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3 votes

Are operators unitary on a real quantum computer?

To a good approximation yes they are unitary. Otherwise experiments like Rabi oscillations would not work. To the extent that they aren't unitary... well, that slight non-unitarity falls under your ...
Craig Gidney's user avatar
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7 votes
Accepted

Are operators unitary on a real quantum computer?

A unitary operator represents a invertible operation that is fully devoid of noise. Similarly to how pure states represent us having perfect knowledge about the quantum state. In realistic scenarios, ...
glS's user avatar
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2 votes

Single-qubit quantum channel from the CNOT gate

First, it is not true that CX leaves the control qubit unaffected. If it did, then it could not create entangled states. In any case, you can convince yourself that the control qubit may be affected ...
Adam Zalcman's user avatar
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1 vote

Single-qubit quantum channel from the CNOT gate

The easiest way to see it, is to write $CX$ as $|00\rangle\langle00| + |01\rangle\langle01| + |10\rangle\langle11| + |11\rangle\langle10|$ (I assumed that the first entry is the system and the second ...
David Dentelski's user avatar
2 votes

Single-qubit quantum channel from the CNOT gate

You calculated $ {\rm tr}_1(\mathrm{CX}(|0\rangle\langle 0|\otimes\rho)\mathrm{CX}) $ which—as you correctly observed—is equal to $\rho$ because the control qubit is in the $|0\rangle$ state so ...
Frederik vom Ende's user avatar
1 vote

How to estimate noisy expectation value in limit of infinite shots?

Depending on your noise model, you should be able to compute the final density matrix that you obtain after your computations, let us denote it $\rho$. Let us assume that the state you intended to ...
Tristan Nemoz's user avatar
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