In Spontaneous Parametric Down Conversion, incoming photons are split into entangled photon pairs by passing through a non-linear optical medium. Since I get a photon pair as an output then how can I say that I am producing single photons?

  • $\begingroup$ Photons are the quanta of the electromagnetic field. It doesn't matter how many photons a state contains (or if the photon number is not even well defined, as in case of thermal radiation), one can always interact with the field by measuring individual photons. $\endgroup$
    – CuriousOne
    Commented Feb 24, 2016 at 6:44
  • $\begingroup$ Then what does "single photon" mean physically? $\endgroup$ Commented Feb 24, 2016 at 9:24
  • $\begingroup$ It means that we detect a quantum of energy with no rest mass, no electric charge, a spin of 1 and without a color charge. If enough of these quanta can be measured, we can recover a classical electromagnetic wave. For a thermal emitter the number of photons in a given volume is not a well defined number, i.e. we can get a distribution of none, one, or two etc. photons. For coherent emitters this distribution will be modulated in time, but we still can't be sure that we have exactly one photon. SPDC can be used to create a pair and one of the photons then indicates the existence of the other. $\endgroup$
    – CuriousOne
    Commented Feb 24, 2016 at 9:50

1 Answer 1


There are indeed some subtle but important differences between an SPDC-based heralded source and a true single photon source. In order to understand these differences, consider what a true single-photon source really means.

A true single-photon source emits a single excitation at a specified frequency when demanded. So mathematically, the output state would be a true Fock state with photon number $n=1$. The probability that the detector would record any other number ($n=0$ or $n=2$ etc.) of photons in an instant is exactly zero. Thus, the probability $p(n)$ that $n$ photons would be detected at a given instant in a detector takes the value unity at $n=1$ and zero for all other values of $n$, implying that the variance $\Delta n$ is zero.

Lets contrast this with a weak coherent source, which is routinely used as an approximation to a single-photon source. A weak coherent source is obtained by attenuating the output from a laser to the point where the mean number of photons per second becomes unity, i.e, $\bar{n}=1$. However, such a source differs in a very important way from a true single-photon source. This is because the light has been attenuated in a way that does not alter the statistical distribution of the photons. The photon statistics of a weak coherent source remain Poissonian, i.e $(\Delta n)^2=\bar{n}$, which is in sharp contrast with the highly sub-Poissonain statistics of a true single photon source with $\Delta n=0$.

The experimental consequences of this difference can be understood by looking at the photon number distribution of a weak coherent source, which is just be the Poisson distribution with $\bar{n}=1$. The salient feature of this distribution is that $p(0)$ takes on a large value, followed by a small value for $p(1)$, and exponentially smaller values for the rest. Clearly, this distribution is sharply different from that of a true Fock state for $n=1$. The large probability $p(0)$ that no photon is detected implies that in an experiment, a large number of detection time windows are wasted without an arrival of a photon. As a result, a large error may be introduced in the measurement due to dark counts.

Now in order to avoid this problem, quite often a single photon source is simulated by means of what is called a heralded single photon source based on SPDC. The idea here is to utilize the strong arrival time correlations of the photon pairs produced from SPDC, which means that one would keep the detector active only conditional to a separate detection of the other photon in the pair. It is then hoped that the ensemble of measurements would approximate the ensemble of measurements from a true single photon source.

However, there are problems with this claim as have been systematically analyzed in Phys.Rev. A 90, 053825 (2014). The essential point becomes clear if you consider the process of SPDC more closely. The Hamiltonian for the process is of the form $H=\epsilon_{0}\chi^{(2)}\hat{a}_{p}\hat{a}^{\dagger}_{s}\hat{a}^{\dagger}_{i}+ c.c$, where the first term signifies the SPDC forward process involving annihilation of the pump photon and the creation of the signal and idler photons, and the second term denotes sum frequency generation, which ensures hermiticity of the Hamiltonian. The time evolved output state $|\psi\rangle$ would then be of the form,


, where the labels $s,i$ denote the signal and idler modes and the pump has been treated as a classical field. Clearly, this output state from SPDC is not just a two-photon Fock state. In fact, the efficiency of the SPDC process is very low (typically less than $10^{-8}$) and consequently, the probability $|c_{0}|^2$ that no photons are produced is almost unity. This is followed by a small probability $|c_{1}|^2$ that a pair of photons is produced, which we use to produce a heralded source. What is crucial however, is that the higher order terms $|c_{2}|^2$, etc. while being small, are certainly not zero. It is precisely these higher order terms that prevent a heralded SPDC-based source from accurately simulating true single-photon source.

Finally, I note that the design and implementation of a true single photon source remains an experimental challenge to this day, and several groups are working actively on this.

  • $\begingroup$ There are no pea-shooters for photons. $\endgroup$ Commented Jun 8, 2019 at 0:56

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