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Is the wave function the invariant thing?

  1. Mar 17, 2015 #1
    Is the wave function of quantum mechanics considered to be the quint-essential invariant object? Is it the wave function that must not change with space, time, gravitational field, etc? It would seem to me that the relative probabilities that things happen is the thing that can not change with perspective (space or time, etc).

    But I am challenged by considering Hawking radiation and Unruh radiation, where things happen at all only for accelerated observers. This would tell me (if I'm understanding it correctly) that even relative probabilities change for different observers. Or is it that these types of radiation always take place, and the relative probability of what kinds of particles are more likely to appear is still the same only more frequent with time for some observers?
     
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  3. Mar 17, 2015 #2
    The discovery of "what is really invariant in physics?" is a puzzling question. The concept of invariance –as everyone knows- has been introduced in physics by A. Einstein. But, for the eyes of a mathematician, this was nothing new since E. B. Christoffel had already explored the question in 1869. The reason why Einstein was led to that concept can reasonably be related to the negative result (the earth has no motion relatively to the presupposed ether) obtained by the Morley and Michelson experiment: the speed of light in vacuum for observers at rest (or with constant speed) is everywhere the same, in extenso: it is an invariant. This was the mile stone allowing the construction of special relativity. From which a whole family of invariants could be deduced; e.g.: the stress-energy tensor.

    Well your question is very interesting because: “what is insuring that we have discovered all invariants quantities in the nature? And what is giving us the insurance that some invariants are not generating some others by the virtue of mathematics?”

    Since the vacuum is unstable (Heisenberg’s uncertainty principle for the pair dE, dt), one is legitimized to ask if, in vacuum, that uncertainty relation is preserved at its quantum limit or not? If yes, this would change our perspective. Now I unfortunately have no published paper illustrating that idea and I ignore how it could be falsified by experiments.

    In our actual understanding, pairs of virtual particles are created (by necessity: the neutral charge of the initial vacuum must be preserved) and they cannot survive a long time since they are forced to annihilate again if they interact together. This point, once more time, remains a mysterious one because we are not actually able to explain why, sometimes, one of the two species survives.

    I am afraid but the opening question is a Pandora box and it will certainly be difficult to close it properly (in extenso: to give a clear answer); or do I miss something important?
     
  4. Mar 17, 2015 #3
    I can understand how the invariance of the wave function can be transferred to the invariance of the Action Integral. If we express the wave function in terms of the Feynman path integral, then the variation of the path integral being set to zero for invariance gets transferred to the invariance of the Action Integral which can be transferred to invariance of the Lagrangian which is where we get the equations of motion. But I'm not sure what would give rise to the invariance of the wave function, except perhaps some invariance of probabilities which I would assume would be the same for all observers. Or maybe not.

    Certain measurements can be teased out of the wave function by applying some operator and getting an eigenvalue as the measurement. But does this mean the state vector itself that's part of the Hilbert space remains invariant before measurement, even though it can be expressed in various bases depending on what measurement you're trying to predict?
     
    Last edited: Mar 17, 2015
  5. Mar 17, 2015 #4
    The question is far over my head and my competences. My answer will only be done at some intuitive level. And what if the equation of motion can be transformed into some differential operator depending on the wave function? And what if that operator can be put into the formalism of a self-adjoint one mimicking a Lorentz transformation acting on the wave function? And what if that transformation is always your proposed "some operator" where ever you are?
     
  6. Mar 17, 2015 #5

    wabbit

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    Sorry, silly question here - isn't the wave function observer-dependent ? I don't understand the sense in which it is invariant - probably missing something obvious..,
     
  7. Mar 17, 2015 #6
    No, you got the right question. We know a measurement requires an observer, right? And we know that the wave function changes (collapses) when a measurement is made. But how is the wave function ( or more precisely the state vector, since the wave function is the state vector in the position basis) - how invariant is the state vector?
     
  8. Mar 17, 2015 #7

    wabbit

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    Actually my question was more basic than that, I was thinkiing of the observer movement. An observer moving relative to a particle and an observer at rest can't see the same wavefunction can they ?
    That s not well defined - maybe better let's say observer A sees the particle in a momentum state p ; observer B moving relative to A will see the particle with a different momentum state - how can the same wave function describe two different momentum states ?

    But you re right we can look at how this translates into A and B s respective measurement's, and their measurements of momentum should differ in proportion to the relative velocity of B to A.

    Actually the same thing happens for position, between A's measurement and B's there s a diifference which is the vector AB. And while it's easy to see how the wave function tranforms from applying this translation, that transformation is not nothing.

    So if we say it is invariant, what is the meaning of this invariance, simce clearly we get a different wave function, not the same ?
     
    Last edited: Mar 17, 2015
  9. Mar 17, 2015 #8
    Yes, that's my question too. Although, I think any observer could only record the probability of outcomes in order to see if the wave function change between observers. My guess is that they would both see the same outcome probabilities. But I don't think the invariance of the probabilities equates to the invariance of the wave function. Since the probability density is equal to the wave function times its complex conjugate, taking the variation of that gives you two terms equated to zero (since it is like the chain rule of derivatives).
     
  10. Mar 17, 2015 #9

    wabbit

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    But I don t see how they can see the same outcome probabilities. Or rather, using the position example (not moving), A will assign probabilty p to position x, and B will assign that same probability p to the different postion x+AB (or is it x-AB)
    In other words
    PsiA(x, t)=PsiB(x-AB, t) which is not the same as PsiB(x, t)
    So PsiB ## \neq ## PsiA
     
    Last edited: Mar 17, 2015
  11. Mar 19, 2015 #10

    haushofer

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    In ordinary QM the wavefunction is not invariant under Galilean boosts, so you're right.
     
  12. Mar 19, 2015 #11

    wabbit

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    Thanks !

    Trying to find a meaning for invariance here I was led to musing about the following ;
    The wave function is a quantum equivalent of a classical trajectory, both are relative to an observer.
    Yet when we talk of a trajectory, we may also have in mind an abstract trajectory, "the set of all pairs ## (A,x_A(.))## related by the appropriate transformations".
    In the same way we could refer to an abstract wavefunction defined as an equivalence class ##{(A,\psi_A)}##.
    I don't know if that is a useful object at all, or whether it is typically discussed in QM?
     
    Last edited: Mar 19, 2015
  13. Mar 19, 2015 #12

    atyy

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    The wave function is not an invariant thing. The probability of the outcomes predicted by the wave function is invariant.
    http://arxiv.org/abs/1007.3977
     
  14. Mar 19, 2015 #13

    wabbit

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    Just had a quick look, this is about delayed choice experiments ?

    Other than that, probability is conserved for uniform motion, but is it still true for accelerated observers ? I suppose it should, just a nagging doubt, maybe just noise from the memory of Unruh radiation which isnt non relativitic QM of course so isnt an issue here...

    Also, I was trying to work out the non-relativistic composition rule for amplitudes, I am stuck - where would I find that ?
    By composition rule I mean I have three observers/particles A B C, both A and B look at C, A looks at B also.
    Classically I have ## x_{AC}=x_{AB}+x_{BC} ## where ## x_{AB} ## is the position (vector) of B from A's viewpoint etc.

    Same situation but QM, ## \psi_{AC}, \psi_{AB}, \psi_{BC}## are the corresponding wavefunctions.

    I can write how the probabilities ## |\psi_{AC}|^2## etc. are related, but i don't know how to compute ## \psi_{AC} ## from ## \psi_{AB} ## and ## \psi_{BC}## . would you have any pointer ? Probabilities compose through convolution, amplitudes compose through..... ?
     
    Last edited: Mar 19, 2015
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