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Setup

I am drawing this circuit:

qc_b = QuantumRegister(2, 'b')
qc_f1 = QuantumRegister(1, 'f1')
qc_a1 = QuantumRegister(1, 'a1')
qc_a2 = QuantumRegister(1, 'a2')
qc_f2 = QuantumRegister(1, 'f2')

qc_0 = QuantumCircuit(qc_a1, qc_f1, qc_b, qc_f2, qc_a2)

qc_0.x(qc_f1)
qc_0.barrier([*range(1, 5)])

for t in range(0, 3):
    if t != 0:
        qc_0.reset(qc_a1)
        qc_0.reset(qc_a2)
    qc_0.unitary(np.eye(2**4), range(1, 5), label=h_label)
    qc_0.unitary(np.eye(2**2), [0, 1], label=l_label)
    qc_0.unitary(np.eye(2**2), [4, 5], label=l_label)


qc_0.barrier([*range(1, 5)])
qc_0.unitary(np.eye(2), [1], label='M')
qc_0.unitary(np.eye(2), [2], label='M')
qc_0.unitary(np.eye(2), [3], label='M')
qc_0.unitary(np.eye(2), [4], label='M')

qc_0.draw('latex')

Which gives me this picture:

circuit_with_numbers

Question

How can I remove the 'qubit numbers' (the 0, 1, 2, 3 before every unitary)?

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1 Answer 1

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As far as I understand, there is no built-in method that allows that unfortunately.

But - what you can do is getting the LaTex code by using the command print(qc_0.draw(output = 'latex_source')).Then you can edit the Latex output and remove those qubit numbers (Using print() is necessary otherwise we get a messy LaTex code).

If you add this piece of code at the bottom of your code:

latex_source = qc_0.draw('latex_source')
for number in range(4):
    look_for = "<<<{" + str(number) + "}"
    latex_source = latex_source.replace(look_for,"<<<{}")
print(latex_source)

Then you get this output:

\documentclass[border=2px]{standalone}

\usepackage[braket, qm]{qcircuit}
\usepackage{graphicx}

\begin{document}
\scalebox{1.0}{
\Qcircuit @C=1.0em @R=0.2em @!R { \\
        \nghost{{a1} :  } & \lstick{{a1} :  } & \qw & \qw & \qw & \multigate{1}{\mathrm{D}}_<<<{} & \gate{\mathrm{\left|0\right\rangle}} & \multigate{1}{\mathrm{D}}_<<<{} & \gate{\mathrm{\left|0\right\rangle}} & \multigate{1}{\mathrm{D}}_<<<{} & \qw & \qw & \qw & \qw\\
        \nghost{{f1} :  } & \lstick{{f1} :  } & \gate{\mathrm{X}} \barrier[0em]{3} & \qw & \multigate{3}{\mathrm{H}}_<<<{} & \ghost{\mathrm{D}}_<<<{} & \multigate{3}{\mathrm{H}}_<<<{} & \ghost{\mathrm{D}}_<<<{} & \multigate{3}{\mathrm{H}}_<<<{} & \ghost{\mathrm{D}}_<<<{} \barrier[0em]{3} & \qw & \gate{\mathrm{M}} & \qw & \qw\\
        \nghost{{b}_{0} :  } & \lstick{{b}_{0} :  } & \qw & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \qw & \gate{\mathrm{M}} & \qw & \qw\\
        \nghost{{b}_{1} :  } & \lstick{{b}_{1} :  } & \qw & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \ghost{\mathrm{H}}_<<<{} & \qw & \qw & \gate{\mathrm{M}} & \qw & \qw\\
        \nghost{{f2} :  } & \lstick{{f2} :  } & \qw & \qw & \ghost{\mathrm{H}}_<<<{} & \multigate{1}{\mathrm{D}}_<<<{} & \ghost{\mathrm{H}}_<<<{} & \multigate{1}{\mathrm{D}}_<<<{} & \ghost{\mathrm{H}}_<<<{} & \multigate{1}{\mathrm{D}}_<<<{} & \qw & \gate{\mathrm{M}} & \qw & \qw\\
        \nghost{{a2} :  } & \lstick{{a2} :  } & \qw & \qw & \qw & \ghost{\mathrm{D}}_<<<{} & \gate{\mathrm{\left|0\right\rangle}} & \ghost{\mathrm{D}}_<<<{} & \gate{\mathrm{\left|0\right\rangle}} & \ghost{\mathrm{D}}_<<<{} & \qw & \qw & \qw & \qw\\
\\ }}
\end{document}

If you render the above output in this engine for example, then you get your wish:

enter image description here

The piece of code that i have wrote just erases the numbers in the phrases <<<{QUBIT_NUMBER_HERE} in the LaTex code. I have set the range to of the loop to 4 and it's enough that case - but of course it can be extended to erase as much qubit numbers as needed.

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3
  • $\begingroup$ Thanks for the hint :). At this point I might just go into GIMP and draw over it in the png. $\endgroup$ Commented Aug 18, 2022 at 6:13
  • $\begingroup$ @AndreasBurger I have modified my answer such that a nice loop handles the issue easily now. $\endgroup$
    – Ohad
    Commented Aug 18, 2022 at 10:31
  • $\begingroup$ Thanks for the effort! Works perfectly :) $\endgroup$ Commented Aug 19, 2022 at 0:04

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