# A Coherent state -- Lasers

1. Apr 17, 2016

### LagrangeEuler

I am confused. In harmonic oscillator problem in quantum mechanics eigenstates of operator $\hat{a}$ are called coherent states. So we practically get Gaussian. In problem of quantum linear harmonic oscillator we have phonons $|n \rangle$. In case of laser we have photons $|n \rangle$. Could you explain me what kind of coherence do we have in ground state of harmonic oscillator?

2. Apr 17, 2016

### Paul Colby

In a laser many excited atoms drive (ideally) a single mode of the radiation field. A radiation field mode is an harmonic oscillator. The $a$ and $a^\dagger$ operators are the quantum equivalent of the wave/mode amplitude. With many atoms driving a mode a laser doesn't produce a $\vert n\rangle$ state but rather something much closer to a coherent state, $a\vert \alpha\rangle = \alpha\vert \alpha\rangle$. The reason for this is that all the excited atoms are coupled through the common field mode.

3. Apr 17, 2016

### vanhees71

Yes, and stimulated emission is a coherent process, i.e., you get a pretty coherent em. wave with well defined phase rather than Fock states with a definite photon number. This coherence it is which makes lasers such a great invention!

4. Apr 17, 2016

### LagrangeEuler

For me is hard to understand for example in case of LHO $\Delta \hat{x} \Delta \hat{p}_x=(n+\frac{1}{2})\hbar$. So if we have third exited state then $\Delta \hat{x} \Delta \hat{p}_x=\frac{7}{2} \hbar$ and in case of ground state $\Delta \hat{x} \Delta \hat{p}_x=\frac{1}{2} \hbar$. What means that $\Delta \hat{x} \Delta \hat{p}_x$ is larger in one state then in another?

5. Apr 17, 2016

### vanhees71

Well, it tells you that only the ground state of the harmonic oscillator and its unitary transformations (the coherent states are unitary transformations of the ground state!) lead to the "minimal space-momentum uncertainty", i.e., to $\Delta x \Delta p =\hbar/2$. The uncertainty relation tells you that for any state $\Delta x \Delta p \geq \hbar/2$.

6. Apr 17, 2016

### LagrangeEuler

Ok. But for example, for third exiting state $\Delta \hat{x} \Delta \hat{p}_x$ is larger then in second exited state. What that means physically?

7. Apr 17, 2016

### Paul Colby

Actually, $e^{-\alpha a^\dagger}$ is not unitary.

8. Apr 17, 2016

### vanhees71

The transformation of the ground state to the coherent state $|\alpha \rangle$ is unitary
$$|\alpha \rangle=\exp(\alpha \hat{a}^{\dagger}-\alpha^* \hat{a})|0 \rangle.$$
Note that
$$\exp(\alpha \hat{a}^{\dagger}-\alpha^* \hat{a})=\exp(-|\alpha|^2/2) \exp(\alpha \hat{a}^{\dagger}) \exp(\alpha^* \hat{a}).$$

9. Apr 17, 2016

### Paul Colby

Sorry, my understanding is that a unitary operator, $U$, is one for which, $UU^\dagger = 1$. The above is clearly unitary, I stand corrected.

10. Apr 19, 2016

### Elemental

LagrangeEuler,

The ground state of a harmonic oscillator is not a coherent state, hence does not have coherence (definite phase). If you look at vanshees71's formula,
and expand the exponential you'll get an infinite series in the creation operator acting on the ground state, creating an infinite series of number states (you also get an infinite series in the annihilation operator but this vanishes when operating on the ground state). The resulting state has no resemblance to the ground state.

Elemental

11. Apr 19, 2016

### Paul Colby

It's not just $\vert \alpha\rangle$ where $\alpha=0$?

12. Apr 19, 2016

I think I have what may be a relevant input to this discussion. A very interesting and inconsistent result occurs if you use a photon model where each photon in the same mode has the same phase, and the E-fields from individual photons (may be incorrect to describe an E-field for a single photon) are all added in phase. The energy density U of the E-field , which is proportional to E^2 would then be proportional to N^2 where N is the number of photons. This dilemma is resolved if the phases of the individual photons are considered constant but random. The phasor diagram then becomes a 2-D random walk, with the result amplitude of the N photons proportional to sqrt (N). Thereby the energy of the E-field will be proportional to the photon number. Would anticipate this result has already been studied somewhere...

13. Apr 19, 2016

### Elemental

Paul Colby,

So it is! Tanking vanshee71's formula (which I'm sure is correct), the ground state is a degenerate case of a coherent state. I wouldn't draw too much from this though. I suspect phase coherence emerges as alpha gets larger, because the number uncertainty will grow with alpha. Glauber wrote several authoritative articles on coherent states in the 1960's (I believe); time permitting, I'll see if he addressed any of these questions there.

Elemental

14. Apr 19, 2016

### strangerep

Heh, first stop: grab a copy of Mandel & Wolf...
(And then maybe try Perelemov's book on generalized coherent states.)

(Of course the vacuum can be considered a coherent state, since the others a generated from the vacuum by the action of "displacement" operators similar to Hendrik Van Hees' formula. Such operators form a group and one can usefully define the concept of coherent states in terms of them.)

15. Apr 20, 2016

### Elemental

Yeah, after checking Amazon, Mandel & Wolf looks like a must-buy; I hadn't read it since it was published after the time I was researching quantum optics.

Something I remember from the quantum mechanics course in graduate school is we entirely skipped over coherent states of the harmonic oscillator. We only studied stationary (number) states. Oscillators in stationary states don't move back and forth like the (macroscopic) oscillators I knew anything about, leaving me as confused as ever. So later I plowed through coherence theory on my own.

Interesting to note that that coherent states are sometimes referred to as "quasi-classical" states, because they most resemble the behavior of macroscopic oscillators found in nature. But, being eigenstates of the annihilation operator, they are non-orthogonal: a consequence of the annihilation operator being non-Hermitian. The ground (vacuum) state is also an eigenstate of the annihilation operator (generalized to the field operator in quantum field theory), since from the eigenvalue equation a|α⟩ = α|α⟩, but |α⟩ = |0⟩ when α = 0 (see above), in which case a|α⟩ = 0 = 0|α⟩. So undoubtedly the ground state is a coherent state.

Elemental

16. Apr 20, 2016

### Paul Colby

Actually, if I might amplify Elemental's comments, coherent states also appear as a good description of thermal sources when suitably averaged. If I understand the calculation correctly, number states behave very differently than coherent states when subjected to a beam splitter. Consider a photon counting experiment with two detectors counting the two outputs of a beam splitter. For number states each photon detected either goes to port 1 and not port 2 or it goes to port 2 and not port 1. This is quite intuitive because macroscopic particles like machine gun bullets (and machine guns) behave this way in that there is an anti-correlation between bullet counts. For a machine gun our beam splitter could be a blade deflecting the bullet up or down. This anti-correlation for bullets is independent of the number of machine guns or their firing rate. For photons and the vast majority of photon sources the correlation is 0. For coherent states the statistics of the two outputs of the beamsplitter are completely uncorrelated.