What is the energy of an atom in a magnetic trap as a function of position?

[Malamala]

Hello! I am a bit confused about trapping using magnetic traps. In a simplified version, assuming we have 2 anti-Helmholtz coils, the magnetic field in between them (assume that the trapping regions is much smaller than the radius of a coil, as well as much smaller than the distance between the coils) along the axial direction is given by:

$$B_z = az$$
where $a$ depends on the geometry. Now the energy of an atom in the trap, with spin $S$ (assume it is 1/2) is given by:

$$E = -g\mu_BS_zB_z = -ag\mu_BS_zz \equiv \alpha zS_z$$
Now I understand that there are high and low field seeking states. But I am not sure how does it work. Say we are in a state with $S_z = +1/2$. Then the energy is given by $E = \alpha/2 \times z$, which means that the atom won't be stable around 0, but will try to move away in a negative direction (assuming $\alpha > 0$). Similarly, for $S_z=-1/2$ the atom will move in the positive direction. In either case, there doesn't seem to be a value of $z$ for which the atoms will have a minimum of energy, which means that they won't get trapped. What am I doing wrong? Thank you!

 

[vanhees71]

This magnetic field doesn't exist in Nature, since $\vec{\nabla} \cdot \vec{B} \neq 0$.

It's rather much more worthwhile studying the Penning trap. Which for obvious reasons is sometimes called an "artificial atom" or "geonium", and you can get very far with analytical solutions.

L. S. Brown and G. Gabrielse, Geonium Theory: Physics of a
Single Electron or Ion in a Penning Trap, Rev. Mod. Phys. 58,
233 (1986), https://doi.org/10.1103/RevModPhys.58.233

 

[DrClaude]

Hello! I am a bit confused about trapping using magnetic traps. In a simplified version, assuming we have 2 anti-Helmholtz coils, the magnetic field in between them (assume that the trapping regions is much smaller than the radius of a coil, as well as much smaller than the distance between the coils) along the axial direction is given by:

$$B_z = az$$
where $a$ depends on the geometry.

That's incorrect. While the magnitude of the magnetic field of a quadrupole trap is approximately linear near the center of the trap, its goes as $B = B' (x^2 + y^2 + 4z^2)^{1/2}$. It is minimum at the center of the trap an increases in all directions.

It's rather much more worthwhile studying the Penning trap.

But that would be changing the subject. The OP is about traps for neutral atoms, not for charged particles.

 

[Malamala]

This magnetic field doesn't exist in Nature, since $\vec{\nabla} \cdot \vec{B} \neq 0$.

It's rather much more worthwhile studying the Penning trap. Which for obvious reasons is sometimes called an "artificial atom" or "geonium", and you can get very far with analytical solutions.

L. S. Brown and G. Gabrielse, Geonium Theory: Physics of a
Single Electron or Ion in a Penning Trap, Rev. Mod. Phys. 58,
233 (1986), https://doi.org/10.1103/RevModPhys.58.233

I am not sure I understand what you mean. People use magnetic traps to trap neutral atoms. I know about Penning traps, but these are used for charged particles. Also what you mean by it doesn't exists in Nature? I just used Maxwell's equations to derive that and did a Taylor expansion around $z=0$. You mean I made a mistake in the derivation?

 

[Malamala]

That's incorrect. While the magnitude of the magnetic field of a quadrupole trap is approximately linear near the center of the trap, its goes as $B = B' (x^2 + y^2 + 4z^2)^{1/2}$. It is minimum at the center of the trap an increases in all directions.But that would be changing the subject. The OP is about traps for neutral atoms, not for charged particles.

Thanks for this. But that formula is for the magnitude of the field and I agree with it. However, for the energy, don't we need the vectorial form of B, in order to take the dot product with the magnetic dipole moment? My question is basically what is the energy as a function of $z$ (assuming $x=y=0$).

 

[vanhees71]

I am not sure I understand what you mean. People use magnetic traps to trap neutral atoms. I know about Penning traps, but these are used for charged particles. Also what you mean by it doesn't exists in Nature? I just used Maxwell's equations to derive that and did a Taylor expansion around $z=0$. You mean I made a mistake in the derivation?

I thought, you just want to have an example for traps. Of course, you can't use a Penning trap for trapping neutral atoms. For that you use indeed magnetic traps, but of course you can only use magnetic fields that exist in Nature, i.e., they must obey Maxwell's equations, i.e., it must obey Gauss's Law for the magnetic field, $\vec{\nabla} \cdot \vec{B}=0$. Here's a review

https://doi.org/10.1103/RevModPhys.79.235

 

[DrClaude]

Thanks for this. But that formula is for the magnitude of the field and I agree with it. However, for the energy, don't we need the vectorial form of B, in order to take the dot product with the magnetic dipole moment? My question is basically what is the energy as a function of $z$ (assuming $x=y=0$).

The field changes direction as you cross the origin,
$$
\mathbf{B} = B' (x,y,-2z)
$$
See https://commons.wikimedia.org/wiki/File:VFPt_anti-helmholtz_coil.svg