I'm having some trouble with successfully working out a rotation about the $z$-axis on the Bloch sphere. Now, I know how this is performed, in principle. A rotation of the Bloch-sphere around an axis $\boldsymbol n$ by an angle $\theta$ is given by $$ R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2} $$ where $\boldsymbol \sigma$ is the vector of Pauli-matrices.

So for example if I start in \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}

And I want to rotate by pi/2 around the x-axis, I simply compute the product with

\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} \\ \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix}

And I end up with \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{-i}{\sqrt{2}} \\ \end{pmatrix}

However, now I want to rotate this around the z-axis, by an angle theta. Now from what I can tell this simply corresponds to the matrix multiplication with \begin{pmatrix} e^{\frac{-i\theta}{2}} & 0 \\ 0 & e^{\frac{i\theta}{2}} \\ \end{pmatrix}

so that I end up with the vector

\begin{pmatrix} \frac{e^{\frac{-i\theta}{2}}}{\sqrt{2}} \\ \frac{-ie^{\frac{i\theta}{2}}}{\sqrt{2}} \\ \end{pmatrix}

However, here things go wrong. For example, I know that if I rotate by $2\pi$, I should end up back at the same spot, but instead I end up at \begin{pmatrix} \frac{-1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} \\ \end{pmatrix}

which is not correct. Could someone point out the obvious error?


1 Answer 1


The rotation operator normalization you have chosen, $ R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2} $, means that for a rotation by $2\pi$ about the $z$ axis, $$ R_{\hat z}(2\pi)=e^{-i\pi \sigma_z} =(-1)^{\sigma_z}=-1,$$ because $(-1)^1=(-1)^{-1}=-1$. Thus, your rotation operator has indeed rotated you back to the same point on the Bloch sphere (because the operator is scalar) but you have accumulated a global phase. There isn't, strictly speaking, an error, because global phases can be ignored.

On the other hand, this 'global' phase is tremendously useful if you are in contact with some other degree of freedom. It is an example of a geometric phase (i.e. if you rotate around some path in the Bloch sphere and return to the same path, you will accumulate a phase of $$\text{(area enclosed by the path in steradians)}/2,$$ where the total area to be had is of course $4\pi$). It is the foundation of what are called phase gates, which are very popular basic entangling gates between two qubits. The idea there is that you perform a controlled rotation: leave qubit A alone if B is in the down state, and rotate it once around the Bloch sphere if B is in the up state. Either way, you come back to the same state, but in one of the two paths you have accumulated a phase, which can be used to generate entanglement.

  • $\begingroup$ Hm, I hadn't thought about it that way. That makes sense, although it is slightly counterintuitive from the perspective of ending up at the same spot. But your explanation makes sense, and I'll give it some thought. Thank you. $\endgroup$
    – user129412
    Feb 24, 2014 at 16:51
  • 1
    $\begingroup$ Yeah, I guess it is counter-intuitive in a way. The thing to keep in mind is that the spot in the Bloch sphere and the actual state are only equivalent up to a global phase, and that sometimes the Bloch sphere does miss important information from there. This is quite a common paradigm when geometric phase is concerned and it takes some getting used to. See the Wikipedia page for examples, or this paper for an accessible explanation. $\endgroup$ Feb 24, 2014 at 21:37
  • $\begingroup$ That is a very, very nice paper. Certainly enlightening. Thanks a bunch! $\endgroup$
    – user129412
    Feb 25, 2014 at 11:15
  • $\begingroup$ Yeah, you're completely correct, which is why I removed the comment. I apologize. $\endgroup$
    – user129412
    Feb 25, 2014 at 12:42
  • $\begingroup$ No worries.$\quad$ $\endgroup$ Feb 25, 2014 at 12:54

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