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I started with Qiskit today and find it very exciting. As a first question I want to understand how to measure an arbitrary state $|\Psi\rangle$ not in the basis of Z ($|1\rangle$, $|0\rangle$) but in the basis of Y ($|R\rangle$, $|L\rangle$).

Since measurements on qubits in Qiskit are carried out in the Z-basis (I would understand it in that way), I have to transform the state by a transformation

$|\Psi'\rangle = U |\Psi\rangle$

into a new state, on which the measurement is done in the Z-basis. U is to be determined. So I want a "mapping" from $|R\rangle \rightarrow |0\rangle$ and $|L\rangle \rightarrow |1\rangle$

The requirement on U is, therefore:

  • the amplitude of $|R\rangle$ in $|\Psi\rangle$ equals to the amplitude of $|0\rangle$ in $U|\Psi\rangle$

  • the amplitude of $|L\rangle$ in $|\Psi\rangle$ equals to the amplitude of $|1\rangle$ in $U|\Psi\rangle$

So I have

  • $\langle 0|U|\Psi\rangle = \langle R|\Psi\rangle$
  • $\langle 1|U|\Psi\rangle = \langle L|\Psi\rangle$

or

  • $U^\dagger|0\rangle = |R\rangle$
  • $U^\dagger|1\rangle = |L\rangle$

Therefore, the Z-base matrix-representation is

$U^\dagger = \begin{pmatrix}1 & 1& \\ i &-i\end{pmatrix} $

$U = \begin{pmatrix} 1 & -i& \\ 1 &i\end{pmatrix} $

This is a well known result: Actually, it can be written as

$U = H S^\dagger$

The circuit in Qiskit is as follows

enter image description here

and works as expected:

enter image description here

I think that's all right so far, but I dislike the my clumsy way of bringing up U. Is there a better, more direct approach? It has nothing to with Qiskit, but the way in general.

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    $\begingroup$ For $\langle Y \rangle$ you should note that $(SH)Z(HS^\dagger) = Y $ $$\langle \psi |Y| \psi \rangle = \langle \psi | (SH)Z (HS^\dagger) | \psi \rangle = \langle \psi SH | Z | H S^\dagger \psi \rangle $$ So you want to apply $S^\dagger$ follow by the Hadamard gate $H$ before measurement in computational basis $\endgroup$
    – KAJ226
    Oct 14 at 21:06
  • $\begingroup$ This is much more elegant, but how should I see the identity Y = ... ? For me this identity is not very intuitive. $\endgroup$
    – MichaelW
    Oct 15 at 12:44
  • $\begingroup$ See Adam's answer. You should from there see that $\frac{1}{\sqrt2}\begin{bmatrix}1 & -i \\ 1 & i\end{bmatrix} = HS^\dagger $ $\endgroup$
    – KAJ226
    Oct 15 at 15:58
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Yes, there is a simple general formula for the unitary $U$ that maps orthonormal basis $|u_1\rangle, |u_2\rangle, \dots, |u_n\rangle$ to orthonormal basis $|v_1\rangle, |v_2\rangle, \dots, |v_n\rangle$

$$ U = |v_1\rangle\langle u_1|+|v_2\rangle\langle u_2|+\dots+|v_n\rangle\langle u_n|=\sum_{i=1}^n|v_i\rangle\langle u_i|. $$

It is easy to check that $U$ is indeed unitary and that it maps $|u_k\rangle$ to $|v_k\rangle$ for each $k=1,\dots,n$.


For example, a $U$ transforming the $Y$ eigenbasis to $Z$ eigenbasis is

$$ \begin{align} U &= |0\rangle\langle R| + |1\rangle\langle L| \\ &= \begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt2} & -\frac{i}{\sqrt2}\end{bmatrix} + \begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt2} & \frac{i}{\sqrt2}\end{bmatrix} \\ &= \frac{1}{\sqrt2}\begin{bmatrix}1 & -i \\ 1 & i\end{bmatrix} \end{align} $$

as expected.

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