# Action of quantum Fourier transform on two-fermion states

In section 2.2 of the paper https://arxiv.org/abs/1807.07112, there appears a Fourier transformation named $$F_k^n$$ that comes out of a matrix called $$F_2$$,

$$F_2 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0\\ 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0\\ 0&0&0&-1 \end{pmatrix}$$

This matrix seems to be explained in the appendix of https://doi.org/10.1103/PhysRevA.79.032316, but I don't understand where the $$c^\prime s$$ appeared and how it is related to the Fourier transformation.

## 1 Answer

It's the beamsplitter unitary (a.k.a. the QFT in two dimensions, a.k.a. the Hadamard gate), represented via its action on two-mode Fermion states.

With two modes, there are four possible fermionic states: $$|11\rangle\equiv c_1^\dagger c_2^\dagger |\text{vac}\rangle$$ (one fermion per mode), $$|01\rangle\equiv c_2^\dagger |\text{vac}\rangle$$ (a fermion in the second mode), $$|10\rangle\equiv c_1^\dagger |\text{vac}\rangle$$ (a fermion in the first mode), and $$|00\rangle\equiv|\text{vac}\rangle$$ (no fermion at all).

A beamsplitter will act on these states as follows: \begin{align} |00\rangle &\to |00\rangle, \\ |10\rangle &\to \frac{1}{\sqrt2}(|10\rangle+ |01\rangle), \\ |01\rangle &\to \frac{1}{\sqrt2}(|10\rangle- |01\rangle), \\ |11\rangle &\to -|11\rangle. \end{align} To see this you just need to consider that the beamsplitter acts on the fermionic modes as $$c_1^\dagger\to\frac{1}{\sqrt2}(c_1^\dagger+c_2^\dagger), \qquad c_2^\dagger\to\frac{1}{\sqrt2}(c_1^\dagger-c_2^\dagger),$$ so that for example $$|11\rangle\equiv c_1^\dagger c_2^\dagger |\text{vac}\rangle \to \frac12(c_1^\dagger+c_2^\dagger)(c_1^\dagger-c_2^\dagger)|\text{vac}\rangle \to -c_1^\dagger c_2^\dagger|\text{vac}\rangle.$$ The rest of the rules is similarly derived. The matrix reported in the paper is simply the matrix representation of these rules.

This is a special case of a more general problem: given a unitary $$U$$, how does it act on many-fermion states? The general result is that the scattering amplitude between an $$n$$-fermion, $$m$$-mode input $$|r_1,...,r_m\rangle\equiv c_1^{r_1\dagger}\cdots c_n^{r_n \dagger}|\text{vac}\rangle$$ and an output $$|s_1,...,s_m\rangle$$ is given by the determinant of the matrix obtained from $$U$$ by taking its first column $$r_1$$ times, its second column $$r_2$$ times, etc., and similarly taking rows according to the occupation numbers in $$|s_1,...,s_m\rangle$$. For example, applying this to the above case with $$U=H$$ and $$|r_1,r_2\rangle=|11\rangle$$, we get that the probability amplitude of $$|11\rangle$$ evolving into $$|11\rangle$$ is the determinant of $$H$$ itself, which is $$-1$$, consistently with what we found by direct analysis before.

• And how does the beamsplitter connect to the Fourier transformation? – Vicky Mar 18 at 15:51
• @Vicky if you work out the qft matrix in 2 dimensions you will notice that it's the hadamard, which is the matrix describing a beamsplitter. The equivalence breaks for higher dimensions though – glS Mar 18 at 16:01
• Yeah, but how does it connects with the $F_k^n$ and that to the Eq. (9) in arxiv.org/abs/1807.07112 ? – Vicky Mar 18 at 18:11
• @Vicky I'm not sure what is the justification for that terminology. The gate $F_k^n$ seems to be the two-fermion unitary corresponding to a beamsplitter with an added phase. In other words, is what you get if you follow the procedure I oulined in the answer but using $\frac1{\sqrt2}\begin{pmatrix}1 & e^{2\pi ik/n} \\ 1 & -e^{2\pi ik/n}\end{pmatrix}$ instead of $H$ – glS Mar 18 at 19:19
• My problem is that I'm not being able to match all the parts in section 2.2 of that paper, so I don't know the reasons and connetions among the different steps – Vicky Mar 18 at 22:36