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I Eigenvalue of coherent states

  1. Jun 27, 2016 #1

    dyn

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    Hi. I don't understand what is meant by the eigenvalue α of a coherent state where a | α > = α | α >. The eigenket |α > is an infinite superposition of the number states , ie | α > = ∑ cn | n > and for each number state a | n > = √n | n-1 >. So for each number state the eigenvalue of the lowering operator is just a number , √n but how is the eigenvalue α arrived at when it is related to an infinite series ?
    Thanks
     
  2. jcsd
  3. Jun 27, 2016 #2
    Just apply the annihilation operator and then shift the Summation index. You should get an extra factor of ##\alpha##.
     
  4. Jun 29, 2016 #3

    dyn

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    Thanks for your reply but i'm still confused. Its easy to see what happens when the lowering operator acts on a number state eg. a|3>=√3|2> but | α > is an infinite superposition of number states so I can't see what the eigenvalue α is
     
  5. Jun 29, 2016 #4

    Paul Colby

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    Coherent states are in a sense "the most classical" states of the field. The ##\alpha## is an eigenvalue of the ##a## operator which in classical terms is the coefficient of a plane wave mode (although one need not use plane waves). ##\alpha## is a phase and amplitude for that component of the field.
     
  6. Jun 29, 2016 #5

    A. Neumaier

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    a applied to a linear combination of number states equals the linear combination of a times the number states. This allows you to evaluate ##a|\alpha\rangle##.
     
  7. Jun 29, 2016 #6

    dyn

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    So its never going to be a simple number like the integer values when the lowering operator is applied to an individual number state ?
     
  8. Jun 29, 2016 #7

    Paul Colby

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    As an undergraduate I learned QM by reading "Radiation and Noise in Quantum Electronics" By W. Louisell. It's a very useful and practically minded book. The author spends quite a bit on ##a## and ##a^\dagger## algebra including ##\alpha##-states. The physics of ##\alpha##-states is quite important. They provide a much better model of virtually all light sources one will find. This is something I only learned quite recently. The more common Fock states are quite strange at least as far as their optical properties. For example when an ##n=3## Fock state entering a 50/50 beam splitter, all the photons either go out port 3 or all go out port 4. This is not at all what an ##\alpha##-state or thermal light will do. Also, speaking more to your question, the value of ##\alpha## may be quite smaller than one. So, even though the admixture of ##n>0## states are quite small, they effect the statistics of the profoundly. For ##\alpha##-states as for thermal light, the photons in a beam splitter are uncorrelated.
     
  9. Jun 29, 2016 #8
    Yes, because the number states are not eigenstates of ##a## except the vacuum state ##|0\rangle ## (with eigenvalue 0).

    Edit: After a second look, I'm not sure what the question was. You apply ##a## to every number states in the infinite sum. At the end you will get the coherent state back, as others have told you.
     
  10. Jun 29, 2016 #9

    blue_leaf77

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    As far as I can remember, there is no restriction at all on ##\alpha## - it can be any real or complex number. Moreover, coherent states with different ##\alpha##'s are not orthogonal, which is to be expected from the fact that ##a## is not a Hermitian operator.
     
  11. Jul 5, 2016 #10
    There is no such meaning to an eigenvalue here. Also here the eigenvalue happens to be an eigenvalue of a non-hermitian operator and so has no "physical interpretation". Although it's a characterization of coherent states. You can take a look at Cohen-Tannoudji, there they have started out with conditions of quasi-classical states and from there arrived at the eigenvalue equation ##a |\alpha>=\alpha |\alpha>##.
     
  12. Jul 5, 2016 #11

    A. Neumaier

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    Of course it has, though it cannot be the Born interpretation. In case of photon coherent states, there is one coherent state ##|A\rangle## for each mode (analytic signal) ##A## of the free Maxwell field, and ##a(x)|A\rangle=A(x)|A\rangle## describes the preservation of the mode under observation of photons. It means that one can extract an arbitrary number of identically prepared photons from a beam prepared in a coherent state.
     
    Last edited: Jul 5, 2016
  13. Jul 5, 2016 #12

    Paul Colby

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    The creation and annihilation operators are the coefficients in a modal expansion of the E and B fields. Clearly I agree with you that the observable values of these operators have a very direct interpretation and meaning corresponding to the classical quantities they replace. Much in the same way as the eigenvalues of position and momentum operators for a particle. Why do you rule out the "Born interpretation" for ##\alpha## in particular? Seems oddly selective.
     
  14. Jul 5, 2016 #13

    A. Neumaier

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    Because the Born interpretation universally assumes a self-adjoint operator to be measured. If the operator is not self-adjoint (or at least normal, which is not the case here), there are no orthogonal projectors.

    The case of coherent states is very special as there is a replacement for it - since they form an overcomplete set with a nice analytic resolution of unity. A correct description of the measurement of analytic signals requires POVMs rather than orthogonal projectors.
     
  15. Jul 5, 2016 #14

    A. Neumaier

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    Indeed, and the coherent states are in this case eigenstates of the annihilation operator ##a=q+ip## (up to a constant factor) with eigenvalue ##\alpha = q+ip##. This shows that measuring ##\alpha## is a joint measurement of ##q## and ##p##, which is not possible with arbitrary accuracy. This shows that one cannot have Born measurements, but only POVM measurements.
     
  16. Jul 5, 2016 #15

    Paul Colby

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    Agreed. Measurement in quantum optics always proceeds via some form of photon counting as the ultimate measured quantity. This is true even in heterodyne detection systems.
     
  17. Jul 5, 2016 #16
    And a POVM measurement can be interpreted as a Born measurement of a greater system.

    Add a test particle in a harmonic oscillator ground state, described by operators ##p_0,q_0##, and measure for the two-particle system ## q-q_0## and ##p+p_0##. They commute, thus, can be measured together. The ground state of the test particle describes, informally, the error of the measurement.
     
  18. Jul 5, 2016 #17

    A. Neumaier

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    But only of a fictitious larger system. This is not how one actually measures systems described by a POVM.
     
  19. Jul 5, 2016 #18
    Of course, in real measurments much more than a single test particle is involved, one needs large, macroscopic measurement devices.
     
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