I'm looking at some lecture notes for electron scattering taking place at a ferromagnet-superconductor junction. The idea is to start from a tight binding model, and eventually obtain the BdG equation.

However, I have some problem with the algebra. In the ferromagnet the exchange term in the Hamiltonian reads $$H_{ex} = -\frac{\Delta_{xc}}{2}\sum_{i,\sigma} \vec{M} \cdot\vec{\sigma}c^{\dagger}_{i\sigma} c_{i \sigma}$$ where $\Delta_{ex}$ is just a scalar number, $\mathbf{M}=(\sin\theta \cos\phi, \sin\theta\sin\phi,\cos\theta)$ is the magnetization vector of the ferromagnet expressed in spherical coordinates, $\vec{\sigma}$ is a vector containing the Pauli matrices, and $c_{i,\sigma}$ annihilates an electron on site $i$ with spin $\sigma$.

In the lecture notes they write that this becomes

$$H_{ex} = -\frac{\Delta_{xc}}{2}\sum_{i,\sigma} \cos\theta c^{\dagger}_{i\sigma}c_{i\sigma}+ \sin\theta e^{-i\phi} c_{i,-\sigma}^{\dagger}c_{i\sigma} + \sin\theta e^{i\phi} c^{\dagger}_{i\sigma}c_{i,-\sigma} - \cos\theta c^{\dagger}_{i,-\sigma} c_{i,-\sigma}.$$

My question is how does one calculate this explicitly?

My own attempt

Clearly one must expand $$\vec{M}\cdot \vec{\sigma} = \sigma_x\sin\theta\cos\phi + \sigma_y \sin\theta \sin\phi + \sigma_z\cos\theta$$

and then act with this on the operator product $c_{i \sigma}^{\dagger}c_{i\sigma}$. However, I do not understand how the Pauli matrices act on the creation operators.

For instance how can you calculate

$$\sigma_x c_{i\sigma}^{\dagger}c_{i\sigma} = ?$$

I'm sure I can do the calculation above if only I knew how one handles one of these terms.

  • $\begingroup$ Are you sure the creation operator is not to the left of $\vec M \cdot \vec\sigma= \begin{pmatrix} \cos \theta & \sin \theta e^{-i\phi}\\ \sin \theta e^{i\phi} & -\cos \theta \end{pmatrix} $? $\endgroup$ Commented Jun 6, 2019 at 0:08
  • $\begingroup$ Yes this is probably correct. Still how would you evaluate $c^{\dagger}_{i \sigma}(\vec{M}\cdot \vec{\sigma})c_{i \sigma}$? $\endgroup$
    – MOOSE
    Commented Jun 6, 2019 at 7:50

1 Answer 1


I strongly suspect you are merely looking at quadratic forms of Dirc oscillators, tallied in garbled aspirational notation.

In the conventional basis of Pauli matrices, $$ \vec M \cdot \vec \sigma = \begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta &-\cos\theta \end{pmatrix}. $$ In this basis, this matrix acts on the 2-vectors $$ v= \begin{pmatrix} c_\sigma \\ c_{-\sigma}\end{pmatrix}, $$ where now the bimodal variables $\sigma=\pm$ have been fixed to a direction, here z, and the sum over them is merely indicated in the bilinear form, $v^\dagger \vec M \cdot \vec \sigma v$, for each i index, still summed over, which I omit.

You might well be overthinking an inept notation.

  • $\begingroup$ That seems reasonable, but let me ask one more simple question just to clarify. So from the start the exchange interaction should be written as $\sum_{i} v^{\dagger} \vec{M}\cdot \vec{\sigma} v$ without a $\sigma$-summation? and the end-result should also be without a $\sigma$-summation? $\endgroup$
    – MOOSE
    Commented Jun 6, 2019 at 14:37
  • $\begingroup$ Indeed: this would be the more elegant/economical/sensible presentation. $\endgroup$ Commented Jun 6, 2019 at 14:41

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