Quantum linear optics Boson sampling

I am currently taking a course in Quantum Optics and I am trying to understand Boson sampling. We were given a formula for Boson sampling: if the input is given by: $$|{\psi_{in}}\rangle=a_1^\dagger a_2^\dagger...a_n^\dagger|{0}\rangle$$ then the output is: $$|{\psi_{out}}\rangle=\sum_{k_1,s_1;...k_n,s_n; k_1+k_2+...k_n=N}\frac{\text{Perm}(\hat{U}_{k_1s_1,...k_ns_n})}{\sqrt{k_1!...k_n!}}|k_1,...k_n\rangle$$ where $$\hat{U}_{k_1s_1,...k_ns_n}$$ is column matrix of $$k_1$$ times the $$s_1$$ row vector and so on of the unitary transformation. But as far as I know this formula holds only for the particular state stated above. What if there are different number of photons in different modes? I had to do an exercise for the three mode Hong Ou Mandel effect for the input state $$|3,0,0\rangle$$ If I then transform the creation operators I get 27=3^3 terms. There must be a simpler way to do it.

• I'm not sure if that's what you are asking, but you can write the general scattering amplitude between two $n$-boson states as the permanent of a specific matrix built out of $U$. See e.g. scottaaronson.com/papers/optics.pdf
– glS
Oct 4 '20 at 11:46
• @glS the bosons have to be exactly indistinguishable to get a permanent (in this case identical arrival times at the interferometer and exactly identical pulse shape). See Fearn, H. and Loudon, R., 1989. Theory of two-photon interference. JOSA B, 6(5), pp.917-927 (illustrated for $n=2$). Oct 4 '20 at 17:29
• @ZeroTheHero I'm not sure what you mean. That formula gives you the transition amplitude to go from $|s_1,...,s_n\rangle$ to $|r_1,...,r_m\rangle$ with $s_i$ the number of bosons in the $i$-th mode of the input, and same for $r_i$. Of course the real amplitude equals that only if the bosons are indistinguishable, but I don't see in the question an assumption of noise being introduced. It's like the HOM effect, of course it only happens if the inputs are indistinguishable bosons
– glS
Oct 4 '20 at 18:59
• @glS if the bosons do not arrive simultaneously at the interferometer there are terms beyond the permanent that are functions of the differences in arrival times to account for the temporal distinguishability. See Brańczyk, A.M., 2017. Hong-ou-mandel interference. arXiv preprint arXiv:1711.00080. See also this experiment: Tillmann, M. et al 2015. Generalized multiphoton quantum interference. Physical Review X, 5(4), p.041015. Oct 4 '20 at 19:23

No and yes. The matrix $$U$$ will contain repeated columns (or rows depending on your you set it up) so that some simplifications are possible. Moreover, since your input state is in the irrep $$(3,0,0)$$ of $$U(3)$$, you can only get an output that is in this irrep. The irrep $$(3,0,0)$$ is of dimension $$10$$ and you can generate the basis states in the form \begin{align} \vert n_1 n_2 n_3\rangle = \frac{(a^\dagger_1)^{n_1}(a^\dagger_2)^{n_2}(a^\dagger_3)^{n_3}}{\sqrt{n_1!n_2!n_3!}}\vert 0\rangle\, ,\qquad n_1+n_2+n_3=3 \end{align} so the output can only be a combination of those states.

The key point is that you cannot gets states in $$(2,1,0)$$ of $$U(3)$$ or in $$(1,1,1)$$ of $$U(3)$$ as they necessarily involve partially symmetric states, which need to be constructed using at least another degree of freedom to distinguish states in the same output ports. As an example one of the states in $$(2,1,0)$$ is of the form $$\left\vert \begin{array}{cc} a^\dagger_{1+}& a^\dagger_{1-}\\ a^\dagger_{2+}& a^\dagger_{2-}\end{array}\right\vert a^\dagger_{1+} \tag{1}$$ i.e. a determinant of creation operators multiplied by another creation operator. By antisymmetry of the determinant you see you need a second label (here $$\pm$$) to distinguish different bosons but in your problems your bosons are indistinguishable so don't have such a second label, which eliminates are states of the form (1).

Thus in principle \begin{align} U\vert 3,0,0\rangle = \sum_{n_1n_2n_3}\vert n_1n_2n_3\rangle U_{n_11}U_{n_21}U_{n_31}\, .\tag{2} \end{align} Now if you detect bosons in some specific output channels - say $$(a,b,c)$$ then your detecting process is modelled as \begin{align} \Pi=\frac{\vert a,b,c\rangle\langle a,b,c\vert}{a! b!c!} \end{align} and thus only a limited number of terms in (2) will survive the projection associated with detection.

• To be honest I don't know much about irreps. So there is no way other than calculating brute force all 27 combinations out of which some will cancel? Doesn't the answer depend not only depend on the input state but also the unitary transformation of the creation operators? Oct 2 '20 at 13:47
• You're NOT gonna get $27$ states. There are only $10$ states of the type $\vert n_1n_2n_3\rangle$. And the answer of course will depend on $U$ as your input state will have in principle overlap with all the $10$ output states. The number of possible overlaps might be further reduced if you specify where the particles are detected. I added some bits to that effect. I think this is enough to get you to finish the calculation. Oct 2 '20 at 13:56
• ok I think i get it now. thanks for the help Oct 2 '20 at 17:34
• out of curiosity, did you see this way of handling many-boson amplitudes via representation theory somewhere? I would be interested in a reference on this
– glS
Oct 4 '20 at 11:48
• @glS see Khalid, A. et al 2018. Permutational symmetries for coincidence rates in multimode multiphotonic interferometry. Physical Review A, 97(6), p.063802, and other papers by that group, and the canonical background paper Yurke, B. et al 1986. SU (2) and SU (1, 1) interferometers. Physical Review A, 33(6), p.4033. Oct 4 '20 at 13:43