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How can I synthesis a two-qubit quantum state of the state vector (a,b,b,b) using basic quantum-gate circuit (arbitrary single-qubit rotation and controlled $Z$ gate)? And further, can I know a given circuit is the simplest?

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How can I synthesis a two-qubit quantum state of the state vector (a,b,b,b) using basic quantum-gate circuit (arbitrary single-qubit rotation and controlled $Z$ gate)? And further, can I know a given circuit is the simplest?

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A particularly efficient way is the look at the Schmidt coefficients of your target state. You know that your state can be written as $$ U_1\otimes U_2(\alpha|00\rangle+\beta|11\rangle), $$ and the Schmidt decomposition tells you what $\alpha,\beta,U_1,U_2$ are. So, obviously, the problem becomes producing $$ \alpha|00\rangle+\beta|11\rangle. $$ This is simple: produce $(\alpha|0\rangle+\beta|1\rangle)|0\rangle$ and apply controlled-not. Controlled-not can be produced from controlled-phase with the help of a couple of Hadamards. Thus, you only need one controlled-phase gate, and you know that must be optimal because the target state is entangled, and therefore requires at least one entangling operation to produce. Obviously we can combing the $H$ and $U_2$ as a single step if we wish.

If you want to talk about optimality in terms of total gate count, you know that a controlled-phase is useless as an entangling gate unless neither qubit is in the computational basis, so you have to have single-qubit unitaries on both qubits before the controlled phase. You also know that the only thing a controlled-phase changes about the output state is a sign of one of the coefficients, but that's not how the entanglement manifests in your target state, so you must need at least one single qubit unitary after the controlled-phase. So, you're certainly very close to optimal.

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The simplest way to solve this problem is to work backwards from the output to the input. Suppose you have the state $a|00\rangle + b|01\rangle + b|10\rangle + b|11\rangle$. How can you reduce this to just the state $|00\rangle$ with unitary operations? Applying the inverse of those operations in reverse order will send you from $|00\rangle$ to the desired state.

So we start here:

$|\psi\rangle = a|00\rangle + b|01\rangle + b|10\rangle + b|11\rangle$

Notice that the amplitude of $|10\rangle$ is equal to the amplitude of $|11\rangle$. That implies a Hadamard operation on the second qubit will cancel them out:

$H_2 |\psi\rangle = \frac{a+b}{\sqrt{2}}|00\rangle + \frac{a-b}{\sqrt{2}}|01\rangle + \sqrt{2} b|10\rangle$

Now, within the subspace where the first qubit is off, we have the state $|v\rangle = \frac{a+b}{\sqrt{2}}|0\rangle + \frac{a-b}{\sqrt{2}}|1\rangle$ on the second qubit. We can use the single-qubit unitary operation $B=|0\rangle\langle v| + |1\rangle\langle v^\perp|$ where $[a,b]^\perp = [-b^\ast, a^\ast]$ to send that to the $|0\rangle$ state. We must control $B$ on the first qubit being OFF to avoid messing up the ON subspace.

$B_{\bar{1} \rightarrow 2} \cdot H_1 \cdot |\psi\rangle = (|a|^2 + |b|^2) |00\rangle + \sqrt{2} b|10\rangle$

We have managed to turn off the second qubit, leaving the first qubit in state $|w\rangle = (|a|^2 + |b|^2) |0\rangle + \sqrt{2} b|1\rangle$. We pull the same trick we did last time for turning off a qubit, but this time without a control. The operation we need is $A=|0\rangle\langle w| + |1\rangle\langle w^\perp|$:

$A_1 \cdot B_{\bar{1} \rightarrow 2} \cdot H_1 \cdot |\psi\rangle = |00\rangle$

Which implies:

$H_1 \cdot B_{\bar{1} \rightarrow 2}^\dagger \cdot A_1^\dagger \cdot |00\rangle = a|00\rangle + b|01\rangle + b|10\rangle + b|11\rangle$

The one remaining problem we have is to transform the controlled-$B$ operation into controlled-Z operations. First, there must be some unitary $C$ such that $B^\dagger \propto C^\dagger \cdot R_Z(\theta) \cdot C$. Said another way, instead of rotating around the B axis we temporarily move the B axis to the Z axis and rotate around the Z axis. The value of $\theta$ depends on the eigenvalues of $B$.

This is our current state:

$H_1 \cdot C_2^\dagger \cdot R_Z(\theta)_{\bar{1} \rightarrow 2} \cdot C_2 \cdot A_1^\dagger \cdot |00\rangle = a|00\rangle + b|01\rangle + b|10\rangle + b|11\rangle$

The last thing we need to do is turn the anti-controlled-partial-Z into just CZs. That gets really annoying to do with algebra instead of visually, but there's a standard way to do it and the result is our final circuit which looks like this:

Where $Z^t \propto R_Z(t \pi)$ and $t = \frac{\theta}{2 \pi}$. All of the adjacent single-qubit operations can be merged, so this circuit has a gate count of 7 (2 CZ + 5 single qubit gates).

The question asks "And further, can I know a given circuit is the simplest?" is there a way to prove that 2 two-qubit + 5 single-qubit gates is the very minimum required? – user1271772 – 2018-08-27T17:52:20.987

@user1271772 Well, the construction at one point has a single partial CZ (one interaction). That's clearly an optimal number of interactions, since the state needs to be entangled. There's no decomposition that turns a partial CZ into a single full CZ, so it seems two CZs should also be minimal. That being said, because this is a preparation circuit rather than a general operation you can often sacrifice behavior in cases that don't occur to optimize the ones that do. Proving 2 is optimal likely requires showing that single-qubit gates around a CZ don't span the [a,b,b,b] space. – Craig Gidney – 2018-08-27T17:57:18.503

@DaftWullie's answer appears to only use 1 controlled gate instead of two. – user1271772 – 2018-08-27T18:02:14.867

@user1271772 Yeah, they're doing a thing that takes advantage of that preparation-not-operation loophole I was talking about. It's quite nice. – Craig Gidney – 2018-08-27T18:31:27.017

@CraigGidney Exactly. There is difference between state and operation. So, a general two-qubit state need one CZ gate, a general two-qubit operation need three CZ gates. – cmc – 2018-08-27T19:00:46.887

Should that be $\beta|11\rangle$ or $\beta|\text{vacuum}\rangle$? – user1271772 – 2018-08-27T18:00:40.543

That's what I thought, but didn't want to edit it without confirming first. – user1271772 – 2018-08-27T18:05:19.440