Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutativity of the Born Rule

  1. Mar 28, 2013 #1
    The born rule, written in the following way:

    P(ψ/[itex]\varphi[/itex])=|<ψ|[itex]\varphi[/itex]>|^2

    As a consequence,

    P(ψ/[itex]\varphi[/itex])=P([itex]\varphi[/itex]/ψ)

    I dont 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!
     
  2. jcsd
  3. Mar 28, 2013 #2

    strangerep

    User Avatar
    Science Advisor

    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.
     
  4. Mar 29, 2013 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    ##\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|##.
     
    Last edited: Mar 29, 2013
  5. Mar 29, 2013 #4
    Thanks for your answers, but I still cant 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 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
     
  6. Mar 29, 2013 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  7. Mar 29, 2013 #6

    strangerep

    User Avatar
    Science Advisor

    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.
     
  8. Mar 29, 2013 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  9. Mar 29, 2013 #8

    strangerep

    User Avatar
    Science Advisor

    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...
     
    Last edited: Mar 29, 2013
  10. Mar 29, 2013 #9
    Id love if the answer were "its because life respects time reversal" but, doesnt weak interaction violate time reversal? So, what do we do?

    Thanks for all the replies!
     
  11. Mar 29, 2013 #10

    strangerep

    User Avatar
    Science Advisor

    Use CPT invariance instead? :biggrin:
     
  12. Mar 30, 2013 #11
    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)
     
  13. Mar 30, 2013 #12

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  14. Mar 30, 2013 #13

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor


    [itex]\vert \langle x \vert y \rangle \vert = \vert \langle y \vert x \rangle \vert [/itex]

    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 [itex]\vert y \rangle[/itex] and project it along a different vector [itex]\vert x \rangle[/itex], you don't get a probabilistic result of 0 or 1, you get a deterministic result, that the projection has length [itex]\dfrac{\vert \langle x \vert y \rangle \vert}{\sqrt{\langle x \vert x \rangle}}[/itex].
     
  15. Mar 30, 2013 #14

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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##.
     
    Last edited: Mar 30, 2013
  16. Mar 30, 2013 #15

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  17. Mar 30, 2013 #16

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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
    [tex]P(\psi|\phi)=|\langle \psi|\phi \rangle|^2[/tex]
    doesn't make sense, except, I misinterpret the symbols:

    Usually the left-hand side denotes a conditional probability that "and event [itex]\psi[/itex]" occurs, given some "condition [itex]\phi[/itex]". 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 [itex]|\psi \rangle[/itex], or equivalently by the projection operator [itex]\hat{R}_{\psi}=|\psi \rangle \langle \psi |[/itex] 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 [itex]A[/itex] is represented by a self-adjoint operator [itex]\hat{A}[/itex]. The possible values this observable can take are given by the spectral values [itex]a[/itex] of this operator. Further let [itex]|a \rangle[/itex] 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 [itex]A[/itex] and [itex]B[/itex] are compatible if the corresponding operators have a complete set of (generalized) common eigen vectors [itex]|a,b \rangle[/itex]. One can show that then the operators necessarily commute [itex][\hat{A},\hat{B}]=0.[/itex]

    (4) A set of compatible independent observables [itex]\{A_{i} \}_{i=1,\ldots n}[/itex] is complete, if any common (generalized) eigenvector [itex]|a_1,a_2,\ldots,a_n \rangle[/itex] 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 [itex]|\psi \rangle[/itex], 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 [itex]|\psi \rangle[/itex], then the probability (density) to find the common values [itex](b_1,\ldots,b_n)[/itex] of (the same or another) complete set of compatible observables [itex]\{B_i \}_{i=1,\ldots,n}[/itex] is given by Born's rule,
    [tex]P(b_1,\ldots,b_n|\psi)=|\langle b_1,\ldots,b_n|\psi \rangle|^2.[/tex]

    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 [itex]\psi[/itex].

    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, [itex]|\psi \rangle[/itex] 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
    [tex]\mathrm{i} \hbar \partial_t |\psi,t \rangle = \hat{H} |\psi,t \rangle,[/tex]
    where [itex]\hat{H}[/itex] 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,
    [tex]\frac{\mathrm{d} \hat{A}(t)}{\mathrm{d} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H} ].[/tex]

    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,
    [tex]\mathrm{i} \hbar \partial_t |\psi,t \rangle = \hat{H}_1 |\psi,t \rangle[/tex]
    [tex]\frac{\mathrm{d} \hat{A}(t)}{\mathrm{d} t}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}_2][/tex]
    with an arbitrary decomposition of the Hamiltonian as
    [tex]\hat{H}=\hat{H}_1+\hat{H}_2.[/tex]
    A commonly used Dirac picture is the "interaction picture", where the observable operators evolve as for free particles, i.e., [itex]\hat{H}_1[/itex] is the kinetic energy only, and the states according to the interaction part [itex]\hat{H}_2[/itex] 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!
     
  18. Mar 30, 2013 #17
    Sorry Vanshees, Im reading 4 and 5 from your post but I dont 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 dont 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
     
  19. Mar 30, 2013 #18
    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 dont want to use the phrase "physical reason" but perhaps it helps you to find what Im asking) in order to explain why the probability to go from [itex]\psi[/itex] to [itex]\varphi[/itex] and the probability to go the other way round are the same.

    Thanks all the same
     
  20. Mar 30, 2013 #19

    atyy

    User Avatar
    Science Advisor

    I'm not sure this is right, but isn't one vector an eigenvector of an observable, while the other is an arbitrary vector?
     
  21. Mar 30, 2013 #20
    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)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutativity of the Born Rule
  1. Born's rule (Replies: 2)

  2. Born rule derivation? (Replies: 2)

Loading...