So starting with an up particle: $$ \lvert \uparrow \rangle = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} $$ My understanding is that you can measure $\lvert \uparrow \rangle$ in $X$ and have a 50% chance of getting $\lvert \leftarrow \rangle$ and a 50% chance of getting $\lvert \rightarrow \rangle$ where: $$ \lvert \rightarrow \rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} \text{ and } \lvert \leftarrow \rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{bmatrix} $$ but you could also operate in $X$ which would be the equivalent of passing it though a NOT gate: $$ X \cdot \lvert \uparrow \rangle = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \lvert \downarrow \rangle $$

I've read that measuring qubit spin is more or less equivalent to measuring the orientation of a photon which can be done by passing it through polarized filters. If the photon is measured to be in one orientation and then is measured in a different orientation it has a certain probability of snapping to be in that other direction.

So in this example, that would be equivalent to measuring an up qubit in $X$. But how does operating in $X$ come into the equation? I understand the mathematical effect it has, but what does this mean for the physical qubit? Is it also being passed through a filter and how is that different from measuring?


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    $\begingroup$ you can rotate spin using magnetic field; I don't remember how exactly to do it, but it should be explained in good QM courses. $\endgroup$
    – kludg
    Oct 16, 2019 at 10:49

1 Answer 1


In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it).

What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (reversibly) states into other states according to some rule.

In the particular case of $X$, you can actually think of it as an evolution, because $X$ is both Hermitian (thus an observable) and unitary (thus an evolution). What this evolution means physically depends on the implementation. If you are dealing with the polarisation of a photon, it can be representing the action of a phase shifter, which rotates the polarisation.

But it's important to distinguish between "acting on a state with the evolution $X$" and "measuring the state with the observable $X$". To "measure $X$" means physically to apply a measurement which makes the state collapse into one of the eigenstates of $X$. This involves doing something like acting on the state with a Hadamard gate and then measuring in the computational basis, so you can see how this is rather different than evolving with $X$.


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