# 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 May 1 '20 at 8:18
• Just a question though. Why is the normal measurement a Pauli Z mesurement May 27 '20 at 13:27
• The Pauli Z matrix is diagonal in the standard measurement basis. 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

$$$$|+i\rangle = |0\rangle + i |1\rangle \qquad |-i\rangle = |0\rangle - i |1\rangle$$$$

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 May 27 '20 at 13:21