Can everything in QM be described with degrees instead of matrices and vectors?

Question

I found this explanation.

"The Hadamard gate can also be expressed as a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis. So $H=XY^{1/2}H = X Y^{1/2}H=XY^{1/2}$."

Can everything in QM be explained with degrees instead of matrices and vectors?

Answer 1

In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates".

For example, you can write an arbitrary 3-modes pure state as $$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\rangle,$$ for $\alpha,\beta,\theta,\phi\in\mathbb R$.

It should be noted, however, that things get much with spaces of dimension larger than $2$. For example, it's harder to interpret arbitrary unitary gates as rotations in this larger space. By this, I mean that even if it is always true that for any given unitary $U$ there is some (and in general an infinity of) Hermitians $H$ such that $e^{-iHt}=U$ at some time $t$, whether these should be considered "rotations" is arguable.

On the one hand, if you represent $H$ as a point in the Bloch representation (you can always do this because the Bloch representation associates a point to any Hermitian matrix, even though this point will in general fall outside of the region representing the set of physical states), then you can think of $H$ as the "axis" of the rotation, as the direction associated with this point will be fixed by the rotation.

On the other hand, in general $t\mapsto e^{-iHt}$ is not a rotation, in the sense that it doesn't "loop back" as rotations do in the Bloch sphere. By this I mean that in general there is no $t>0$ such that $e^{-iHt}=I$, which is what you would expect from a rotation. An easy example of this is: $$H\equiv \begin{pmatrix}\alpha&0&0\\0&\beta&0\\0&0&\gamma\end{pmatrix},\quad\alpha,\beta,\gamma\in\mathbb R.$$ Then, $e^{-iHt}=\mathrm{diag}(e^{-i\alpha t},e^{-i\beta t},e^{-i\gamma t})$, and $e^{-iH t}=I$ if and only if $e^{-i\alpha t}=e^{-i\beta t}=e^{-i\gamma t}=1$, which aren't simultaneously satisfiable for incommensurable $\alpha,\beta,\gamma$.

Answer 2

The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not hold for all quantum systems.

Answer 3

Short answer: Yes, except for measurement.

There two postulates in play here:
1- the evolution of a closed QM system can always be described by a unitary matrix.
2- measurement operators (observable operator) are always hermitian

Hermitian operators can't (always) be described as a rotation in any space.

Unitary operators can always be considered as rotation around some axis since they always preserve the inner products. However the dimensions of the space where the rotation happens increases as the dimension of the system increase:

• operation on 1 qubit is a rotation on the 2D surface of a 3D sphere (Bloch sphere).
• operation on m qubits is a rotation on the $2^{m+1}$ − 2 dimensions surface of some $2^{m+1}$ − 1 sphere.

Which makes this visualization (degrees of rotation) not very helpful when applied to a multi-qubit system.