What does it mean that a spin 1/2 particle needs two full rotations?

[vijayantv]

TL;DR

what does it mean that spin 1/2 particle needs two full rotations to returns to the previous state.

I know that we can change the spin orientation of a spin 1/2 particle up or down and test it in the Stern Gerlach apparatus.
And the spin 1/2 particles need two full rotations to return to the previous state.

Questions:

1). what does state mean?

2). Is, Changing spin orientation to up or down, known as one full rotation?
(if a spin 1/2 particle is in spin-up state and it will become Spin-down after one full rotation And another full rotation returns it back to the spin-up state. is that correct?)

 

[vanhees71]

You ask the right questions to realize that the claim in your Summary is wrong.

Ad (1): In non-relativistic quantum mechanics you can describe a pure state by a wave function. In your case of a spin-1/2 particle it is a socalled Pauli spinor, which is a function $\psi(t,\vec{x}) \in \mathbb{C}^2$. It is characterized by its behavior under rotations. The rotation means you change the position vector $\vec{x}$ to $\vec{x}' = \hat{R}_{\vec{n}}(\phi) \vec{x}$, where $\hat{R}_{\vec{n}}(\phi)$ is an SO(3) matrix (i.e., a $3 \times 3$ real matrix that is orthogonal, i.e., for which $\hat{R}^{\text{T}}=\hat{R}^{-1}$ and with $\mathrm{det} \hat{R}=1$), that describes a rotation around an axis with direction $\vec{n}$ by an angle $\phi$ (according to the right-hand rule as usual).

Now a Pauli spinor also changes under rotations with a unitary matrix given by
$$\hat{U}_{\vec{n}}(\phi)=\exp(\mathrm{i} \phi \vec{n} \cdot \hat{\vec{\sigma}}/2),$$
i.e., under a rotation the Pauli-spinor valued wave function transforms as a spinor field,
$$\psi(t,\vec{x}) \rightarrow \psi'(t,\vec{x}') = \hat{U}_{\vec{n}}(\phi) \psi(t,\vec{x}) = \hat{U}_{\vec{n}} \psi(t,\hat{R}_{\vec{n}}(\phi)^{-1} \vec{x}').$$
Now it is indeed true that for any $\vec{n}$
$$\hat{U}_{\vec{n}}(\phi=2 \pi)=-\hat{1},$$
i.e., the wave function changes its sign when you apply a rotation by $2 \pi$ on it, but now saying "the wave function represents a (pure) quantum state" is too sloppy! Everything physical you describe with the state does not change when you multiply the wave function by a phase factor. This means a quantum state is described uniquely by the wave function modulo a phase factor, and this implies that the state does not change under rotations by an angle of $2 \pi$, because the wave function is only changed by a factor of $(-1)$, i.e., a phase factor $-1=\exp(\mathrm{i} \pi)$.

Ad 2): What one means by "spin flip" is that if you have a particle prepared at time $t=0$ with the spin component $s_z$ pointing "up" (having eigenvalue $+\hbar/2$), i.e., in a state described by a spinor of the form $\psi=(\psi_{1/2},0)$ and the particle is interacting, e.g., with a magnetic field (not in $z$ direction), then there is some probability that when measuring $s_z$ you get $-\hbar/2$. The change of the spin state is indeed described by a rotation for the spinor-wave function due to the Hamiltonian
$$\hat{H}=-\frac{g q}{2mc} \vec{B} \cdot \hat{\vec{s}}=-\frac{g q}{2 mc} \vec{B} \cdot \frac{\hbar}{2} \hat{\vec{\sigma}}.$$
A spin flip does not necessarily mean that the spinor underwent a rotation by $2 \pi$.

 

[PeroK]

Summary:: what does it mean that spin 1/2 particle needs two full rotations to returns to the previous state.

I know that we can change the spin orientation of a spin 1/2 particle up or down and test it in the Stern Gerlach apparatus.
And the spin 1/2 particles need two full rotations to return to the previous state.

Questions:

1). what does state mean?

2). Is, Changing spin orientation to up or down, known as one full rotation?
(if a spin 1/2 particle is in spin-up state and it will become Spin-down after one full rotation And another full rotation returns it back to the spin-up state. is that correct?)

Here's a simple answer. The spin of the particle is described by a state vector. If we rotate the physical system by $2\pi$ then the state is multiplied by $-1$. This is physically equivalent to the original state, in the sense that the results of any measurement on either state will get the same results with the same probabilities.

Nevertheless, it's not the same state. And, if you can engineer a superposition of the two states, then you will get destructive quantum interference.

 

[vijayantv]

Thanks.

Somehow it makes sense to me.

how to observe this phenomenon practically.

Can you please explain in a physical scenario using electrons, and Stern Gerlach apparatus.?

 

[PeroK]

Thanks.

Somehow it makes sense to me.

how to observe this phenomenon practically

There's a reference in Modern Quantum Mechanics by JJ Sakurai to a neutron interferometry experiment that confirms this prediction if QM.

 

[vanhees71]

Here's a simple answer. The spin of the particle is described by a state vector. If we rotate the physical system by $2\pi$ then the state is multiplied by $-1$. This is physically equivalent to the original state, in the sense that the results of any measurement on either state will get the same results with the same probabilities.

Nevertheless, it's not the same state. And, if you can engineer a superposition of the two states, then you will get destructive quantum interference.

It's not a different state as you argued yourself before. The pure state is represented by a ray in Hilbert space, not the Hilbert space vector itself or equivalently the statistical operator $|\psi \rangle \langle \psi$.

The fact that a $2\pi$ rotation changes the ket by a factor -1 for half-integer spins and by 1 for integer spins rather implies a superselection rule forbidding the superposition of half-integer and integer spin states, because otherwise indeed a $2\pi$ rotation would change the state described by such a superposition.

Nevertheless the sign change of half-integer spinors under a $2\pi$ rotation can be observed in, e.g., neutron-interferometry experiments:

https://www.google.com/url?sa=t&sou...8QFnoECAMQAQ&usg=AOvVaw08QtoJ9XxoSUUzS0BOv6GO

 

[PeterDonis]

It's not a different state as you argued yourself before.

It is if you have a superposition of the two, as in the neutron interferometry experiments you reference.