The traditional way to formulate the measurement of the intensity in an optical field is by using the number operator $$ \hat{n} = a^{\dagger} a . $$ In effect one can say that the intensity, which is proportional to the average number of photons in the field, is obtained from the number operator due to the bosonic enhancement associated with the detection of a single photon. If $\hat{\rho}$ is the density operator for the quantum state of the optical field, then the avergae number of photons in the field is given by $$ \langle n \rangle=\text{tr}\{\hat{n}\hat{\rho}\} . $$

There is another way one can formulate the measurement of intensity quantum mechanically. One can use projective measurements to obtain the probability to measure $n$ photons and then use it to compute the expectation value for the number of photons $$ \langle n \rangle = \sum_n n\ \text{tr}\{\hat{P}_n\hat{\rho}\} = \sum_n n\ \langle n|\hat{\rho}|n\rangle . $$

These two approaches give the same result provided that one excludes other degrees of freedom and only consider the particle-number degrees of freedom. However, when one includes spatiotemporal degrees of freedom the results differ.

The detailed calculation to show the difference is rather complicated. However, one can understand where such a difference comes from by noting that a multi-photon state would contain multiple factors (tensor products) of a single-photon wave function $\psi(\mathbf{x})$. The localized number operator will produce a factor of $|\psi(\mathbf{x})|^2$ with the expectation value for the number of particles in the state, regardless of the number of photons in the state. On the other hand, the projection operator would produce $|\psi(\mathbf{x})|^{2n}$ for every Fock state with $n$ photons in the state.

For example, consider a coherent state that is parameterized by $\alpha\psi(\mathbf{x})$, where $\alpha$ is an arbitrary complex constant. A local observation modeled by the number operator gives $$ \langle\alpha|\hat{n}(\mathbf{x})|\alpha\rangle = |\alpha\psi(\mathbf{x})|^2 , $$ while the observation modeled by the projection operator gives $$ \sum_n n \langle\alpha|\hat{P}_n(\mathbf{x})|\alpha\rangle = |\alpha\psi(\mathbf{x})|^2\exp\left(-|\alpha|^2+|\alpha\psi(\mathbf{x})|^2\right) . $$

My question is: which one of these two approaches should one use? The first approach seems to be what one usually finds in the literature, but I would like to have a deeper understanding why that is the way it should be done.

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    $\begingroup$ Can you elaborate on where you see the differences between the two schemes, given that $\hat n = \sum_n n \hat P_n$? $\endgroup$ – Norbert Schuch Oct 22 at 6:25
  • $\begingroup$ @NorbertSchuch: Please see the additional paragraph $\endgroup$ – flippiefanus Oct 22 at 10:33
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    $\begingroup$ I don't think I can see your point. (And what is wrong with the argument in my comment above?) If $P_n$ projects onto the $n$-photon sector of a specific mode (or a set of modes), and $\hat n$ counts the photons in that mode (or those modes), there will be no difference. --- Maybe it would help if you could add the simplest possible example where those differ? $\endgroup$ – Norbert Schuch Oct 22 at 14:53
  • $\begingroup$ @NorbertSchuch: please see the example. May I point out that I have worked through the analysis and found the results to be different. If you are implying that I must have made a mistake somewhere, then I would appreciate it if you can provide the correct analysis showing that when the spatiotemporal degrees of freedom are included one still obtains the same result. $\endgroup$ – flippiefanus Oct 23 at 4:27
  • $\begingroup$ What exactly is $\psi(\mathbf{x})$? Is it a first-quantized wavefunction? A second-quantized one? What is $\mathbf{x}$? --- Also, you don't explain the example - you just state results, so it is unclear how you derive them. $\endgroup$ – Norbert Schuch Oct 23 at 17:49

Actually, I figured out the answer on my own. As it turns out, the two operators have different eigenstates (which is perhaps obvious, given the observed behaviour).

To understand what the difference is between the two approaches, we'll consider a simplified scenario. We'll fix the time at a specific value $t$, and divide the space into discrete points so that we can write an arbitrary single-photon state as \begin{equation} |\psi\rangle = \sum_m |\mathbf{x}_m\rangle \psi(\mathbf{x}_m) , \end{equation} where $\psi(\mathbf{x})=\langle\mathbf{x}|\psi\rangle$ is the single-photon wave function. We should point out that the position basis does not represent position eigenstates of the photon, but simply the coordinates where a photon is detected at time $t$. (If it bothers you that I'm using position space basis elements, then you are welcome to Fourier transform everything to momentum space where everything is well-defined.) A multi-photon state with $N$ photons where all photons have the same wave function is then given by \begin{equation} |N_{\psi}\rangle = \sum_{m..p} |\mathbf{x}_m\rangle...|\mathbf{x}_p\rangle \mathcal{N}_{m..p} \psi(\mathbf{x}_m) ...\psi(\mathbf{x}_p) , \end{equation} where $\mathcal{N}_{m..p}$ is the appropriate combinatoric factor. Note that the order of the ket-vectors has no meaning, as the state would be symmetrized. Say the detector is located as $\mathbf{x}=\mathbf{x}_0$, then we see that the multi-photon state will contain terms with a varying number of photons located at $\mathbf{x}_0$, ranging from $0$ to $N$. Each term will have a complimentary number of photons located at all the other possible locations.

We'll assume that the detector has a 100% quantum efficiency. So, if a term has a certain number of photons located at the detectors, we'll assume that all those photons are detected, giving a photon-current proportional to that number of photons. All the other photons located at other points are not detected. Therefore, we can trace over them. The result is a mixed state given by \begin{equation} \hat{\rho}_N(\mathbf{x}_0)= \text{tr}_{\mathbf{x}\neq\mathbf{x}_0} \left\{ |N_{\psi}\rangle \langle N_{\psi}| \right\} = \sum_{n=0}^N |n(\mathbf{x}_0)\rangle \frac{N!}{n!(N-n)!} |\psi(\mathbf{x}_0)|^{2n} \left[1-|\psi(\mathbf{x}_0)|^2\right]^{N-n} \langle n(\mathbf{x}_0)| , \end{equation} where $|n(\mathbf{x}_0)\rangle$ is a Fock state with $n$ photons located at $\mathbf{x}_0$ and we used the normalization of the wave function to say that \begin{equation} \sum_{\mathbf{x}\neq\mathbf{x}_0} |\psi(\mathbf{x})|^2 = 1-|\psi(\mathbf{x}_0)|^2 . \end{equation} One can check that the mixed state is properly normalized.

We see that, even if the number of photons is fixed for the entire state, the number of photons detected at the location of the detector can vary. This is not what would be obtained from projection operators, unless all the photons are located at the detector. Therefore, the projection operator approach cannot give the correct result.

In general, the state would not have a fixed number of photons, but would instead be a superposition of different Fock states. Therefore, a more general situation is where the multi-photon pure state, with all photons having the same spectral function, is given by \begin{equation} |\varphi\rangle = \sum_N |N_{\psi}\rangle C_N . \end{equation} After the trace operation over this more general pure state, we obtain \begin{align} \hat{\rho}(\mathbf{x}_0) = & \text{tr}_{\mathbf{x}\neq\mathbf{x}_0} \left\{ |\varphi\rangle \langle\varphi| \right\} = \sum_N |C_N|^2 \hat{\rho}_N(\mathbf{x}_0) \nonumber \\ = & \sum_{N=0}^{\infty} |C_N|^2 \sum_{n=0}^N |n(\mathbf{x}_0)\rangle \frac{N!}{n!(N-n)!} |\psi(\mathbf{x}_0)|^{2n} \left[1-|\psi(\mathbf{x}_0)|^2\right]^{N-n} \langle n(\mathbf{x}_0)| . \end{align} One can now apply the number operator to the reduced density operator to obtain \begin{equation} \text{tr}\left\{\hat{n}(\mathbf{x}_0) \hat{\rho}(\mathbf{x}_0)\right\} = \sum_{N=0}^{\infty} |C_N|^2 \sum_{n=0}^N n \frac{N!}{n!(N-n)!} |\psi(\mathbf{x}_0)|^{2n} \left[1-|\psi(\mathbf{x}_0)|^2\right]^{N-n} = \sum_{N=0}^{\infty} |C_N|^2 N |\psi(\mathbf{x}_0)|^2 . \end{equation}

Consider, for example, the case where the state is a coherent state, so that \begin{equation} C_N = \exp\left(-\tfrac{1}{2}|\alpha|^2\right) \frac{\alpha^N}{\sqrt{N!}} . \end{equation} Then \begin{equation} \text{tr}\left\{\hat{n}(\mathbf{x}_0) \hat{\rho}(\mathbf{x}_0)\right\} = \sum_{N=0}^{\infty} \exp\left(-|\alpha|^2\right) \frac{|\alpha|^{2N}}{{N!}} N |\psi(\mathbf{x}_0)|^2 = |\alpha\psi(\mathbf{x}_0)|^2 . \end{equation} This result agrees with what I previously found with the localized number operator.

Based on this understanding, one can now propose a modified projection operator that would give the correct result. Instead of requiring all photons to be at the same location, the projection operator can be separated into two parts, one part operates on those photons that are found at the detector location and the other part operates on those that are everywhere but at the detector. Each part is associated with a specific number of photons. Since this goes beyond the question I've asked, I won't bore you with the details though.

Hope somebody finds this useful.

  • $\begingroup$ Shouldn't the wavefunction be symmetrized (i.e. the first formula where $\mathcal N$ appears)? $\endgroup$ – Norbert Schuch Oct 28 at 8:31

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