What are the principal results of the decoherence in QM?

In summary: H1, the |psi> state will have the form |psi> = |H1eigenstate>|. If you choose the eigenbasis of H2, the |psi> state will have the form |psi> = |H2eigenstate>|. The rule is that the |psi> state is the unique state that is the product of the eigenstate of the system under study and the eigenstate of the measurement apparatus.
  • #1
Terra Incognita
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Hi all,

1)What are the principal results of the decoherence in QM?
2) Is there a general theorem we can use to determinate the states of macroscopic bodies (huge number of particles)?
3) if yes, what are the known limitations?

Terra Incognita
 
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  • #2
Terra Incognita said:
Hi all,

1)What are the principal results of the decoherence in QM?
2) Is there a general theorem we can use to determinate the states of macroscopic bodies (huge number of particles)?
3) if yes, what are the known limitations?

A very good review in my opinion is given in quant-ph/0312059

The main result of decoherence is the following: when you take into account the (tiny) interactions with the environment, and consider unitary evolution describing this, then the overall wavefunction evolves very quickly into a sum of states:

|psi> = |someenvironmentstate> |pointerstate1> |sys1> + |someenvironmentstatebis> |pointerstate2> |sys2> +...

where "pointerstate" corresponds to a classical state of the measurement apparatus, and "sys1", "sys2"... correspond to the eigenstates of the system under study, of the operator implemented by the measurement apparatus.

As such, decoherence solves a part of the "measurement problem", namely the "preferred basis" problem.

cheers,
Patrick.
 
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  • #3
vanesch said:
A very good review in my opinion is given in quant-ph/0312059

Ok, seems a good paper. I've quiclky read it. I think they do not explain the preferred basis selection.

vanesch said:
The main result of decoherence is the following: when you take into account the (tiny) interactions with the environment, and consider unitary evolution describing this, then the overall wavefunction evolves very quickly into a sum of states:

|psi> = |someenvironmentstate> |pointerstate1> |sys1> + |someenvironmentstatebis> |pointerstate2> |sys2> +...

where "pointerstate" corresponds to a classical state of the measurement apparatus, and "sys1", "sys2"... correspond to the eigenstates of the system under study, of the operator implemented by the measurement apparatus.

As such, decoherence solves a part of the "measurement problem", namely the "preferred basis" problem.

cheers,
Patrick.

I don't understand very well. Any |psi> state may be decomposed like that (we just select the eigenbasis of each hilbert space and write down the |psi> state). Therefore I do not understand why a particular eigenbasis for the |psi> vector is selected or if you prefer the particular eigenbasis of |pointerstate> for example (what makes this particular selection?: the rule).

If one can select the eigenbasis as one likes, what additional information does decoherence add relatively to the collapse postulate?

Thanks in advance,

Terra Incognita :frown:
 
  • #4
Terra Incognita said:
Ok, seems a good paper. I've quiclky read it. I think they do not explain the preferred basis selection.

You cannot have read a paper of 40 pages in 5 minutes :-)

cheers,
Patrick.
 
  • #6
vanesch said:
You cannot have read a paper of 40 pages in 5 minutes :-)

cheers,
Patrick.

You are right. But I already knew it (I've read it one year ago I think) :tongue2: . This paper explains very well the current status on the decoherence program (end 2003), better than zureck papers (external view, even if it is also a biased view). However it does not explain the rule to choose a particular eigen basis.
In other words, if we have an hamiltonian H= HSO+HEO+Hint (where HSO is the free system hamiltonian, HEO the free environment and Hint the interaction). I do not know what eigen basis to select (the eigen basis of H, the one of HSO etc ...).

Terra Incognita
 
  • #7
Edgardo said:
Hello Terra Incognita,

Maybe you will find this article useful:
http://physicsweb.org/articles/world/18/3/5/1

It's about interference experiments with large molecules.

Thanks edgardo. This one I haven't already read it. I will check it and give a feed back.

Terra Incognita
 
  • #8
Edgardo said:
Hello Terra Incognita,

Maybe you will find this article useful:
http://physicsweb.org/articles/world/18/3/5/1

It's about interference experiments with large molecules.

Ok, after a brief review. I have the same problem (what selects the preferred basis).
In this paper, the implicitly selected eigenbasis (the preferred basis) to study the decoherence is mainly the position (e.g. the position of the photons or molecules on the screen on the double slit experiment). However, No explanation is given on why the observed results of the experiment are in this basis (the preferred basis problem).
If you prefer, if I construct a quantum system experiment, I would like to know, a priori, what basis I will see during my observation: the collapse postulate does not tell what the basis is.

I may understand the entanglement and the unitary time evolution. However, to infer experimental results, if I have no rule to select the eigen basis, I stay with the collapse postulate of QM and I do not see what additional information brings decoherence.

Terra Incognita :frown: .
 
  • #9
Terra Incognita said:
I don't understand very well. Any |psi> state may be decomposed like that (we just select the eigenbasis of each hilbert space and write down the |psi> state).

That is not true for an arbitrary basis. If you have a tensor product of two hilbert spaces, H = H1 x H2, and if {|a_n>} is a basis of H1, and {|b_m>} is a basis of H2, then a basis of H is of course {|a_n> x |b_m> }. This means that any ket |psi> of H can be written as nxm terms. But this means, for instance, that a particular |a_k> appears several times in the decomposition (in principle it can appear m times). However, there is only ONE basis (apart from degeneracy) in H1 and H2 so that each |a_n> only appears ONCE. That is the theorem of Schmidt decomposition. This basis is dependent on |psi> of course.

cheers,
Patrick.
 
  • #10
Terra Incognita said:
Ok, after a brief review. I have the same problem (what selects the preferred basis).

That's exactly the problem that decoherence takes up!

In this paper, the implicitly selected eigenbasis (the preferred basis) to study the decoherence is mainly the position (e.g. the position of the photons or molecules on the screen on the double slit experiment). However, No explanation is given on why the observed results of the experiment are in this basis (the preferred basis problem).

Well, that should follow from the interaction hamiltonian of the measurement apparatus with the environment: only (rough) position states are robust against "mixing" under this hamiltonian (meaning: under this evolution, the Schmidt decomposition of |psi(t)> remains roughly the same).

In more detail:

if at t0, we have a certain psi and a Schmidt decomposition:

|psi(t0)> = |environment state1> |pointer POSITION state1> + |environment state2> |pointer POSITION state 2> +...

then at a later time, evolving under the interaction hamiltonian of measurement system and environment, we have of course another state |psi(t1)> with ANOTHER Schmidt decomposition |environment state B1> |pointer POSITION state B1> +...

Well, it turns out that the pointer position state B1, although different from state1 (and maybe almost orthogonal in Hilbert space if there are many pointer positions), is still classically very close to it. And you will not find states that look like, say, pure momentum states of the pointer. The Schmidt decomposition of |psi(t)>, once the interaction hamiltonian with the environment is taken into account, is dominated by almost classical and almost stable position states for the pointer (in fact highly localized wave packets).

This is in fact a property of the interaction hamiltonian with the environment, which is roughly diagonal in the position wavepacket basis.

I have to say I'm myself still trying to understand all the details of this, but I can feel in my bones that this is a plausible explanation...

cheers,
Patrick.
 
  • #11
vanesch said:
That is not true for an arbitrary basis. If you have a tensor product of two hilbert spaces, H = H1 x H2, and if {|a_n>} is a basis of H1, and {|b_m>} is a basis of H2, then a basis of H is of course {|a_n> x |b_m> }. This means that any ket |psi> of H can be written as nxm terms. But this means, for instance, that a particular |a_k> appears several times in the decomposition (in principle it can appear m times). However, there is only ONE basis (apart from degeneracy) in H1 and H2 so that each |a_n> only appears ONCE. That is the theorem of Schmidt decomposition. This basis is dependent on |psi> of course.

cheers,
Patrick.

Thanks Patrick, but I have problems to understand.

For every |psi>, you can write |psi>= sum_nm c_nm|a_n>|b_m>= sum_n |a_n>(sum_m cnm|b_m>)= sum_n |a_n>|c(a_n)>.
Therefore, for each |a_n> (measurement result a_n), I know I have the associated state |c(a_n)>.
If I choose another eigenbasis, I will have other possibilities (freedom of the collapse postulate).
Therefore, what eigenbasis is the preferred one and why? Why do we need to choose the particular schmidt decomposition you proposed and not my decomposition:
If I choose the double slit experiment, I think the interference pattern is not in the schmitt decomposition you proposed but in one of the decompositions, I have shown.

Terra Incognita :frown:
 
  • #12
vanesch said:
Well, it turns out that the pointer position state B1, although different from state1 (and maybe almost orthogonal in Hilbert space if there are many pointer positions), is still classically very close to it. And you will not find states that look like, say, pure momentum states of the pointer.

Therefore, you are saying that we are not measuring a position eigenbasis but another different eigenbasis that is "close" to the position eigenbasis. Ok. But why this eigenbasis and not another one? (collapse postulate always work, whatever eigenbasis we select).


vanesch said:
The Schmidt decomposition of |psi(t)>, once the interaction hamiltonian with the environment is taken into account, is dominated by almost classical and almost stable position states for the pointer (in fact highly localized wave packets).

This is in fact a property of the interaction hamiltonian with the environment, which is roughly diagonal in the position wavepacket basis.

This is not the case for the double slit experiment, where the impacts of the photons on the screen are the energy of the photons time the sreen position.
Why do we select the energy photon eigenbasis and the position screen eigenbasis? :frown:

Once you have selected your eigenbasis, with the collapse postulate you have no more problems (what I understand). However, I do not know where one shows what eigenbasis is selected in a measurement experiment (if there exists a rule of selection).
I may accept that QM does not describe this selection procedure (external from the theory): we have to learn (a posteriori) from the different experiments in order to know what eignen basis is selected during a particular experiment. If this is the case, frankly, I think QM has to be amended with an external rule to select this basis (if it is possible).

Terra Incognita
 
  • #13
Terra Incognita said:
Thanks Patrick, but I have problems to understand.

For every |psi>, you can write |psi>= sum_nm c_nm|a_n>|b_m>= sum_n |a_n>(sum_m cnm|b_m>)= sum_n |a_n>|c(a_n)>.
Therefore, for each |a_n> (measurement result a_n), I know I have the associated state |c(a_n)>.

You puzzled me for about 10 minutes, I felt woossshh !
I reread the statement of the Schmidt decomposition theorem (which states that the decomposition is UNIQUE) and didn't find directly where your argument fails.

I think I found it. There is no guarantee that |c(a_k)> and |c(a_l)> are ORTHOGONAL states.

The Schmidt decomposition tells us that, requiring orthogonal states (a BASIS) in both spaces, you have a unique decomposition (except if the coefficients of certain terms are identical, in which case you can play your game ; that's the case of degeneracy).

cheers,
Patrick.
 
  • #14
Edgardo said:
Hello Terra Incognita,

Maybe you will find this article useful:
http://physicsweb.org/articles/world/18/3/5/1

It's about interference experiments with large molecules.
Maybe I should start another thread for this question.

From the article: "To avoid the seemingly decisive role played by the observer, physicists put forward many alternative theories and interpretations. Often this was done at the price of introducing as-yet-unobserved quantities into quantum mechanics called hidden variables.

Decoherence theory, in contrast, is based firmly on the conventional framework of quantum mechanics. ..."

Why can't hidden variables be introduced/included in Decoherence theory?

Regards
Don
 
  • #15
vanesch said:
You puzzled me for about 10 minutes, I felt woossshh !

Sorry, It wasn't my intention :tongue2:

vanesch said:
I think I found it. There is no guarantee that |c(a_k)> and |c(a_l)> are ORTHOGONAL states.

Why do we need individual orthogonal states? (is there a good explanation, or just a common guess as the use of the position eigen basis as a preferred basis in most of the experiments).
Collapse postulate just requires the global state to be orthogonal <a_k|<c(a_k)|a_l>|c(a_l)>= delta(k,l). And we have this result, with my proposed decompostion.

vanesch said:
The Schmidt decomposition tells us that, requiring orthogonal states (a BASIS) in both spaces, you have a unique decomposition (except if the coefficients of certain terms are identical, in which case you can play your game ; that's the case of degeneracy).
cheers,
Patrick.

Let's suppose a moment this case is the general case. What eigenbasis is selected in such experiments?

Terra Incognita :frown:
 
  • #16
dlgoff said:
Maybe I should start another thread for this question.

From the article: "To avoid the seemingly decisive role played by the observer, physicists put forward many alternative theories and interpretations. Often this was done at the price of introducing as-yet-unobserved quantities into quantum mechanics called hidden variables.

Decoherence theory, in contrast, is based firmly on the conventional framework of quantum mechanics. ..."

Why can't hidden variables be introduced/included in Decoherence theory?

Regards
Don

Reformulated with other words, we can say this is the problem of the eigenbasis selection (what I would like to know).
If we know, a priori, using QM theory, what eigenbasis will be selected by a given experiment, no hidden variable is required (what eigenbasis we select to apply the collapse postulate).
If QM theory is really independant of the eigenbasis selection (the collapse postulate), we therefore are free to imagine an external eigenbasis ad hoc selection to reflect what is given by all the possible experiments. This can be done through hidden variables (e.g. trhough the labelling of all the past, present and future experiments: experiment(a) => eigenbasis(a) or through more compact procedures: may be what decoherence tries to do - my current questions).

TI.
 
  • #17
Terra Incognita said:
Why do we need individual orthogonal states? (is there a good explanation, or just a common guess as the use of the position eigen basis as a preferred basis in most of the experiments).
Collapse postulate just requires the global state to be orthogonal <a_k|<c(a_k)|a_l>|c(a_l)>= delta(k,l). And we have this result, with my proposed decompostion.

Well, I'm also of the opinion that SOME extra postulate is necessary, and I think that the Schmidt decomposition enters in that postulate.
Warning: this is my personal view (but I think I'm in the company of people like Wigner and Stapp).
I think there needs to be an extra postulate that 1) requires us to write the state of the universe as a Schmidt decomposition in H_mybody x H_restofuniverse, 2) makes us experience consciously only ONE of those terms and 3) with a probability given by the Born rule.

But what is interesting is that we do not have to postulate a particular basis by hand: requiring Schmidt decomposition naturally leads to terms which correspond to "classical" states, which most of the time come down to "position" states (wave packets).

I'm also of the opinion that this is an interesting result but doesn't solve the entire issue, as some (like Zurek) seem to claim.

cheers,
Patrick.

EDIT: however, if you somehow accept the Born rule, even in your basis you will then find a result, which is probably the most tangible result of decoherence: if you calculate the reduced density matrix of the measurement apparatus (by tracing out the states of the environment), you are NATURALLY lead to a Schmidt decomposition, because it is in the corresponding basis for the measurement apparatus that this reduced density matrix is diagonal. It is then very natural to consider that reduced density matrix as describing an (improper) mixture. If you work in another basis, your reduced density matrix will not be diagonal.
 
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  • #18
vanesch said:
Well, I'm also of the opinion that SOME extra postulate is necessary, and I think that the Schmidt decomposition enters in that postulate.
Warning: this is my personal view (but I think I'm in the company of people like Wigner and Stapp).
I think there needs to be an extra postulate that 1) requires us to write the state of the universe as a Schmidt decomposition in H_mybody x H_restofuniverse, 2) makes us experience consciously only ONE of those terms and 3) with a probability given by the Born rule.

But what is interesting is that we do not have to postulate a particular basis by hand: requiring Schmidt decomposition naturally leads to terms which correspond to "classical" states, which most of the time come down to "position" states (wave packets).

I'm also of the opinion that this is an interesting result but doesn't solve the entire issue, as some (like Zurek) seem to claim.

cheers,
Patrick.

Thanks a lot Patrick. But it is not a very satisfactory reply. (it is a sort of "do the experiment first then use the selected basis displayed by the experiment to make other predictions"). :frown:

For example, can we build by hand an interaction between a system and a human being where in the schmitt basis decomposition, the human position is no more the the preferred basis?

Or, don't you think that this schmitt decomposition may lead to false predictions on some particular huge systems (where macroscopic quantum effects are viewed)?

TI.
 
  • #19
Terra Incognita said:
... (it is a sort of "do the experiment first then use the selected basis displayed by the experiment to make other predictions"). :frown:
TI.
My thought was, maybe that's it. The hidden varable somehow being the act of experimenting?

Does this sound as weird to you as it does to me?
 
  • #20
Terra Incognita said:
For example, can we build by hand an interaction between a system and a human being where in the schmitt basis decomposition, the human position is no more the the preferred basis?

I think you have to consider the environment versus the human body, not the system under study and the human body, if the human body is to be part of the measurement apparatus.
I think that what you get out of decoherence is that if that human body is in a room where there is air, EM radiation corresponding to a temperature of 20 degrees, some light etc... then the Schmidt decomposition in H_room x H_body+system very quickly leads to states where the body is in localised states. These are not pure position states. These are wavepackets that define position and momentum as accurately as possible (often called coherent states).

Have you seen my edit concerning the reduced density matrix in my previous post ?

cheers,
Patrick.
 
  • #21
vanesch said:
Have you seen my edit concerning the reduced density matrix in my previous post ?

cheers,
Patrick.

Thanks for this clarification, I've have taken a coffee break, so I haven't noticed your modifications. I will check it.
The current point of discussion is close to the post of dlgof. So it will try to reply to both at the same time.

TI. :frown:
 
  • #22
dlgoff said:
My thought was, maybe that's it. The hidden varable somehow being the act of experimenting?

Does this sound as weird to you as it does to me?

Well, once we accept that a theory does not describe the whole thing (independance, in order to get coherent results), the labelling of the experiments becomes the hidden variable of the eigenbasis selection (this is the a posteriori selection). Therefore you call this label "the act of experimenting number a" as long as there is no possible confusion with another act of experimenting.

In fact it is not as weird (or stupid) as we can think. In classical mechanincs, we have mainly 2 choices to describe the universe:
a) using the f=m.a and the initial conditions to descibe the paths of the particles (very dense formula) or
b) labelling all the paths of all the particles in the universe (i.e. the experiment label "a" just diplays a path of a particle).

I think egyptians point of view were closer to point b) while modern occidental point of view is closer to a), a compact form (~reductionism). Both have their advantages.

But this does not explain if decoherence can tell us or not the preferred basis.

T.I :frown:
 
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  • #23
Terra Incognita said:
But this does not explain if decoherence can tell us or not the preferred basis.
Well I don't feel as stupid thanks to your reply.

vanesch said:
I'm also of the opinion that this is an interesting result but doesn't solve the entire issue, as some (like Zurek) seem to claim.
Does this mean that there were einstates back at the time of the big band that are still fixed in our universe/reality?
 
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  • #24
vanesch said:
EDIT: however, if you somehow accept the Born rule, even in your basis you will then find a result, which is probably the most tangible result of decoherence: if you calculate the reduced density matrix of the measurement apparatus (by tracing out the states of the environment), you are NATURALLY lead to a Schmidt decomposition, because it is in the corresponding basis for the measurement apparatus that this reduced density matrix is diagonal. It is then very natural to consider that reduced density matrix as describing an (improper) mixture. If you work in another basis, your reduced density matrix will not be diagonal.

Nevertheless, why (the reason) do we have to choose this particular eigenbasis (where the densiy matrix is almost diagonal). If I take the double slit experiment, after the slits I have:
|psi_photon+slitplate>= |slits_plate>(|left_photon_state>+|right_photon_state>).
In this basis <left_photon_state|right_photon_state>=/=0.
Your Schmitt basis decomposition simply does not work (observation of interferences) in this case.
Therefore, I do not see the natural way to select one eigen basis versus another one.


vanesch said:
I think you have to consider the environment versus the human body, not the system under study and the human body, if the human body is to be part of the measurement apparatus.
I think that what you get out of decoherence is that if that human body is in a room where there is air, EM radiation corresponding to a temperature of 20 degrees, some light etc... then the Schmidt decomposition in H_room x H_body+system very quickly leads to states where the body is in localised states. These are not pure position states. These are wavepackets that define position and momentum as accurately as possible (often called coherent states).

I have brought the human into this subject in order to show why, for me, it is difficult to see where/why there is an eigenbasis selection.
Just suppose, we can model the human thought by a quantum state, we call |human_mind> (if you prefer the state of the brain or of the soul :biggrin: ).
We may thus model the interaction of the human thought with the rest of the world through a "simple" hamiltonian (very coarse simplification):

H= H_ROW+H_mind+ Hint.

(Simpler form than yours: we avoid taking into account unusefull information)

Where Hint describes an unknown interaction between the mind/thought and the rest of the world (including the human body, the eyes, the system, the experiment etc ...).

If I use the collapse postulate, I will say |human_mind>=|a photon at position x at the screen at time t> for example.
When I say that, I have implicitely selected an eigen basis that defines automatically the state of the ROW (through the interaction).
However, I am free to select any |human_mind> state: they are just possible thoughts of a human mind. They can belong to any basis . Therefore with the collapse postulate, I have no preferred eigen basis for the |human_mind> states.
However, practically, most of the times we notice that an experiment has a preferred basis of states in |human_mind>: Only the collapse postulate applied to an adequate eigenbasis for a given experiment seems to work.
I just do not understand well, how a schmitt decomposition reflect this result.

TI. :frown:
 
  • #25
Terra Incognita said:
If I use the collapse postulate, I will say |human_mind>=|a photon at position x at the screen at time t> for example.
When I say that, I have implicitely selected an eigen basis that defines automatically the state of the ROW (through the interaction).
However, I am free to select any |human_mind> state: they are just possible thoughts of a human mind. They can belong to any basis . Therefore with the collapse postulate, I have no preferred eigen basis for the |human_mind> states.
However, practically, most of the times we notice that an experiment has a preferred basis of states in |human_mind>: Only the collapse postulate applied to an adequate eigenbasis for a given experiment seems to work.
I just do not understand well, how a schmitt decomposition reflect this result.

TI. :frown:

However, my example hilights the limit of the collapse postulate (and the born rules) and QM in general:
What can we say about the statistics of an object if we just have a single instance of this object (what statistics we may compute)?

(To verify the statistics of |human_mind> I must have identical humans !)

TI. :frown:
 
  • #26
Terra Incognita said:
However, my example hilights the limit of the collapse postulate (and the born rules) and QM in general:
What can we say about the statistics of an object if we just have a single instance of this object (what statistics we may compute)?

(To verify the statistics of |human_mind> I must have identical humans !)

TI. :frown:
Is it possible that the eigenbasis selection is also imprinted on our being? That is, the thought process has a preferred direction? This is beginning to sound more Philosophical.

Regards
Don
 
  • #27
dlgoff said:
Is it possible that the eigenbasis selection is also imprinted on our being? That is, the thought process has a preferred direction? This is beginning to sound more Philosophical.

Regards
Don

This is far from beeing evident (my point of view). Especially when we consider the "thought observable" may have degenerated eigenvalues.

To illustrate this and my previous post let's use the optical illusions: depending on the context, we observe different results.

Case 1: moving patches with or without grid:

http://www.michaelbach.de/ot/mot_feet_lin/index.html

(choose colour of, for a maximal effect)

We can see that the apparent motion (the observation) of the two moving blocks depend on the context. What the |human_mind> see in this case depends on what is behind (the grid) the 2 moving patches:

|human_mind1(t)>|patches(t)>|no_grid>
|human_mind2(t)>|patches(t)>|grid>

While the state |patches(t)> does not change.

We may wrongly interpret the observation |human_mind1(t)> and |human_mind2(t)> as the virutally observed states "|patches1(t)>" (continuous displacement) and |patches2(t)> (discrete displacement).

Case 2: the shadow.

http://psylux.psych.tu-dresden.de/i1/kaw/diverses Material/www.illusionworks.com/html/shadow.html

The light-colored check in the middle of the shadow is the same shade of gray as the dark checks outside the shadow.

In this case, we may say that the colour interpretation of the "light-colored check" in the middle depends on the presence of this special shadow (the shadow does not change the colour of the light-colored check in the middle).

Case 3 (joint document): the young/hold woman.

In this case, what we see depends on what we want to see. The same image may be interpreted at least in 3 ways:
1) |human_mind1> sees a young woman
2) |human_mind2> sees a hold woman
3) |human_mind3> sees simply a strange picture

The 3rd one may be interpreted by the thought as the "observation" of a superposition of the states 1 and 2.

The choice between the 3 states depends on what we want to see (and the way our brain has been educated) and not on what it is (the external context).

TI.
 

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  • #28
I see what you mean Terra. But human beings are a lot alike.

Regards
 
  • #29
Terra Incognita said:
If I take the double slit experiment, after the slits I have:
|psi_photon+slitplate>= |slits_plate>(|left_photon_state>+|right_photon_state>).
In this basis <left_photon_state|right_photon_state>=/=0.

I don't understand this. I would think that |left_photon_state> and |right_photon_state> are (almost) orthogonal ! (as wavepackets centered around different positions).
Or do you mean by "left photon state" the photon state from the left hole ?
Even in that case, they are orthogonal (or almost so, no ?)

But this has nothing to do with the Schmidt decomposition: you have to apply that one after the plate has interacted with the photons and with the environment. And then you're supposed to find something like:

|someenv> |brightleftscreen>|leftphotonwavepacket> + |otherenv> |brightrightscreen>|rightphotonwavepacket>

With |leftphotonwavepacket> about (|lefthole> + |righthole) and |rightphotonwavepacket> about (|lefthole> - |righthole>),

and |someenv> orthogonal to |otherenv> ; |brightleftscreen> orthogonal to |brightrightscreen> ...

Your Schmitt basis decomposition simply does not work (observation of interferences) in this case.
Therefore, I do not see the natural way to select one eigen basis versus another one.

However, you are right that apparently the Schmidt decomposition in a strict sense does not always give us the correct basis if there is (near) degeneracy. I wasn't aware of this, just read up on it last night after this recent exchange of posts. In that case one also has to resort to some kind of stability criterion. That doesn't invalidate the "coarse" Schmidt decomposition (in which the reduced density matrix is almost diagonal). But there are some difficulties here I wasn't aware of.

However, I am free to select any |human_mind> state: they are just possible thoughts of a human mind. They can belong to any basis . Therefore with the collapse postulate, I have no preferred eigen basis for the |human_mind> states.
However, practically, most of the times we notice that an experiment has a preferred basis of states in |human_mind>: Only the collapse postulate applied to an adequate eigenbasis for a given experiment seems to work.
I just do not understand well, how a schmitt decomposition reflect this result.

Well, the point was that you need somehow *A* basis for your brainstates and that the Schmidt decomposition gives you one (an essentially unique one, except for problems of degeneracy), when you consider H_universe = H_brain x H_restofuniverse.

So this is a natural way of obtaining such a basis, through the interaction of your brain with the rest of the universe. But, as you point out, without any extra specification, nothing tells us what we should DO with that basis, or why it should be the one that is to be used when applying the Born rule. That's where I think AN EXTRA POSTULATE is necessary (but which comes down to the traditional Born rule when we, intuitively use the correct basis: in that, when we buy a "momentum measurement apparatus" (because the vendor told us so, and it is written on the documentation), that we use the eigenstates of the P operator, and if we use an "angular momentum measurement apparatus" (again, because the salesman told us so), that we use the eigenstates of the L operator. In fact, if we are only given the interaction hamiltonian of the apparatus with the system and with the environment, we are not naturally led to consider P or L, unless we introduce an extra postulate, for instance, which postulates that we should take, as preferred basis, what comes out of the Schmidt decomposition. But we can also work in the opposite way: we can say that we constructed quantum mechanics the way it is, because it works, when the salesman tells us that we have a momentum measurement apparatus, that things work out correctly when we do use the P eigenbasis etc... So maybe salesmen have a divine inspiration which is not reducible to physical laws: they _define_ our physical laws :tongue2:

cheers,
Patrick.
 
  • #30
vanesch said:
I don't understand this. I would think that |left_photon_state> and |right_photon_state> are (almost) orthogonal ! (as wavepackets centered around different positions).
Or do you mean by "left photon state" the photon state from the left hole ?
Even in that case, they are orthogonal (or almost so, no ?)

Sorry, I was talking about "first order" interference (i.e. based on single photon events): left and right is the wave packet extension on the different areas of the same photon (approximated by spherical waves relatively to the slits sources).
In this case they are not orthogonal because we can see the interference results when they reach the screen. In fact it hilights the problem:
a) at the vicinity of the holes we have <left_photon_state|right_photon_state>= 0
b) at longer distances :
<left_photon_state|right_photon_state>=/=0 (inteferences)

Just by choosing the distance between the screen and the slits plate, I may have or not orthogonal states. Hence, I have problems with the Schmidt decomposition rule.


vanesch said:
But this has nothing to do with the Schmidt decomposition: you have to apply that one after the plate has interacted with the photons and with the environment. And then you're supposed to find something like:

|someenv> |brightleftscreen>|leftphotonwavepacket> + |otherenv> |brightrightscreen>|rightphotonwavepacket>

With |leftphotonwavepacket> about (|lefthole> + |righthole) and |rightphotonwavepacket> about (|lefthole> - |righthole>),

and |someenv> orthogonal to |otherenv> ; |brightleftscreen> orthogonal to |brightrightscreen> ...

I may simplify the problem:

H= H_slit_plate + H_photon + Hint

Where H_slit_plate describes the state evolution of the slit plate and Hint the interaction between the plate and the photons (we can replace photons by electrons if it is simpler to understand).
Now just suppose that we have [h_slit_plate, Hint]= 0.
In the case of electrons, Hint is a simple quantum wall potential with holes (the slits) at given positions.

If we now have @ to |psi(to)>= |slit_plate(to)>|photons(to)>

Due to this Hamiltonian, we will have at any t> to:

|psi(t)>= |slit_plate(t)>|photons(t)>

Where |photons(t)> is the state described in previous posts after the slits.
And |slit_plate(t)> is, for example, one eigenvalue of the Hamiltonian H_slit_plate.

=> In this case, we have no entanglement of the slit plate with the photons. You may notice that this example just highlights the implicit hypothesis used in the double slit experiment to show the QM results of collapse postulates and so on: no entanglement with the environment is required to get the known results.


vanesch said:
However, you are right that apparently the Schmidt decomposition in a strict sense does not always give us the correct basis if there is (near) degeneracy. I wasn't aware of this, just read up on it last night after this recent exchange of posts. In that case one also has to resort to some kind of stability criterion. That doesn't invalidate the "coarse" Schmidt decomposition (in which the reduced density matrix is almost diagonal). But there are some difficulties here I wasn't aware of.

Very interesting. May you detail?

My problem: I don’t see (or I don’t understand) how the Schmitt basis has been brought (the logical reasons) into the decoherence problem. And may be this is the source of my problems. Saying that the Schmitt decomposition is good just because it gives a single solution to the eigenbasis problem seems a very week argument (the same as saying the position is the preferred basis in most of problems). If this the case, I prefer to say that we only know the preferred eigenbasis after the experiment.

I would like to know how an observer infers a single eigenbasis for the observed results. Or if the eigenbasis is not important (independence): it’s only a matter of view (like the optical illusions analogy in my previous post). In this case, how to explain the different point of views an observer may have of an experiment (especially the ones concerning incompatible observables).

vanesch said:
That's where I think AN EXTRA POSTULATE is necessary (but which comes down to the traditional Born rule when we, intuitively use the correct basis: in that, when we buy a "momentum measurement apparatus" (because the vendor told us so, and it is written on the documentation), that we use the eigenstates of the P operator, and if we use an "angular momentum measurement apparatus" (again, because the salesman told us so), that we use the eigenstates of the L operator.

Do you know (or someone else) an apparatus that really measures the momentum p ?
I only know apparatuses that do position measurements (where we apply the born rules) and therefore deduce the momentum through the positions results (e.g. trough relations like <p>=d<r>/dt).
I also do not understand very well this point. Can we directly measure the momentum in QM?


vanesch said:
In fact, if we are only given the interaction hamiltonian of the apparatus with the system and with the environment, we are not naturally led to consider P or L, unless we introduce an extra postulate, for instance, which postulates that we should take, as preferred basis, what comes out of the Schmidt decomposition. But we can also work in the opposite way: we can say that we constructed quantum mechanics the way it is, because it works, when the salesman tells us that we have a momentum measurement apparatus, that things work out correctly when we do use the P eigenbasis etc... So maybe salesmen have a divine inspiration which is not reducible to physical laws: they _define_ our physical laws :tongue2:

cheers,
Patrick.
So what you are saying is « we just know the eigen basis after the experiment results ». If this is the case, you are saying that QM theory does not explain, a priori, why we see experimental results in a given eigenbasis. Rather it is an a posteriori selection (the salesperson has already tested its product: he knows what eigenbasis its product selects and we have to learn how its product works in order to “see” the results in the given eigenbasis ).

TI. :frown:
 
  • #31
Terra Incognita said:
Sorry, I was talking about "first order" interference (i.e. based on single photon events): left and right is the wave packet extension on the different areas of the same photon (approximated by spherical waves relatively to the slits sources).
In this case they are not orthogonal because we can see the interference results when they reach the screen.

I don't understand this reasoning. If f(x) = sin(x) and g(x) = cos(x), then I can see interference (look at |f(x) + g(x)|, it is not equal to |sin(x)| + |cos(x)| ), but f(x) and g(x) are orthogonal, no ?


In fact it hilights the problem:
a) at the vicinity of the holes we have <left_photon_state|right_photon_state>= 0
b) at longer distances :
<left_photon_state|right_photon_state>=/=0 (inteferences)

I think you have a strange idea of what orthogonality is ?
Once two states are orthogonal, unitary propagation cannot change that (conservation of in -product), so how can they be orthogonal "at the holes" and "not orthogonal" after having evolved to the screen ?

cheers,
Patrick.
 
  • #32
vanesch said:
I don't understand this reasoning. If f(x) = sin(x) and g(x) = cos(x), then I can see interference (look at |f(x) + g(x)|, it is not equal to |sin(x)| + |cos(x)| ), but f(x) and g(x) are orthogonal, no ?

Yes, you are right and I am wrong :yuck: . I have confused the position eigenbasis orthogonality with respect to the state with the orthogonality in general:

I need <x|f>=/0 and <x|g> =/=0 to be able to see an interference in the x position eingenbasis for the state |f>+|g>.

I need P(x)= |<x|f+g>|^2=/=|<x|f>|^2+|<x|g>|^2= Pf(x)+Pg(x) in order to see an interference in the x position relatively to the states |f> and |g> in the double slit experiment.
(while I was wrongly working with <psi|psi>=<f|f>+<g|g>=1 which is completely stupid!)

Therefore you are right:

just after the slits we have <left_photon_state(to)|right_photon_state(to)>= 0 => <left_photon_state(t)|right_photon_state(t)>= 0 for all t > to

We have (at to):
<x| left_photon_state(to)> <x| right_photon_state(to)>=0 for all x

While for t> t1>to we have:

<x| left_photon_state(t)> <x| right_photon_state(t)>=/=0 on the area of the interference.

(and int_x {<x| left_photon_state(t)> <x| right_photon_state(t)>}=0)


In other words if we select a measurement eingenbasis "orthogonal" to the vectors ( |left_photon_state(t)>, | right_photon_state(t)>), we have no interference (or if we put the screen too close to the slits plate).

vanesch said:
I think you have a strange idea of what orthogonality is ?
Once two states are orthogonal, unitary propagation cannot change that (conservation of in -product), so how can they be orthogonal "at the holes" and "not orthogonal" after having evolved to the screen ?

cheers,
Patrick.

It is your fault! :yuck: . I was just confusing orthogonality with interference (due to the non orthogonality of the measurement eigenbasis vectors versus the vector state of the measured system). This is due to your Schmitt basis decomposition, I’m trying to understand, that makes me move towards the dark side of the force (the Schmitt side :biggrin: ).

Now, with this new point of view (thanks to your remarks in this post), we see that the interference pattern depends on the selected eigenbasis of the measurement and not on the Schmitt decomposition of the state.

=> I am lost with the utility of the Schmitt decomposition :confused: . (a priori no connection with the measurement eigenbasis). And I think I do not understand at all decoherence (while at the beginning of this thread, I was thinking I could reach a better comprehension.

TI. :frown:
 
  • #33
Terra Incognita said:
=> I am lost with the utility of the Schmitt decomposition :confused: . (a priori no connection with the measurement eigenbasis). And I think I do not understand at all decoherence (while at the beginning of this thread, I was thinking I could reach a better comprehension.

Come to me, Luke ! Come...

Well, I have to say I also feel like I know less about it than I thought :grumpy:

What the Schmidt (or let's say, the decoherence) eigenbasis is supposed to do is - as far as I understand - the following: when you apply the Born rule, you have to apply it with respect to a certain basis. Now of course, when you look at the "wave function of this chunk of the universe", you can expand that in any basis you like - as you pointed out. However, we seem to observe a rather classical world which means that the application of the Born rule must somehow be in a basis of "this chunk of the universe" which corresponds to classical-like states, and the question is: why ?
The way I understand decoherence is that you have to split "this chunk of the universe" into two parts: namely "you + whatever you care to observe" and "the rest" (= the environment), and work in a Schmidt-like decomposition following this splitting of this part of the universe into two things (namely you and what you observe on one hand, and the environment on the other), and that it is in THIS SCHMIDT-LIKE DECOMPOSITION THAT YOU HAVE TO APPLY THE BORN RULE. Now, as I say, this is a bit personal, I never saw that explicitly written down. Whether this is the "natural" thing to do, or whether you postulate this explicitly (I think you have to do the latter) doesn't really matter, if at the end of the day, somehow you have to apply the Born rule *in that basis*.
And - as far as I understood the spirit of decoherence - this Schmidt-like decomposition then results in classical-like states for the "you + whatever you observe" basis. It didn't have to ! But this seems to be a consequence of the typical interaction hamiltonian between you and the environment.

This means that, if after observation (= application of the Born rule) you + whatever you observe is in a rather classical state (one term in the Schmidt-like decomposition), that your measurement apparatus you are looking at is in a classical-like state (which is called a "pointer state") ; for instance, the pointer on a voltmeter scale is in a rather well-defined position (and not in a superposition of positions).
This then defines a particular basis (the pointer basis) of all measurement apparatus you can observe, and from the interaction hamiltonian (and hence the evolution operator) between the apparatus and the system under study, you can then arrive at what basis for the system is being measured.
As such, the interaction hamiltonian of the apparatus (depending on the way it is constructed) then uniquely defines what is the corresponding hermitean measurement operator of the system that corresponds to it. So you don't have to take the vendor's word for it: you can look at how exactly the apparatus works, and what are its macroscopic classical states (because these are the ones you will be obtaining in the Schmidt decomposition of the state of this chunk of the universe). From these classical states, you can work back to the system and find out what states they are.

At least that's how I think I understand the decoherence program.

cheers,
Patrick.
 
  • #34
vanesch said:
Come to me, Luke ! Come...
No! No! I want to breathe like anybody.
vanesch said:
Well, I have to say I also feel like I know less about it than I thought :grumpy:

Welcome to the padawan club :biggrin: .

vanesch said:
What the Schmidt (or let's say, the decoherence) eigenbasis is supposed to do is - as far as I understand - the following: when you apply the Born rule, you have to apply it with respect to a certain basis. Now of course, when you look at the "wave function of this chunk of the universe", you can expand that in any basis you like - as you pointed out. However, we seem to observe a rather classical world which means that the application of the Born rule must somehow be in a basis of "this chunk of the universe" which corresponds to classical-like states, and the question is: why ?
I think the problem is not why we live in a classical word, but rather why we choose a basis rather than another one. I explain what I mean.
In statistical classical mechanics, we always select the phase space basis to give the observed results of statistics. We have a Schrödinger equation for the evolution of the classical system (instead of having the configuration Hilbert space H=Hilbert(q) of QM, we have a phase Hilbert space H=Hilbert(q)(x)Hilbert(p)). This the Koopman-von Neumann approach (for example, the first page of arXiv:quant-ph/0305063 document). We always give the results in the |q>|p> phase position basis and not, for example in the phase momentum basis |P_q>|P_p>. This is a kind of superselection rule: why?

The first evidence I have to try to build an answer is almost evident. The classical Hamiltonian commutes with the observables q and p. Therefore, if something is located at [qo,po,to], I know with certainty (before doing another measurement), it is at position [q(t), p(t),t]. A classical Hamiltonian does not provide an interaction that is able to shift the classical system from an initial state [q(t), p(t),t] to a state “outside” the same eigenbasis.

However, this is not enough. With the collapse postulate, I may study, formally, the outcomes of a classical system in the momentum eigenbasis |P_q, P_p> from an initial state |qo,po>. But I know (or may I don’t know how to do), I cannot build a real experiment where I may get these particular statistics. Why?

This is the second evidence, in a classical world where all the interactions commute with the q,p observables, I am only able to see statistics in the |q,p> eigenbasis. I can apply the born rules in other bases, but I am only able to build experiments with results in the |q,p> eigenbasis. How may I explain that in a coherent way (by inference)?


If I were able to explain why the collapse postulate in classical mechanics should be applied in the |q,p> eigenbasis, I think I would be able to explain why we have to choose one versus another eingenbasis in QM.

I think this is the core of the preferred basis problem in QM.




vanesch said:
The way I understand decoherence is that you have to split "this chunk of the universe" into two parts: namely "you + whatever you care to observe" and "the rest" (= the environment), and work in a Schmidt-like decomposition following this splitting of this part of the universe into two things (namely you and what you observe on one hand, and the environment on the other), and that it is in THIS SCHMIDT-LIKE DECOMPOSITION THAT YOU HAVE TO APPLY THE BORN RULE. Now, as I say, this is a bit personal, I never saw that explicitly written down. Whether this is the "natural" thing to do, or whether you postulate this explicitly (I think you have to do the latter) doesn't really matter, if at the end of the day, somehow you have to apply the Born rule *in that basis*.
And - as far as I understood the spirit of decoherence - this Schmidt-like decomposition then results in classical-like states for the "you + whatever you observe" basis. It didn't have to ! But this seems to be a consequence of the typical interaction hamiltonian between you and the environment.

I think this is not always true (the schmidt basis is the preferred basis of the measurement) especially for the cases we are able to observe interferences . I have used the example of the double slit experiment to highlight this fact. Double slits results (the interference pattern) may be explained without having the environment interacting with the photons. This has been done in post #30 H= H_slit_plate + H_photon + Hint (where [h_slit_plate, Hint]= 0).
-The orthogonality wasn’t correct, but it wasn’t important for the example-
The photons are not entangled with the plate neither with the environment.

We have |psi(t)>= |slit_plate(t)>|photons(t)> before and after the wave packet has crossed the double slit plate.

Where |photons(t)> is the wave packet carrying the interference pattern. We can say the slit plate has done a degenerated x-position measurement of the photons without requiring any entanglement with the environment or with the plate.

The x-position measurement of the slit plate may be given by the projector P_plate=|x1><x1|+|x2><x2|: if we do a second position measurement just after the slits, we will always measure photons at either one of the slit locations (we assume measurement uncertainty large enough to get this result).

Note: |x1> represents the area of one slit and |x2> the area of the other one.

We have therefore a result: the slit plate is a measurement with projector P_plate=|x1><x1|+|x2><x2| if “we observe” the photons in the area after the slit plate=> The projector of this external observation must commute with the P_plate projector in order not to be independant of the P_plate measurement result.

=> we always have an additional observation in order to transform a system described by an Hamiltonian (H= H_slit_plate + H_photon + Hint) into a measurement apparatus+system. We can apply this simple result to stern-gerlach measurement apparatuses or others. Without an extra observation, we cannot get the projector of the measurement.

Therefore, we always come to the same, damned question, what is the eigenbasis of this extra observation? Why should we take one that commutes with the interaction (connection with the classical mechanics part of this post).

We note that this extra observation is purely a thought experiment (it commutes with the P_plate projector: it may be done before or after the wave packet has interacted with the plate, if I am right). We just have an interaction Hamiltonian (H_slit_plate + H_photon + Hint). Only the conjunction of this Hamiltonian with this hypothetic extra observation (on the good eigenbasis) gives the good measurement projector (and the good eigenbasis) and the born rules.

We can think this extra observation that commutes with the P_plate projector, is in fact the initial condition of the system under study! The context (hamiltonian) dependant hidden variable. :yuck:

:uhh: Ooooh, dark side of the force, you are so strong ! Obi wan where are you? :cry:

In a next post, I will try to better separate the decoherence results for the eigen basis problem in order to help my understanding of the problem (if I can): the second part of your post.

TI.
 
Last edited:
  • #35
Terra Incognita said:
I think the problem is not why we live in a classical world, but rather why we choose a basis rather than another one. I explain what I mean.

But isn't that the same issue ? If somehow you only apply the Born rule in a basis which is classical, you will observe a classical world ! Simply because you decided that the outcomes will always belong to one of the basis states (you pick ONE by the Born rule). So if somehow you impose on those basis states to be classical, you will observe a classical world, and if you pick a basis which is strongly non-classical, and pick one term according to the Born rule, you will observe a world which is non-classical.

In statistical classical mechanics, we always select the phase space basis to give the observed results of statistics. We have a Schrödinger equation for the evolution of the classical system (instead of having the configuration Hilbert space H=Hilbert(q) of QM, we have a phase Hilbert space H=Hilbert(q)(x)Hilbert(p)). This the Koopman-von Neumann approach (for example, the first page of arXiv:quant-ph/0305063 document). We always give the results in the |q>|p> phase position basis and not, for example in the phase momentum basis |P_q>|P_p>. This is a kind of superselection rule: why?

I didn't look yet at that paper ; however I've been playing myself also with H_q x H_p stuff for fun (not thinking any serious scientist would do something like that :-). What I found was that the dynamics is trivial: each |p0,q0> state is an eigenstate of the hamiltonian and hence nothing ever happens. P and Q commute with each other, and with the hamiltonian, so in the Heisenberg picture, because [P,H] = [Q,H] = 0, dP/dt = 0 = dQ/dt. Because all variables are supposed to be functions of P and Q, all variables are constants of motion. Nothing moves.

But what I wanted to say is this: we sloppily say "classical position basis" but actually we mean a coherent state which is localised as much as possible in PHASE space. A wave packet (gaussian - style) which has a rather well-defined p and q value.

The first evidence I have to try to build an answer is almost evident. The classical Hamiltonian commutes with the observables q and p. Therefore, if something is located at [qo,po,to], I know with certainty (before doing another measurement), it is at position [q(t), p(t),t]. A classical Hamiltonian does not provide an interaction that is able to shift the classical system from an initial state [q(t), p(t),t] to a state “outside” the same eigenbasis.

I even think it stays at [q0,p0,t]...

However, this is not enough. With the collapse postulate, I may study, formally, the outcomes of a classical system in the momentum eigenbasis |P_q, P_p> from an initial state |qo,po>. But I know (or may I don’t know how to do), I cannot build a real experiment where I may get these particular statistics. Why?

You mean, after a canonical transformation ? I'd think that a coherent state (localized p and q values in a wave packet) can "survive" a canonical transformation (p,q) -> (P,Q) ; but I admit never having looked deep into this.

This is the second evidence, in a classical world where all the interactions commute with the q,p observables, I am only able to see statistics in the |q,p> eigenbasis. I can apply the born rules in other bases, but I am only able to build experiments with results in the |q,p> eigenbasis. How may I explain that in a coherent way (by inference)?

I'm not sure you recover classical mechanics in a |q,p> eigenbasis... at least it didn't work out when I tried.

If I were able to explain why the collapse postulate in classical mechanics should be applied in the |q,p> eigenbasis, I think I would be able to explain why we have to choose one versus another eingenbasis in QM.

I think this is the core of the preferred basis problem in QM.

I agree with you. As I said, I'm convinced this must in one way or another be postulated.

We have |psi(t)>= |slit_plate(t)>|photons(t)> before and after the wave packet has crossed the double slit plate.

I don't think so if you illuminate the slit plate with a light beam that also covers the plate outside of the holes! There is clearly an interaction between the beam and the plate in that case, which will lead us to an entangled state.

Consider the EM modes to be resolved in 3 (orthogonal) eigenspaces of position: E1 (modes that go through hole 1), E2 (modes that go through hole 2), E3 (modes that are completely absorbed by the plate).

So if the field state before the plate is decomposed on these 3 eigenspaces:

|photon> = a |e1> + b |e2> + c |e3>

and the plate is "cold" |cold>

So the state before the beam hits the plate is:
|cold> (a |e1> + b |e2> + c |e3>)

then after the plate you have:
|cold> (a |e1> + b |e2> ) + c |hot> |0>

(where |0> is the absorbed photon state: vacuum state) and the plate got heated by the absorbed photon.

Where |photons(t)> is the wave packet carrying the interference pattern. We can say the slit plate has done a degenerated x-position measurement of the photons without requiring any entanglement with the environment or with the plate.

Well, I don't think so...

But this thing by itself is not really a measurement. A measurement is a computer that prints out a certain result on paper or something like that.


Therefore, we always come to the same, damned question, what is the eigenbasis of this extra observation? Why should we take one that commutes with the interaction (connection with the classical mechanics part of this post).

Well, I don't think that putting a plate with holes in a beam is a measurement. You can DO a measurement now, like looking whether a photon got through it or not. But you haven't done any yet, except if you consider that the heating of the plate by the absorption of the e3 photons is sufficiently observable. It could be considered so for gamma rays, if the plate was instrumented.


We note that this extra observation is purely a thought experiment (it commutes with the P_plate projector: it may be done before or after the wave packet has interacted with the plate, if I am right).

I don't think that this extra observation is superfluous: it is the core of the measurement if you want this plate to be a measurement apparatus, no ?

We just have an interaction Hamiltonian (H_slit_plate + H_photon + Hint). Only the conjunction of this Hamiltonian with this hypothetic extra observation (on the good eigenbasis) gives the good measurement projector (and the good eigenbasis) and the born rules.

Yup, because only when you add that something, you have a real observation (a computer that prints out a result on a sheet of paper).

cheers,
Patrick.
 

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