Commutativity of the Born Rule

[the_pulp]

The born rule, written in the following way:

P(ψ/\varphi)=|<ψ|\varphi>|^2

As a consequence,

P(ψ/\varphi)=P(\varphi/ψ)

I don't see it as an obvious fact from real life, do you? Why does it happen? Is there any intuitive reasoning / experience behind this commutativity of states?

Thanks!

 

[strangerep]

I don't see it as an obvious fact from real life, do you? Why does it happen? Is there any intuitive reasoning / experience behind this commutativity of states?

It's not "commutativity". More like time reversal invariance, which is usually appears in theories based on Galilean/Poincare symmetry unless specific techniques are employed to avoid it.

 

[Fredrik]

$\left|\langle x,y\rangle\right|$ is the angle between the lines through 0 (i.e. 1-dimensional subspaces) that contain x and y.

The Cauchy-Schwartz inequality says that for any two vectors x,y, we have $\left|\langle x,y\rangle\right|\leq\|x\|\|y\|$. If we had been dealing with a real vector space, this would have implied that
$$-1\leq\frac{\langle x,y\rangle}{\|x\|\|y\|}\leq 1,$$ and this would have allowed us to define the angle θ between the two vectors by
$$\cos\theta=\frac{\langle x,y\rangle}{\|x\|\|y\|}.$$ Since we're dealing with a complex vector space, this doesn't quite work. So we can't define the angle between the vectors, but we can define the angle between the 1-dimensional subspaces that contain x and y as
$$\frac{\left|\langle x,y\rangle\right|}{\|x\|\|y\|}.$$ If x and y are unit vectors, i.e. if $\|x\|=\|y\|=1$, this of course reduces to $\left|\langle x,y\rangle\right|$.

 

[the_pulp]

It's not "commutativity". More like time reversal invariance, which is usually appears in theories based on Galilean/Poincare symmetry unless specific techniques are employed to avoid it

|⟨x,y⟩| is the angle between the lines through 0 (i.e. 1-dimensional subspaces) that contain x and y.

Thanks for your answers, but I still can't see it. What I'm trying to say is that, with this axiom, it Will always happen that, given two experiments. E1 And E2, with eigenstates ψ₁ⁱ And ψ₂^j the two following experiments:

1 prepare the system in order to be in the state 1i, perform the experiment E2 And see how likely is 2j to happen

2 prepare the system in order to be in the state 2j, perform the experiment E1 And see how likely is 1i to happen

have the same probability.

I mean, it is an implicit axiom of the theory, but, before starting reading about QM And QFT, I was not expecting that to be obvious to happen. So, is it an experimental fact? Was it obvious in 1920 that laws of physics should imply this? Is it obvious to happen? Is it right to say "its obvious that the probability of situation 1 And situation 2 should be the same"?

Thanks

 

[Fredrik]

I don't know a reason to think that P(x|y)=P(y|x) should hold in all theories. It is however pretty obvious that it must hold in QM, since P(x|y) is just the angle between the subspaces that contain x and y.

 

[strangerep]

1 prepare the system in order to be in the state 1i, perform the experiment E2 And see how likely is 2j to happen

2 prepare the system in order to be in the state 2j, perform the experiment E1 And see how likely is 1i to happen

have the same probability.

Aren't these just time-reversed versions of each other?

Early in Weinberg vol 1, he uses assumptions of time-reversal invariance to derive the anti-linear nature of the time-reversal operator.

 

[Fredrik]

Aren't these just time-reversed versions of each other?

I find it hard to answer that due to the the non-deterministic nature of measurements. But at least now I see what you had in mind.

I think Weinberg's approach makes it clear that a quantum theory possesses a certain type of invariance if and only if it contains an operator corresponding to that invariance. For example a theory is invariant under translations in the x direction of space if and only if it contains an "x component of momentum" operator. So a quantum theory should be time-reversal invariant if and only if it contains a time-reversal operator. But the result discussed in this thread holds even if there's no time-reversal operator in the theory.

 

[strangerep]

But the result discussed in this thread holds even if there's no time-reversal operator in the theory.

Maybe it's simpler to appeal to Bayes' theorem, aka the principle of "inverse probability".
Cf. Ballentine, p31.
$$
P(B|A\&C) ~=~ \frac{P(A|B\&C) \; P(B|C)}{P(A|C)}
$$
(and specialize to the case where $C$ is a certainty).

Of course, this is a consequence of assuming $A\&B = B\& A$ .

Edit: Actually this is a bit fuzzy so I probably need to go revise some (quantum) propositional logic...

 

[the_pulp]

Id love if the answer were "its because life respects time reversal" but, doesn't weak interaction violate time reversal? So, what do we do?

Thanks for all the replies!

 

[strangerep]

Id love if the answer were "its because life respects time reversal" but, doesn't weak interaction violate time reversal? So, what do we do?

Use CPT invariance instead?

 

[the_pulp]

Use CPT invariance instead?

Thats the real reason or its just a guess? Sounds promising, can you expand a little or give some reference that shows the link between the two of them? (I have the intuition that it may be the answer, but I'm not sure)

 

[Fredrik]

Id love if the answer were "its because life respects time reversal" but, doesn't weak interaction violate time reversal? So, what do we do?

What makes you think that something needs to be done? The theory doesn't need to be time-reversal invariant for $\left|\langle x,y\rangle\right|=\left|\langle y,x\rangle\right|$ to hold.

 

[stevendaryl]

What makes you think that something needs to be done? The theory doesn't need to be time-reversal invariant for $\left|\langle x,y\rangle\right|=\left|\langle y,x\rangle\right|$ to hold.

\vert \langle x \vert y \rangle \vert = \vert \langle y \vert x \rangle \vert

makes perfect sense as a claim about vectors, but as the original poster said, it's weird as a statement about probabilities. As a statement about vectors, if you start with a vector \vert y \rangle and project it along a different vector \vert x \rangle, you don't get a probabilistic result of 0 or 1, you get a deterministic result, that the projection has length \dfrac{\vert \langle x \vert y \rangle \vert}{\sqrt{\langle x \vert x \rangle}}.

 

[Fredrik]

\vert \langle x \vert y \rangle \vert = \vert \langle y \vert x \rangle \vert

makes perfect sense as a claim about vectors, but as the original poster said, it's weird as a statement about probabilities. As a statement about vectors, if you start with a vector \vert y \rangle and project it along a different vector \vert x \rangle, you don't get a probabilistic result of 0 or 1, you get a deterministic result, that the projection has length \dfrac{\vert \langle x \vert y \rangle \vert}{\sqrt{\langle x \vert x \rangle}}.

Right, but if x is a unit vector, the denominator is 1. If we for some reason choose to not work with unit vectors the way we usually do, the probability formula is
$$P(x|y)=\frac{\left|\langle x,y\rangle\right|^2}{\|x\|^2\|y\|^2}$$ so we still have $P(x|y)=P(y|x)$, which is what he asked about. The intuitive way of looking at this is that $P(x|y)$ is the square of the angle between the 1-dimensional subspaces $\mathbb Cx$ and $\mathbb Cy$.

 

[stevendaryl]

Right, but if x is a unit vector, the denominator is 1. If we for some reason choose to not work with unit vectors the way we usually do, the probability formula is
$$P(x|y)=\frac{\left|\langle x,y\rangle\right|}{\|x\|\|y\|}$$ so we still have $P(x|y)=P(y|x)$, which is what he asked about. The intuitive way of looking at this is that $P(x|y)$ is given by the angle between the 1-dimensional subspaces $\mathbb Cx$ and $\mathbb Cy$.

My point is not about the vector space result. As I said, it's the interpretation of the result as a probability that is strange.

 

[vanhees71]

I've been already puzzled by the question, because it doesn't make any sense to me. The more I'm puzzled by the answers given so far. So, here are my 2 cents:

First of all the conception to write something like
P(\psi|\phi)=|\langle \psi|\phi \rangle|^2
doesn't make sense, except, I misinterpret the symbols:

Usually the left-hand side denotes a conditional probability that "and event \psi" occurs, given some "condition \phi". The right-hand side takes two Hilbert-space vectors and the modulus of its scalar product squared (tacitly assuming that the vectors are normalized to 1, as I'll do also in the following). This resembles vaguely Born's rule, but it's not really Born's rule!

One has to remember the very basic postulates of quantum theory to carefully analyze the statement of Born's rule, which partially is interfering with a lot of philosophical ballast known as "interpretation". Here, I follow the minimal statistical interpretation, which is as free as possible from this philosophical confusion.

(1) A completely determined state of a quantum system is given by a ray in Hilbert space, represented by an arbitrary representant |\psi \rangle, or equivalently by the projection operator \hat{R}_{\psi}=|\psi \rangle \langle \psi | as the statistical operator. Such a state is linked to a system by a preparation procedure that determines a complete set of compatible observables (a notion which is explanable only with the following other postulates).

(2) Any observable A is represented by a self-adjoint operator \hat{A}. The possible values this observable can take are given by the spectral values a of this operator. Further let |a \rangle denote the corresponding (generalized) eigenvectors (I don't want to make this posting mathematically rigorous by introducing formally the spectral decomposition a la von Neumann or the more modern but equivalent formal introduction of the "rigged Hilbert space").

(3) Two observables A and B are compatible if the corresponding operators have a complete set of (generalized) common eigen vectors |a,b \rangle. One can show that then the operators necessarily commute [\hat{A},\hat{B}]=0.

(4) A set of compatible independent observables \{A_{i} \}_{i=1,\ldots n} is complete, if any common (generalized) eigenvector |a_1,a_2,\ldots,a_n \rangle is determined up to a (phase) factor and if any of these observables cannot be expressed as the function of the others. If a system is prepared such that such a complete set of independent compatible observables take a determined value, then the system is prepared in the corresponding state.

In the following, I assume that in the case that some of the observables have a continuous part in the spectrum the corresponding eigenvectors are somewhat smeared with a square-integrable weight such that one has a normalizable true Hilbert-space vector |\psi \rangle, which is also a mathematical but pretty important detail, which we can discuss later, if necessary.

(5) If the system is prepared in this sense in a state, represented by the normalized vector |\psi \rangle, then the probability (density) to find the common values (b_1,\ldots,b_n) of (the same or another) complete set of compatible observables \{B_i \}_{i=1,\ldots,n} is given by Born's rule,
P(b_1,\ldots,b_n|\psi)=|\langle b_1,\ldots,b_n|\psi \rangle|^2.

Now the whole construct makes sense, because on the left-hand side we have the probability to measure certain values for a complete compatible set of independent observables under the constraint that the system has been prepared in the state, represented by \psi.

BTW, note that it's the modulus squared not the modulus, which is also very important, because otherwise one wouldn't get a well-definied probability distribution in the sense of Kolmogorov's axioms!

It is important to note that we have in some sense an asymmetry here: The one vector in Born's rule, |\psi \rangle represents the state of a system, which is linked to the system by a preparation procedure to bring it into this state, and the other vector is the (generalized) common eigenvector of a complete set of compatible independent observables, referring to the measurement of this set of observables.

One should note that the clear distinction of the two vectors in Born's rule is crucial for the whole formalism to work also mathematically. Particularly, the dynamics of these vectors is described differently. The mathematical time dependence of both kinds of vectors is determined only up to a time-dependent unitary transformation, which freedom is often made use of in practical calculations.

Usually, in non-relativistic quantum mechanics one starts with the Schrödinger picture, where the state kets carry the full time evolution, according to the equation
\mathrm{i} \hbar \partial_t |\psi,t \rangle = \hat{H} |\psi,t \rangle,
where \hat{H} is the Hamilton operator of the system. The operators and generalized eigenvectors of observables that are not explictly time dependent, are independent of time.

The other extreme is the Heisenberg picture, where the state ket is time-independent, but the Operators representing observables and thus the eigenvectors carry the full time dependence through the equation of motion,
\frac{\mathrm{d} \hat{A}(t)}{\mathrm{d} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H} ].

The most general case is the Dirac picture, where both the state vector and the observables are time dependent, according to the equations of motion,
\mathrm{i} \hbar \partial_t |\psi,t \rangle = \hat{H}_1 |\psi,t \rangle
\frac{\mathrm{d} \hat{A}(t)}{\mathrm{d} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}_2]
with an arbitrary decomposition of the Hamiltonian as
\hat{H}=\hat{H}_1+\hat{H}_2.
A commonly used Dirac picture is the "interaction picture", where the observable operators evolve as for free particles, i.e., \hat{H}_1 is the kinetic energy only, and the states according to the interaction part \hat{H}_2 of the Hamiltonian.

Of course at the end all these descriptions must lead to the same physical results. Most importantly all the probabilities must be independent of the picture of time evolution, i.e., independent of the split of the Hamiltonian into the two parts within the Dirac picture (of which Schrödinger and Heisenberg picture are only special cases). As one can easily check this works out only when making clearly the distinction between state vectors and the generalized eigenvectors of the observable operators!

So there is an asymmetry in the two vectors involved in Born's rule, and it's a physically very meaningful asymmetry!

 

[the_pulp]

Sorry Vanshees, I am reading 4 and 5 from your post but I don't see the diference from my formula. In fact, if you make b1,b2...bn=phi, then you have my formula right?, then you can interchange phi and psy and you get the same probability and I don't see a clear reason of why it should be like this...

Sorry I insist but I really did not get the reason why what you write can invalidate my question?

Nevertheless I appreciate your answers very much

 

[the_pulp]

Fredrik, as stevendaryl says, my question is not about the math (I understand it). My question is if there is any reason that can be stated in a few words (I don't want to use the phrase "physical reason" but perhaps it helps you to find what I am asking) in order to explain why the probability to go from \psi to \varphi and the probability to go the other way round are the same.

Thanks all the same

 

[atyy]

I'm not sure this is right, but isn't one vector an eigenvector of an observable, while the other is an arbitrary vector?

 

[the_pulp]

I'm not sure this is right, but isn't one vector an eigenvector of an observable, while the other is an arbitrary vector?

As I see it, the two of them can be regarded as eingenvectors of an observable (or a couple of observables to be more general)

 

[atyy]

As I see it, the two of them can be regarded as eingenvectors of an observable (or a couple of observables to be more general)

What if I prepare a state which is a sum of eigenvectors?

 

[the_pulp]

What if I prepare a state which is a sum of eigenvectors?

You mean a mixed state, right? Well, in that case its different. If you start with a pure state and you finnish with a pure state, then there is no difference, right? (This thread was intended to that case)

 

[atyy]

You mean a mixed state, right? Well, in that case its different. If you start with a pure state and you finnish with a pure state, then there is no difference, right? (This thread was intended to that case)

No, I mean a pure state. I suppose my question is whether every state is an eigenvector of some observable.

 

[the_pulp]

I don't see why that could happen. Anyway, let's stick please to the easy example because it seems as if we are going off topic. I repeat my first question: We have two states \varphi and \psi (lets suppose they describe the same system in two different situations and there are observables associated with these states that allow us to detect exactly if we are in one of them).

Then:

P(ψ|ϕ) = P(ϕ|ψ)

Besides the fact that the math implies this relation in an easy way, the question is, what is the physical reason that forces this to happend?

I will paste below another way in which I wrote it, perhaps it helps you to see my doubt:

What I'm trying to say is that, with this axiom, it Will always happen that, given two experiments. E1 And E2, with eigenstates ψ1i And ψ2j the two following experiments:

1 prepare the system in order to be in the state 1i, perform the experiment E2 And see how likely is 2j to happen

2 prepare the system in order to be in the state 2j, perform the experiment E1 And see how likely is 1i to happen

have the same probability.

I mean, it is an implicit axiom of the theory, but, before starting reading about QM And QFT, I was not expecting that to be obvious to happen. So, is it an experimental fact? Was it obvious in 1920 that laws of physics should imply this? Is it obvious to happen? Is it right to say "its obvious that the probability of situation 1 And situation 2 should be the same"?

Thanks

 

[atyy]

Would your question be equivalent to asking why projective measurements are projective?

 

[Fredrik]

Fredrik, as stevendaryl says, my question is not about the math (I understand it). My question is if there is any reason that can be stated in a few words (I don't want to use the phrase "physical reason" but perhaps it helps you to find what I am asking) in order to explain why the probability to go from \psi to \varphi and the probability to go the other way round are the same.

That follows immediately from the Born rule, so I guess you're asking why the Born rule holds. The Born rule is part of the definition of QM, so what you're really asking is why QM is a good theory (or rather, a good framework in which we can define theories of matter and interactions). The only thing that can answer that is a better theory.

 

[atyy]

That follows immediately from the Born rule, so I guess you're asking why the Born rule holds. The Born rule is part of the definition of QM, so what you're really asking is why QM is a good theory (or rather, a good framework in which we can define theories of matter and interactions). The only thing that can answer that is a better theory.

That too is my understanding of what is being asked. I suppose the proposals for deriving the Born rule are:

1) Bohmian mechanics - works for non-relativistic cases, no consensus on relativistic cases

2) Many-worlds - widely believed to work, but I don't understand this derivation

3) Gleason's theorem - uncontroversial, but mystifying

4) Zurek's envariance - pretty new attempt, and no consensus yet

 

[Fredrik]

I'm not sure this is right, but isn't one vector an eigenvector of an observable, while the other is an arbitrary vector?

This is true, but even the arbitrary vector is an eigenvector of some observable.

 

[atyy]

This is true, but even the arbitrary vector is an eigenvector of some observable.

Is there a short proof, or can you point me to a reference? (Anyway, I do think it's not very relevant to the question, since I know this is true in many cases.)

 

[micromass]

Is there a short proof, or can you point me to a reference? (Anyway, I do think it's not very relevant to the question, since I know this is true in many cases.)

A nonzero vector $|u>$ is an eigenvector of $|u><u|$.

 

[the_pulp]

I don't want you to explain the interpretation of the born rule. Just of a little part of it, the apparent symmetry between the inicial And final state in the calculation of probabilities. I thought that it may have a clear analogy in real life, but, perhaps I was wrong And it is just as obscure as all the rest of the Born rule.

Thanks all the same for your usual help!

Ps I know that the ultimate justification of QM is its prdictive power. I thought that this particular part of QM may have a proper interpretation.

 

[Fredrik]

I've been already puzzled by the question, because it doesn't make any sense to me. The more I'm puzzled by the answers given so far. So, here are my 2 cents:

First of all the conception to write something like
P(\psi|\phi)=|\langle \psi|\phi \rangle|^2
doesn't make sense, except, I misinterpret the symbols:Usually the left-hand side denotes a conditional probability that "and event \psi" occurs, given some "condition \phi". The right-hand side takes two Hilbert-space vectors and the modulus of its scalar product squared (tacitly assuming that the vectors are normalized to 1, as I'll do also in the following). This resembles vaguely Born's rule, but it's not really Born's rule!

The OP spoke of "the Born rule, written in the following way", so I assumed that the notation P(x|y) should be interpreted as "the probability that the measurement will have the result that leaves the system in the state x, given that the state before the measurement is y". This makes sense if we're talking about a measuring device that's represented by an operator such that x is one of its eigenvectors, and the eigenspace corresponding to the eigenvalue of x is the 1-dimensional subspace that contains x. So I assumed that we are talking about a measurement with such a measuring device.

BTW, note that it's the modulus squared not the modulus

Oops. I see that I forgot the square in post #14. I just fixed that in an edit. Also, in post #5 I said that P(x|y) is the angle between the 1-dimensional subspaces $\mathbb Cx$ and $\mathbb Cy$. I can't edit that anymore, so I'll just say here that it's the square of the angle, not the angle itself.

It is important to note that we have in some sense an asymmetry here:

I don't follow this argument, and I don't think that there is an asymmetry, at least not one that's severe enough to cause any problems. A measurement that doesn't destroy the system is a preparation procedure, because it leaves the system in a known state. So both the "before" and "after" states have been produced by preparation procedures.

I suppose you could nitpick that statement by insisting that only a procedure that leaves the particles in the same state every time should be considered a preparation procedure, but that is only a matter of discarding the particle every time we get the "wrong" result.

 

[atyy]

A nonzero vector $|u>$ is an eigenvector of $|u><u|$.

Are all eigenvalues of $|u><u|$ real?

 

[atyy]

I don't want you to explain the interpretation of the born rule. Just of a little part of it, the apparent symmetry between the inicial And final state in the calculation of probabilities. I thought that it may have a clear analogy in real life, but, perhaps I was wrong And it is just as obscure as all the rest of the Born rule.

Thanks all the same for your usual help!

Ps I know that the ultimate justification of QM is its prdictive power. I thought that this particular part of QM may have a proper interpretation.

But as Fredrik has been saying, this seems to follow from the idea that when you measure, you make a projection. So your question is like asking why a projection?

 

[Fredrik]

Are all eigenvalues of $|u><u|$ real?

That's a projection operator, so it only has eigenvalues 0 and 1.

Note that if $P^2=P$ and $Px=\lambda x$ where $\lambda$ is a complex number and $\|x\|\neq 0$, we have $P^2x=P(Px)=P(\lambda x)=\lambda Px=\lambda^2x$, and also $P^2x=Px=\lambda x$. So $\lambda(\lambda-1)x=0$, and this implies that $\lambda$ is 0 or 1.

 

[atyy]

That's a projection operator, so it only has eigenvalues 0 and 1.

Note that if $P^2=P$ and $Px=\lambda x$ where $\lambda$ is a complex number and $\|x\|\neq 0$, we have $P^2x=P(Px)=P(\lambda x)=\lambda Px=\lambda^2x$, and also $P^2x=Px=\lambda x$. So $\lambda(\lambda-1)x=0$, and this implies that $\lambda$ is 0 or 1.

Ok, thanks. I have to say that it seems very unintuitive in the case of the wave function, but perhaps it's just very hard to measure such an observable, although it exists. (It seems quite intuitive to me for spin).

 

[strangerep]

I don't see why that could happen. Anyway, let's stick please to the easy example because it seems as if we are going off topic.

Yes, let's get back to your question, since the deeper I think about it, the more trouble I have with it...

I repeat my first question: We have two states \varphi and \psi (lets suppose they describe the same system in two different situations and there are observables associated with these states that allow us to detect exactly if we are in one of them).

Then: P(ψ|ϕ) = P(ϕ|ψ)

OK, part of the difficulty is in the way you're writing and interpreting this, as Hendrik hinted.

Let's go back to classical probability theory. Suppose we have a set of events, which I'll denote as U
(meaning the universal set). Let A,B,C be subsets of U. The definition of conditional probability is:
$$
P(A|B) ~:=~ \frac{P(A\cap B)}{P(B)}
$$hence $$
P(A|B) ~\ne~ P(B|A)
$$in general, unless $P(A) = P(B)$.

To relate this to QM properly requires a lot more explanation than I can reproduce here. Please grab a copy of Ballentine, and review ch9, especially sect 9.6 on "joint and conditional probabilities", and the subsection therein describing its application to cascaded spin measurements. If one is willing to work through all this carefully (as I've just spent the past few hours re-doing), it reveals quite strikingly how time ordering is indeed part of QM. Fredrik objected to this (at least, I think it was an objection, but I might be wrong), however this part of Ballentine shows how it enters in practice.

I'll end this post with a cautionary quote from Ballentine:

Although these examples are quite simple, they serve as a warning against formal axiomatic theories of measurement that do not explicitly take the dynamical action of the apparatus into account.

(Edit: BTW, the "CPT" thing earlier was just me wondering out loud how to extend Ballentine's non-relativistic example to a relativistic context, but let's stick to the non-relativistic case for now.)

 

[vanhees71]

Good that you mention this, strangerep. I forgot to mention this very important point! Often, people make up a problem with the direction of time, because almost any systems considered are described by a Hamiltonian which is time-reversal invariant, but in our macroscopic experience we clearly have a sense of the direction of time. There is a qualitative difference between past and future.

This view overlooks the fact that from the very beginning of all physics we make the assumption that nature behaves according to causal laws. This becomes very clear in the basic fundamentals of quantum theory as I tried to summarize in my previous posting: To link a state as a description of the conditions of a given system to this system, we have to prepare the system in this state by determining a complete set of compatible observables (I left out the more common case of incomplete state determination, but this is not so difficult to generalize; you just have to substitute a statistical operator for an appropriate mixed state, given the information about the system due to the incomplete preparation) before we make any measurements of the same or other observables. If you measure the same observables, according to quantum theory you get back the values you have the system prepared in, which makes the idea of state preparation consistent with the statistical interpretation of the states according to Born's Law. All other observables which are incompatible with the determined set have no determined values, and we can tell only the probabilities for finding a certain possible value when measuring this observable.

However, there is clearly an order of time in this description: You first have to prepare the system before you are able to make predictions about future! In this way there is already a tacitly assumed direction of time in the very foundation of quantum theory. The very same arguments can also be made in classical mechanics or field theory, which all are described by causal laws.

Time-reversal invariance is just a property of many Hamiltonians (Lagrangians). In fact, as far as we know today, it's only weekly broken by the weak interaction, but it's no problem for the fact that we have an orientation of the time direction, because that's already inherent in the very foundations of physics. It's a postulate, which is not explanable by another simpler foundation so far.

To understand the meaning of time-reversal invariance, one must keep in mind, how to test it in practice. Of course, we cannot really reverse the direction of time, according to the fundamental postulate about the direction of time, explained above. What is really implied by this very special symmetry is better described as "invariance under the reversal of motion". For a time-reversal invariant Hamiltonian by definition there exists a well-defined symmetry which corresponds to "time reversal" in the following sense: There exists an antiunitary transformation for all states and observables of the system, which can best be described by thinking about filming a dynamical process and then letting the movie run backwards and then interpret what you observe as a real process possible in nature according to the causal flow of time.

What you then observe obviously is the evolution of a system from an initial state which corresponds to the final state of the process captured by the movie, but with time-reversed values of the observables that determine the final state, i.e., a complete set of observables that have been prepared in the time-reversed values of the state of the real event fixed on the movie. Then time-reversal invariance means that after the same time interval you end up with the state, corresponding to the initial state of the orignal event with again time-reversed values for the observables that are determined when preparing the system in this state.

This also shows, why time-reversal invariance is practically absent for macorscopic objects: To observe it would mean that you had to prepare the system at some moment in time with the time-reversed state and then observe that it dynamically evolves back the the time reversed initial state you prepared it some time before. This is practically impossible as soon as you have to deal with a sufficiently large system, and macroscopic systems are always huge (\sim 10^{24} particles).

This shows that the often cited "macroscopic arrow of time", defined as the direction of growing entropy, is identical with the "causal arrow of time" already assumed at the very foundations of any physical theory. This becomes also very clear when deriving macroscopic equations like the Boltzmann equation from classical (or also quantum) theory: At some step there is always an assumption already hidden in the derivation of this equation. In the usual way to derive the Boltzmann equation from microscopic Hamiltonian dynamics (according to the original derivation by Boltzmann or the more modern one by Bogoliubov), it's at the point, where you assume that possibly present higher-order correlations at some point of time get lost in the evolution to later times, again implicitly assuming already an arrow of time (what I call the "causal arrow of time"), and this implies an increase of the entropy by the very definition of entropy as the measure of missing information: We already assume in the derivation of the Boltzmann equation that probably available information (parametrized in terms of higher-order many-body correlation functions) gets lost with the evolution of the system forward in time (in the sense of the causal arrow of time). Last but not least this assumption is only justified by the common experience and observations of nature, as any deeper natural law has finally to be justified empirically.

 

[Fredrik]

Let's go back to classical probability theory. Suppose we have a set of events, which I'll denote as U
(meaning the universal set). Let A,B,C be subsets of U. The definition of conditional probability is:
$$
P(A|B) ~:=~ \frac{P(A\cap B)}{P(B)}
$$hence $$
P(A|B) ~\ne~ P(B|A)
$$in general, unless $P(A) = P(B)$.

It's important to keep in mind that this definition is intended for the case when P is a probability measure on a σ-algebra. In QM, we're dealing with a probability measure on a lattice. σ-algebras are lattices, but lattices aren't σ-algebras. QM is not probability theory, it's a generalization of probability theory.

To relate this to QM properly requires a lot more explanation than I can reproduce here. Please grab a copy of Ballentine, and review ch9, especially sect 9.6 on "joint and conditional probabilities", and the subsection therein describing its application to cascaded spin measurements.

I have only had a quick look. So far I don't see what this has to do with the_pulp's question. Ballentine explains how to calculate probabilities of sequences of measurement results, but the_pulp only asked about the formula for the probability of a single measurement.

 

[the_pulp]

However, there is clearly an order of time in this description: You first have to prepare the system before you are able to make predictions about future! In this way there is already a tacitly assumed direction of time in the very foundation of quantum theory.

Vanshees, although I think that your post does not state a clear position about why in QM P(ψ|ϕ)=P(ϕ|ψ), which is the aim of this thread, let me give you 2 facts that, perhaps you already know, go totally against the idea of "arrow of time" in foundational physics.

1) Fluctuations Theorems: As I understood, they seem to say that given any situation of entropy, if you move a "microsecond" backward or forward, you have the same probability of increasing or decreasing entropy (stating, in a way, that we see growth of entropy just because in the past there was very little of it)

2) Time Symmetric Quantum Mechanics: Its an interpretation of QM, very nice, that states a total symmetry between the future and the past.

From these two facts, I don't see a big difference between the future and the past. To me, we experience the arrow of time because we need to live in the increasing entropy order. Nevertheless, it´s just a matter of interpretation.

However, my question is not a matter of interpretation. It's something that seems too foundational to say that "it's just the way that Born rule is" (but perhaps that's the sad answer we have right now, I thought that there could be something). Again, my question is:

P(ψ|ϕ)=P(ϕ|ψ), Why? (not mathematically, but physically why?)

I thought that it may be something like time reversal, but weak interactions are against them. Then I thought that it may be CPT, but I can't see that very clear. It could be anything else (or nothing).

Thanks again to all for sharing your knowledge

 

[Fredrik]

P(ψ|ϕ)=P(ϕ|ψ), Why? (not mathematically, but physically why?)

What's the difference? It's especially hard to see a difference in QM, where it's unclear if anything in the theory should be thought of as a description of what's actually happening.

I thought that it may be something like time reversal, but weak interactions are against them. Then I thought that it may be CPT, but I can't see that very clear.

Since $\left|\langle x,y\rangle\right|=\left|\langle y,x\rangle\right|$ for all x,y regardless of whether the specific quantum theory we're dealing with includes a time-reversal operator, this can't have anything to do with any of those things.

You are really just asking why QM is a good theory. As I said, only a better theory can answer that.

So neither the general framework of QM, nor any specific theory in that framework, can answer the question. There could perhaps exist a theory-independent answer, something that follows from an abstract definition of "theory" that's general enough to include all the classical theories and all the quantum theories. But would you even consider such an answer, or would you dismiss it as "mathematical"?

 

[vanhees71]

Ok, now I understood your question (although it's still not put in the wrong way, as explained in my very first posting to this thread). We can argue about the arrow of time later, because this is obviously off topic with respect to your question, although due to my misunderstanding of your question, I thought you indeed have something about time reversal in mind. As I understood it you misinterpret somehow Born's law as giving some kind of transition-probility (like an S-matrix element in scattering theory) which clearly is not the case. It deals with the probability for finding certain values at one instant of time, given the state of the system at this very time.

What you are asking now is, whether there is a significance in the fact that indeed |\langle \phi|\psi \rangle|^2 is symmetric under exchange of the vectors \phi and \psi. In my opinion there is none:

Again, please note that the two vectors are not symmetric in their physical meaning within Born's rule, as explained in my previous posting! Note that also formally the state \psi must be described by a true Hilbert-space vector |\psi \rangle which must be normalizable to 1, i.e.,
\langle \psi|\psi \rangle=1.
The other vector refers to the measured complete set of compatble observables and may also be a generalized eigenvector of the corresponding self-adjoint operators. It's in the dual space of the nuclear space of the operators, i.e., the dense subspace of the Hilbert space, where the operators are defined (see Ballentine for details about this important point). Thus you can have something like generalized eigenvectors for momentum |\phi \rangle=|p_x,p_y,p_z \rangle of a particle, which are normalized to a \delta distribution, and then the probability density for measuring a certain momentum is given by
P(p_x,p_y,p_z|\psi)=|\langle p_x,p_y,p_z|\psi \rangle|^2=|\langle \psi|p_x,p_y,p_z \rangle|^2.
Of course, in this case it doesn't make sense to ask about the "exchange" of \phi and \psi, because here |\phi \rangle can never ever describe a state of the particle!

This clearly shows that the objects in the (generalized!) scalar product in Born's rule are physically different and also nowhere anything is said about time evolution, transition amplitudes, or time-reversal invariance.

 

[atyy]

2) Time Symmetric Quantum Mechanics: Its an interpretation of QM, very nice, that states a total symmetry between the future and the past.

I thought that it may be something like time reversal, but weak interactions are against them. Then I thought that it may be CPT, but I can't see that very clear. It could be anything else (or nothing).

strangerep's time reversal observation seems in the spirit of Time Symmetric QM.

Do the weak interactions really violate time reversal symmetry in a way that renders Time Symmetric QM inapplicable?

 

[vanhees71]

I don't know what time-symmetric quantum theory is.

Recently it has been experimentally shown directly that indeed the time-reversal invariance is violated by the weak interaction. For quite some time we know that parity is violated (Wu experiment 1956) and that CP symmmetry is violated (Cronin&Fitch 1964). Then due to the CPT theorem of local microcausal quantum field theory also T should be violated, but the direct experimental proof of this has been found only very recently:

http://physicsworld.com/cws/article...direct-measurement-of-time-reversal-violation

 

[Fredrik]

Time-symmetric QM replaces the Born rule with another rule called the ABL rule (Aharonov, Bergmann, Lebowitz) that gives you the probability of a measurement result given a state before the measurement and a state after the measurement. It can be derived from the Born rule using Bayes theorem. It makes QM at least look like it doesn't favor a direction in time.

I don't think this replacement changes any of the predictions of the theory, so it doesn't have anything to do with CPT invariance or anything like that.

 

[atyy]

http://physicsworld.com/cws/article/news/2012/nov/21/babar-makes-first-direct-measurement-of-time-reversal-violation[/url]

The OP requires that both states are pure. Does T-violation by the weak interaction involve mixed states?

I came across
http://arxiv.org/abs/1208.0773
http://philosophyfaculty.ucsd.edu/faculty/wuthrich/PhilPhys/EarmanJohn2002IntStudPhilSci_TRI.pdf

Both talk about T-violation. The former mentions a state that is filtered (para after Eq 3), the latter gets T-violation by a mixed state (p259)

Edit 1: No, I don't think it has to do with mixed states. In the first reference http://arxiv.org/abs/1208.0773, Eq (25) there is no time reversal violation for Δt=0, which seems closest to the condition of the OP.

Edit 2: It also seems that the time-reversal symmetry violation by the weak interaction is consistent with Time Symmetric QM. In http://arxiv.org/abs/1208.0773, the time reversal violating Eq (21, 24, 26, 27) are derived from Eq (1) and Eq (17-20), which are just normal QM. Time Symmetric QM is, I believe, just normal QM rewritten http://arxiv.org/abs/0706.1232. The reformulation was useful because it helped the development of "weak measurements", but "weak measurements" can also be arrived at from normal QM.

 

[the_pulp]

As I understood it you misinterpret somehow Born's law as giving some kind of transition-probility (like an S-matrix element in scattering theory) which clearly is not the case. It deals with the probability for finding certain values at one instant of time, given the state of the system at this very time.

I understand from the Born rule, exactly what you said, a transition probability. Indeed, after the measurement, the system transits to a state where it has the values observed, right?

The other vector refers to the measured complete set of compatble observables and may also be a generalized eigenvector of the corresponding self-adjoint operators. It's in the dual space of the nuclear space of the operators, i.e., the dense subspace of the Hilbert space, where the operators are defined (see Ballentine for details about this important point).

I will re-read it right now, nevertheless, the dual space of a hilbert space is isomorphic to the original space and, so, it is indeed a hilbert space. For every element in the original space there is one in the dual space with the same properties.

Thus you can have something like generalized eigenvectors for momentum |ϕ⟩=|px,py,pz⟩ of a particle, which are normalized to a δ distribution, and then the probability density for measuring a certain momentum is given by
P(px,py,pz|ψ)=|⟨px,py,pz|ψ⟩|2=|⟨ψ|px,py,pz⟩|2.

Of course, in this case it doesn't make sense to ask about the "exchange" of ϕ and ψ, because here |ϕ⟩ can never ever describe a state of the particle!

But in this case, px,py,pz> is indeed a state (or the dual of states but, from what I said, it has all the properties of the original state). It is the state where the system has px,py and pz momentum. So?

Nevertheless, suppose you are right and, in this case, what you put on the left is something different from what you put on the right... But there are cases that, what you put on the right can be put on the left (example, px,py and pz momentum as phy and x,y, and z position as psi). So, in thes cases what I said still happens... we can switch the states and the probabilitie will remain the same...

I guess your explanation of the significance of this will remain the same:

In my opinion there is none

But, because of what I said, I can't agree with your arguments, sorry :(

Ps: I am going to re-read Ballentine

 

[atyy]

Suppose we have a set of events, which I'll denote as U
(meaning the universal set). Let A,B,C be subsets of U. The definition of conditional probability is:
$$
P(A|B) ~:=~ \frac{P(A\cap B)}{P(B)}
$$hence $$
P(A|B) ~\ne~ P(B|A)
$$in general, unless $P(A) = P(B)$.

Is this another way of saying that the intuition that P(x|y) ≠ P(y|x) in general doesn't obviously hold in the first place for quantum states?

Usually, Bayes's rule starts from

P(x,y)=P(x|y)P(y)=P(y|x)P(x)

which leads to

P(x|y) = P(y|x)P(x)/P(y).

However, there is no joint probability of P(x,y) in the first place in QM where x and y are quantum states.

 

[Fredrik]

I understand from the Born rule, exactly what you said, a transition probability. Indeed, after the measurement, the system transits to a state where it has the values observed, right?

I don't see why this view would be wrong either. $\left|\langle x,y\rangle\right|$ is the probability that a system in state y before the measurement will be in state x after the measurement, at least if we're talking about a measurement that doesn't destroy the system.

I will re-read it right now, nevertheless, the dual space of a hilbert space is isomorphic to the original space and, so, it is indeed a hilbert space. For every element in the original space there is one in the dual space with the same properties.

This is correct. However, vanhees71 wasn't talking about the dual of the Hilbert space. He was talking about the dual of one of its vector subspaces. If $K\subset H$, then $H^*\subset K^*$. The dual of the subspace is larger than the dual of the Hilbert space. It's a topological vector space, but not a Hilbert space. Members of that space may not have a norm for example. This trick allows us to assign a meaning to "eigenstates" of operators like position and momentum. They can be defined as members of $K^*-H^*$.

(Not sure if this has anything to do with the topic of this thread. I'm just making an observation about the mathematics).

 

[strangerep]

It's important to keep in mind that this definition is intended for the case when P is a probability measure on a σ-algebra. In QM, we're dealing with a probability measure on a lattice. σ-algebras are lattices, but lattices aren't σ-algebras. QM is not probability theory, it's a generalization of probability theory.

Yes, notions of probability can be formulated in various ways. Kolmorgorov (measure on an event set), Cox (axiomatically -- which is Ballentine's starting point), and also the lesser known Whittle (generalized by Arnold Neumaier in arXiv:0810.1019, see sect 8.2, 8.4, etc, in which states are mappings on an algebra of observables, hence more akin to the algebraic approach to quantum theory).

But in my previous post, I was trying to tease apart the more intuitive (naive?) notion of conditional probability from the the subtle modifications thereof required in QM. Ballentine delays discussion of Cox's 4th axiom in the context of QM until sect 9.6, as it involves some tricky ideas about the notion of joint probability $P(A\& B)$, and the subtleties of noncommutative observables. Resorting to filter-type measurements to produce a new state, he eventually arrives at regarding the 4th axiom, i.e.,
$$P(A\& B|C) ~=~ P(A|C) \; P(B|A\& C)$$ as a definition of the joint probability on the left hand side in the case of QM, but times of events A and B must be such that $t_A<t_B$. (See middle of p248.)

IOW, he does indeed generalize the concepts involved in Cox's formulation of probability to the QM case, but in doing so (i.e., dealing with noncommutativity) he is forced to introduce time ordering in an important way.

I have only had a quick look. So far I don't see what this has to do with the_pulp's question. Ballentine explains how to calculate probabilities of sequences of measurement results, but the_pulp only asked about the formula for the probability of a single measurement.

I was trying to highlight the difficulties and pitfalls of relating the QM squared-modulus formula to the classical notion of conditional probability. To discuss the latter properly, one must also discuss joint probability, which is problematic in QM due to noncommutativity. The_pulp was asking about physical motivations, and I think this is only possible through detailed examination of the dynamics of real measurements, which is one of the main areas in Ballentine's ch9, hence deserves careful study.

Edit 1: I now see there's been lots of posts since I started composing this one.

Edit 2: @atyy, as you can probably guess from the above, my answer to your post #48 is, broadly, "yes".