# Why can I apply $HS^\dagger$ and then measure in the computational basis to measure $Y$?

I come from a CS background

I was reading Neven and Farhi's paper ("Classification with Quantum Neural Networks on near Term Processors"), and I am trying to implement the subset parity problem using Qiskit, and solve it using a quantum Neural Network.

There is one thing that doesn't make sense to me though. In the paper, they measure "the Pauli Y gate on the readout qubit" (perhaps this phrasing is wrong, as I have to admit that whenever one does not measure in the computational basis, the whole thing doesn't make sense to me anymore). In one of the questions I already asked on this site, I was told that measuring in a basis other than the computational basis is simply the same as applying a matrix to the qubit and then measuring it in a computational basis.

Through various research, I was able to determine that, for this problem "to measure the Pauli Y gate the readout qubit", I had to apply $$HS^{\dagger}$$ and then measure in the computational basis in order to obtain the same result. It works, but I don't understand why it has to be this matrix in particular (is there any mathematical proof that shows that this is indeed this matrix ?)

Your normal measurement is a pauli-$$Z$$ measurement. If you apply a unitary $$U$$ just before measurement, this transforms the $$Z$$ measurement into $$U^\dagger ZU$$. So, any $$U$$ that transforms $$U^\dagger ZU=Y$$ will do the job. One convenient way of doing this is $$\frac{Y+Z}{\sqrt{2}},$$ but your choice will also work: $$SHZHS^\dagger=SXS^\dagger=-iS^2X=-iZX=Y$$

If you want to know why it's the transformation $$U^\dagger ZU$$, well think about a circuit with input $$|\psi\rangle$$ that has a unitary $$U$$ enacted upon it, and then it's measured in the standard basis. The probability of getting the 0 answer is $$|\langle 0|U|\psi\rangle|^2,$$ which is the same as the probability that $$|\psi\rangle$$ is in the state $$U^\dagger|0\rangle$$. This corresponds to a measurement projector $$U^\dagger |0\rangle\langle 0|U$$, so you can see that transformation starting to come out.

• Got it, thank you – Skyris May 1 '20 at 8:18
• Just a question though. Why is the normal measurement a Pauli Z mesurement – Skyris May 27 '20 at 13:27
• The Pauli Z matrix is diagonal in the standard measurement basis. – DaftWullie May 27 '20 at 13:50

Measurement in $$Y$$ basis means that we want to measure is the qubit in $$|+i\rangle$$ state or $$|-i\rangle$$ state which are eigenbasis vectors for $$Y$$ gate. Because they are eigenbasis vectors we can express any $$|\psi_1 \rangle$$ state in this form:

$$| \psi_1 \rangle = \alpha_{+i} |+i\rangle + \alpha_{-i} |-i\rangle$$

where $$|\alpha_{+i}|^2$$ is the probability of measuring $$|+i\rangle$$ state and $$|\alpha_{-i}|^2$$ is the probability of measuring $$|-i\rangle$$. And

$$\begin{equation} |+i\rangle = |0\rangle + i |1\rangle \qquad |-i\rangle = |0\rangle - i |1\rangle \end{equation}$$

Now when we apply $$HS^{\dagger}$$ to $$|\psi_1 \rangle$$ state, we will obtain:

$$| \psi_2 \rangle = \alpha_{+i} |0\rangle + \alpha_{-i} |1\rangle$$

Then, with $$|\alpha_{+i}|^2$$ we will measure $$|0\rangle$$ (the same probability that we had for $$|+i \rangle$$ measurment in the initial $$|\psi_1\rangle$$), and with with $$|\alpha_{-i}|^2$$ we will measure $$|1\rangle$$ (the same probability that we had for $$| -i \rangle$$ measurment in the initial $$|\psi_1\rangle$$). For any gate that will do $$U |+i\rangle = e^{i \varphi_1} |0\rangle$$ and $$U |-i\rangle = e^{i \varphi_2}|1\rangle$$ mapping (where $$\varphi_1$$ and $$\varphi_2$$ are some phases that will not have any influence on probabilities), we will have this correspondence. For example, if I understand this Riggeti's code right, they are doing $$Y$$ basis measurement by applying firstly $$U = R_x(\pi /2)$$ gate that maps $$R_x(\pi /2) |+i\rangle = |0\rangle$$ and $$R_x(\pi /2) |-i\rangle = -i|1\rangle$$.

The other thing is to measure the expectation value of $$Y$$ operator:

$$\langle \psi_1 | Y | \psi_1 \rangle = |\alpha_{+i}|^2 - |\alpha_{-i}|^2$$

that can easily be calculated after enough measurements in the $$Y$$ basis. Here we took into accout that $$Y|+i\rangle = (+1)|+i\rangle$$ and $$Y|-i\rangle = (-1)|-i\rangle$$. $$|\alpha_{+i}|^2 = \frac{N_{+i}}{N}$$ and $$|\alpha_{-i}|^2 = \frac{N_{-i}}{N}$$, where $$N$$ is the number of measurements, $$N_{+i}$$ is the number of $$| +i \rangle$$ measurements, and $$N_{-i}$$ is the number of $$| -i \rangle$$ measurements.

I guess in the paper they mean expectation value of $$Y$$ operator, not just one simple measurement in the $$Y$$ basis, because of this line "Our predicted label value is the real number between $$−1$$ and $$1$$... which is the average of the observed outcomes if $$Y_{n+1}$$ is measured in multiple copies of...".

• You're right, in the paper, they compute the expectation value of Y. Since I am implementing this on a simulator, I run the circuit multiple times and compute the probabilities just like you mentioned. However, the simulator only does measurements in the computational basis, hence my question – Skyris May 27 '20 at 13:21