- #1

stone

- 41

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Any help would be wonderful!

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- Thread starter stone
- Start date

- #1

stone

- 41

- 0

Any help would be wonderful!

- #2

stone

- 41

- 0

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

- #3

saaskis

- 66

- 0

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.

- #4

stone

- 41

- 0

Let us consider a simple example of a tight binding model with two bands with energies

E=E

How do we proceed from here?

Thanks in advance......

- #5

saaskis

- 66

- 0

Okay, so the relevant quantum numbers are the lattice momentum k and the spin s. The free Hamiltonian is

Let us consider a simple example of a tight binding model with two bands with energies

E=E_{0}+_{-}2tcos(ka)

How do we proceed from here?

Thanks in advance......

[tex]

H=\sum_{k,s} \epsilon_k c_{ks}^{\dagger} c_{ks},

[/tex]

and now you want to add a magnetic field h to the system, say along the z-axis. What is [tex]hS^z_{tot}[/tex] written in terms of [tex]c_{ks}^{\dagger} [/tex] and [tex]c_{ks}[/tex]? 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.

- #6

stone

- 41

- 0

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 [tex] h [/tex], isn't it?

Thanks again for your time and patience!

- #7

saaskis

- 66

- 0

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?

- #8

stone

- 41

- 0

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!

- #9

hiyok

- 109

- 0

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

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