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How do we perform a measurement of an arbitrary 1-qubit quantum state in any arbitrary orientation?

Quantum Computing Asked on January 17, 2021

Let’s imagine we have an arbitrary 1-qubit quantum system $alpha vert 0 rangle + beta vert 1 rangle$ Making a measurement in the +/- basis is equivalent to performing a Hadamard gate and then making a measurement in the standard computational basis.

Can we extend this notion generally for measurement of an arbitrary 1-qubit state in any arbitrary orientation? So, if we want to make a measurement of an arbitrary 1-qubit quantum state in some arbitrary orientation in the 2-dimensional Bloch sphere, can we perform some unitary transformation $U$ (supposing such $U$ can be composed theoretically) that transforms the basis state into orthonormal states of that orientation and then perform a measurement in the computational basis?

Let’s say we have unitary transformation $U$ that maps $vert 0 rangle$ to $vert a rangle$ and $vert 1 rangle $ to $vert b rangle$ with $langle a vert b rangle = 0$ (where $vert a rangle$ and $vert b rangle$ have certain orientation with respect to the standard basis states) the same way $H$ transforms $vert 0 rangle$ to $vert + rangle$
and $vert 1 rangle $ to $vert – rangle$ state. One difference could be that $H$ is Hermitian too while $U$ may not be.
Another sub question is whether $U$ needs to be Hermitian also to make this generalization.

One Answer

Yes, this observation can be generalized. To start with, let's notice why is Hadamard the transformation required to measure a state $| psi rangle$ in the $sigma_{x}$ basis. This is because it is the ``unitary intertwiner'' connecting the $sigma_{x}$ basis to the $sigma_{z}$ basis (a.k.a. computational basis). Recall that the $sigma_{x}$ eigenvectors are ${ | + rangle, | - rangle }$ and the $sigma_{z}$ eigenvectors are ${ | 0 rangle, | 1 rangle }$. The unitary operator connecting these basis is: $$ mathcal{U}_{sigma_{x} rightarrow sigma_{z}} = | 0 rangle langle + | + | 1 rangle langle - | = H. $$

Let's take a moment to interpret the action of this intertwiner: when it acts on the $| + rangle$ state, it sends it to the $| 0 rangle$ state and when it acts on the $| - rangle$ state, it sends it to the $| 1 rangle$ state, thereby connecting the basis elements (and by linearity, any other vector expressed in these bases). Therefore, measuring $| psi rangle$ in the $sigma_{z}$ basis is the same as applying $mathcal{U}_{sigma_{x} rightarrow sigma_{z}}$ and then measuring in the $sigma_{z}$ basis.

For connecting to an arbitrary basis, we simply replace ${ | pm rangle }$ with the new basis vectors, say $ mathbb{B} = { | phi_{+} rangle, | phi_{-} rangle }$, giving us, $$ mathcal{U}_{mathbb{B} rightarrow sigma_{z}} = | 0 rangle langle phi_{+} | + | 1 rangle langle phi_{-} | $$

In the most general case, where you want to connect a basis $mathbb{B}_{0} = { phi_{j} }$ with $mathbb{B}_{1} = { chi_{j} }$, the intertwiner is defined as, $$ mathcal{U}_{mathbb{B}_{0} rightarrow mathbb{B}_{1}} = sumlimits_{j=1}^{d} | chi_{j} rangle langle phi_{j} | . $$

Update: No, the unitary doesn't need to be hermitian as well, it just so happens in the specific case because we're dealing with Pauli matrices, which are both unitary and hermitian. Given any two bases, there always exists a unitary connecting them (that is essentially what I construct above).

Correct answer by keisuke.akira on January 17, 2021

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