By Manishearth

2014-01-30 13:20:58 8 Comments

I just went through Observation of Dirac monopoles in a synthetic magnetic field.

What exactly has been observed?

More importantly, are these monopoles localized inside the apparatus (no stray monopole field lines coming out), or can they be used to create tangible monopoles? (By tangible monopoles I mean an object that is overall a monopole without any external influence. It may contain a condensate or similar inside)


@Jim 2014-02-04 20:40:07

After reading this paper, I wracked my brain trying to come up with the perfect analogy. Suffice it to say I failed, so here is my less than ideal answer.

The monopole created and referred to in this article is not a true Dirac monopole. It is no more a real monopole than a thermal vacuum testing chamber is outer space. That is, it is an artificially created object that exhibits most of the desirable properties of a Dirac monopole, specifically the properties that many scientists really want to experiment on. Although I am not a particle physicist and don't work much with theories involving monopoles, it came across to me like the monopole was localized; there was no net flux of magnetic field into or out of the laboratory. Instead, they often refer to it being monopolar within the scope of a derived synthetic magnetic field, not a true magnetic field as jinawee pointed out in the comments.

That said, this is still a major breakthrough. This represents that scientists have the ability to create something in a lab that, when used in certain experiments, would behave just like we expect a real Dirac monopole would.


I should probably have pointed out that they only describe producing one type of pole at a time (specifically North poles I believe), which is why if they were true monopoles, there would be a net flux of magnetic field into/out of the laboratory.

@Mitchell Porter 2014-02-05 08:43:18

Why is it a major breakthrough?

@Jim 2014-02-05 14:02:49

@MitchellPorter performing experiments on something that mimics the properties of Dirac monopole (even if only in a synthetic magnetic field) will still give us great insight into real Dirac monopoles, which may help us determine how to make them, where to find them, or if they would exist in nature at all. Among a host of many other possible applications. That, at least to me, constitutes a breakthrough

@Emilio Pisanty 2014-02-04 22:53:22

Let me make quite clear that the recent experiment does NOT imply the detection of a true magnetic monopole. Somehow, in all the excitement, the word "synthetic" was dropped rather quickly from the phrase "synthetic magnetic field".

A synthetic magnetic field is a physical quantity that obeys the same equations as a magnetic field, typically realized in things like the velocity field of the atoms inside a BEC. The recent paper claims the observation of the analogue of a Dirac monopole in such a quantity, and it is a significant result in the field of quantum simulation. However, their experiment is made completely out of atoms, and therefore of protons, neutrons and electrons. It cannot have any magnetic flux into or out of any region within or around it, and it does not violate the laws of electromagnetism. Indeed, it's built within them.

Synthetic magnetic fields are explained nicely in this review:

Colloquium: Artificial gauge potentials for neutral atoms. Jean Dalibard et al. Rev. Mod. Phys. 83 no. 4, 1523–1543 (2011). arXiv:1008.5378.

They describe a simple toy model which captures, I think the basics of the recent experiment.

Consider a two-level atom which is at the influence of an external field which couples its two internal states. Its hamiltonian can be written as $$ H=\left(\frac{P^2}{2m}+V\right)+U, $$ where the part in brackets does not couple to the internal state, and $$ U=\frac{\hbar\Omega}{2}\begin{pmatrix}\cos\theta&e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix} $$ in the ground - excited state basis $\{|g\rangle,|e\rangle\}$. In such a scheme $\Omega$ is the Rabi frequency and determines the strength of the coupling; $\phi$ is the phase of the laser that's used to couple the states; and $\theta$ is called the mixing angle and determines whether the coupling acts more like a Stark shift (at $\theta=0$), more like a pure coupling (at $\theta=\pi/2$), or somewhere in between. The distinction between those regimes is dependent on the detuning to the atomic transition, and it can vary with the atom's position $\mathbf{r}$.

The game here is to have the atom move around slowly enough that it will always stay in the ground state of the laser-induced atomic hamiltonian $U$. This is the first of the two eigenstates $$ |\chi_1\rangle =\begin{pmatrix}\cos(\theta/2) \\ e^{i\phi}\sin(\theta/2)\end{pmatrix}, \quad |\chi_2\rangle =\begin{pmatrix}-e^{-i\phi}\sin(\theta/2)\\ \cos(\theta/2)\end{pmatrix}, $$ which depend on the position $\mathbf{r}$ through the laser parameters $\theta$ and $\phi$. If you vary these parameters slowly enough, you will stay in the same eigenstate as you started in, without any nonadiabatic transitions.

More precisely, because the states $|\chi_j\rangle$ are a basis, one can always write the atom's wavefunction as $$ |\Psi(\mathbf{r},t)\rangle=\langle \mathbf{r}|\Psi(t)\rangle = \sum_j\psi_j(\mathbf{r},t)|\chi_j(\mathbf{r})\rangle. $$ (Note that this is the position-dependent internal state, which can be obtained from the full state $|\Psi(t)\rangle$ by projecting on a position state $|\mathbf{r}\rangle$.) If the atom's velocity spread is small enough, it will never transition to $|\chi_2\rangle$ and one can describe the system simply in terms of a single Schrödinger equation for $\psi_1(\mathbf{r},t)$.

To get that equation, you first have to work rigorously with the full state, and then afterwards neglect the possibility of transitions. Thus, if you act with the momentum $\mathbf{P}=-i\hbar\nabla$ on the component $\psi_j(\mathbf{r},t) |\chi_j(\mathbf{r})\rangle$, you will get contributions from both factors: $$ \nabla\left[\psi_j(\mathbf{r},t)|\chi_j(\mathbf{r})\rangle\right] =\left(\nabla\psi_j(\mathbf{r},t)\right)|\chi_j(\mathbf{r})\rangle +\psi_j(\mathbf{r},t)\nabla_\mathbf{r}|\chi_j(\mathbf{r})\rangle. $$ Multiplying on the left with the completeness relation $\sum_i|\chi_i\rangle\langle\chi_i|$, we can obtain a nice expression for the action of the momentum: $$ \mathbf{P}\psi_j(\mathbf{r},t)|\chi_j(\mathbf{r})\rangle =\sum_i|\chi_i\rangle \left[\delta_{ij}\mathbf{P}-\mathbf{A}_{ij}\right] \psi_j(\mathbf{r},t), $$ where $\mathbf{A}_{ij}=i\hbar\langle\chi_i|\nabla_\mathbf{r}|\chi_j\rangle$ is, broadly speaking, the rate at which nonadiabatic transitions can occur. Dotting this again with the momentum operator, one obtains $$ P^2\psi_j(\mathbf{r},t)|\chi_j(\mathbf{r})\rangle =\sum_{l,i}|\chi_l\rangle(\delta_{li}\mathbf{P}-\mathbf{A}_{li}) (\delta_{ij}\mathbf{P}-\mathbf{A}_{ij})\psi_j(\mathbf{r},t), $$ as long as one can neglect a term in $\nabla\cdot\mathbf{A}_{ij}=i\hbar\nabla\cdot\langle\chi_i|\nabla_\mathbf{r}\chi_j\rangle$. This corresponds to saying that the potential's variation is only relevant to first order at the low momenta encompassed by $\Psi$.

One can then neglect the possibility of transitions, and simply ignore all the terms that have $|\chi_2\rangle$ in them. (Or one might not, in which case you can try and build non-Abelian gauge fields instead.) If you do this, all the sums go away, and you get the simple Schrödinger equation $$ i\hbar\frac{\partial\psi_1}{\partial t} = \left[ \frac{1}{2m}(\mathbf{P}-\mathbf{A})^2+V+\frac{\hbar\Omega}{2}+W. \right]\psi_1 $$

This equation is identical in form to that of a single particle under the action of the old potential $V$, an 'electrostatic potential' $$W=\frac{\hbar^2}{2m}|\langle\chi_2|\nabla\chi_1\rangle|^2$$ which measures virtual transitions from $|\chi_1\rangle$ to $|\chi_2\rangle$ and back, and a magnetic field with vector potential $$\mathbf{A}=i\hbar\langle\chi_1|\nabla_\mathbf{r}|\chi_1\rangle.$$ This magnetic field will be nonzero whenever $\theta$ and $\phi$ both have significant spatial dependence with non-collinear gradients.

Most importantly, this 'magnetic field' depends on experimentally controllable parameters through $\theta(\mathbf{r})$ and $\phi(\mathbf{r})$. If one can build a sufficiently clever experiment, the 'field' $\mathbf{B}$, which is really the vector field $$\nabla\times\langle\chi_1|\nabla\chi_1\rangle,$$ can exhibit monopolar behaviour. One can create 'coupled' monopoles (i.e. separated by a finite Dirac string) or even take one of those out of the cloud to make a 'true' Dirac monopole as far as the cloud is concerned. It is my understanding that this is the case for the recent experiment. This does not, of course, imply that any net real magnetic flux is coming out of that apparatus.

@Steve Byrnes 2014-02-09 19:24:25

In a true magnetic monopole, the Dirac string is completely unobservable, even in theory. The string is not a real thing, just a way of describing some of the mathematical manipulations. Is that the case here too?

@Emilio Pisanty 2014-02-10 15:09:48

@SteveB I'm unsure whether that's the case, but I would tend to think it won't be observable.

@lancbert 2014-02-04 19:05:13

It is not a monopole, is just an artifact. Note that any addition of dipolar contributions of magnetic field (spin, coils, magnets…) must be dipolar or of a higher order magnetic field, in no case it could be monopolar. It is to say, there is not possible to construct a monopolar field as a superposition of dipolar (or higher order) contributions. The artifact the authors claim as a Dirac monopole is not that. It is simply a collection of atomic spin that invert the direction when the external field rotates. Intimately joined with every spin magnet there are the magnetic closed field lines that the author did not draw. Is perhaps the pression to publish papers under this kind of papers?

@Tom-Tom 2014-02-04 20:32:52

Welcome to the physics.SE! You wrote: "in any case it could be monopolar" I think you meant that in no case it could be monopolar.

@lancbert 2014-02-04 21:37:12

Yes, I'm sorry. There is not possible to construct a monopolar field as a superposition of dipolar (or higher order) contributions.

@Tom-Tom 2014-02-04 21:38:01

In that case, you should edit your answer, please. Thanks.

@Everett You 2014-02-05 07:01:53

I do not think you have gone through the paper carefully. The monopole observed this time is not constructed by a superposition of dipoles. It is a single separable monopole of the synthetic gauge field.

@lancbert 2014-02-05 22:56:31

No, I explain here with more detail my comment. It is impossible to construct a true magnetic monopole (net source or sink of B-lines of force) in this manner. The authors start from conventional point magnets (dipoles) and coils (no point sources, higher order n-poles). As you know the dipole cancels the monopolar contribution, the 4-pole cancels the 2-pole contribution and so on. You can remember the conventional multipolar expansion as appear in all books.

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