Conjugate transpose of a U-gate

Question

I want to be able to create the circuit depicted below, but running the code below results in an empty circuit when viewing the job afterwards.

backend = provider.backends.ibmq_armonk


qreg_q = QuantumRegister(1, 'q')
creg_c = ClassicalRegister(1, 'c')
circuit = QuantumCircuit(qreg_q, creg_c)

circuit.u(pi/2, pi/2, pi/2, qreg_q[0])
circuit.x(qreg_q[0])
circuit.y(qreg_q[0])
circuit.x(qreg_q[0])
circuit.y(qreg_q[0])
circuit.u(pi/2, pi/2, pi/2, qreg_q[0]).inverse()

circuit.measure(qreg_q[0], creg_c[0])


qobj = assemble(transpile(circuit, backend=backend), backend=backend)
job = backend.run(qobj)
retrieved_job = backend.retrieve_job(job.job_id())

Answer 1

Indeed, your circuit can be reduced to an identity circuit as @KAJ226 nicely explains here. The transpilation process notices that and reduces your gates to idle during circuit depth optimization:

qreg_q = QuantumRegister(1, 'q')
creg_c = ClassicalRegister(1, 'c')

circuit = QuantumCircuit(qreg_q, creg_c)
circuit.u(pi/2, pi/2, pi/2, qreg_q[0])
circuit.x(qreg_q[0])
circuit.y(qreg_q[0])
circuit.x(qreg_q[0])
circuit.y(qreg_q[0])
circuit.u(pi/2, pi/2, pi/2, qreg_q[0]).inverse()

circuit.measure(qreg_q[0], creg_c[0])

transpile(circuit, backend=backend).draw('mpl')

You can avoid the optimization by reducing the optimization level to $0$:

transpile(circuit, backend=backend, optimization_level=0).draw('mpl')

In your example, you need to modify the transpile call like this:

qobj = assemble(transpile(circuit, backend=backend, optimization_level=0), backend=backend)

Answer 2

Just want to add a small additional detail here that might be useful for future encounter.

The reason why you see the circuit with only measurement after the transpilation process is because $(XY)^\dagger = Y^\dagger X^\dagger = -XY$. Since

$$ XY = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix} = \begin{pmatrix} -i & 0\\ 0 & i \end{pmatrix}$$ and $$ (XY)^\dagger = Y^\dagger X^\dagger = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix}$$

Now note that $(XY)(XY)^\dagger =I $ because of definition of Unitary, and so $ XYXY = -XY(XY)^\dagger = -I $ but in quantum computing, global phase doesn't matter, so $-I = I$... since there is no experiment you can do to distinguish $|0\rangle$ from $-|0\rangle$.

Now, looking at your circuit, we have: $$UXYXYU^\dagger = UXY[-(XY)^\dagger] U^\dagger = -[UXY(XY)^\dagger U^\dagger] = -[U I U^\dagger] = -[UU^\dagger] = -I \equiv I $$
During the transpilation step, it recognizes this and made the reduction to save the possible number of quantum operations needed...hence the reason why your circuit only consists of the measurement procedure.