# Measuring qubit's state in a real experiment

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?

• What kind of qubit? – Norbert Schuch Jul 3 '20 at 14:11
• Why not just rotate your measurement setup by 90 degrees? – probably_someone Jul 3 '20 at 14:28
• @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? – TheDorkSide Jul 4 '20 at 5:14
• @probably_someone , will that work? – TheDorkSide Jul 4 '20 at 5:15
• @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. – probably_someone Jul 4 '20 at 17:23

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 $$|-\rangle$$.
• 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? – TheDorkSide Jul 4 '20 at 7:24