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_{n}| 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

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Thanks

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Paul Colby

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A. Neumaier

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Paul Colby

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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.

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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.

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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.the eigenvalue happens to be an eigenvalue of a non-hermitian operator and so has no "physical interpretation".

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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.Of course it has, though it cannot be the Born interpretation.

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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.Why do you rule out the "Born interpretation" for α in particular?

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.

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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.Much in the same way as the eigenvalues of position and momentum operators for a particle.

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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.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.

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And a POVM measurement can be interpreted as a Born measurement of a greater system.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.

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.

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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.a POVM measurement can be interpreted as a Born measurement of a greater system.

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