# Hamiltonian on qubit: necessarily induces rotation. What about the reflections? Are they impossible to implement?

## 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.

$$U(2) \to U(2)/U(1) = SO(3)$$
• 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 ? Commented Nov 25, 2020 at 0:35
• 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. Commented Nov 25, 2020 at 0:36
• @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. Commented Nov 25, 2020 at 1:45
• 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 ? Commented Nov 25, 2020 at 9:55
• @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. Commented Nov 25, 2020 at 22:32