# I Eigenvalue of coherent states

1. Jun 27, 2016

### dyn

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. Jun 27, 2016

### Mr-R

Just apply the annihilation operator and then shift the Summation index. You should get an extra factor of $\alpha$.

3. Jun 29, 2016

### dyn

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

4. Jun 29, 2016

### Paul Colby

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.

5. Jun 29, 2016

### A. Neumaier

a applied to a linear combination of number states equals the linear combination of a times the number states. This allows you to evaluate $a|\alpha\rangle$.

6. Jun 29, 2016

### dyn

So its never going to be a simple number like the integer values when the lowering operator is applied to an individual number state ?

7. Jun 29, 2016

### Paul Colby

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.

8. Jun 29, 2016

### Truecrimson

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.

9. Jun 29, 2016

### blue_leaf77

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.

10. Jul 5, 2016

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 $a |\alpha>=\alpha |\alpha>$.

11. Jul 5, 2016

### A. Neumaier

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
12. Jul 5, 2016

### Paul Colby

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.

13. Jul 5, 2016

### A. Neumaier

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.

14. Jul 5, 2016

### A. Neumaier

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.

15. Jul 5, 2016

### Paul Colby

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.

16. Jul 5, 2016

### Ilja

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.

17. Jul 5, 2016

### A. Neumaier

But only of a fictitious larger system. This is not how one actually measures systems described by a POVM.

18. Jul 5, 2016

### Ilja

Of course, in real measurments much more than a single test particle is involved, one needs large, macroscopic measurement devices.