# POVM three-qubit circuit for symmetric quantum states

I have been reading this paper but don't yet understand how to implement a circuit to determine in which state the qubit is not for a cyclic POVM. More specifically, I want to implement a cyclic POVM with $$m=3$$;

Update: I came across having to implement this unitary matrix: $$M= \frac{1}{\sqrt{2}}\left[ {\begin{array}{cc} 1 & 1 \\ 1 & w \\ \end{array} } \right]$$ Where $$w$$ is a third root of unity using rotations, after which I am stuck.

• That's not a unitary matrix unless w=-1.. – Craig Gidney Mar 6 at 19:49

This is not the unitary that you have to implement: you need a two-qubit unitary $$\frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 & \omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array}\right),$$ where $$\omega=e^{2i\pi/3}$$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.

I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.

Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly... Here, I'm using $$Z^r$$ to denote $$\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\pi r} \end{array}\right),$$ and $$V=\frac{1}{\sqrt{3}}\left(\begin{array}{cc} 1 & \sqrt{2} \\ -\sqrt{2} & 1 \end{array}\right).$$

• Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something? – chubakueno Mar 4 at 16:09
• @chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3\times 3$ matrix which we must embed into a $4\times 4$ matrix if we're using qubits. – DaftWullie Mar 4 at 16:11
• I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know. – chubakueno Mar 4 at 16:26
• @chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit. – DaftWullie Mar 4 at 16:48
• @xbk365 once i’m done with the evening’s childcare responsibilities... – DaftWullie Mar 4 at 18:21

Check this
$$\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ \end{array}\right) = \left(\begin{array}{cc} H & 0 \\ 0 & 1 \\ \end{array}\right) \cdot \frac{1}{\sqrt{3}}\left(\begin{array}{ccc} \sqrt{2} & 0 & 1 \\ 0 & \sqrt{3} & 0 \\ 1 & 0 & -\sqrt{2} \\ \end{array}\right) \cdot M_{3}$$

$$M_{3} = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}}i\omega^2 & \frac{1}{\sqrt{2}}i\omega \\ 0 & -\frac{1}{\sqrt{2}}\omega^2 & -\frac{1}{\sqrt{2}}\omega \\ \end{array}\right)$$

$$\frac{1}{\sqrt{2}}\left(\begin{array}{ccc} i\omega^2 & i\omega \\ -\omega^2 & -\omega \\ \end{array}\right) = X \cdot S \cdot X \cdot Z \cdot H \cdot \left(\begin{array}{ccc} \omega^2 & 0 \\ 0 & \omega \\ \end{array}\right) \cdot Z$$