Why does a rotation of $\pi/2$ on $|1\rangle$ yield $i|1\rangle$?

It seems $$V(\pi/2, \qr{1}) = i \qr{1}.$$ I didn't expect that. To me $$\qr{1}$$ points up because $$\qr{0}$$ points to the right. So rotating $$\qr{1}$$ by $$\pi/2$$ should yield $$-\qr{0}$$. What am I missing here?


When people talk about single qubit rotations, the rotations they are visualising are generally rotations on the Bloch Sphere. In this picture, orthogonal states are represented by vectors that point in opposite directions, so $$|0\rangle$$ points up while $$|1\rangle$$ points down. Since the Bloch sphere is a 3 dimensional shape, there is a lot of freedom to pick the axis that is being rotated about. Since you don't define $$V$$ or describe where it comes from, I can only guess, but my guess would be that you're talking about a rotation about the Z axis, which preserves the states $$|0\rangle$$ and $$|1\rangle$$, up to phases. You're probably thinking (more or less) about a rotation about the X axis.

• What's the X axis in this picture? – R. Chopin Oct 18 '18 at 16:21
• @R.Chopin the one marked by + – DaftWullie Oct 18 '18 at 16:35
• And where would be $i\langle 1 \rangle$? – R. Chopin Oct 18 '18 at 18:55
• @R.Chopin the i makes no difference (global phases don’t change states). The 1 is at the bottom (South Pole). – DaftWullie Oct 18 '18 at 20:14

First off, you may have some misunderstanding regarding the placement of Bloch vectors on the sphere. The placement of a state on the sphere is dictated by the following parametrisation of this state:

$$|\psi\rangle=\cos\theta/2\,|0\rangle+e^{i\phi}\sin\theta/2\,|1\rangle.$$

The parameters ($$\theta, \phi$$) are then taken as the tangential and azimuthal angles for the Bloch vector representation. As such, the $$|0\rangle$$ and $$|1\rangle$$ vectors, parametrised by $$(0, 0)$$ and $$(2\pi, 0)$$ respectively, are parallel to one another on the sphere – as confusing as it may seem – since $$|0\rangle$$ points up (by convention) and $$|1\rangle$$ points down.

Next, it seems that you are rotating about the Z axis, instead of about the X or Y axis. The exponentiated operator $$e^{-i\alpha/2\,{\rm Z}}=:{\rm R}_{\rm Z}(\alpha)$$, where $$\alpha\in(-2\pi, 2\pi]$$, is a rotation about the Z axis, and has the following matrix form: $${\rm R}_{\rm Z}(\alpha)= \begin{bmatrix}e^{-i\alpha/2}&0\\ 0&e^{i\alpha/2}\end{bmatrix}.$$ You can check that this indeed corresponds to a Z rotation of a Bloch vector by the same angle.

Assuming you took $$V(\beta,|\psi\rangle)={\rm R}_{\rm Z}(2\beta)|\psi\rangle$$, apply this operator to the $$|1\rangle$$ state with $$\beta=\frac\pi2$$ and you get $${\rm R}_{\rm Z}(\pi)|1\rangle=e^{i\pi/2}|1\rangle=i|1\rangle.$$

The X and Y rotation operators, on the other hand, look like this:

$${\rm R}_{\rm X}(\alpha)=e^{-i\alpha/2\,{\rm X}}= \begin{bmatrix}\cos\alpha/2&-i\sin\alpha/2\\ -i\sin\alpha/2&\cos\alpha/2\end{bmatrix},$$

$${\rm R}_{\rm Y}(\alpha)=e^{-i\alpha/2\,{\rm Y}}= \begin{bmatrix}\cos\alpha/2&-\sin\alpha/2\\ \sin\alpha/2&\cos\alpha/2\end{bmatrix}.$$

Following the same reasoning, let us apply $${\rm R}_{\rm X}(\pi)$$ and $${\rm R}_{\rm Y}(\pi)$$ to the $$|1\rangle$$ state:

$${\rm R}_{\rm X}(\pi)|1\rangle=-i\sin\pi/2\,|0\rangle+\cos\pi/2\,|1\rangle=-i|0\rangle,$$

$${\rm R}_{\rm Y}(\pi)|1\rangle=-\!\sin\pi/2\,|0\rangle+\cos\pi/2\,|1\rangle=-|0\rangle.$$

Long story short, it depends on the axis about which you rotate (as is always the case with rotations). In this case you were expecting a Y rotation, but you did a Z rotation.

For more information, I highly recommend you check out this excellent lecture by Ian Glendinning on Pauli rotations.