In this paper it is explained how an SPDC source can have its emission inhibited by destructive interference with emission from a previous time.

I have had a (maybe incorrect?) general intuition that destructive interference of wavefunctions can never make things "disappear." For instance, while destructive interference of a photons' possible paths can cause a particular path to vanish, the photon itself can only have its pathway altered. (Likewise in two-photon interference, like HOM interference, the output possibilities are interfering destructively)

In this example (as illustrated from the picture from the paper below), mirrors are set up such that photon pairs created by SPDC interfere upon reflection with a subsequent spdc event.

It appears as though when two creation events are prepared to have opposite phase-relations, the creation event is inhibited, as shown here: $$|\Psi\rangle = \alpha|1\rangle_{s_1}|1\rangle_{i_1} + e^{\mathrm{i}\phi_p}\alpha|1\rangle_{s_2}|1\rangle_{i_2}, $$

When the photons pairs are made to be identical, then these events are seen to destructively interfere. $$ I_i = I_s = 2I_0 \{ 1 + \cos(\phi_i+\phi_s-\phi_p) \},$$ where $I_0\propto |\alpha|^2$ is the rate of photon emission into either mode without mirrors present.

My question is: Where does the probability "go" in this case? In the case of a quantum mach-zender interferometer, for instance, to have a wavefunction interfere with itself you need to construct a unitary operator in which probability amplitudes can split between different pathways. When there is destructive interference, there's constructive interference in another pathway. So where is the constructive interference that's happening in parallel in this case?

Maybe other "leftover" modes are enhanced (in other spatial modes, for example) enhanced as a result of this inhibition due to interference?

Or maybe the total state could be expressed in a form describing a larger system, such as: $|\text{pair creation}\rangle + |\text{no pairs}\rangle$ And something constructive is happening there?

  • $\begingroup$ If you explain SPDC and HOM it would make the question more self-contained. $\endgroup$
    – my2cts
    Sep 29, 2019 at 10:43
  • 1
    $\begingroup$ 'destructive interference of wavefunctions can never make things "disappear." ' Correct but it will prevent things from appearing in the first place. $\endgroup$
    – my2cts
    Sep 29, 2019 at 10:45
  • $\begingroup$ I think I found this on arxiv, arxiv.org/abs/1909.03513 ( I do not have easy access to phys rev). It is not my field , it needs a quantum optics specialist, but I have found very useful in accepting the behavior of photons in interference the Mit video youtube.com/watch?v=RRi4dv9KgCg . It explains for the shown experimental setup where the energy goes (let alone the probabilities) : one has to consider the whole setup , the energy goes back to the laser. In this case crystal and mirrors and what ever generates the waves,, should be in one quantum mechanical solution. $\endgroup$
    – anna v
    Sep 29, 2019 at 10:52

1 Answer 1


SPDC sources create states that contain mostly vacuum, and a small amount of light. The full output state of an SPDC source is:

$$ |\Psi\rangle = \mathrm{exp}[-\frac{i}{\hbar}(\kappa \hat{a}_s^{\dagger}\hat{a}_i^{\dagger}+ \kappa^* \hat{a}_s\hat{a}_i)] = \sum_n \frac{1}{(i\hbar)^n}\frac{(\kappa \hat{a}_s^{\dagger}\hat{a}_i^{\dagger}+ \kappa^* \hat{a}_s\hat{a}_i)^n}{n!}|0,0\rangle_{s,i} $$

Here $\hat{H} = \kappa \hat{a}_s^{\dagger}\hat{a}_i^{\dagger}+ \kappa^* \hat{a}_s\hat{a}_i$ is the interaction Hamiltonian of the process, $\kappa$ is a complex number representing the classical pump field, and the second term corresponds to sum-frequency generation (SFG). Usually the sum is truncated after the first non-vacuum term. Physically this is motivated when the probability of generating more than a single photon pair is very low, and much smaller than the probability of generating a single pair. The SFG part of the Hamiltonian acting on vacuum produces no output, and therefore the truncated sum is:

$$ |\Psi\rangle \approx \alpha_0 |0,0\rangle_{s,i} + \alpha_1 |1,1\rangle_{s,i}, $$

with $|\alpha_0|^2 >> |\alpha_1|^2$. The paper you referenced did not explicitly write out this vacuum term. If we once again neglect higher order terms, then the state after the second crystal reads:

$$ |\Psi\rangle \approx \alpha_0 |0,0\rangle_{s,i} + \alpha_1 |1,1\rangle_{s,i} + e^{i\phi}\alpha_1 |1,1\rangle_{s,i}. $$

Here we've assumed that $\alpha_0 \approx 1$, so that it is left unchanged by the down-conversion process. If $\phi = \pi$ then the photon-pair terms cancel, and the probability "goes" back to the vacuum. While that is indeed where the probability goes, this description is clearly flawed since the amplitude $\alpha_0$ doesn't depend on the phase $\phi$, i.e. whether the two down-conversion process actually cancel or not. In practice we'd also have to include the SFG process $|1,1\rangle_{s,i} \to |0,0\rangle_{s,i}$, and all higher order combinations of SPDC and SFG processes. It is these terms that contain the constructive interference with the $|0,0\rangle_{s,i}$ term. The problem is that we're trying to add two infinite sums that are truncated.

One way to see that the two emission processes do in fact cancel perfectly is by considering two different interaction Hamiltonians that only differ by a phase:

$$ \hat{H}_1 = \kappa \hat{a}_s^{\dagger}\hat{a}_i^{\dagger}+ \kappa^* \hat{a}_s\hat{a}_i,\quad \hat{H}_2 = e^{i\phi}\kappa \hat{a}_s^{\dagger}\hat{a}_i^{\dagger}+ e^{-i\phi}\kappa^* \hat{a}_s\hat{a}_i. $$

Setting $\phi = \pi$ we have $\hat{H}_2 = -\hat{H}_1$ and clearly $[\hat{H}_1,\hat{H}_2] = 0$. We can therefore write the state after the two crystals as:

$$ |\Psi\rangle = e^{-\frac{i}{\hbar}\hat{H}_2}e^{-\frac{i}{\hbar}\hat{H}_1}|0,0\rangle_{s,i} = e^{-\frac{i}{\hbar}(\hat{H}_2+\hat{H}_1)}|0,0\rangle_{s,i} = e^0|0,0\rangle_{s,i} = |0,0\rangle_{s,i}. $$

More generally, this type of interference between two nonlinear processes corresponds to an $\mathrm{SU}(1,1)$ interferometer, and is better analyzed in the Heisenberg picture. In this type of interferometer the beam-splitters are replaced by the nonlinear process through which the modes couple. Such an interferometer transforms the creation operators $\hat{a}_0$ and $\hat{b}_0$ of the two input modes as:

$$ \hat{a}_2 = a\hat{a}_0 - b\hat{b}_0 $$ $$ \hat{b}_2 = e^{i\theta}(a\hat{b}_0 - b\hat{a_0}), $$ where $\theta$ is the phase difference between the two modes $a,b$ acquired between the two processes, and where

$$ a=\cosh r_1 \cosh r_2 + e^{i(\phi_2-\phi_1-\theta)}\sinh r_1 \sinh r_2 $$ $$ b=e^{i\phi_1}\sinh r_1 \cosh r_2 + e^{i(\phi_2-\theta)}\cosh r_1 \sinh r_2. $$

Here $r_k e^{i\phi_k}$ is the squeezing parameter of the corresponding process. Two identical down-conversion processes that are out of phase correspond to setting $\theta = 0$, $\phi_2 - \phi_1 = \pi$ and $r_1 = r_2$, and doing this gives:

$$ a = \cosh^2 r - \sinh^2 r = 1 $$

$$ b = e^{i\phi_1}(\sinh r \cosh r - \cosh r \sinh r) = 0, $$ and hence $\hat{a}_2 = \hat{a}_0$, $\hat{b}_2 = \hat{b}_0$, meaning that the input state is the same as the output state. In the case of SPDC this is just the vacuum state. For a reference see:

Effect of losses on the performance of an SU(1,1) interferometer


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