\"Linear anisotropy\" of magnetic susceptibility?

[Yevgeny Kats]

Suppose there is a line of atoms X and Y. For atom Y, S = L = 0, but not for the atom X. The arrangement is like this:

--X-Y---X-Y---X-Y---X-Y-- (periodical)

Ferromagnetic exchange interaction exists between atoms X. Atoms X have a spin-orbit coupling. Atoms X and Y are close enough to each other so that their orbitals may overlap.

Will the magnetic moment depend on whether the magnetic field is applied to the right or to the left?

[Edited on 9/2/04 by Yevgeny Kats]

[roch]

The problem is to determine if there is a mirror symetry for the hamiltonian of the sytem:

First, the atom X are regularly spaced and from the point of vue of the exchange interaction, the hamiltonian will be symetric.
so, left or right, that should make no difference.

However, I assume that the way you have written your atoms chain

--X-Y---X-Y---X-Y---X-Y--

means that \"---\" is a distance tree time larger than \"-\". In this case, there is no mirror symetry from the crystallographic point of vue:
Through a mirror transformation, the chain becomes

--Y-X---Y-X---Y-X---Y-X--

which cannot be superimpose to the original chain.

Maybe, anotherway of putting it is that the action of atom Y on X is different from the right than from the left:

...Y---X-Y...

Therefore the mirror reflection which transfroms into:

...Y-X---Y...

is not a symetry of the system.

Now the shape of the X atom orbitals will reflect this NON-symetry since the X and Y orbitals are mixing together. In other words, not only the shape of the X atom orbitals must be axisymetric (the direction of the chain) but also they must have a preferential orientation, towards the left or towards the right.

If you add that the spin of atoms X couples (interaction energy) with the orbitals of the atoms X, then you see immediately that there should be a preferential direction AND orientation for the spin of atom X in term of energy. In other words, the spin distinguish between their right and their left.

Hence, the susceptibility of the system will depends wether you apply a magnetic field to the right or to the left. :)

Does this make sens to you ?

[Yevgeny Kats]

This picture sounds reasonable.

However, as far as I know, in real materials the magnetization is the same in both directions.

I asked two of my friends, and very quickly they came up with the answer.

Moshe G. answered that looking at the time-reversed picture shows that the magnetization in both directions MUST BE THE SAME (when the direction of time is reversed, the directions of the magnetic field and the magnetization are reversed, while the crystal remains unchanged; and it is known that the relevant laws of physics are symmetric in time).

Vadim K. answered that looking at the mirror reflection, with the mirror perpendicular to the line of atoms, also shows that the magnetization in both directions MUST BE THE SAME (in the mirror, the direction of the crystal is reversed, but the magnetic field and the magnetization keep their directions; and it is known that the relevant laws of physics are symmetric under mirror reflection).

I added then that if the mirror is parallel to the line of atoms, it also proves this statement in a similar way.

:beer: :beer: :beer:

[roch]

Originally posted by Yevgeny Kats
This picture sounds reasonable.

However, as far as I know, in real materials the magnetization is the same in both directions.

Could you give any exemple of a \"REAL\" material? I have never heard of such an atom chains as the one you described in your problem. Give me a reference to an article in which they did measurement on such a system. I thought this problem was more like an school exercise.

I asked two of my friends, and very quickly they came up with the answer.

Moshe G. answered that looking at the time-reversed picture shows that the magnetization in both directions MUST BE THE SAME (when the direction of time is reversed, the directions of the magnetic field and the magnetization are reversed, while the crystal remains unchanged; and it is known that the relevant laws of physics are symmetric in time).

Vadim K. answered that looking at the mirror reflection, with the mirror perpendicular to the line of atoms, also shows that the magnetization in both directions MUST BE THE SAME (in the mirror, the direction of the crystal is reversed, but the magnetic field and the magnetization keep their directions; and it is known that the relevant laws of physics are symmetric under mirror reflection).

I added then that if the mirror is parallel to the line of atoms, it also proves this statement in a similar way.

:beer: :beer: :beer:

I do not understand the argument of your two friends. What is important to know is: if you keep you external field fixed and do a mirror symmetry (a mirror perpendicular to the chain axis which distinguish between right and left) then have you change the hamiltonian of the system? If you do, the system is anisotropic, if you do not, the system is isotropic.

My argument was that, while the exchange interaction remains unchanged, the spin-orbit interaction has changed: because the orbitals are not symetric for that mirror transformation. Now your friends should give a bit more explaination on how the symetry they apply change or not the details of the system.

Now, you are right about the fact that if the mirror is parallel to the chain axis, the system remains invariant for such a transformation. Therefore, whether the magnetic field is up or down, the amplitude of the magnetisation (in other words the susceptibility) should be the same. In fact, you can go further and say that any rotation about the chain axis leave the system invariant. Therefore the susceptibilty is the same for any field applied perpendicular to the chain.

But I have not been convinced by your friends that the susceptibility of your chain is isotropic from left to right.

[Yevgeny Kats]

Could you give any exemple of a \"REAL\" material? I have never heard of such an atom chains as the one you described in your problem. Give me a reference to an article in which they did measurement on such a system. I thought this problem was more like an school exercise.

There are many 3-dimensional crystalline materials whose basic unit includes more than one kind of atom (I asked about the 1-dimensional case just to make things simpler). In all such materials, the susceptibilities with oppositely applied fields are the same. People don\'t even bother to check this. (Anyone who has a touch with measurements of magnetization in materials, please comment on this!)

I do not understand the argument of your two friends. What is important to know is: if you keep you external field fixed and do a mirror symmetry (a mirror perpendicular to the chain axis which distinguish between right and left) then have you change the hamiltonian of the system? If you do, the system is anisotropic, if you do not, the system is isotropic.

OK, I will describe the solution in a more detailed way. I will describe the solution of Vadim:

1. The laws which govern the magnetization in our system (the electromagnetic interaction, the spin-orbit coupling, etc.) are symmetric with respect to mirror reflection. The meaning of this statement is that if you found a solution, then the mirror reflection of everything (both the external conditions and the internal configuration) is also a solution. (In many places the term \"parity\" is used instead of speaking about mirror reflection). Do you agree with this statement?

2. Put a mirror perpendicularly to the line of atoms. The magnetic field will keep its direction (think about the currents which create the field), and the magnetization will also remain the same (think about the rotating charges which create it), with the same magnitude. However, the direction of the chain will reverse.

3. The situation which you see in the mirror is exactly what will happen if instead of changing the direction of the field (in the original experiment) you put the chain oppositely. This is, of course, the same experiment (because the space is isotropic).

Hope now it is more clear.

Yevgeny.