I'm reading Linear Optics Quantum Computation: an Overview.

On page 16 we're given a matrix

$$U = \begin{pmatrix}1-\sqrt{2}&\frac{1}{\sqrt{\sqrt{2}}}&\sqrt{\frac{3}{\sqrt 2}-2}\\\frac{1}{\sqrt{\sqrt 2}}&\frac{1}{2}&\frac{1}{2}-\frac{1}{\sqrt 2}\\\sqrt{\frac{3}{\sqrt{2}}-2}&\frac{1}{2}-\frac{1}{\sqrt 2}&\sqrt 2 - \frac{1}{2}\end{pmatrix}$$

and this exercise:

What transformation does this implement on the input state $(\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle)|10\rangle$? What is the transformation if we measure $|10\rangle$ in modes $2$ and $3$? This will be important later!

I know that I should write states using the creation operator and $|00\rangle$, and then let $U$ act on creation operators. Yet this approach does not seem to get me anywhere.

  • $\begingroup$ Thank you @SanchayanDutta for the edit. I hope someone can shed some light on this $\endgroup$
    – Bashir
    Jun 20 '19 at 19:43

I know that I should write states using the creation operator and $|00⟩$, and then let $U$ act on creation operators.

This is exactly right.

Writing the state in terms of creation operators gives $$(\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle)|10\rangle = \left(\alpha + \beta a_1^\dagger + \gamma a_1^\dagger a_1^\dagger\right)a_2^\dagger\left|000\right>.$$

We know from page 10 that the unitary transformation matrix acts on the creation operators as


That is, overall it gives $$\left(\alpha + \beta \sum_mU_{m1}a_m^\dagger + \gamma \sum_{m, l}U_{m1}U_{l1}a_m^\dagger a_l^\dagger\right)\sum_jU_{j2}a_j^\dagger\left|000\right>.$$ The reason that this looks like a mess is simply because it is (at this point anyway) - 3 photons in 3 modes is considerably more complicated than 3 photons in 1 mode or 1 photon in 3 modes, so naively, the only option is to persevere and see what comes out the other end. Making use of the Reck decomposition (equation 15, which looks like $U = D^\dagger T^\dagger_{2, 1} T^\dagger_{3, 1} T^\dagger_{3, 2}$ for 3 modes, as on page 15) can sometimes be used to make things easier to handle, but in this case, it doesn't particularly look like it would help. What might make it easier is splitting it into chunks, then taking the sum: So, taking the $\alpha$ component, then the $\beta$ component, then the $\gamma$ component and summing them together. The occasional term simplifies down a bit but overall, this (part) seems to be one of those questions where the purpose is to give the reader practise in doing these sorts of calculations, so actually writing it down here in this answer won't help that.

However, for the latter part of the question, there is a bit of simplification that can be done. This lies in realising that the $U$ is symmetric: $U^\dagger = U$ and so, going 'backwards' is the same evolution as going 'forwards'. The above considers writing the output in terms of the input $\left(a_l^\dagger\left|0\right>\rightarrow\sum_mU_{ml}a_m^\dagger\left|0\right>\right)$. However, the input can also be written in terms of the output and we know both the input and two modes of the output, so we're only looking at outputs of the form $$(\alpha'|0\rangle+\beta'|1\rangle+\gamma'|2\rangle)|10\rangle = \left(\alpha' + \beta' a_1^\dagger + \gamma' a_1^\dagger a_1^\dagger\right)a_2^\dagger\left|000\right>.$$

Indeed, taking the $\alpha'$ component gives that, the only single photon term that survives is $\frac{1}{2}\alpha\left|010\right>$ as the only possible way to go from an input of $a_2^\dagger$ to $a_2^\dagger$ is by the operation $U_{22}$. Similarly, the $\beta$ term gives the only two photon term that survives as $$\left(U_{11}U_{22} + U_{12}U_{21}\right)\beta\left|110\right> = \left(\frac{1}{2}\left(1-\sqrt{2}\right) + \frac{1}{\sqrt{2}}\right)\beta\left|110\right> = \frac{1}{2}\beta\left|110\right>,$$ corresponding to the possible transformations of $a_1^\dagger a_2^\dagger \rightarrow a_1^\dagger a_2^\dagger$. The $\gamma'$ term then gives $$\left(\left(1-\sqrt{2}\right)^2\frac{1}{2} + 2\left(1-\sqrt{2}\right)\frac{1}{\sqrt{2}}\right)\gamma\left|210\right> = -\frac{1}{2}\gamma\left|210\right>,$$ giving the overall output if we measure $\left|10\right>$ in modes 2 and 3 as $$(\alpha|0\rangle+\beta|1\rangle-\gamma|2\rangle)|10\rangle,$$ where the factor of 1/2 shows that the probability of obtaining this state is 1/4 and is then factored out for normalisation.

If the former part of the question is fully calculated, then doing the latter part is a relatively simple matter of looking for the terms of the correct form, however, the above shows an easier way of performing this latter part without having to do the former and transforms a brute force 'calculate all the terms and look for the ones you want' into a much smaller combinatorics-type problem of 'what are the ways that this input can be transformed into this output?' and multiplying and summing the relevant matrix elements.

This method is essentially an adaption of that used in e.g. Ralph's Linear Optical CNOT Gate in the Coincidence Basis, which will hopefully offer more insight into what's happening.

  • $\begingroup$ Thanks a lot @Mithrandir24601 for the detailed and excellent answer. The second method (i.e. not the brute force method) is much better... $\endgroup$
    – Bashir
    Jun 22 '19 at 20:10

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