# Hydrogen atom in a magnetic field

1. Feb 28, 2017

### Silviu

1. The problem statement, all variables and given/known data
A beam of neutral hydrogen atoms in their ground state is moving into the plane of the page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, in how many beams was the original beam deflected?

2. Relevant equations

3. The attempt at a solution
The electron and proton has both spin 1/2 so the total spin of the hydrogen atom, by the addition of spin and angular momentum can be 1 or 0, which means that the beam would not be deflected at all (spin 0) or would be deflected in 3 beams. However the answer seems to be that the beam will be deflected in 2 beams. Can someone explain to me why?
Thank you!

2. Feb 28, 2017

### kuruman

In a strong magnetic field, the hyperfine coupling between the electron and proton is broken.

3. Mar 5, 2017

### John Park

The magnetic moment of the proton is much smaller than that of the electron. But the angular momentum of any spin-1/2 particle can take up one of two orientation is a magnetic field.

4. Mar 5, 2017

5. Mar 5, 2017

### kuruman

I am sure as long as we we are in the strong field regime stipulated by the problem.
Quoting from the Feynman reference, which you so kindly provided, just above Fig. 12-3
That means two beams after the hyperfine interaction is broken. For good measure, I extended Feynman's Fig.12-3 to higher values of μB/A (see below). The argument is that in a strong external field there are two states, the low energy state with both magnetic moments parallel to the field and the high energy state with both magnetic moments antiparallel to the field. As the plot shows, that's the situation when the Zeeman energy μB is much larger than the hyperfine coupling energy A.

6. Mar 6, 2017

### phyzguy

kuruman - Thanks for that. But what you are showing is the energy levels. The question asks about the deflection in the inhomogeneous magnetic field. This depends on the spin state, not the energy, doesn't it? The true ground state has spin zero, and should not be deflected at all.

7. Mar 6, 2017

### Staff: Mentor

As @kuruman said, in a strong magnetic field, coupling of the individual magnetic moments with the field is stronger than the hyperfine coupling, so the total spin is not relevant.

Think back to the original Stern-Gerlach experiment. Are there significant differences between Ag and H in the context of this experiment?

8. Mar 6, 2017

### kuruman

To elaborate on what DrClaude wrote. Indeed the true ground has spin zero and should not be deflected at all. However, once the external magnetic field is turned on, the true ground state is no longer "true" but acquires a magnetic moment from admixtures from the other states. As you can see from the energy level diagram, at high μB/A the energy of the ground state decreases linearly which means the ground state has a magnetic moment and that would be close to μelectron + μproton. The next state up close to the ground state would have magnetic moment μelectron - μproton. Likewise, for the higher energy pair of lines we have moments -μelectron + μproton and -μelectron - μproton.

The two pairs of lines should be resolved with a Stern-Gerlach experiment. I don't think that lines within pairs can be experimentally resolved because μelectron ≈1000 μproton. The forces acting on the dipoles are proportional to the magnetic moments so if the difference in force is small so would the deflection. In theory, there should be 4 deflected beams, but in practice there will be 2 detected beams.

9. Mar 6, 2017

### phyzguy

What I struggle to understand is the following. I have a beam of H atoms in the ground state - the red line in kuruman's figure from post #5. As they enter the strong B field, the energy of that state is decreased, but how is the spin state changed? If the original beam were a mixture of all 4 states, then I can see that it would be deflected into multiple beams, but if all of the atoms are in the ground state, what causes them to change state? Is it that interaction with the B-field induces them to populate the other states?

10. Mar 6, 2017

### kuruman

Why do you think that all the atoms are in the ground state? The zero-field splitting energy is of order of Feynman's parameter A. If the overall splitting equal to 4A corresponds to the 21 cm line, then A is of order 10-29 Joules. The thermal energy kBT at room temperature is of order 10-21 Joules, 8 orders of magnitude above A. It is safe to conclude that all levels are equally populated. The assumption here is that the atoms in the beam are in thermal equilibrium with the room.

11. Mar 6, 2017

### phyzguy

OK. That wasn't clear from the original question. It is not stated what the thermal state of the beam is. But your explanation makes sense. Thanks.

12. Mar 6, 2017

### kuruman

Actually, I can see where you are coming from. According to the statement of the problem (emphasis mine)
Which ground state? Probably the electronic ground state with equal populations of spin up and spin down electrons. In other words, we are talking about a Stern-Gerlach experiment with hydrogen instead of silver atoms and a negligible hyperfine interaction. The hyperfine interaction interpretation of the question was OP's suggestion.

13. Mar 6, 2017

### phyzguy

Yes. I see it would be very difficult to prepare a beam of H atoms all in the true ground state, because as you pointed out the energy difference is so small. It would require that the beam's effective thermal excitation be in the microKelvin range.