I have hamiltonian for fermionic field as $${\cal H}_F=E_0+\int dx[\frac{v}{2}(\Psi^\dagger\frac{\partial \Psi^\dagger}{\partial x}-\Psi\frac{\partial \Psi}{\partial x})+\Delta\Psi^\dagger\Psi]\tag{1}$$ And partition function is

$$\mathcal{Z}=Tre^{-\frac{H_F}{T}}=\int D\Psi D\Psi^{\dagger}e^{-\int_0^{1/T}dx^0 dx^1\mathcal{L}}. \tag{2}$$


$$\mathcal{L}=\frac{-\dot\iota}{2}\Big(\Psi^\dagger\frac{\partial\Psi}{\partial x^0}+\Psi\frac{\partial\Psi^\dagger}{\partial x^0}+\Psi^\dagger\frac{\partial\Psi^\dagger}{\partial x^1}-\Psi\frac{\partial\Psi}{\partial x^1}\Big)$$

The text says, by using Grassmann variables, one can write the partition function given above. I read about Grassmann variable (Grassmann Algebra), but still, I don't know how to write the partition function given above.

We get Hamiltonian above after Jordan-Winger and Bogoliubov transformation of the quantum Ising model in a transverse field.

Text also talks about Majorana fermions. How to relate above field as Majorana field fermion?

  • 1
    $\begingroup$ I think it means that you have to interpret $\Psi$ and $\Psi^\dagger$ as grassmann variables in the integral formulation of $\mathcal{Z}$. $\endgroup$ Commented Mar 24, 2020 at 17:11
  • $\begingroup$ @Raskolnikov I tried, but couldn't solve RHS of equation (2) to verify that it is indeed partition function. $\endgroup$ Commented Mar 27, 2020 at 16:51

1 Answer 1


As you can see the action is a sesquilinear form in $\Phi, \Phi^\dagger$. This means that you can write

$$ S = \Phi^\dagger G \Phi. $$

Now you realize that the partition function is a gaussian (infinite dimensional) integral over Grassmann variables, i.e.

$$ Z = \int D\Phi D\Phi^\dagger e^{- \Phi^\dagger G \Phi }. $$

The formula for Gaussian integral over Grassmann variables is similar to the one for complex variables with an additional (-1).

The formula reads

$$ Z = \det(G). $$

The problem is reduced to the computation of that (infinite dimensional) determinant.

Note In the post there is some confusion about naming the fields $\Phi$ vs $\Psi$. Also the Hamiltonian has a mass term $\Delta$ which is absent in the Lagrangian.

Edit 04/06/2020

I overlooked one important detail. The Hamiltonian (and Lagrangian) contains pair creation terms $\Psi^\dagger \partial_x \Psi^\dagger$ (and the corresponding pair annihilation). In this case in order to write the action as a quadratic form you must first pass to so called Nambu spinor:

$$ \Phi = \left ( \begin{array}{cc} \Psi \\ \Psi^\dagger \end{array} \right ), $$

taking care of anticommutation relations.

  • $\begingroup$ Sorry. I have rectified mass term in Langrangian. Any way you can show for $\Delta=0$ because for Majorana fermions are massless. IDK know how to write action as a matrix $G$? Perhaps, a hint would be enough. $\endgroup$ Commented Mar 28, 2020 at 3:12

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