# Interpretation of rotating a quantum state

I'm currently taking an introductory quantum mechanics course, where the last covered topic was spin and identical particles. During some reading online, I stumbled across the term "rotating a quantum state" (it was related to the fact that rotating a spin 1/2 state by $$2\pi$$ let's the state pick up a minus sign). To get to my question: what is meant by "rotating a quantum state"? Is there any intuitive interpretation?

Yes, it means exactly the same thing it means in classical mechanics.

For example, suppose you describe the state of a classical particle with its position $$\mathbf{r}$$ and $$\mathbf{p}$$. Then after a rotation, these quantities will be modified by a rotation matrix, $$\mathbf{r} \to R \mathbf{r}, \quad \mathbf{p} \to R \mathbf{p}$$ where the two $$R$$'s here have to be the same, because both are vectors. Different quantities can transform in different ways. For example, the energy of particle just stays the same, $$E \to E.$$ On the other hand, the moment of inertia of a rigid body is a matrix, and it turns out to transform as $$I \to R^T I R.$$ To perform a rotation on a classical state means to take it to another classical state, the state it would be in if you physically rotated the entire system.

In quantum mechanics, everything works the same way. For example, a point particle might be specified by a position, so its state could be $$|\mathbf{r} \rangle$$. Under a rotation, $$|\mathbf{r} \rangle \to |R \mathbf{r} \rangle.$$ That's what we mean by "rotating the state". It's just the result of a physical rotation.

The point at which this departs from classical mechanics is that in quantum mechanics, there are things that can transform very weirdly under rotations, even weirder than $$I$$ in classical mechanics. For example, many courses spend a lot of time talking about the rotation properties of spin $$1/2$$ particles. But you shouldn't forget that the principle is the same: we are just cataloging what can happen when you pick up the system and physically rotate it. It's just that in quantum mechanics, there are more and stranger possibilities.

The simplest way to think about “rotating a quantum state” is to think about rotating the physical apparatus used to measure the quantum state. (For example, you might change the direction of a magnetic field that you are using to measure an electron’s spin.) This is a “passive” rotation rather than an “active” one, but they are completely equivalent, and you don’t have to think about things like “how do I rotate an electron?”

A previous question asks about qubits as the Bloch sphere and I point out that a spin-1/2 system in a state $$|\psi\rangle = \cos(\theta/2) |0\rangle + e^{i\phi}\sin( \theta/2)|1\rangle$$ is indeed in an eigenstate of an operator that works out to be $$\sin\theta\cos\phi ~\sigma_x + \sin\theta\sin\phi~\sigma_y + \cos\theta~\sigma_z,$$ so these are spherical coordinates $$(\theta, \phi)$$ of the direction that the spin-1/2 system is spinning in.

Now let us apply a Hamiltonian proportional to $$\sigma_z$$ or $$|0\rangle\langle 0| - |1\rangle\langle 1|$$. The Schrödinger evolution of this is exceptionally simple, it maps for some $$\eta$$, \begin{align}|0\rangle &\mapsto e^{i\eta t}|0\rangle,\\ |1\rangle &\mapsto e^{-i\eta t}|1\rangle.\end{align} In turn the state maps to $$|\psi\rangle \mapsto e^{i\eta t}\left[ \cos\left(\frac\theta2\right) ~|0\rangle + e^{i(\phi - 2\eta t)} \sin\left(\frac\theta2\right)~|1\rangle\right].$$

The key thing to look at here is this expression $$\phi - 2\eta t$$, which will return to the same rotation $$\phi - 2\pi$$ when $$\eta t = \pi$$ due to the periodicity of $$e^{i\theta}.$$ You can see that under this Hamiltonian the qubit simply precesses about the $$z$$-axis. Something similar happens for $$\sigma_{x,y}$$ but the math is a little more complicated so I cheated and took the easiest case.

However we do see that at this time there is also a global prefactor $$e^{i\eta t} = -1$$ and the state has been mapped to its own negative under this transformation. I mentioned in the comments to that answer to that previous question, such a global prefactor is unobservable in quantum mechanics. But that is true only if it is truly a global prefactor. This setup we have discussed could be applied to a two-qubit system to change either of the two qubits' phase prefactors independently, and that change is indeed observable. So the fact that the one spin-1/2 system maps to its negative when we rotate it by $$2\pi$$ does indeed have physical consequences and marks that system as a spinor.

Let $$\left\{J_i\right\}_{i=1,2,3}$$ a collection of operators such that $$[J_i, J_j] = \epsilon_{ijk}J_k$$ and let $$\mathfrak{D}(\theta) = \textrm{exp}\big(-i\theta\mathbf{J}\cdot{\mathbf{n}}\big)$$; we refer to such an operator as to "rotations". Given a quantum state $$|\psi\rangle$$, the state $$|\psi\rangle' = \mathfrak{D}(\theta)|\psi\rangle$$ is the $$\theta$$-rotated quantum state, given $$|\psi\rangle$$. By applying the above using the Pauli matrices you end up with a minus sign.

• That does not answer my question. My question was how a rotation of a quantum state is to be understood in the first place. – pjHart1000 Mar 19 '19 at 20:35
• What do you mean by " is to be understood"? There is nothing to understand, it is just another state obtained applying a linear transformation to the original one. Most you can say is that rotations preserve the norm, but given that as states we take the equivalence classes of rays in the Hilbert space, the norm is unitary anyway. – gented Mar 19 '19 at 20:38
• @gented You could at least add some vocabulary to the answer, s.t. people can look stuff up. Like Lie group/algebra, or that the angular momentum and spin operators fulfill the Lie bracket constraint. In the end his question boils down to that of "In which space does the rotation happen?", while the reason for the behavior of the spin is SU(2) being a double cover of SO(3) (rotation group), I guess. – Paul G. Mar 19 '19 at 20:58
• @Paul well, that still wouldn't make the rotation "more understandable", would it? :p – gented Mar 19 '19 at 21:12
• @gented Well to me it does, at least up to a point. A rotation in 3D position space is sth. one can easily imagine. A rotation in the inner 3D spin space not as easy, but still possible. – Paul G. Mar 19 '19 at 21:22

I think he means, is the particle rotating, or is it really a point particle? What is the intuitive notion behind a spinning quantum particle?

Some might say, the system is not really rotating, but there have been a number of scientists which aren't or haven't been sure about this topic - certainly the behaviour of particles with angular momentum suggests they could be spinning. The thing is, there is no issue, in classical physics - classical physics pretty much predicted a small enough ball would interact like a point. Points don't spin, if an electron could for instance (assuming it really was a point) would have to spin 720 degrees to reach its original orientation - instead of believing that may indicate there is something wrong with viewing an electron as a point particle, we seem to have embraced it, tagging it an ''intrinsic property.'' Interestingly, some scientists attempted to measure the shape of an electron - where there is shape, there is area... they found it was pretty spherical. This spherical picture they built up was in fact a measure of its charge distribution, so maybe electrons, one of the smallest particles in the standard model, may not be pointlike. Likewise, evidence from rotating molecules, called rotational bands suggests that there are real rotation going on for atoms at least.