How do we determine the direction of a time evolution $e^{-itH}$ on the Bloch sphere?

I have a question about the direction of time evolution on a Bloch sphere: suppose I'm performing a unitary time-evolution $$\exp(-iHt/\hbar)$$ for a single qubit, then on the Bloch sphere it corresponds to a rotation around an axis $$\hat n$$, which is an eigenvector of the Hamiltonian. My question is how do we determine if the time evolution corresponds to a clockwise or counterclockwise rotation of $$\hat n$$? Is the direction dependent on the sign of energy eigenvalue? Thanks!

• I believe that Bloch rotations are assumed counter-clockwise by convention (assuming a right-handed coordinate system) Mar 30 '21 at 17:10
• It's basically just a convention because you could easily state the vector to be $\vec{n}$ or $-\vec{n}$, and the corresponding direction would be opposite in the two cases. May 14 '21 at 6:44

Let $$H$$ be a $$2\times 2$$ Hermitian matrix.

You can always choose a basis with respect to which $$H$$ is represented as a diagonal matrix, so let us fix this basis. We can also assume, without loss of generality, that the first eigenvalue is zero (because changing the Hamiltonian by a constant factor doesn't affect the physics).

Then, $$H=\operatorname{diag}(0,\lambda)$$ for some $$\lambda\in\mathbb R$$, and $$e^{-iHt} = \begin{pmatrix}1 & 0 \\ 0 & e^{-i\lambda t}\end{pmatrix}.$$ Let $$|\psi\rangle=\cos(\theta/2)|0\rangle+\sin(\theta/2)e^{i\phi}|1\rangle$$, with $$\theta,\phi\in\mathbb R$$, be an arbitrary pure state. You can easily verify that, in the Bloch sphere, this (or more precisely, the corresponding projection) has coordinates $$|\psi\rangle\!\langle\psi| = \frac12\left(I + \cos(\theta)Z + \sin(\theta)\cos(\phi) X + \sin(\theta)\sin(\phi) Y\right).$$ At the same time, $$e^{-iHt}|\psi\rangle = \cos(\theta/2)|0\rangle +e^{i(\phi-\lambda t)}\sin(\theta/2)|1\rangle,$$ which in Bloch coordinates reads $$e^{-iHt}|\psi\rangle\!\langle \psi| e^{iH t} = \frac12[I + \cos(\theta)Z + \sin(\theta)\cos(\phi-\lambda t)X + \sin(\theta)\sin(\phi-\lambda t) Y].$$ In other words, upon rotation, the $$z$$ coordinate of $$|\psi\rangle$$ remains unchanged, whereas $$x$$ and $$y$$ coordinates transform as $$x \to x \cos(\lambda t) + y \sin(\lambda t), \\ y \to y \cos(\lambda t) - x \sin(\lambda t).$$ It follows that, if $$\lambda>0$$, then the rotation follows the right-hand rule (i.e. it is counterclockwise around the axes corresponding to the Hamiltonian's eigenvector).

To state this more precisely: given the standard conventions on how the Bloch representation is defined, the rotation is counterclockwise with respect to the direction going from the origin to the first eigenvector of $$H$$, assuming the phase difference between second and first eigenvalues of $$H$$ is positive.

• Thanks so much for the great explanation:) If $\lambda$ is negative, should the rotation be clockwise instead?
– ZR-
Mar 30 '21 at 19:59
• @ZR- yes. Or if $t<0$, or if you define the evolution operator as $e^{iHt}$ rather than $e^{-iHt}$. These are all purely conventional choices
– glS
Mar 30 '21 at 20:20