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Phase in quantum mechanics

  1. Mar 1, 2009 #1
    Hi everyone.
    I am now learning the perturbation theory in QM.
    and I have encountered something that puzzles me.

    from

    http://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics [Broken])

    says,

    "...Since the overall phase is not determined in quantum mechanics, without loss of generality, we may assume <n(0)|n> is purely real. Therefore, <n(0)|n(1)>=-<n(1)|n(0)>, and we deduce..."

    What do you mean by "the overall phase is not determined in quantum mechanics" ?

    I especially emphasize "in quantum mechanics". Does this mean that we cannot find a specific value for the phase by quantum calculation?

    I appreciate your help!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 1, 2009 #2
    sorry, you cannot jump from the hyperlink above.
    make sure copy the last ")" as well.
     
  4. Mar 1, 2009 #3

    malawi_glenn

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    What we observe is never the wavefunction but the wavefunction modulus square. You can multiply to every wavefunciton a phase factor:

    exp(i \theta)

    where theta is a real number. This phase factor will never change the outcome of measurements and observables, since what you observe is always the wavefunction modulus square:

    |psi|^2 = exp(-i theta) Psi* exp(i theta) Psi = 1 psi*psi = |psi|^2
     
  5. Mar 1, 2009 #4
    Thank you malawi_glenn!

    I have finally understood it. It was much easier than I thought.
     
  6. Mar 1, 2009 #5

    f95toli

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    I am not quite sure I agree with that description. I think it is better to say that there is no such thing as absolute phase in QM (or in any other field in physics) meaning we are free to choose whatever value we want as a starting point; there are certainly experiments where we can see the effect of a e.g. a difference in phase between two trajectories; the Aharonov-Bohm effect would be an obvious example.
    In this example it simply means that we choose to use [itex]\theta=0[/itex] as our staring point; i.e. n(0)|n> is purely real.

    Note that I am not disagreeing with what Malalwi_Glenn wrote; it is just that it is important to understand that phase is something "real" in quantum mechanics; it is NOT just mathematical tool that always "disappears" at the end of a calculation.
     
  7. Mar 2, 2009 #6

    malawi_glenn

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    I tried to be as simple as possible ;-)

    Then we can impose LOCAL phase invariance of the fields, and then we have the starting point for the standard model of particle physics.
     
  8. Mar 2, 2009 #7
    As the phase evolves in a time (and ergo space) dependant way depending on the energy (faster rate of change for higher energy) we have that changing the phase of *every* state by a constant factor is equivalent to choosing a new zero for the energy. Since it is only ever the differences in energies that have physical meaning, this poses no problems. This choice of zero energy is the same in classical mechanics, but I'm not sure that phases exist there..
     
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