# How do you rotate an electron 360 degrees?

1. Feb 28, 2014

### dsoodak

I went through an undergraduate physics program and I've internalized the results of the Stern–Gerlach experiment to the point where it makes more sense to me than the classical version.

However, one thing I somehow missed (or maybe it wasn't taught) is what "must rotate an electron 720 degrees to return it to its original state" actually means. Its not like you can just pick it up and spin it. Do you just make a series of spin measurements at regularly spaced angles?

I would really appreciate a description of or link to an experiment that shows the effect of the negative sign added to an electron's wavefunction after a 360 degree rotation.

thanks,

Dustin Soodak

2. Feb 28, 2014

### Bill_K

The neutrons in one arm of an interferometer were made to rotate 360 degrees by means of a magnetic field. Lo and behold, this changed the interference pattern.

3. Feb 28, 2014

### hilbert2

To OP:

In elementary quantum mechanics, the state of an electron is described by a wave function $\psi$. We can also form a unitary operator $U_{\theta}$ that rotates the coordinate system with respect to some axis by an angle $\theta$. For rotations around z-axis, the unitary operator is $U_{\theta}=\exp\left(-\frac{i\theta J_{z}}{\hbar}\right)$, where $J_{z}$ is the z-component of angular momentum operator.

If you operate on a wavefunction with the rotation operator corresponding to 360 degrees rotation, i.e., with $U_{2\pi}$, you get the original wavefunction back.

However, in relativistic QM of electrons and other fermions, the dynamical equation is the Dirac equation instead of the usual time-dependent Schrodinger equation, and the state of the electron is described with a four-component spinor instead of a one-component wave function. If you form the unitary rotation operator in Dirac theory, and perform a 360 degrees rotation on a spinor, you get back the negative of the original spinor. This is what this means mathematically.

4. Feb 28, 2014

### WannabeNewton

Hey Dustin. Recall that the $j = \frac{1}{2}$ representation of the rotation generators $\hat{J^i}$ gives rise to the Pauli spin matrices $\sigma^i$ which act on normalized linear combinations $|\psi \rangle$ of the two-component basis spin states $|\uparrow \rangle$ and $|\downarrow \rangle$. Now a rotation through an angle $\theta$ about an axis $\hat{n}$ will transform such a state according to $|\psi' \rangle = e^{i\theta \hat{n}\cdot \hat{J}/\hbar}|\psi\rangle = e^{i\frac{\theta}{2} \hat{n}\cdot \sigma}|\psi\rangle$. Say the rotation is about the $y$ axis then expanding out the exponential in a power series in $\theta$ and using the known Taylor series expansions of sine and cosine it is easy to see that $|\psi' \rangle = \begin{pmatrix} \cos(\theta/2) &-\sin(\theta/2) \\ \sin(\theta/2)&\cos(\theta/2) \end{pmatrix}|\psi\rangle$. Note that a rotation by $2\pi$ yields $|\psi' \rangle = -|\psi\rangle$ whereas a rotation by $4\pi$ yields $|\psi' \rangle = |\psi\rangle$.

5. Feb 28, 2014

### strangerep

Umm,.... Bill,... is that really the reference you intended?
(It also talks about nucleon structure in terms of counter-rotating vortices and "etherons". Also look at the abstract on p1 of Winfield's book. I won't quote any of it here -- for obvious reasons.)

To the OP: the description of neutron interferometry experiments is essentially ok: one can use magnetic fields to rotate an elementary fermion that has a magnetic moment. Applying such a magnetic field to just one side of an interferometer allows one to observe the effects of (e.g.,) 360deg rotation on fermions since they interfere destructively with the unrotated side.

Last edited: Feb 28, 2014
6. Mar 1, 2014

### ChrisVer

Thank the invariance of the Dirac Equation under the Lorentz Group...

7. Mar 1, 2014

### WannabeNewton

Thank it for what?

8. Mar 1, 2014

### ChrisVer

Coz it's what causes spin-1/2 particles needing 4pi rotations to return to their initial states....

9. Mar 1, 2014

### strangerep

Actually, one doesn't need the Dirac eqn or special relativity to derive those features. They're already present in the nonrelativistic (Galilean) case. Cf. Ballentine, ch7.

10. Mar 1, 2014

### WannabeNewton

In addition to what strangerep said:

Please correct me if I'm wrong but I'm not seeing the relevance of the Dirac equation. We define a Dirac field in terms of left-handed and right-handed Weyl fields which are themselves defined in terms of how they transform under Lorentz transformations e.g. $\psi_L'(x') = e^{(-i\theta - \eta)\cdot \sigma/2}\psi_L(x)$ and the rotational properties of $\psi_L$ come out of that.

11. Mar 2, 2014

### ChrisVer

what I said had nothing to do with left or right mover fields...
If you try to impose Lorentz invariance on the Dirac equation, you will get that your field will change:
$\psi'(x')= S(Λ) \psi(x)$
Under rotations the matrix $S(Λ)$ has that property....
What I mean is that you don't even need left-right movers to clear it out...
On the other hand what you wrote is exactly what I said- invariance under lorentz transformation needs the 4pi rotations (in your case the $iθσ/2$ in the exponential)

Also for the non relativistic case, yes- but the Dirac equation brings from within the properties of spin1/2 particles , something that the non relativistic QM didn't... If I recall well, we imposed the spins in QM, while for the dirac case we just take it out of the theory...

12. Mar 2, 2014

### strangerep

We don't need to "impose" the spins in nonrelativistic QM. They're a consequence of representing the rotation group unitarily on Hilbert space. Ref: Ballentine ch7.

There's also this paper by Levy-Leblond which (among other things) constructs wave equations for nonrelativistic particles of any spin. (That Springer url is a paywall, but if you search on Google Scholar, it seems easily downloadable via researchgate.net.)

13. Mar 3, 2014

### dsoodak

Thanks for all the replies.
The description of the experiment referenced in the linked book says that they rotated neutrons with a magnetic field (neutrons having a magnetic moment is another thing I missed or forgot).
I also appreciate the summaries of the pertinent equations.

14. Mar 3, 2014

### dsoodak

One last thing...They basically use Larmor precession formulas to calculate what kind of magnetic field will rotate it exactly 360 degrees, right?

15. Mar 3, 2014

Yes.