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In theory, it is fairly simple to measure a qubit's state. Let us consider a Ramsey experiment, where the qubit is in the ground state $|0\rangle$ initially. Then by applying a Hadamard gate the qubit becomes a superposition state $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$. After waiting for a while the state evolves to $|\psi\rangle=\alpha|+\rangle+\beta|-\rangle$. Finally, we will measure the state to be in $|+\rangle$ with the probability $|\alpha|^2$.

The question is how to measure the qubit to be in $|+\rangle$ real experiment? For example, a very well known way to know if the qubit is in an excited state or in ground state is to apply a dispersive readout, where the resonator's frequency shifts depending on the state of the qubit. However, it seems like it is applicable for $|\sigma_z\rangle$ basis only. Is this method applicable for $|\pm\rangle$ basis as well?

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  • $\begingroup$ What kind of qubit? $\endgroup$ – Norbert Schuch Jul 3 '20 at 14:11
  • $\begingroup$ Why not just rotate your measurement setup by 90 degrees? $\endgroup$ – probably_someone Jul 3 '20 at 14:28
  • $\begingroup$ @Norbert Schuch, what if superconducting qubits. I don't have much experience with experiments, so I don't know whether the type of qubit matters. But the question is how does one measure the state of the qubit along a general axis, for example here $\sigma_x$ axis? $\endgroup$ – TheDorkSide Jul 4 '20 at 5:14
  • $\begingroup$ @probably_someone , will that work? $\endgroup$ – TheDorkSide Jul 4 '20 at 5:15
  • $\begingroup$ @TheDorkSide The + and - states are the spin-up and spin-down eigenstates along the x-axis. Whether a spin measurement along the x-axis is possible depends on the specific mechanical details of your experiment. $\endgroup$ – probably_someone Jul 4 '20 at 17:23
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Firstly, apply Hadamard gate. If the qubit is in state $|+\rangle$, it becomes $|0\rangle$. In case it is in state $|-\rangle$, it becomes $|1\rangle$. Now, you can measure the qubit in z-basis - if outcome is $|0\rangle$, the qubit was formely in state $|+\rangle$, if $|1\rangle$, the qubit was in state $|-\rangle$.

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  • $\begingroup$ Thank you. The qubit will be in $|0\rangle$ with $|\alpha|^2$ probability, that makes sense. So, there is no direct way that will collapse and leave the state of the qubit in $|+\rangle$ state after the measurement? And do you know if it is possible in an experiment to measure the state of the qubit in an arbitrary basis, not only z or x-basis? $\endgroup$ – TheDorkSide Jul 4 '20 at 7:24
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    $\begingroup$ @TheDorkSide: First question: In some implementation of quantum computing it is possible to measure in other basis directly, but I am not sure how widespread these implementations are (IBM Q Experience, QuTech Quantum Inspire use z-basis as default). Second question: you can measure in any basis you want. The way how to do so is to apply an inverse circuit to one transforming z-basis to desired basis and then do measurement in z-basis. Based on mapping between these bases you can infer the results. $\endgroup$ – Martin Vesely Jul 5 '20 at 5:00

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