is not needed here. If you use the SG apparatus to spatially separate the different spin particles, you end up with a mixed state $\rho_{SG} = \left|x_1,\uparrow\right>\left<x_1,\uparrow\right|+\left|x_2,\downarrow\right>\left<x_2,\downarrow\right|$ (the environment has already been traced out and the small off-diagonal terms have been neglected). Assume you want to do scattering experiments with the spin up particles by a potential $V(x)$, which is supported in a bounded region $R$. You would arrange the SG apparatus in such a way that the spin down particles end up in a different region ($x_2\notin R$), while $x_1\in R$. Now you would choose a basis for $L^2(R)$ and calculate the partial trace $\rho_R=\mathrm{Tr}_R\rho_{SG}=\left|x_1,\uparrow\right>\left<x_1,\uparrow\right|$. If you only want to measure observables in $R$, the states $\rho_{SG}$ and $\rho_R$ are indistinguishable for you. The whole system is still in a mixed state (or even in a pure state, if you include the environmental degrees of freedom) and no collapse has ever happened, but you have isolated a state $\rho_R$ that is indistinguishable from a hypothetically collapsed, pure state $\left|x_1,\uparrow\right>$ for experimenters who only measure in the region $R$.

 

[vanhees71]

I only fight against the word "collapse" and want to substitute by what's really done. You shoot a particle in an arbitary (perhaps unknown) state into an SG apparatus for the spin-z component. Experimental evidence tells us (since 1923 when Stern and Gerlach did their experiment for the first time) that it is deflected in such a way that the spin-z component has a well defined value in the corresponding region. Then I let all particles with the unwanted spin states run into the beam dump and I've either no particle or one with the desired spin-z value. That's what's called, in my understanding a state preparation by a filter (von Neumann) measurement. There's nothing in this setup making it necessary to invoke some strange "mechanism", claiming quantum theory to be invalid and call it "collapse".

 

[Demystifier]

There's nothing in this setup making it necessary to invoke some strange "mechanism", claiming quantum theory to be invalid and call it "collapse".

Of course. To repeat the words of Goldstein, "Either the Schrodinger equation is not right or wave function is not all." The collapse hypothesis corresponds to the Schrodinger-equation-not-right option, while you choose (even if you are not always aware* of it) the other, wave-function**-not-all option.

(*Here is the place where you are aware of it: When Quantum Mechanics is thrashed by non-physicists #1 )

(**In this context, the "wave function" means not only wave function in the narrow non-relativistic-QM sense, but also wave functional in QFT, or more abstractly, a ray in the physical Hilbert space.)

 

[atyy]

I think vanhees is right. The projection postulate is not needed here. If you use the SG apparatus to spatially separate the different spin particles, you end up with a mixed state $\rho_{SG} = \left|x_1,\uparrow\right>\left<x_1,\uparrow\right|+\left|x_2,\downarrow\right>\left<x_2,\downarrow\right|$ (the environment has already been traced out and the small off-diagonal terms have been neglected). Assume you want to do scattering experiments with the spin up particles by a potential $V(x)$, which is supported in a bounded region $R$. You would arrange the SG apparatus in such a way that the spin down particles end up in a different region ($x_2\notin R$), while $x_1\in R$. Now you would choose a basis for $L^2(R)$ and calculate the partial trace $\rho_R=\mathrm{Tr}_R\rho_{SG}=\left|x_1,\uparrow\right>\left<x_1,\uparrow\right|$. If you only want to measure observables in $R$, the states $\rho_{SG}$ and $\rho_R$ are indistinguishable for you. The whole system is still in a mixed state (or even in a pure state, if you include the environmental degrees of freedom) and no collapse has ever happened, but you have isolated a state $\rho_R$ that is indistinguishable from a hypothetically collapsed, pure state $\left|x_1,\uparrow\right>$ for experimenters who only measure in the region $R$.

The projection postulate is only needed for sequential measurements. What you are doing here is instead of sequential measurements, you have the system interact with an ancilla, then take first and second measurements at late and delayed time. It is a principle http://en.wikipedia.org/wiki/Deferred_Measurement_Principle that one can always push all measurements to the final step of the analysis, so that there are no sequential measurements, which is why collapse is not needed if one thinks of the SG experiment in this way.

But can one do this in the Bell test?

 

[vanhees71]

Why shouldn't one be able to do that for a Bell test? Usually you just count entangled biphotons in coincidence experiments. So why can't I describe this in this way?

Of course, I'm silent about how the probabilities come about. For me that's just an independent postulate of the formalism, and within quantum theory I cannot answer the question of, how this occurs, i.e., I don't know of any more simple system of postulates that make the Born postulate to a derivable theorem rather than an independent postulate. Weinberg gives in his newest textbook, "Lectures of Quantum Mechanics" a pretty convincing line of arguments that the Born rule indeed cannot be derived but is an independent postulate.

Of course, that doesn't exclude the possibility that somebody finds a more comprehensive theory of some kind that "explains" it from "more simple" postulates, but as long as I don't see such a theory, I'm inclined to live with quantum theory.

 

[atyy]

Why shouldn't one be able to do that for a Bell test? Usually you just count entangled biphotons in coincidence experiments. So why can't I describe this in this way?

It cannot be done in a Bell test, because the measurements are spacelike separated. So if the measurements are simultaneous in one frame, they will be sequential in another frame.

 

[vanhees71]

In the simple measurement a la Zeilinger Alice and Bob just measure the polarizations of their photons and compare measurement protocols, leading to a violation of Bell's inequality if they adjusted their polarization directions appropriately (see DrChinese's website for the details).

Ok, this reminds me that I still owe you the corresponding calculation within QED...

 

[atyy]

In the simple measurement a la Zeilinger Alice and Bob just measure the polarizations of their photons and compare measurement protocols, leading to a violation of Bell's inequality if they adjusted their polarization directions appropriately (see DrChinese's website for the details).

Well, as far as I know, if we use the Schroedinger picture, the initial state is |00> + |11>, and if Alice measures and gets the result "0", the state collapses to |00>. The collapse makes a successful prediction, since if Bob now measures, he will also get "0". For all measurement choices, the nonlocal correlations are consistent with collapse.

Ok, this reminds me that I still owe you the corresponding calculation within QED...

You only need it if you use the Schroedinger picture. It is that picture which has the collapse. The collapse can be hidden in some forms of the Heisenberg picture, eg. Eq 37 of http://arxiv.org/abs/quant-ph/0209123.

 

[rubi]

But can one do this in the Bell test?

In the SG apparaturs, I'm assuming that decoherence has approximately transformed the initial state $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow\right>+\left|0\downarrow\right>\right)$ into $\rho_{SG} = \frac{1}{2}\left(\left|x_1\uparrow\right>\left<x_1\uparrow\right|+\left|x_2\downarrow\right>\left<x_2\downarrow\right|\right)$. The rest of the argument is just the mathematical version of the claim that FAPP $\rho_{SG}$ is just as good as a collapsed state. So the crucial point is not the partial trace, but decoherence.

If you want to know whether this is also possible for the Bell test, you are effectively asking whether decoherence can approximately (up to small off-diagonal terms) transform the Bell state $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow0\downarrow\right>+\left|0\downarrow0\uparrow\right>\right)$ into $\rho = \frac{1}{2}\left(\left|x_1\uparrow x_2 \downarrow\right>\left<x_1\uparrow x_2 \downarrow\right|+\left|x_1\downarrow x_2\uparrow\right>\left<x_1\downarrow x_2\uparrow\right|\right)$. Taking partial traces afterwards is not problematic. So the question is: Can decoherence accomplish this? Unfortunately, I don't know, but my feeling is that it should be possible (The non-local correlations would then be hidden in the neglegted off-diagonal terms and the environmental degrees of freedom). Maybe someone else knows?

 

[atyy]

So the question is: Can decoherence accomplish this? Unfortunately, I don't know, but my feeling is that it should be possible (The non-local correlations would then be hidden in the neglegted off-diagonal terms and the environmental degrees of freedom). Maybe someone else knows?

I mean avoid collapse in the Bell test, and also predict the nonlocal correlations (ie. calculate the quantum result for the violation of the Bell inequalities) using the Schroedinger picture.

 

[rubi]

Maybe I'm wrong, but I don't see why an apparent collapse is not good enough for explaining the Bell test. Can you explain what goes wrong?

 

[atyy]

Maybe I'm wrong, but I don't see why an apparent collapse is not good enough for explaining the Bell test. Can you explain what goes wrong?

Let me think about it. However my intuition is to read your equation that $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow0\downarrow\right>+\left|0\downarrow0\uparrow\right>\right)$ evolves into $\rho = \frac{1}{2}\left(\left|x_1\uparrow x_2 \downarrow\right>\left<x_1\uparrow x_2 \downarrow\right|+\left|x_1\downarrow x_2\uparrow\right>\left<x_1\downarrow x_2\uparrow\right|\right)$ in a different way. Here you mean the variables |AaBb> to mean that A and B are the positions of Alice's and Bob's electrons. However, one can also read A and B as representing Alice's and Bob's measurement ancillae. If that is the case, the state vector is complete, and there is no decoherence since the measurement apparatus is included. However, this means that a pure state has turned into a mixed state, which would mean non-unitary evolution.

 

[billschnieder]

It cannot be done in a Bell test, because the measurements are spacelike separated. So if the measurements are simultaneous in one frame, they will be sequential in another frame.

I know you don't like hearing this but I've told you repeatedly how it is done in a Bell test. It is called "post-processing" using time-tags, aka "coincidence counting". The time tags are local to each station, but they are compared with each other "globally" in order to select events. The time tags have to be transformed to the same frame in order to compare them, so simultaneity is not a problem at all.

 

[atyy]

I know you don't like hearing this but I've told you repeatedly how it is done in a Bell test. It is called "post-processing" using time-tags, aka "coincidence counting". The time tags are local to each station, but they are compared with each other "globally" in order to select events. The time tags have to be transformed to the same frame in order to compare them, so simultaneity is not a problem at all.

I understand that what you are saying may be correct: in such a scenario, the Bell inequalities are not violated at spacelike separation. However, for the time being, we are accepting that quantum mechanics predicts the violation of the Bell inequalities at spacelike separation, so please, give your point a rest in this thread (until vanhees71 brings it up again - I think the two of you are the only local realists around here).

 

[atyy]

Maybe I'm wrong, but I don't see why an apparent collapse is not good enough for explaining the Bell test. Can you explain what goes wrong?

Just to add a bit to my comment in post #112. The reason for my alternative reading of |AaBb> as including the apparatus is that if we hope to obtain the nonlocal correlations, we cannot trace out the apparatus, because when we include the apparatus, it is observables on the apparatuses that we measure.

Perhaps one could make an even bigger wave function including the environment so that the total state is |EAaBb>, and maybe that would trace out to |AaBb> evolving from $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow0\downarrow\right>+\left|0\downarrow0\uparrow\right>\right)$ to $\rho = \frac{1}{2}\left(\left|x_1\uparrow x_2 \downarrow\right>\left<x_1\uparrow x_2 \downarrow\right|+\left|x_1\downarrow x_2\uparrow\right>\left<x_1\downarrow x_2\uparrow\right|\right)$. I'm not sure this will happen, but it does seem plausible, as you say. If so, then we should be able to consider a frame in which Alice and Bob measure simultaneously, and we should get the nonlocal correlations.

However, one would still be left with the question of what is happening for sequential measurements, where we consider a frame in which Alice measures before Bob. The Born rule (without collapse) only gives the probabilities of outcomes for a single measurement, not the joint probabilities for two measurements separated in time.

Actually, the point is really calculating the joint probabilities in sequential measurements. If one makes sequential measurements in the SG case, one also has the same problem, since one is measuring one observable followed by another observable. The reason for going to a Bell test to illustrate this is that in the SG case, one could conceivably deny that the two observables are measured at different times, and say that there really only are observables measured at the same time. In a Bell test, this is not possible because of the spacelike separation, which makes simultaneous measurements in one frame become sequential measurements in a different frame.

 

[rubi]

Just to add a bit to my comment in post #112. The reason for my alternative reading of |AaBb> as including the apparatus is that if we hope to obtain the nonlocal correlations, we cannot trace out the apparatus, because when we include the apparatus, it is observables on the apparatuses that we measure.

You are right, in a careful analysis, we should not trace out the apparatus. However, the $x_i$ were really meant to be the positions of the particles, which we should keep as well, if we want to model the behaviour of the particles after the mesaurements (in analogy to what happened with the SG apparatus above).

Perhaps one could make an even bigger wave function including the environment so that the total state is |EAaBb>, and maybe that would trace out to |AaBb> evolving from $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow0\downarrow\right>+\left|0\downarrow0\uparrow\right>\right)$ to $\rho = \frac{1}{2}\left(\left|x_1\uparrow x_2 \downarrow\right>\left<x_1\uparrow x_2 \downarrow\right|+\left|x_1\downarrow x_2\uparrow\right>\left<x_1\downarrow x_2\uparrow\right|\right)$. I'm not sure this will happen, but it does seem plausible, as you say. If so, then we should be able to consider a frame in which Alice and Bob measure simultaneously, and we should get the nonlocal correlations.

That's what I was thinking when I wrote this. The full system+environment wave-function evolves unitarily, but the restriction to the system yields a mixed state. If it is possible, I expect that the correlations are still there, at least in the full wave-function. In order to calculate it, we need to pick a model for the system-environment interaction and just do the computation. It's probably a bit involved, but maybe I find some time to do it on the weekend.

However, one would still be left with the question of what is happening for sequential measurements, where we consider a frame in which Alice measures before Bob. The Born rule (without collapse) only gives the probabilities of outcomes for a single measurement, not the joint probabilities for two measurements separated in time.

Actually, the point is really calculating the joint probabilities in sequential measurements. If one makes sequential measurements in the SG case, one also has the same problem, since one is measuring one observable followed by another observable. The reason for going to a Bell test to illustrate this is that in the SG case, one could conceivably deny that the two observables are measured at different times, and say that there really only are observables measured at the same time. In a Bell test, this is not possible because of the spacelike separation, which makes simultaneous measurements in one frame become sequential measurements in a different frame.

If we have a relativistic theory and a description of the process in one frame, then we should be able to use the representation of the Lorentz group to transform the whole physics (i.e. the wave-function and also the observables) to the other frame and the physics should be invariant under this transformation. The measurements themselves are events and therefore all observers should agree on them. Maybe the reduced density matrix would not transform covariantly, but that would not be a problem since taking a partial trace is not a physical process. However, writing down actual models for this gets more and more complicated, because we not only need to model the decoherence, but we also need to make sure that we model it in a Lorentz-invariant way. If decoherence is a physically viable theory, then it should be possible to do this in principle, since the environment is nothing but a lot of particles that should obey the laws of physics by themselves. I strongly believe that we would not measure a violation of Lorentz invariance if we compared the results of Bell tests from the POV of different observers and that's why I tend to opt out for an apparent collapse incuded by decoherence. Unfortunately, the bets are still on.

 

[atyy]

Unfortunately, the bets are still on.

But if the bets are still on as to whether decoherence alone can do the trick, I hope we agree that if one uses collapse, although the time evolution of the wave function is not covariant, the probabilities for all events remain Lorentz invariant if collapse is used. The non-covariance of the wave function doesn't matter since (in Copenhagen) we are agnostic about the physicality of the wave function. So collapse is viable in the postulates of relativistic quantum theory. ?

 

[kith]

If you want to know whether this is also possible for the Bell test, you are effectively asking whether decoherence can approximately (up to small off-diagonal terms) transform the Bell state $\left|\psi\right>=\frac{1}{\sqrt{2}}\left(\left|0\uparrow0\downarrow\right>+\left|0\downarrow0\uparrow\right>\right)$ into $\rho = \frac{1}{2}\left(\left|x_1\uparrow x_2 \downarrow\right>\left<x_1\uparrow x_2 \downarrow\right|+\left|x_1\downarrow x_2\uparrow\right>\left<x_1\downarrow x_2\uparrow\right|\right)$. Taking partial traces afterwards is not problematic. So the question is: Can decoherence accomplish this? Unfortunately, I don't know, but my feeling is that it should be possible (The non-local correlations would then be hidden in the neglegted off-diagonal terms and the environmental degrees of freedom). Maybe someone else knows?

I too would like to know how decoherence works in Bell tests. The last time I googled, I only found papers where decoherence was related to disturbances and errors and not used as a fundamental part in the modelling of the measurement process.

 

[rubi]

But if the bets are still on as to whether decoherence alone can do the trick, I hope we agree that if one uses collapse, although the time evolution of the wave function is not covariant, the probabilities for all events remain Lorentz invariant if collapse is used. The non-covariance of the wave function doesn't matter since (in Copenhagen) we are agnostic about the physicality of the wave function. So collapse is viable in the postulates of relativistic quantum theory. ?

That depends on whether each observer carries his own wave-function independently of the other observers and it only represents the information that each observer has. However, if the wave-function of one observer is related to the wave-function of another observer by a unitary transformation, even after a collapse, then the collapse can't be consistent with relativity, because the collapse changes the wave-function on a $t=const$ hyperplane. But different observers don't agree on these hyperplanes, so a collapse in one reference frame will appear as an influence on the past for a different observer.

As far as i can tell, people using relativistic quantum physics, like for example at colliders, never make sequential measurements, whereas people who make sequential measurements such that collapse plays a role, like people working in quantum optics, always work in the same frame of reference. So there seems to be no experimental evidence that could help us answer such quenstions.

I too would like to know how decoherence works in Bell tests. The last time I googled, I only found papers where decoherence was related to disturbances and errors and not used as a fundamental part in the modelling of the measurement process.

I'm also not aware of any such research. I think we could learn a lot about quantum theory by doing such calculations.

 

[atyy]

That depends on whether each observer carries his own wave-function independently of the other observers and it only represents the information that each observer has. However, if the wave-function of one observer is related to the wave-function of another observer by a unitary transformation, even after a collapse, then the collapse can't be consistent with relativity, because the collapse changes the wave-function on a $t=const$ hyperplane. But different observers don't agree on these hyperplanes, so a collapse in one reference frame will appear as an influence on the past for a different observer.

As far as i can tell, people using relativistic quantum physics, like for example at colliders, never make sequential measurements, whereas people who make sequential measurements such that collapse plays a role, like people working in quantum optics, always work in the same frame of reference. So there seems to be no experimental evidence that could help us answer such quenstions.

My understanding is that the quantum optics and the collider people are using the same theory, just that the collider people never test the collapse postulate, and the quantum optics people never test all the decays. So the wave function and collapse is different for each frame of reference. The wave functions of different observers are not related by unitary transformation after a collapse. However, collapse is consistent with Lorentz invariance, since all predictions of the theory are Lorentz invariant. For example:
http://arxiv.org/abs/quant-ph/9906034
http://arxiv.org/abs/1007.3977

These are just some recent references, but it was figured out quite some time ago that collapse doesn't cause any problems for relativity. So my understanding is that the standard model of particle physics does include the collapse postulate (I think one has to use Landau-Lifshitz-Weinberg style quantum mechanics, unless one uses consistent histories, because there is no consensus whether BM works for the standard model because of the chiral fermion problem, and I think there is still no consensus on whether MWI works even for non-relativistic QM in all technical details).

 

[atyy]

I think vanhees is right. The projection postulate is not needed here. If you use the SG apparatus to spatially separate the different spin particles, you end up with a mixed state $\rho_{SG} = \left|x_1,\uparrow\right>\left<x_1,\uparrow\right|+\left|x_2,\downarrow\right>\left<x_2,\downarrow\right|$ (the environment has already been traced out and the small off-diagonal terms have been neglected). Assume you want to do scattering experiments with the spin up particles by a potential $V(x)$, which is supported in a bounded region $R$. You would arrange the SG apparatus in such a way that the spin down particles end up in a different region ($x_2\notin R$), while $x_1\in R$. Now you would choose a basis for $L^2(R)$ and calculate the partial trace $\rho_R=\mathrm{Tr}_R\rho_{SG}=\left|x_1,\uparrow\right>\left<x_1,\uparrow\right|$. If you only want to measure observables in $R$, the states $\rho_{SG}$ and $\rho_R$ are indistinguishable for you. The whole system is still in a mixed state (or even in a pure state, if you include the environmental degrees of freedom) and no collapse has ever happened, but you have isolated a state $\rho_R$ that is indistinguishable from a hypothetically collapsed, pure state $\left|x_1,\uparrow\right>$ for experimenters who only measure in the region $R$.

Before tackling decoherence in the Bell case, I would first still like to understand how sequential measurements are treated here. In the scheme you outlined, there is a first measurement that takes place on a measurement apparatus that is later traced out. The measurement of the local observable on the apparatus will produce the same statistics p(M at t1) as if the measurement had taken place on the electron. Then the apparatus and other regions are traced out to show that a second local measurement will indeed produce the same statistics p(N at t2) on the state, as if collapse had taken place. But how do you treat the correlations between the first and second measurement? If the first measurement detects that a spin is up, the second measurement will also detect that the spin is up. For this we need something like p(M at t1, N at t2) or p(N at t2|M at t1), but the Born rule doesn't say anything about what happens for measurements at different times.

The problem can be evaded by saying that since the apparatus only briefly interacts with the electron, we can measure later on the apparatus p(M at t2) = p(M at t1), which is what I meant by deferring the measurement so that there are no sequential measurements. My reason for going to the Bell test was to force a sequential measurement in some frame. But if we accept that we can measure M at t1, then I don't understand how the Born rule without collapse can produce p(M at t1, N at t2).

 

[rubi]

My understanding is that the quantum optics and the collider people are using the same theory, just that the collider people never test the collapse postulate, and the quantum optics people never test all the decays. So the wave function and collapse is different for each frame of reference. The wave functions of different observers are not related by unitary transformation after a collapse. However, collapse is consistent with Lorentz invariance, since all predictions of the theory are Lorentz invariant. For example:
http://arxiv.org/abs/quant-ph/9906034
http://arxiv.org/abs/1007.3977

These are just some recent references, but it was figured out quite some time ago that collapse doesn't cause any problems for relativity. So my understanding is that the standard model of particle physics does include the collapse postulate (I think one has to use Landau-Lifshitz-Weinberg style quantum mechanics, unless one uses consistent histories, because there is no consensus whether BM works for the standard model because of the chiral fermion problem, and I think there is still no consensus on whether MWI works even for non-relativistic QM in all technical details).

My point is that if wave-functions of individual observers aren't related by unitary transformations, then the postulate of having a unitary representation of the Poincare group on the Hilbert space is not really justified, because our reason to postulate it in the first place is to make the theory independent of the choice of frame. I feel that the success of that postulate suggests that we should seek to describe the collapse in such a way that unitary equivalence is always obeyed. That would also make the quantum state a coordinate-independent object like a tensor in general relativity. (As a person coming from the quantum gravity camp, I'm basically obliged to prefer such a point of view. :D)

Before tackling decoherence in the Bell case, I would first still like to understand how sequential measurements are treated here. In the scheme you outlined, there is a first measurement that takes place on a measurement apparatus that is later traced out. The measurement of the local observable on the apparatus will produce the same statistics p(M at t1) as if the measurement had taken place on the electron. Then the apparatus and other regions are traced out to show that a second local measurement will indeed produce the same statistics p(N at t2) on the state, as if collapse had taken place. But how do you treat the correlations between the first and second measurement? If the first measurement detects that a spin is up, the second measurement will also detect that the spin is up. For this we need something like p(M at t1, N at t2) or p(N at t2|M at t1), but the Born rule doesn't say anything about what happens for measurements at different times.

The problem can be evaded by saying that since the apparatus only briefly interacts with the electron, we can measure later on the apparatus p(M at t2) = p(M at t1), which is what I meant by deferring the measurement so that there are no sequential measurements. My reason for going to the Bell test was to force a sequential measurement in some frame. But if we accept that we can measure M at t1, then I don't understand how the Born rule without collapse can produce p(M at t1, N at t2).

I would calculate all the individual probabilities using $\mathrm{Tr}(\rho(t)P)$, where $\rho(t)$ is the density matrix for the different times and $P$ is the projection you are interested in. Then I would calculate the conditional probabilities using the rule for conditional probabilities from probability theory. Note that I use the projection operators only as a tool to calculate the probabilities. I'm not actually applying it to the state afterwards.

 

[atyy]

I would calculate all the individual probabilities using $\mathrm{Tr}(\rho(t)P)$, where $\rho(t)$ is the density matrix for the different times and $P$ is the projection you are interested in. Then I would calculate the conditional probabilities using the rule for conditional probabilities from probability theory. Note that I use the projection operators only as a tool to calculate the probabilities. I'm not actually applying it to the state afterwards.

But how do I get the conditional probabilities or joint probabilities between observables that are measured at different times? The Born rule only gives probabilities and conditional probabilities between observables measured at the same time.

 

[rubi]

But how do I get the conditional probabilities or joint probabilities between observables that are measured at different times? The Born rule only gives probabilities and conditional probabilities between observables measured at the same time.

I probably don't understand what the problem is, but here is an example: 50% of all particles leave the SG apparatus as spin up at time $t_1$ and maybe 25% of the particles leave the apparatus as spin up at $t_1$ and are scattered into a solid angle $\Omega$ at time $t_2$. Then the conditional probability would be $\frac{1}{4}/\frac{1}{2} = \frac{1}{2}$. The Born rule is not modified if I reject the projection postulate.

 

[atyy]

I probably don't understand what the problem is, but here is an example: 50% of all particles leave the SG apparatus as spin up at time $t_1$ and maybe 25% of the particles leave the apparatus as spin up at $t_1$ and are scattered into a solid angle $\Omega$ at time $t_2$. Then the conditional probability would be $\frac{1}{4}/\frac{1}{2} = \frac{1}{2}$. The Born rule is not modified if I reject the projection postulate.

Can the Born rule calculate that 25% of the particles leave the apparatus as spin up at $t_1$ and are scattered into a solid angle $\Omega$ at time $t_2$? That uses two different times, so if a probability is calculated using $\mathrm{Tr}(\rho(t)P)$, what $t$ is being plugged into the formula: $t_1$ or $t_2$?

The problem I'm having is that if I have:
$P(M(t_1)) = \mathrm{Tr}(\rho(t_1)M)$
$P(N(t_2)) = \mathrm{Tr}(\rho(t_2)N)$,

how can I calculate $p(N(t_2)|M(t_1))$?

From Bayes's rule, I have $P(N|M) = P(M,N)/P(M)$, but although $P(M)$ is known, neither $P(N|M)$ nor $P(M,N)$ are known.

 

[strangerep]

Try Ballentine, section 9.6.

 

[atyy]

Try Ballentine, section 9.6.

Ballentine assumes the collapse hypothesis in his Eq 9.28.

 

[strangerep]

Ballentine assumes the collapse hypothesis in his Eq 9.28.

No, he doesn't. He's using a filtering operator which maps one state into another. But we've been over this before, so I won't retrace that journey.

 

[Jazzdude]

[...]. I feel that the success of that postulate suggests that we should seek to describe the collapse in such a way that unitary equivalence is always obeyed. That would also make the quantum state a coordinate-independent object like a tensor in general relativity. (As a person coming from the quantum gravity camp, I'm basically obliged to prefer such a point of view. :D)

I think this will not work out, even if you don't have any form of collapse. The multi-particle spaces used in quantum theory are not unitarily equivalent for different rest frames. A simple example like a 2-electron Hilbert space already demonstrates the problem, and introducing a general Fock space doesn't make it any simpler.

So before we discuss the general covariance of a possible collapse mechanism, we'd have to find an alternative for the state space.

 

[Dale]

This thread has well outlived its usefulness