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Any Hamiltonian dynamic on a qubit is necessarily a rotation.

I consider a two level system having a Hamiltonian $H$. Decomposing this Hamiltonian on the Pauli basis and using the fact it is Hermitian, it is easy to show that it can be rewritten as: $$H=\frac{\hbar \Omega}{2}\overrightarrow{n}.\overrightarrow{\sigma}$$ This Hamiltonian implements the following unitary when evolved to a time $t$:

$$U_t = e^{-\frac{i \Omega t}{2} \overrightarrow{n}.\overrightarrow{\sigma}}$$

Such transformation induces a rotation in the Bloch-sphere around the axis $\overrightarrow{n}$ of an angle $\theta=\Omega t$.

Then, any (time-independent) Hamiltonian necessarily performs a rotation in the Bloch sphere. Why not.

Reflexion is a "valid" operation to do on a Bloch vector.

On the other hand, a qubit density matrix can be written as:

$$ \rho = \frac{1}{2} \left( \sigma_0 + \overrightarrow{p}.\overrightarrow{\sigma} \right)$$

Seeing it as a column vector $(1,\overrightarrow{p})$ (its decomposition in the Pauli basis), its evolution for a trace-preserving process can be computed through the Pauli-transfer matrix that has the following shape:

$$M=\begin{pmatrix} 1 & \overrightarrow{0} & \\ \overrightarrow{t} & A \end{pmatrix}$$

Where $A$ is a $3$x$3$ matrix, there are $0$'s on the first line for trace preservation. In the particular case $\overrightarrow{t}=0$ (unital map), $A$ is simply a matrix acting on $\overrightarrow{p}$. I could imagine $A$ being a reflexion: $A^T A= I$ but $det(A)=-1$

And this is where I get confused: We saw that a time-independent Hamiltonian necessarily induces a rotation. But as we can reason in $\mathbb{R}^3$ when seeing a qubit, reflections are perfectly valid operations to perform on it.

Does that imply that reflections are:

  • Not unitary operations when "translated" on the Bloch sphere
  • Are unitary operations, but cannot be implemented by a time-independent Hamiltonian (but I guess my example could be generalized as well to those cases...)

I would also be interested in a generalisation of this in higher dimension.

Note: I think this is very closely related to some notions of group theory, like the relationship between $U(n)$ and $O(n)$. But I have extremely bad memories of all that. So if an answer uses those notions, could you please remind me the very basic properties to know about this.

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The reflection isn't unitary. The action of unitaries on the density matrix is by conjugation, which is trivial for the scalar matrices, giving a reduction

$$U(2) \to U(2)/U(1) = SO(3)$$

meaning there are only rotations, not reflections.

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  • $\begingroup$ Thank you for your answer. Naive questions from a big beginner of group theory. $U(2)/U(1)$ is the quotient group. Physically here it means you "remove" the global phase from the description (you identify to a unique element all unitary differing from a global phase). Is that correct ? The equality with $SO(3)$ means isomorphic I guess. Do you have a reference for this ? $\endgroup$
    – StarBucK
    Commented Nov 25, 2020 at 0:35
  • $\begingroup$ Also. Is there an equivalent property in higher dimensions ? Up to a global phase $U(N)/U(1)$ is isomorphic to $SO(N+1)$ ? I am assuming I was correct in my previous comment. $\endgroup$
    – StarBucK
    Commented Nov 25, 2020 at 0:36
  • $\begingroup$ @StarBucK To your first comment: yes exactly. Two matrices in $U(2)$ which differ by $e^{i \theta}$ are identified. The isomorphism with $SO(3)$ is actually special to this case. For larger $N$ $U(N)/U(1)$, denoted $PU(N)$ for the projective unitary group, is just its own thing. To prove the isomorphism one can simply calculate that if we conjugate the matrix $\rho = 1 + a \sigma^x + b \sigma^y + c \sigma^z$ by $e^{i \theta \sigma_x}$, the effect is a rotation of the vector $(a,b,c)$ by an angle $2\theta$ about x. This gives us the map $U(2) \to SO(3)$, then we just need to find the kernel. $\endgroup$
    – Ryan Thorngren
    Commented Nov 25, 2020 at 1:45
  • $\begingroup$ Thank you. I will try to see the proof on my side with your suggestion. So just to be sure: a $N>3$ dimension reflexion in $R^N$ could have a corresponding unitary matrix ? $\endgroup$
    – StarBucK
    Commented Nov 25, 2020 at 9:55
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    $\begingroup$ @StarBucK For general density matrices the action is given by the adjoint representation of $U(N)$ acting by conjugation on Hermitian matrices. This gives a map $U(N) \to SO(N^2-1)$. A simple reason it cannot contain a reflection is because $U(N)$ is connected, while $O(N^2-1)$ is disconnected. $\endgroup$
    – Ryan Thorngren
    Commented Nov 25, 2020 at 22:32

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