I don't get how measuring in different bases works in experiments. From an experimental point of view, what do we do to measure in $|\pm \rangle$ basis?

If I'm getting it right measuring in the computational basis means just recording a click on the detector behind the PBS. Whereas when we want to measure in $|\pm \rangle$ basis we put a half-wave plate between the PBS and the detector.

  • 1
    $\begingroup$ For photonic qubits it just works exactly as you explained changing the orientation of a polarizer, but again this is platform dependent. In some qubits it just means to change the orientation of a magnetic field for example. $\endgroup$
    – Mauricio
    May 19 '21 at 15:05

This fully depends on the experimental implementation and the context.

Generally speaking, $|\pm\rangle$ is a basis like any other, so asking "how does measuring in the $|\pm\rangle$ basis works" has the same answer as the question "how does measuring in the $|0\rangle,|1\rangle$ basis works?". However, when asking this question you probably implicitly mean that you want to measure in a basis that is rotated with respect to the "computational basis" used to express the rest of the protocol.

In other words, you probably mean to ask something along the lines of: assuming my experimental apparatus "naturally operates" in the computational basis, how do I measure in the $|\pm\rangle$ basis?

At a formal level, this amounts to applying an Hadamard gate before the measurement: the Hadamard implements the operations $|0\rangle\mapsto|+\rangle$ and $|1\rangle\mapsto|-\rangle$, and thus measuring in the computational basis after an Hadamard amounts to measuring in the $|\pm\rangle$ before it.

For example, suppose you are encoding a qubit in the polarisation degree of freedom of a light beam. A way to measure the polarisation state is then to put a detector after a polarisation filter, or equivalently measuring the position after a PBS. Then, to measure in the $|\pm\rangle$ simply means to rotate the polarisation before said measurement, e.g. using a quarter waveplate.

If instead you encode a qubit in the position of, say, a photon, then applying the Hadamard amounts to evolving the state through a balanced beamsplitter and measuring after it.

You can also imagine more complicated examples. For example, you can encode a qubit in the Fock degree of freedom of a photon. In other words, you can consider the presence of a photon in a single mode as the $|1\rangle$ state and its absence as the $|0\rangle$ state. Then measuring on the rotated basis would become significantly more complicated as it would require a nonlinear operation.

  • $\begingroup$ So by measurement, we don't always mean a click on the detector. I mean when we send our photon with a known polarization state through a Hadamard gate we're still making a measurement even if our circuit includes other elements after that because we're telling our qubit to either give 1 or 2 with a hadamard gate $\endgroup$
    – pantinah93
    May 19 '21 at 9:44
  • $\begingroup$ @pantinah93 well, in optics at the end of the day you still usually measure clicks on some detectors, yes, but there's plenty of experimental platforms which have nothing to do with photons or light, and then measurements are performed in different ways $\endgroup$
    – glS
    May 19 '21 at 9:52
  • $\begingroup$ I understand that. I just don't find a way around this to understand it well. Thank you for your answers. It was very helpful. $\endgroup$
    – pantinah93
    May 19 '21 at 10:00

Lets assume that resuls of some calculation on a quantum computer is either state $|+\rangle$ or $|-\rangle$. If you measure these states in computational basis, in both cases you end up with probability of state $|0\rangle$ and $|1\rangle$ to be 50 %. This means that you are not able to distinguish between $|+\rangle$ and $|-\rangle$.

However, if you put Hadamard gate before the measurement, state $|+\rangle$ is transformed to $|0\rangle$ and $|-\rangle$ to $|1\rangle$. So, after measurement in Hadamard basis (i.e. with $H$ before the measurement in z-basis), you end up with state $|0\rangle$ with 100 % probability if you measure $|+\rangle$ or with state $|1\rangle$ with 100 % probability if you measure $|-\rangle$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.