Editor’s note: The following is a guest post contributed by Anshul Kogar.  Anshul is a postdoctoral researcher in experimental condensed matter physics who currently splits his time between Argonne National Laboratory and the University of Illinois at Urbana-Champaign.  He maintains the blog This Condensed Life, which discusses conceptual ideas and recent developments in condensed matter physics.

This may surprise you, but the answer to the question in the title leads to some profound quantum mechanical phenomena that fall under the umbrella of the Berry or geometric phase, the topic I’ll be addressing here. There are also classical manifestations of the same kind of effects, a prime example being the Foucault pendulum.

Let’s now return to the original question. Ultimately, whether or not one returns to oneself after walking in a closed loop depends on what you are and where you’re walking, as I’ll describe below. The geometric phase is easier to visualize classically, so let me start there. Let’s consider a boy, Raj, who is pictured below: Raj

Now, Raj, for whatever reason, wants to walk in a rectangle (a closed loop). But he has one very strict constraint: he can’t turn/twist his body while he’s walking. So if Raj starts his walk in the top right corner of the rectangle (pictured below) and then walks forward normally, he has to start side-stepping when he reaches the bottom right corner to walk to the left. Similarly, on the left side of the rectangle, he is constrained to walk backwards, and on the top side of the rectangle, he must side-step to the right. The arrow in the diagram below is supposed to indicate the direction which Raj faces as he walks. When Raj returns to the top right corner, he ends up exactly in the same place that he did when he started — not very profound at all! But now let’s consider the case where Raj is not walking on a plane, but walking on the surface of a sphere: Again, the arrow is supposed to indicate the direction that Raj is facing. This time, Raj starts his trek at the north pole, heads to Quito, Ecuador on the equator, then continues his walk along the equator and heads back up to the north pole. Notice that on this journey, even though Raj obeys the non-twisting constraint, he ends up facing a different direction when he returns to the north pole! Even though he has returned to the same position, something is slightly different. We call this difference anholonomy.

Why did anholonomy result in the spherical case and not the flat rectangle case? Amazingly, it turns out to be related to the hairy ball theorem (don’t ask me how it got its name). Crudely speaking, the hairy ball theorem states that if you have a ball covered with hairs, you can’t comb the hairs straight without leaving at least one little bald spot or tuft. In the image below you can see a little tuft on both the top and bottom of the sphere: The sphere which Raj traversed had a little bald spot at the north pole, leading to the anoholonomy.

Now before moving onto the Foucault pendulum, I want to explicitly state the items that were critical in obtaining the anholonomy for the spherical case: (i) The object that is transported must have a direction (i.e. be a vector); (ii) The object must be a transported on a surface on which one cannot properly comb hair.

Now, how does the Foucault pendulum get tied up in all this? Well, the Foucault pendulum, swinging in Paris, does not oscillate in the same plane after the earth makes a full 24-hour rotation. This difference in angle between the original and the next-day plane of oscillation is also an anholonomy, except the earth rotates instead of the pendulum taking a walk. Check out this great animation from Wikipedia below: If the pendulum was at the north (south) pole, the pendulum would come back to itself after a 24-hour rotation of the earth, $\theta=360^o$ ( $\theta=-360^o$). If the pendulum was at the equator, the pendulum would not change its oscillation plane at all ( $\theta=0^o$). Now, depending on the latitude in the northern hemisphere where a Foucault pendulum is set up, the pendulum will make an angle between $0^o$ and $360^o$ after the earth makes its daily rotation. The north pole case, in light of the animation above, can be inferred from this image by imagining the earth rotating about its vertical axis: The concept on anholonomy in the quantum mechanical case can actually be pictured quite similarly to the classical case described above. In quantum mechanics, we describe particles using a wavefunction, which in a very basic sense is also a vector. The vector does not exist in “real space” but what physicists refer to as “Hilbert space”. Nonetheless, the geometrical game I played with classical anholonomy can also be played in this abstract “Hilbert space”. The main difference is that the anholonomy angle becomes a phase factor in the quantum realm. The correspondence is as so:

(Classical) $\theta \rightarrow \mathrm{e}^{i\theta}$     (Quantum)

The expression on the right is precisely the Berry/geometric phase.  In the quantum case, regarding the two criteria above, we already have (i) a “vector” in the form of the wavefunction — so all we need is (ii) an appropriate surface on which hair cannot be combed straight. It turns out that in quantum mechanics, there are many ways to do this, but the most famous is undoubtedly the case of the Aharonov-Bohm effect.

In the Aharonov-Bohm experiment, one prepares a beam of electrons, splits the beam and passes them on either side of the solenoid, recombining them on the other side. A schematic of this experimental steup is shown below: In the image, B labels the magnetic field and A is the vector potential. While many readers are probably familiar with the magnetic field, the vector potential may not be as household a concept. The vector potential was originally thought up by Maxwell, and considered to be a mathematical oddity. He realized that one could obtain B by measuring the curl of A at each point in space, but A was not given any physical meaning.

Now, it doesn’t immediately seem like this experiment would give one a geometric phase, especially considering the fact that the magnetic field outside the solenoid is zero. But let’s take look at the pattern for the vector potential outside the solenoid (top-down view): Interestingly, the vector potential outside the solenoid looks like it would have a tuft in the center! Criterion number (ii) may therefore potentially be met. The one question left to be answered is this: can the vector potential actually “rotate” the electron wavefunction (or “vector”)? The answer to that question deserves a post to itself, and perhaps Brian or I can fill that hole in the future, but the answer seems to be emphatically in the affirmative.

The equation describing the relationship between the anholonomy angle and the vector potential is: $\theta = \alpha \oint \textbf{A}(\textbf{r})\cdot d\textbf{r}$

where $\theta$ is the rotation (or anholonomy) angle, the integral is over the closed loop of the electron path, and $\alpha$ is just a proportionality constant.

The way to think about the equality is as so: $d\textbf{r}$ is an infinitesimal “step” that the electron takes, much in the way that Raj took steps earlier. At each step the wavefunction is rotated a little compared to the previous step by an amount $\textbf{A}(\textbf{r})d\textbf{r}$, dictated by the vector potential. When I add up all the little rotations caused by $\textbf{A}(\textbf{r})$ over the entire path of the electrons, I get the integral around the closed loop.

Now that we have the anholonomy angle, we need to use the classical $\rightarrow$ quantum relation from above. This gives us a phase difference of $\mathrm{e}^{i\theta}$ between the electrons that go to the right and left of the solenoid. Whenever there is a non-zero phase difference, one should always be able to measure it using an interference experiment — and this is indeed the case here.

An experiment consisting of an electron beam fired at double-slit interference setup coupled with a solenoid demonstrates this interference effect most profoundly. On the setup to the left, the usual interference pattern is set up due to the path length difference of the electrons. On the setup to the right, the entire spectrum is shifted because the extra phase factor from the anholonomy angle. Again, let me emphasize that there is no magnetic field in the region in where the electrons travel. This effect is due purely to the geometric effect of the anholonomy angle, a.k.a. the Berry phase, and the geometric effect arises in relation to the swirly tuft of hair!

So next time you’re taking a long walk, think about how much the earth has rotated while you’ve been walking and whether you really end up where you started — chances are that something’s just a little bit different.

1. September 10, 2015 8:22 pm

Fantastic post 🙂 Thank you..

• September 11, 2015 9:23 am

Thanks!

2. September 11, 2015 6:35 am

Superb post but still m confused if magnetic field doesn’t contribute to anholonomy.. Then y u used solenoid?
You could have use just a cylinder

• September 11, 2015 9:22 am

Thanks — I appreciate the kind words.

The solenoid is necessary because while there is no magnetic field outside, the solenoid causes there to be a non-zero vector potential outside with the desired pattern. This desired pattern is of course the one with the tuft in the middle.

A cylinder would not necessitate the non-zero vector potential outside it.

That the vector potential affects the electron beam at all is probably one of the most stunning quantum mechanical effects we know of today. In classical electrodynamics, the vector potential does not have any physical meaning.

3. September 11, 2015 9:57 am

By the way, my own personal favorite example of (classical) Berry phase is the following:

Imagine if you had an expensive globe (like, say, this one: http://www.1worldglobes.com/images/GemstoneGlobes/mopfullcutgoldtri2.jpg), mounted on a perfectly frictionless axis and sitting in your office. Then, when you leave the room, someone comes in and picks up the globe, turns it around 360 degrees, and puts it back down again. Would you be able to tell when you returned to the office that someone had been playing with the globe?

If the axis of the globe were completely vertical, then you wouldn’t. When the person who played with the globe spun it around, the frame of the globe would spin around 360 degrees, but the globe itself would remain stationary with respect to the rest of the room. Then the globe would have rotated 360 degrees with respect to the frame, and everything would be back to the way it looked before. The Berry phase here is 360 degrees.

On the other hand, if the axis of the globe were completely horizontal, the whole thing would rotate rigidly when it is spun around: globe and frame would turn 360 degrees together. But the globe would not revolve at all — the Berry phase would be zero degrees. And it again would be impossible to tell that anything had happened.

If, on the other hand, the angle $\theta$ of the axis with respect to the vertical is anything other than $0$ or $90$ degrees (or $180$ degrees), then the rotation of the globe on its axis will be incomplete when the globe is put back down, and you’ll be able to tell that someone spun it around. In particular, the rotation of the globe on its axis — the Berry phase — is $2 \pi \cos \theta$.

This is all essentially identical to Anshul Foucault pendulum example above, but I like it because it allows you to imagine spinning things around in your hands.

4. September 11, 2015 10:35 am

It’s a little disingenuous to say the anholonomy on the sphere walk is “caused by” the HBT. It’s directly caused by the fact that the surface of the sphere is curved; the anholonomy around a closed path is (basically) the integral of the curvature over the portion of the surface contained within.

The HBT itself is a consequence of the Atiyah-Singer Index Theorem and the fact that the sphere has Euler characteristic 2, so all vector fields must have index 2, and thus at least one zero.

The Gauss-Bonnet formula (another consequence of ASIT) tells us that a surface of characteristic 2 cannot be intrinsically flat, so it must be curved, and this is where the anholonomy comes from. So the HBT and the anholonomy are both “caused by” ASIT, but neither one directly causes the other.

• September 11, 2015 10:59 am

Thanks for your comment, John. I totally agree with what you’re saying and of course you are completely correct.

However, for the sake of communicating my point in the post to a more general audience, I thought that these less precise but more intuitive arguments to be satisfactory. Describing the Berry curvature of the flat 2D plane in the Aharonov-Bohm problem was something I found particularly difficult to communicate, which is why I chose the description based on the HBT. It allows for a more visual interpretation, and I think the trade-off with respect to precision is justified in some sense.

• September 11, 2015 12:24 pm

Maybe “is related to” would serve the same purpose?

• 5. 