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Properties of controlled $z$-rotations

Physics Asked by Leo Rogers on October 27, 2020

Given $n$ qubit gate of the form c-z-z-z… (shorthand for c-z between qubit 1 and 2 followed by c-z between 1 and 3 and so on up to n qubits) it seems to be possible to find local unitaries which will change this gate into one that preserves phase when flipping all the qubits. That is |1111..1> will have the same phase as |000..0> ans so on.

An example would be c-z. All input states remain unchanged except |11>, which picks up a – sign. However, if you put an S gate on each qubit (a pi/2 phase gate) you will find that |00> and |11> are left unchanged, and |01> and |10> pick up an i. That is, each basis state has the same quantum amplitude after the gate as its bitwise flipped counterpart.

I haven’t proven this, but I have tested it for 2, 3 and 4 qubits. However I haven’t been able to do the same thing for gates with more than one control qubit, such as c-c-z.
Is anyone aware of any reason why this might be, or any even slightly relevant research which might shed light on this?

Btw, if it helps, the reason I ask is that I am looking at using a coherent optical state to mediate entangling interactions between qubits. A coherent state is defined by a complex number, and in this particular system, when it interacts with a qubit, this complex number can be displaced in an arbitrary direction conditional on the state of the qubit, |0> and |1> displace the coherent state in opposite directions.
When you use a series of these controlled displacements to take the coherent state in a closed path in the complex plane, this leaves a a phase on the qubits of the form e^(±iA) where A is the area enclosed by the path, and the sign is dependant on whether the path is travelling clockwise or anti-clockwise.
You can use this effect to build entangling gates, specifically c-not as shown in http://arxiv.org/abs/quant-ph/0509202.

I have been trying to describe a c-c-z gate using fewer qubit-bus interactions than building it out of many 2-qubit gates.

The types of paths I am looking at (if not all paths in general) all seem to have this property where flipping all the qubits does not affect the phase, but it seems not all multiqubit gates have this property (including c-c-z), but I haven’t worked out a way of determining which do and which don’t apart from trial and error.

Any help would be appreciated.

One Answer

I've done a lot more work on this since asking, and now have the answers to my questions. First, I mentioned that it seemed that with single qubit gates you can always change the state given by c-z-z-z... (from the equal superposition) to one with this 'bit-flip' property, but that I hadn't proven it. Turns out this is true for any circuit made up of c-z gates, and this can be proven to be true. It works for one c-z gate, if you start with a circuit producing a state with the bit-flip property, and concatenating another circuit (such as one containing only one c-z) that does this will produce another circuit with this property.

Second, it turns out that there are no single qubit gates that only alter the phases on the qubits to produce the bit-flip property for the c-c-z gate. I only have a proof regarding c-c-z, not any general criteria about which circuits you can do this with and which you can't, except that it is always possible to do it with any circuit using only c-z gates and single qubit gates that commute with c-z.

I'm still working on this. I'll update when I have more general results.

Answered by Leo Rogers on October 27, 2020

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