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The physical meaning of a phase factor

  1. Oct 31, 2011 #1
    A phase factor is [itex]e^{i\phi}[/itex]. Mathematically a multiplication of a wavefunction by phase factor is equivalent to a rotation of the state vector by the angle [itex]\phi[/itex], however the probability amplitudes are unchanged. So what is the physical meaning of it?
     
    Last edited: Oct 31, 2011
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  3. Oct 31, 2011 #2

    f95toli

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    I don't think this is conceptually very different from a classical wave. The phase doesn't matter in a measurement UNLESS you are dealing with some kind of interference. However, since in most "real world" uses of QM interference DOES matter phase is most definitely important. It can also be argued that it is "real" since it under some circumstances can be quantized (it is the conjugate variable to charge in an electronics circuit).

    One obvious example of the effect of the phase in QM is the Aharonov-Bohm effect.
     
  4. Oct 31, 2011 #3
    If we rotate (multiply by a phase factor?) a state vector in a Hilbert space, does the state changes? It seems that yes because if we rotate an eigenvector we can get an superposition state or an another eigenvector perpendicular to the first. So the expectation value of the operator changes. What I'm missing here?
     
  5. Oct 31, 2011 #4

    kith

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    If you multiply an eigenvector by a factor, it remains an eigenvector due to the linearity of the eigenvalue equation.

    The problem here is the term "rotation". A complex phase factor corresponds to a rotation in the complex plane, which does not change expectation values. A rotation in Hilbert space is described by a unitarian transformation, which in general does change expectation values.
     
  6. Oct 31, 2011 #5

    vanhees71

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    A physical (pure) state is represented not by the state vector but to the corresponding ray, i.e., the state doesn't change when you simply multiply the state vector by a phase factor.

    More generally you can represent any state, pure or mixed, by the Statistical Operator [itex]\hat{R}[/itex] of the system. For a pure state, represented by a normalized Hilbert-space vector [itex]|\psi \rangle[/itex], it reads

    [tex]\hat{R}=|\psi \rangle \langle \psi|.[/tex]

    As you see, just multiplying the state with a phase factor [itex]\exp(\mathrm{i} \alpha)[/itex] with [itex]\alpha \in \mathbb{R}[/itex] doesn't change the statistical operator, because multiplication with a scalar commutes with any operator:

    [tex]\exp(\mathrm{i} \alpha) \hat{R} \exp(-\mathrm{i} \alpha)=\hat{R}.[/tex]

    All this, of course, doesn't hold anymore for a general unitary transformation [itex]\hat{U}[/tex] that generally doesn't commute with the Statistical Operator and thus in general maps the original to another state.

    Of course, nothing changes with any observable quantity (expectation values of observables), if you apply the unitary transformation to all states and all observables since the physics content of quantum theory is invariant under unitary transformations.
     
  7. Oct 31, 2011 #6
    Thank you all.

    So, actually, the probability amplitudes are rotating on the complex plane, not the state vectors. The state and the magnitude of the probability amplitude remains the same, so in the most cases this is enough to say that the phase factor does not have any physical meaning. However the relative phases between the components of an superposition state in some basis are important and can be measured (in the double slit experiment etc.).

    My second question is how exactly the interference depends on the relative phases of the components of an superposition state? For example how different are the two states: [itex]a|0\rangle +b|1\rangle=e^{i0}(a|0\rangle +b|1\rangle)[/itex] and [itex]a|0\rangle -b|1\rangle=a|0\rangle +e^{i\pi}b|1\rangle[/itex]. In the first state the basis components are in phase but in the second they are out of the phase.
     
  8. Oct 31, 2011 #7

    Matterwave

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    The most simple way to talk about phase that I can think of is in the 2 slit experiment. If we look at a point P on the screen (where the detector is), then the laws of quantum mechanics say that the amplitude (key word here is "amplitude" and not "probability) to detect a particle at that point, is the sum of the amplitudes for the particle to move through slit 1 and slit 2 to this point. In other words, if the amplitude for the particle to move through slit 1 to point P is [itex]\psi_1[/itex], and the amplitude for the particle to move through slit 2 to point P is [itex]\psi_2[/itex], then the amplitude for a detection at P is simply [itex]\psi_{total}=\psi_1+\psi_2[/itex].

    The PROBABILITY for this detection is then:

    [tex]P=|\psi_{total}|^2=|\psi_1+\psi_2|^2[/tex]

    So, let us take the case where [tex]\psi_1=\psi_2[/tex], then the probability for detection is simply

    [tex]P=4|\psi_1|^2[/tex]

    Now, if [tex]\psi_1=-\psi_2[/tex] (i.e. I rotated one of them 180 degrees out of phase), then it is obvious that now the probability for detection is 0.

    If the two are theta degrees out of phase:
    [tex]\psi_1=e^{i\theta}\psi_2[/tex]

    Then the probability for detection is:

    [tex]P=2|\psi_1|^2+e^{i\theta}|\psi_1|^2+e^{-i\theta}|\psi_1|^2=2|\psi_1|^2(1+\cos\theta)[/tex]



    Therefore, the phase of the state really DOES matter when interference effects are present. (I used the symbol [itex]\psi[/itex] for amplitude)
     
    Last edited: Oct 31, 2011
  9. Oct 31, 2011 #8

    Here is a marvelously clear explanation.

    http://quantum.phys.cmu.edu/CQT/chaps/cqt02.pdf
     
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