# Calculation of Magnetic susceptibility

• stone
In summary: The derivation for pauli susceptibility does not go beyond first order effects. For a more in-depth discussion, you might want to look into e.g. "Quantum Mechanics: An Introduction" by Bloch.I just looked up Ashcroft and Mermin. I think the derivation of pauli susceptibility is quite clear. And yes the final result comes out to be independent of the value of h, due to the taylor expansion we make for the dos, so that magnetisation is directly proportional to the value of h and hence \chi becomes independent of h. But if there is no DOS at the Fermi level, does that mean that the susceptibility is zero (for an insulator, for
stone
I wanted to calculate the magnetic susceptibility for a model system, with no interaction. I know the energy eigenvalues and eigenvectors. How can I proceed?
Any help would be wonderful!

I wanted to use the Pauli susceptibility formula but I am not sure.

Hi!

The simple way: calculate the magnetization <m> for the Hamiltonian H'=H-hm (where H is your original Hamiltonian and m is the magnetization operator, usually sum over spins), differentiate the result with respect to h and set h equal to zero.

You could also determine the susceptibility directly from H by calculating the m-m correlation function. To find out how to do this, see e.g. "Principles of Condensed Matter Physics" by Chaikin & Lubensky.

Thanks a lot...I went through Chaikin and Lubensky but it is still not clears how to do this for a simple hamiltonian!
Let us consider a simple example of a tight binding model with two bands with energies
E=E0+-2tcos(ka)
How do we proceed from here?

stone said:
Thanks a lot...I went through Chaikin and Lubensky but it is still not clears how to do this for a simple hamiltonian!
Let us consider a simple example of a tight binding model with two bands with energies
E=E0+-2tcos(ka)
How do we proceed from here?
Okay, so the relevant quantum numbers are the lattice momentum k and the spin s. The free Hamiltonian is
$$H=\sum_{k,s} \epsilon_k c_{ks}^{\dagger} c_{ks},$$
and now you want to add a magnetic field h to the system, say along the z-axis. What is $$hS^z_{tot}$$ written in terms of $$c_{ks}^{\dagger}$$ and $$c_{ks}$$? You will notice that the new, additional term can be absorbed into the quadratic term (the free Hamiltonian) and that the new eigenvalues become spin-dependent. What is the expectation value of the magnetization in the new ground state? The ground state clearly contains a different number of spin-up and spin-down electrons, so there is a non-zero magnetization. Once you know this, you can write the final expression for the susceptibility. Good luck, just ask if you don't know how to proceed.

I can write $$hS^z_{tot}$$ in terms of the creation and annihilation operators and I also see that the no. of up and down spin electrons are different (because I have the eigenvectors) . But I did not get the part about the additional term getting absorbed into the free hamiltonian!
Sorry if it sounds trivial but I am just coming to grips with the second quantized notation.
Also ultimately after taking the average of magnetization I get the susceptibility in the zero field limit of $$h$$, isn't it?
Thanks again for your time and patience!

Whoops, if you are not that familiar with the second quantization, we of course have to solve the problem without it first :) Earlier you asked about using the Pauli paramagnetism formula, and yes, you are correct, the formula is applicable. But you should really try to understand the derivation: check any classical text (Ashcroft-Mermin/your textbook) on Pauli paramagnetism and make sure that you understand the derivation of the susceptibility. You will notice that the derivation is normally made for a free electron gas in a box ($$\epsilon_p=p^2/2m$$), but that the final result only makes use of the density of states at the Fermi level. This is because the change in the spin-up and spin-down electron densities is directly given by the density of states times the Zeeman splitting. So in your case, the main problem is to calculate the density of states for the tight-binding model.

And yes, the zero-field susceptibility is obtained by setting h=0 in the end. But if you check the derivation, you will notice that the final result for the susceptibility seems to be independent of h! Can you see why that happens?

I looked up Ashcroft and Mermin. I think the derivation of pauli susceptibility is quite clear. And yes the final result comes out to be independent of the value of h, due to the taylor expansion we make for the dos, so that magnetisation is directly proportional to the value of h and hence $$\chi$$ becomes independent of h.
But if there is no DOS at the Fermi level, does that mean that the susceptibility is zero (for an insulator, for instance) or that Pauli susceptibility is not applicable. Then what can we do?

Thanks so much for your help!

Then you have to go to higher order effects, which are non-linear.

## 1. What is magnetic susceptibility?

Magnetic susceptibility is a measure of the degree to which a material can be magnetized in the presence of an external magnetic field. It is a dimensionless quantity that indicates how easily a material can be magnetized or how strongly it responds to an applied magnetic field.

## 2. How is magnetic susceptibility calculated?

Magnetic susceptibility is calculated by dividing the magnetization of a material by the applied magnetic field strength. It can also be expressed in terms of the relative permeability of the material, which is equal to the magnetic susceptibility plus one.

## 3. What factors affect the magnetic susceptibility of a material?

The magnetic susceptibility of a material is influenced by various factors, such as its composition, crystal structure, temperature, and magnetic field strength. In general, materials with unpaired electrons in their atomic or molecular structure tend to have higher magnetic susceptibilities.

## 4. How is magnetic susceptibility used in scientific research?

Magnetic susceptibility is an important parameter in many fields of scientific research, such as geophysics, materials science, and chemistry. It can provide valuable information about the composition and properties of materials, as well as the structure and dynamics of Earth's magnetic field.

## 5. What are some common units for magnetic susceptibility?

The most commonly used units for magnetic susceptibility are dimensionless quantities, as it is a ratio of two magnetic properties. However, it can also be expressed in terms of SI units such as m^3/kg or in cgs units such as cm^3/g. In some cases, it may also be reported in units of parts per million (ppm) or parts per thousand (ppt).

• Other Physics Topics
Replies
0
Views
2K
• Atomic and Condensed Matter
Replies
1
Views
2K
• Atomic and Condensed Matter
Replies
1
Views
1K
• Atomic and Condensed Matter
Replies
2
Views
2K
• Atomic and Condensed Matter
Replies
2
Views
2K
• Classical Physics
Replies
11
Views
477
• Atomic and Condensed Matter
Replies
3
Views
1K
• Atomic and Condensed Matter
Replies
2
Views
4K
• Atomic and Condensed Matter
Replies
1
Views
1K
• Programming and Computer Science
Replies
4
Views
974