Quantum Optics and Coherence Functions

[Niles]

Hi

Classically, we can interpret the coherence functions (1st order) by using the definition of visibility, i.e., when there is complete coherence, the visibily of a fringe is maximum, and when the is complete incoherence, the visibility of a fringe is minimum.

Can one make the same interpretation of the QM coherence functions?


Niles.

 

[Cthugha]

That depends on which order we are talking about.

When changing from classical field correlation functions to quantum field correlation functions, the main change lies in the fact that the field operators do not necessarily commute. However, this is not important for first order coherence as you only have two operators which do not have to be reordered. Therefore, first-order coherence functions are in almost all respects the same in classical and quantum treatments. First-order coherence also does not carry any information about whether the light field is classical or non-classical.

Such information is only contained in higher-order correlation functions. Here you have at least for operators. As measurements of these functions usually measure the normal-ordered correlation functions, the field operators need to be reordered to relate them to photon numbers. In this reordering processes, the differences between classical and quantum correlation functions become apparent (basically the difference is that the detection of a photon destroys it and therefore alters the light field itself).
However, these functions do not relate to the visibility of some interference pattern. They relate to the presence (or not) of deviations from statistical independence of photons.

 

[Niles]

Thanks, that was a very good answer. I have another question related to correlation functions (I wrote coherence in the topic, but meant correlation). In my book, they say that if we have the (classical) first order correlation function

$$\gamma ^{(1)} (x_1 ,x_2 )\frac{{\left\langle {E^* (x_1 )E(x_2 )} \right\rangle }}{{\sqrt {\left\langle {|E(x_1 )|^2 } \right\rangle \left\langle {|E(x_2 )|^2 } \right\rangle } }}$$
then we have complete coherence if the numerator can be written as

$$\left\langle {E^* (x_1 )E(x_2 )} \right\rangle = \left\langle {E^* (x_1 )} \right\rangle \left\langle {E(x_2 )} \right\rangle$$

I cannot see why that is the case, since we in the denominator have amplitudes?Niles.

 

[Cthugha]

You can assume to have fields of the kind (I assume the same frequency for simplicity):
$$E_i(x,t)=\epsilon_i e^{i \omega t} e^{i \phi_i}$$.
Then you will get a $$\gamma^{(1)}$$ that looks like:
$$\frac{\langle \epsilon_1 \epsilon_2 e^{i(\phi_2 -\phi_1)} \rangle}{\epsilon_1 \epsilon_2}$$.

Whether or not the numerator factorizes depends on the phase difference term. For ideally first-order coherent light, this phase difference willl always be the same. In this case the numerator factorizes. The degree of first-order coherence in this case is in some sense a quantity describing how well you can predict the phase at position B given you know the phase at position A.

If the fields at the two positions are completely independent, both phases will fluctuate somewhat randomly, leaving the phase difference to be also totally random and average out to zero. As the phase difference changes constantly, the numerator does not factorize and you are not able to predict the phase at position B by knowing it at position A.

Therefore, you get basically:
$$\gamma^{(1)}=\langle e^{i (\phi_2-\phi_1)} \rangle$$
which amounts to the well known
$$\left| \gamma^{(1)}\right|=1$$ if the phase difference is fixed
and
$$\left| \gamma^{(1)}\right|=0$$ if the phase difference is random.

 

[Niles]

Nice, thank you very much indeed. I just started studying correlation functions, and it was somewhat abstract in the beginning. It is getting better now. I hope you don't mind that I ask some more questions. You seem quite experienced.

In my book it says that the function

$$G^{(1)} (x,x) = Tr\left[ {\hat \rho \hat E^{( - )} (x)\hat E^{( + )} (x)} \right]$$

is the intensity at the space-time point x=(r, t). Here rho is the density matrix, Tr denotes a trace and E are the E-field operators (- for negative frequency, and + for positive). In my book they merely state that the above is the intensity at the photodetector, after showing that the above is proportional to the probability of photodetection. How do they suddenly make that jump?Niles.

 

[Cthugha]

I just started studying correlation functions, and it was somewhat abstract in the beginning. It is getting better now. I hope you don't mind that I ask some more questions. You seem quite experienced.

I had the pleasure/misfortune to measure them for a while. Whether it was a pleasure or misfortune depended on how ell the experiments went. ;)

How do they suddenly make that jump?

I am not quite sure I get which jump you mean, so I will try to explain that stuff in my own words. If I miss your point, please ask again.

$$G^{(1)} (x,x)$$ for equal coordinates is always the intensity at that coordinate.
$$\hat E^{( - )} (x)$$ and $$\hat E^{( + )}$$ are proportional to the photon creation and annihilation operators at the corresponding coordinates. Using the density matrix and the trace is then necessary to sum over all states if the field is not in a pure state.

Basically, $$G^{(1)} (x,x)$$ is the correlation between a field and itself at the same coordinate. Accordingly the amplitudes have to be the same and there can be no phase difference. Putting this into the definition gives you just the intensity.

 

[Niles]

Basically, $$G^{(1)} (x,x)$$ is the correlation between a field and itself at the same coordinate. Accordingly the amplitudes have to be the same and there can be no phase difference. Putting this into the definition gives you just the intensity.

I see, but I guess G⁽¹⁾(x,x) is not the intensity, but something proportional to the intensity (otherwise, I can't get the units to match)?

Furthermore, in my book (Gerry and Knight), they find that the probability of the field undergoing a transition from some state i to some state f summed over all possible f for a general (not necessarily pure) state is given by

$$Tr\left[ {\hat \rho \hat E^{( - )} (x)\hat E^{( + )} (x)} \right]$$

I interpret this as the probability of photodetection, since if the field changes state, so has our atom (= detector). Afterwords, they claim that this is proportional (or equal, I guess it is not super important which one it is in this case) to the intensity. So is it correct to conclude that the intensity at the detector is proportional to the probability of photodetection?Niles.

 

[A. Neumaier]

is it correct to conclude that the intensity at the detector is proportional to the probability of photodetection?

Yes. Photodetection is simply a noisy, binary measurement of the intensity, which becomes more accurate the more detection events are recorded.

 

[Cthugha]

I see, but I guess G⁽¹⁾(x,x) is not the intensity, but something proportional to the intensity (otherwise, I can't get the units to match)?

Usually they are identical. I do not know the definition used in the Gerry/Knight, so I have to read it tomorrow, but usually it is the intensity (given in some photon number per unit volume and/or time).

I interpret this as the probability of photodetection, since if the field changes state, so has our atom (= detector). Afterwords, they claim that this is proportional (or equal, I guess it is not super important which one it is in this case) to the intensity.

Well, this is discussed to some extent in the Milburn/Walls for example.
The transition probability and the intensity are equal for ideal detectors. Then the transition probability is given by:
$$T_{if}=\left|\langle f|E^{(+)}(r,t)|i \rangle\right|^2$$
The total counting rate is then giving by summing over all final states and is identical to the intensity as can be easily seen:
$$I(r,t)=\sum_f T_{fi}=\sum_f \langle i|E^{(-)}(r,t)|f \rangle \langle f|E^{(+)}(r,t)|i \rangle$$
which is equal to
$$\langle i|E^{(-)}(r,t) E^{(+)}(r,t)|i \rangle$$

If you do not assume pure states, but arbitrary ones, you just need to add the density matrix and the trace and arrive at the equation you gave.

 

[Niles]

Thanks guys, good answers. I have a final question for this round. This is regarding 2nd order (quantum) correlation functions. For coherent light, the photons arrive at the detector with random time intervals, we have g⁽²⁾=1 for all times.

If I e.g. look at collisional broadened light with a Lorentzian spectrum, then I will find

$$g^{(2)} = 1+e^{-2|\tau|/{\tau_0}}$$

where tau₀ is the collisional (=coherence) time. For small time intervals, this is seen to be greater than 1, meaning it is bunched light. Around t=tau₀, the correlation function has a value close to 1. Is it correct to interpret this that since the light is coherent for t<tau₀, then photons tend to come in pairs for these time intervals? I.e., that the photons are bunched for small times in this case is just because the light is coherent for these times?Niles.

 

[Cthugha]

Is it correct to interpret this that since the light is coherent for t<tau₀, then photons tend to come in pairs for these time intervals? I.e., that the photons are bunched for small times in this case is just because the light is coherent for these times?

Well, it is difficult to find a good terminology without ambiguities, but your statement is in principle correct. However, "the light is coherent for these times" can be a bit misleading as every kind of light is first-order coherent to some extent, so it will have a certain coherence time, but the light you are talking about is not second- and higher-order coherent. But it is correct to say that photon bunching effects arising due to light being thermal and not second-order coherent will only show up for times smaller than the (first-order) coherence time of the light. For large time delays photon bunching will disappear.

 

[Niles]

But it is correct to say that photon bunching effects arising due to light being thermal ... will only show up for times smaller than the (first-order) coherence time of the light. For large time delays photon bunching will disappear.

I have been thinking about this statement for some time, and I cannot quite see why the following is the case: Why is it only for times satisfying t<t_coherence that photons (from a thermal source) tend to bunch together? There seems to be some link between coherence and bunching that I haven't seen.

 

[Cthugha]

Strictly speaking we are not discussing absolute times, but delay times $$\tau$$.

If you detect a photon now, the probability to detect another one simultaneously ($$\tau$$=0) is enhanced by a factor of two which is the bunching effect. The probability to detect another one five minutes later (large $$\tau$$) should show such an increase. It should be independent on the earlier detection.

Another possible viewpoint is to connect the bunching process to the photon statistics and noise properties. The photon number distribution of thermal light is very noisy and you have large fluctuations around the mean. So if you detect a photon now, the probability is quite high that the momentary intensity is above the mean value. Therefore it is also very probable to detect another one shortly thereafter. If you do not detect a photon, the probability that the momentary intensity is way below the mean intensity is quite high and the probability to detect one shortly thereafter will also be quite small. However, the momentary photon number fluctuates and randomizes on some timescale on the order of the coherence time, so that the bunching effect vanishes for large delays.

 

[Niles]

Strictly speaking we are not discussing absolute times, but delay times $$\tau$$.

If you detect a photon now, the probability to detect another one simultaneously ($$\tau$$=0) is enhanced by a factor of two which is the bunching effect. The probability to detect another one five minutes later (large $$\tau$$) should show such an increase. It should be independent on the earlier detection.

Another possible viewpoint is to connect the bunching process to the photon statistics and noise properties. The photon number distribution of thermal light is very noisy and you have large fluctuations around the mean. So if you detect a photon now, the probability is quite high that the momentary intensity is above the mean value. Therefore it is also very probable to detect another one shortly thereafter. If you do not detect a photon, the probability that the momentary intensity is way below the mean intensity is quite high and the probability to detect one shortly thereafter will also be quite small. However, the momentary photon number fluctuates and randomizes on some timescale on the order of the coherence time, so that the bunching effect vanishes for large delays.

The highlighted part is what made me understand it. Thank you for your help, a very good answer indeed.


Niles.

 

[Niles]

I guess I have a final question

Will the second order correlation function for a QM field always go to 1 for large delay times? In the books I am reading, it isn't shown explicitly, but only for thermal fields with Lorentzian spectra.

 

[Cthugha]

In theory or in reality?

The correlation function returns to 1 on the timescale of the coherence time. A thermal light field with infinite coherence time would never return to one. However, this is more or less a theoretical consideration. In reality any thermal light source has finite coherence time.

 

[Niles]

In theory or in reality?

The correlation function returns to 1 on the timescale of the coherence time. A thermal light field with infinite coherence time would never return to one. However, this is more or less a theoretical consideration. In reality any thermal light source has finite coherence time.

I meant theory. But can one show rigorously that it will always return to 1 for large delays, or is that more of a definition?

 

[Cthugha]

Well, for classical thermal light one can derive the Siegert relation stating that:
$$g^{(2)}(\tau)=1+(g^{(1)}(\tau))^2$$.

Basically it all depends on the coherence time of your emission. If you had a completely single mode thermal field, you would get a value of 2 for all times. In practice you can always increase the coherence time of your light source by spectral filtering. If you use a narrow bandwidth spectral filter, the light passing through it will have longer coherence time, but still be thermal. You can do the same for spatial coherence if you place the source very far away from your detectors. This is why the HBT experiment on light from Sirius B worked. This is by the way also the reason why second-order coherence is a better criterion to distinguish different light sources than first-order coherence alone. You can always increase the coherence time by filtering your signal, but you usually do not alter second-order coherence by such processes unless you scatter your light.

 

[Niles]

You say that "the correlation function returns to 1 on the timescale of the coherence time". For single mode thermal light, we have that g⁽²⁾(tau) = 2 for all delays. This makes sense, since it has infinite coherence time, so it never returns to 1. For a single mode coherent field we have g⁽²⁾(tau) = 1 for all delays. This is 1 for all times. So this brings about the question: Why is it more probable to detect coincident photons of a single mode thermal light than in coherent light?

Also, I cannot see how it even make sense to have g⁽²⁾(tau) > 1 for some fields. Wasn't g⁽²⁾(tau) a probability?Niles.

 

[Cthugha]

Why is it more probable to detect coincident photons of a single mode thermal light than in coherent light?

This is a consequence of the noise (variance of the underlying photon number distribution. You can describe the momentary photon number as $$\langle n \rangle +\Delta$$, the sum of the mean photon number and a fluctuation. After reordering the operators, $$g^{(2)}(0)$$ is defined as:
$$g^{(2)}(0)=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}=\frac{\langle \langle n \rangle+\Delta (\langle n \rangle +\Delta-1) \rangle}{\langle n \rangle^2}$$
which is
$$\frac{\langle \langle n \rangle^2 + 2 \Delta \langle n \rangle \Delta^2 -\langle n \rangle -\Delta \rangle}{\langle n \rangle^2}$$
The expectation values of all terms linear in $$\Delta$$ vanish as the expectation value of the deviation from the mean value must be zero on average. What is left is:

$$g^{(2)}(0)=\frac{\langle n\rangle^2 +\langle \Delta^2 \rangle -\langle n \rangle}{\langle n \rangle^2}=1-\frac{1}{\langle n \rangle }+\frac{\langle \Delta \rangle^2}{\langle n \rangle^2}$$

Now the interesting term is $$\langle \Delta^2 \rangle$$ which is the variance of the photon number distribution. For coherent light this is a Poissonian distribution, while it is a Bose-Einstein distribution for thermal light. Now the variances of these distributions are well known. You get $$\langle \Delta^2 \rangle=\langle n\rangle$$ for a Poissonian distribution and $$\langle \Delta^2 \rangle=\langle n \rangle^2 +\langle n\rangle$$ for the Bose-Einstein distribution. Inserting those into the equation gives the well-known values of 1 and 2, respectively.

Also, I cannot see how it even make sense to have g⁽²⁾(tau) > 1 for some fields. Wasn't g⁽²⁾(tau) a probability?

It is a relative or normalized probability. In detail, it is the probability to detect a secoond photon after a time delay $$\tau$$ after the first photon was detected, compared to the same probability if the photons were emitted statistically independent of each other. In other words: If you take some snapshots (integration time much shorter than the coherence time) and count the total photon number inside them and get on average one photon every 6 pictures, you would expect to detect a pair of photons every 36 pictures on average. If you have the same mean photon count rate, but detect a pair every 18 pictures on average (and accordingly more pictures without any photons inside), you have a twofold increase which would be the signature of thermal light.

 

[Niles]

Thanks, that was very kind of you. I feel really confident now! Now, on to first order But they seem somewhat less abstract.

Best (and thanks again),
Niles.