# What does coherence mean?

1. Dec 22, 2009

### pellman

Coherent states are defined as eigenstates of the annihilation (lowering) operator for the quantum harmonic oscillator.

So why are these called "coherent"? What does "coherence" generally mean in physics? And what does it have to do with what is known as "quantum de-coherence?"

Edit:
I just realized there is an article on Coherence in physics generally. http://en.wikipedia.org/wiki/Coherence_(physics [Broken]) But I don't see the connection with the eigenstates of the annihilation operator.

Also, the article on plain coherence states in its section on quantum coherence: "The quantum description of perfectly coherent paths is called a pure state," so if you can elaborate on the relation between pure states and coherent states as well, that would be appreciated.

Last edited by a moderator: May 4, 2017
2. Dec 22, 2009

### Mentz114

I'd like to know this also. For what it's worth my
impression is that there's some overloading of the term 'coherent' and it's derivatives. In quantum optics the states called 'coherent' referred to above should probably be called 'most coherent' because they have the miniumum uncertainty between conjugates that is allowed by the HUP. So does 'coherence' in this context refer to a property ? Like the magnitude of the uncertainty ?

In the wider context, 'decoherence' seems to be the process that causes 'wave-function collapse' and has no obvious connection to the HUP.

3. Dec 22, 2009

### Cthugha

Great question. First of all, one should make clear that optical coherence has a special meaning and is very different from the usage of the term "coherence" in e.g. spin coherence or decoherence. What makes it even worse is the fact that there are even two meanings of coherence in optics: the classical one describing more or less the ability of light to interfere in a Michelson interferometer or a double slit experiment (which is a bit closer to the meaning of coherence in other branches of physics) and the meaning in quantum optics. The latter context is not that easy to grasp. So let me start explaining it in simple terms.

One might have the opinion that the classical meaning of coherence is enough to describe light sources e.g. by just determining their coherence time. Coherence time is the Fourier transform of the spectral power density of the source and more or less a measure of how sharp the mode is. This is why a laser is classically very coherent and sunlight is not. However, this would also mean that if you pass sunlight through an extremely narrow spectral filter so that it becomes comparable to the laser linewidth, you would get light with laser properties. Obviously there must be something else to laser light.

This extra property is higher order coherence. This can be tested by building a Hanbury Brown and Twiss setup, which is nothing else, but putting a light beam on a beam splitter, putting two photodiodes in the two light paths and do start-stop counting to produce a histogram of coincidence counts. Now the expectation value of coincidence counts $$\langle n^2 \rangle$$ for a classical situation would be easy to describe. Photons should be statistically independent and therefore:
$$\langle n^2 \rangle=\langle n\rangle \langle n\rangle$$
One usually normalizes the counts actually seen in the experiment on the value expected for independently emitted photons and gets:
$$g^{(2)}=\frac{\langle n^2 \rangle}{\langle n \rangle^2}=1$$ in the classical case.

From a semiclassical point of view, the problem is more delicate. It is well known that the detection of a photon changes the light field as there is now one photon less. So you now have:

$$g^{(2)}=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}$$

The momentary n can be described as the sum of the mean value of n and the fluctuations:$$n=\langle n \rangle + \delta n$$ and accordingly

$$g^{(2)}=\frac{\langle (\langle n \rangle + \delta n) (\langle n \rangle + \delta n-1) \rangle}{\langle n \rangle^2}$$

If you do the calculations you will find that all terms containing the expectation value of a single $$\delta n$$ will be 0 and what is left behind is:

$$g^{(2)}=\frac{\langle n \rangle^2 -\langle n \rangle + \langle (\delta n)^2 \rangle}{\langle n \rangle^2}$$
$$=1+\frac{\langle (\delta n)^2 \rangle}{\langle n \rangle^2}-\frac{1}{\langle n \rangle}$$

Sorry for all this math. You now have two main effects, which can influence the number of joint detection: the photon number fluctuations given by $$\langle (\delta n)^2 \rangle$$ (the variance of the photon number distribution) and the $$\frac{1}{\langle n \rangle}$$ term accounting for the change in the light field by detecting one photon.

Now let us consider three easy examples:

1) an ideal Fock state with $$\langle n \rangle=1$$ and $$\langle (\delta n)^2 \rangle = 0$$. Accordingly you get:
$$=g^{(2)}=1-\frac{1}{\langle 1 \rangle}=0$$
This is pretty logical. As only one photon is present, there will be no simultaneous detections. Although the noise is minimal, we are far from the classical limit and this state is of course also not very stable in terms of loss mechanisms.

2) a thermal state. Thermal light follows the Bose-Einstein distribution. Here $$\langle (\delta n)^2 \rangle = \langle n\rangle^2 +\langle n\rangle$$. Accordingly we get:
$$=g^{(2)}=1-\frac{1}{\langle n \rangle}+\frac{\langle n\rangle^2 +\langle n\rangle}{\langle n \rangle^2}=2$$

So the possibility to have simultaneous detections is twice as high as expected for statistically independent photons. This is a clear measure of the noisy photon number distribution and also shows the bosonic nature of photons. Here loss of a photon means tht it is very possible that other photons will be there at the same time, too.

3) laser light. Here photon numbers follow a Poisson distribution and $$\langle (\delta n)^2 \rangle = \langle n\rangle$$. Accordingly we get:
$$g^{(2)}=1-\frac{1}{\langle n \rangle}+\frac{\langle n\rangle}{\langle n \rangle^2}=1$$

So here we recovered the classical limit and photons are indeed emitted statistically independent of each other because of the inherent noise present. Therefore the state is immune to loss and destruction of a photon does not change the mean photon number. This is why a coherent state is an eigenstate of the annihilation operation. This is the "most classical" state one can get in a quantum description.

So now one can distinguish between sunlight and laser light simply by measuring whether there is this bunching effect ($$g^{(2)}=2$$) or photons are statistically independent
($$g^{(2)}=1$$). Now if one does not have a look at simultaneous detection, but detection with a delay it is clear that the value of $$g^{(2)}$$ will go from 2 towards 1 with increasing delay as photons emitted with a large delay should be statistically independent. This drop of $$g^{(2)}$$ usually happens within the (classical) coherence time and therefore this decay time in combination with the classical decay time is a good quantum measure of coherence. For good lasers this decay goes from almost 1 to 1 and therefore the decay time will be extremely long. Of course you can also define "higher quality" coherence by detecting 2,3,4 or more photons simultaneously. However, this is often not extremely interesting.

However, in conclusion there are a lot of effects behind the concept of optical coherence. Coherent states are immune to loss, are states of minimal uncertainty, robust against loss, resemble classical states and coherent states also mean that photons are statistically independent. Of course it also means that photons within the same coherence volume are indistinguishable.

4. Dec 22, 2009

### pellman

First let me say: awesome response, Cthugha!

I made it this far. What are coincidence counts?

5. Dec 23, 2009

### Cthugha

Ah, ok. Coincidence counts are just simultaneous detections of photons. You just have a look at the two photodiodes after the beam splitter and check how often you detect photons simultaneously on both. So you are looking for photon pairs instead of photons.

6. Dec 23, 2009

### hamster143

In quantum physics, coherent states are those that can be described by vectors in Hilbert space. That is the most complete state description allowed within the framework (unless you believe in nonlocal hidden variables).

Sometimes you don't have the complete knowledge of the system, and that means you can't use Hilbert space vectors. The fallback option is to use density matrix instead. The process of information loss that forces you to use density matrix in place of Hilbert vector is called decoherence.

7. Dec 23, 2009

### pellman

Cthugha, I 'm unclear now about what $$n$$ is.

I would very muck like to get this. I think understanding your post would be very helpful.

8. Dec 23, 2009

### Cthugha

n is just the photon number or momentary photon count rate at one detector. Accordingly $$\langle n \rangle$$ is the mean photon count rate at one detector and $$\langle n^2 \rangle$$ is the mean photon pair or simultaneous detection count rate. So if you have e.g. a mean photon count rate of 0.5 photons per millisecond on each detector and switch on the detectors for 1 ms and switch it off again and repeat this several times, you would expect a detection in 50% of those "on-times" for each of the detectors and therefore you would expect a simultaneous detection at both detectors in 25% of those "on-times".

9. Dec 24, 2009

### dede102

State said to be coherence state if expectation value of Energy operator in this state has same value as classical value.....but i still don't get it, why it is called coherence state.....

10. Dec 24, 2009

### pellman

Both laser light (as indicated by Cthugha's post) and coherent states (see the article linked in OP) have a Poisson distribution.

so is laser light an example of a quantum system in a coherent state, in the strict sense of being an eigenstate of the annihilation operator?

11. Dec 24, 2009

### Fredrik

Staff Emeritus
I don't know much about coherent states, or why they're called that. I just want to add that I've heard the term "coherent superposition" used just to emphasize that the system is really in a superposition of eigenstates, and not just in a specific but unknown eigenstate. The term "decoherence" seems to come from that usage. Decoherence is a process that turns a pure state (almost) into a mixed state (i.e. into a specific but unknown eigenstate).

12. Dec 24, 2009

### Cthugha

Yes, it is. It is well known how the annihilation operator works on a Fock state:
$$a|n\rangle=\sqrt n|n-1\rangle$$

A coherent state in optics $$|\alpha \rangle$$ is a weighted sum of Fock states:

$$|\alpha \rangle= e^{-0.5|\alpha|^2} \sum_{n=0} ^\infty \frac{\alpha^n}{\sqrt{n!}}|n \rangle$$

So

$$a|\alpha \rangle= e^{-0.5|\alpha|^2} \sum_{n=0} ^\infty \frac{\sqrt{n} \alpha^n}{\sqrt{n!}}|n-1 \rangle=e^{-0.5|\alpha|^2} \sum_{n=0} ^\infty \frac{\alpha \alpha^{n-1}}{\sqrt{(n-1)!}}|n-1 \rangle=\alpha|\alpha \rangle$$

So by changing the range of the sum you get back the original result times $$\alpha$$, which is the complex amplitude of the coherent state. Laser light is (usually) in such a coherent state as given above. There might be differences in lasers, which offer high beta-values and work with extremely low light levels like VCSELs and other microcavity lasers and also similar differences in few-atom lasers, but those are mainly of academic interest and not what is used for applications.

13. Dec 24, 2009

### pellman

Thank you very much for your illuminating posts, Cthugha.

(no pun intended!)