# Parametric down-conversion - QFT necessary?

In quantum optics, one ususally starts by quantizing the free electric field and obtains an expression for the electric field operators:

$$E(\vec{r},t) = \sum_{\vec{k},p} C_{\vec{k}} \vec{e}_{\vec{k},p} a(\vec{k},t,p) + h.c. \tag{1}$$

where $$e_{\vec{k},p}$$ is the polarization vector, $$a(\vec{k},t,p)$$ are the plane wave operators and $$p$$ indexes the polarization.

When studying parametric down-conversion, we use the same expression and plug it into the interaction Hamiltonian (Gerry & Knight, Appendix D)

$$H = \int d^3r \chi^{(2)}_{ijk} E_i E_j E_k \tag{2}$$

But how can this be justified given that (1) was derived for free fields without polarizations in the materia?

The closest answer I have found in literature is in the book "The quantum theory of nonlinear optics" by Drummond and Hillery, where the derivation is done starting from Maxwell's equations. However, the resulting expressions do not look the same and are more "like quantum field theory". Can you help me with this confusion?

Is there a way to justify combining (1) and (2) this way?

See also for example M. Hillery, "An Introduction to the Quantum Theory of Nonlinear Optics", arXiv:0901.3439.

• Parametric down conversion is an atomic/molecular spectrum effect, is it not? It's awfully hard to calculate that with relativistic QFT from first principles, so at most we are doing either effective field theory here or we would do better writing some explicit electronic states of atoms or molecules in there in an ad-hoc perturbation approach. Whether we want to call this "QFT" is probably a matter of taste, especially since the optical materials break relativity (Lorentzian and Galilean relativity because of momentum non-conservation). Commented Jul 20 at 23:07
• Where is the parameter in your Eq.(2)? After all parametric downconversion depends on a parameter… Commented Jul 20 at 23:33

Your approach is correct. The only issue is: why can we do it? Standard practice in QFT is to quantize the fields as free fields and then add the interactions perturbatively, assuming the couplings are weak. Fortunately, the coupling or nonlinear coefficient in parametric down-conversion is extremely weak. Therefore, the perturbative approach is well justified.

Perhaps you can ask how we can get significant down-conversion if the coupling is so weak. The efficiency of the down-conversion is increased by bosonic enhancement due to the large number of photons in the pump field. That allows us to use another simplification, namely to replace the pump field by a classical field, leading to a semi-classical approximation which works very well in most cases.

Hope that helps.

• Thanks, very helpful! Do you perhaps know some literature which discusses this comparison? Because all I could find was Drummond and Hillery saying that the approach is "incorrect" (or rather "should be avoided"). Commented Jul 21 at 10:04
• Could you perhaps provide a link to that paper? The literature on parametric down-conversion is very large. Commented Jul 22 at 4:03
• I have added the literature I have to my questions. The first two sources I mentioned were books, but I found a paper which contains the main part of the book. Commented Jul 22 at 9:47
• See especially p. 59 in the cited arxiv paper Commented Jul 22 at 9:55
• So I had a look at the paper. It is trying to derive Maxwell's equations within the nonlinear medium and, predictably, comes with a mess, partly because the expressions that only apply for the free field theory part and those that also involve the interactions are mixed in an inconsistent way. The perturbative nature of the process is completely ignored. It does not present a credible argument. Commented Jul 25 at 4:03

The problem of using the free fields inside a dielectric has been discussed elsewhere as well, for example in 'Why you should not use the electric field to quantize in nonlinear optics' by Quesada and Sipe. They present a simple and general argument for why Faraday's law does not hold when quantizing the dielectric Hamiltonian using the free field.

In this case, the nonlinear Hamiltonian of order $$N$$ will be some $$N+1$$ order polynomial of the creation and annihilation operators:

$$H = \mathrm{poly}_{N+1}(\hat{a},\hat{a}^{\dagger})$$

while the electric field is (by construction) a first order polynomial of these operations, and in a non-magnetic material the magnetic field is as well. Faraday's law is

$$\tag{1} \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}.$$

Using Heisenberg's equation of motion the r.h.s. can be written as

$$\frac{\partial \mathbf{B}}{\partial t} = -\frac{i}{\hbar}[\mathbf{B},H] = [\mathrm{poly}_{1}(\hat{a},\hat{a}^{\dagger}),\mathrm{poly}_{N+1}(\hat{a},\hat{a}^{\dagger})]$$

Using the relations $$[\hat{a},f(\hat{a},\hat{a}^{\dagger})] = \frac{\partial f}{\partial \hat{a}}$$ and $$[\hat{a}^{\dagger},f(\hat{a},\hat{a}^{\dagger})] = -\frac{\partial f}{\partial \hat{a}^{\dagger}}$$ we can infer that

$$\frac{\partial \mathbf{B}}{\partial t} = \mathrm{poly}_{N}(\hat{a},\hat{a}^{\dagger})$$

but this is in contradiction with

$$\nabla \times \mathbf{E} = \mathrm{poly}_{1}(\hat{a},\hat{a}^{\dagger})$$

and thus (1) does not hold.

In practice this leads to several problems. A straightforward one, pointed out by Hillery, is that if you go through the trouble of deriving the wave equation for the free fields inside the dielectric you get

$$\nabla^2 \mathbf{E} = \frac{1}{n^2c^2}\frac{\partial^2 \mathbf{E}}{\partial t}$$

and the solutions to this equation are waves that propagate at $$nc > c$$. As discussed by Quesada et al. there are also problems in nonlinear interactions, for example in nonlinear self-interactions.

When is the approach using the free field justified? One could say "never", since it is simply wrong, but people still use it to get correct results. One reason for this is that parametric down-conversion is often described in the interaction picture, and the time-dependence of the field operators is usually introduced explicitly. For example, Loudon (2000) introduces the electric-field operator

$$\hat{E}^+ = i\int_0^{\infty} \left(\frac{\hbar\omega}{4\pi\epsilon_0 c A \eta(\omega)} \right)^{1/2} \hat{a}\ \mathrm{exp}[-i\omega t + ik(\omega)z].$$ This operator corresponds to waves propagating at the correct speed by construction, and the problem of the Hamiltonian actually generating a different time-evolution is swept under the rug. As for the problems with nonlinear interactions, these don't really appear in the low-gain regime of PDC, and it's worth keeping in mind that most perturbative approaches to PDC completely ignore time ordering as well (which does not work when self-interaction becomes significant).

Finally, regarding your question about the QFT formulation, it is not actually necessary to start from a Lagrangian density and quantize the field "properly". An alternative approach is presented in the resources you linked. The correct approach is to quantize the $$D$$-field instead of the $$E$$-field. The reason for this is that the $$E$$-field is not fundamental inside a dielectric medium, and in a macroscopic approach to quantization the fundamental excitations inside the medium are in fact combinations of light and matter excitations.

Hillery goes through the steps of how to re-write the nonlinear dielectric Hamiltonian in terms of the $$D$$-field and the inverse susceptibilities $$\beta$$, though this process is a bit tedious. After introducing the appropriate creation operator, you end up with the field operators

$$\mathbf{D}(\mathbf{r},t) = \sum_{\mathbf{k},\alpha} i\sqrt{\frac{\hbar \epsilon_0 \omega_k }{2 V}} \hat{e}_{\mathbf{k},\alpha} \Big( \hat{a}_{\mathbf{k},\alpha}(t) e^{i\mathbf{k}\cdot \mathbf{r}} - \hat{a}^{\dagger}_{\mathbf{k},\alpha}(t) e^{-i\mathbf{k}\cdot \mathbf{r}}\Big)$$

$$\mathbf{B}(\mathbf{r},t) = \sum_{\mathbf{k},\alpha} i\sqrt{\frac{\hbar}{2 \epsilon_0 \omega_k V}} (\mathbf{k}\times\hat{e}_{\mathbf{k},\alpha}) \Big( \hat{a}_{\mathbf{k},\alpha}(t) e^{i\mathbf{k}\cdot \mathbf{r}} - \hat{a}^{\dagger}_{\mathbf{k},\alpha}(t) e^{-i\mathbf{k}\cdot \mathbf{r}}\Big) \\$$

If you take the corresponding free Hamiltonian $$H = \int \left(\frac{1}{2\mu_0}\mathbf{B}^2 + \frac{1}{2}\beta^{(1)}\mathbf{D}^2 \right)dr^3$$

and apply Heiseberg's equation of motion, you will see that the $$D$$-field obeys

$$\nabla^2 \mathbf{D} = \frac{n^2}{c^2}\frac{\partial^2 \mathbf{D}}{\partial t^2}$$

and the dynamics are therefore correct. In conclusion, you can use the "standard" description by simply choosing to work with the displacement field instead of the electric field. You do also see this approach taken in some of the literature. However, when working in the interaction picture with the field operators introduced by hand, the answer for most experimentally relevant quantities are the same.

• The inconsistencies that you demonstrate is a consequence of the assumptions that you can perform the calculations in the same way that you would do it in standard EM theory and QM without the nonlinearity. Obviously Faraday's law etc. won't work as before. Now with quantizing $\mathbf{D}$ in the nonlinear medium the question is what part of the nonlinearity has been taken care of in the quantization process and what part do you still need to add with interactions? So it leads to ambiguities whenever you do not quantize the field as a free field. Commented Jul 24 at 3:50
• There are two interactions here, one is between the photons and the medium, and the other is between the photons (p-p). In a macroscopic approach to quantization the interaction with the medium is treated as an effective nonlinear susceptibility, and the p-p interaction is described perturbatively. This is true regardless of which field you quantize. When using the free field, however, you get wrong predictions ($v>c$) even for a linear medium (no p-p interaction). When quantizing $D$, the nonlinear part (p-p) is not included, it is accounted for by adding nonlinear terms to the Hamiltonian. Commented Jul 24 at 9:47
• There is only one interaction. There are no interactions between photons that are not mediated by the nonlinear medium. Since D=E+P, it already includes the polarization which include the nonlinear effects. When you quantize the theory as a free field you should obvious use the correct dispersion relation. Then you will the correct speed of light. Commented Jul 25 at 3:43
• "Using the correct dispersion relation" means including the linear response of the medium, as that is what causes the dispersion. In other words you're doing $E\to E+P^{(1)}$. This solves the problem of "free" propagation, but, as discussed in detail in the various resources linked within this question, still results in wrong predictions for nonlinear effects. Commented Jul 25 at 10:08