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.

  • $\begingroup$ Afaik we have no \ket, but we have | and \rangle: $|\alpha \rangle$. $\endgroup$
    – peterh
    Jul 31, 2020 at 18:33
  • $\begingroup$ Thanks for the edit. $\endgroup$ Aug 1, 2020 at 3:58
  • $\begingroup$ @RabinsWosti No problem, and welcome to QCSE. $\endgroup$ Aug 1, 2020 at 5:40
  • $\begingroup$ you can use \ket, you just need to include $\newcommand{\ket}[1]{\lvert #1\rangle}$ to the beginning of the post $\endgroup$
    – glS
    Aug 3, 2020 at 12:07

1 Answer 1


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, $$ \mathcal{U}_{\mathbb{B}_{0} \rightarrow \mathbb{B}_{1}} = \sum\limits_{j=1}^{d} | \chi_{j} \rangle \langle \phi_{j} | . $$

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).

  • $\begingroup$ I actually meant to ask if U needs to be Hermitian(not idempotent) also to make the generalization. Sorry for the mistake. $\endgroup$ Aug 1, 2020 at 5:03
  • $\begingroup$ @RabinsWosti I see. I'm updating my answer now to clarify this. $\endgroup$ Aug 1, 2020 at 5:14

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