# Constructing Lagrangian from Hamiltonian for Majorana fermions

The text gives the Hamiltonian density as $$$${\cal H}=\frac{v}{2}\Big(\psi^\dagger\frac{\partial\psi^\dagger}{\partial x}-\psi\frac{\partial\psi}{\partial x}\Big)+\Delta\Psi^\dagger\Psi$$$$

and the Langrangian density as $$$${\cal L}=\psi^\dagger\frac{\partial\psi}{\partial \tau}+\frac{v}{2}\Big(\psi^\dagger\frac{\partial\psi^\dagger}{\partial x}-\psi\frac{\partial\psi}{\partial x}\Big)+\Delta\Psi^\dagger\Psi$$$$

The text also says that field is related to Majorana fermion when $$\Delta=0$$ whose Langrangian density (from another text) is

$$$${\cal L_M}=\frac{1}{2}\Big(\psi^\dagger\frac{\partial\psi}{\partial x^0}-\psi^\dagger\frac{\partial\psi}{\partial x^1}+\psi\frac{\partial\psi^\dagger}{\partial x^0}+\psi\frac{\partial\psi^\dagger}{\partial x^1}\Big)$$$$ If I go about deriving Langrangian density from Hamiltonian using $${\cal L}=\Pi\dot\Phi-\cal H$$ (by taking $$\Pi=\Psi^\dagger$$, which I am not sure that I am doing right), I get Lagrangian Density as

$${\cal L}_{\text{wrong}}=\frac{v}{2}\psi\frac{\partial\psi}{\partial x}$$

1. How all the Lagrangians are related?
2. Where I am going wrong in finding Langrangian from Hamiltonian?

More particularly

1. How $$\cal L_M = L$$?
2. What should be my $$\Pi$$ in $${\cal L}=\Pi\dot\Phi-\cal H$$?

References:

(text1: http://edu.itp.phys.ethz.ch/fs13/cft/SM2_Molignini.pdf) (page: 28-29)

(text2: Introduction to conformal field theory, R Blumenhagen and E Plauschinn) (page: 56-57)

• Related post by OP: physics.stackexchange.com/q/538770/2451 Mar 28, 2020 at 6:31
• Minor comment (v4): Langrangian should be Lagrangian. Mar 28, 2020 at 13:22
• Okay. I will rectify that. Mar 28, 2020 at 13:50

## 1 Answer

Those who work in the field of cond. matt. phy. must have already recognized that the Hamiltonian is Majorna Fermions. [https://arxiv.org/abs/cond-mat/0010440]

For field theorist, we encounter this in a different way. In CFT, the Lagrangian for the Majorana fermions (fermions whose anti-particle is particle itself) is $${\cal L_M}=\frac{1}{2}\bar\Psi (\dot\iota\Gamma^\mu\partial_\mu-m)\Psi$$ Where $$\Psi=\begin{pmatrix}\psi\\ \psi^\dagger\end{pmatrix}$$ We can construct the Hamiltonian as $${\cal H}=\Pi\dot\Psi-{\cal L}$$ Conjugate field $$\Pi$$ is $$\Pi=\frac{\partial\cal L}{\partial\dot\Psi}=\frac{\dot\iota}{2}\bar\Psi\Gamma^0$$ And the Hamiltonian is $${\cal H}=\frac{\dot\iota}{2}\bar\Psi\Gamma^0\dot\Psi-\frac{\dot\iota}{2}\bar\Psi \Gamma^0\partial_0\Psi-\frac{\dot\iota}{2}\bar\Psi \Gamma^1\partial_1\Psi+\frac{1}{2}m\bar\Psi\Psi$$ $${\cal H}=-\frac{\dot\iota}{2}\bar\Psi \Gamma^1\partial_1\Psi+\frac{1}{2}m\bar\Psi\Psi$$ $${\cal H}=-\frac{\dot\iota}{2}\begin{pmatrix}\psi^\dagger &\psi\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\partial_1\begin{pmatrix}\psi\\ \psi^\dagger\end{pmatrix}+\frac{m}{2}\begin{pmatrix}\psi^\dagger &\psi\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}\psi\\ \psi^\dagger\end{pmatrix}$$ $${\cal H}=-\frac{\dot\iota}{2}\begin{pmatrix}\psi^\dagger &-\psi\end{pmatrix}\partial_1\begin{pmatrix}\psi^\dagger\\ \psi\end{pmatrix}+\frac{m}{2}\begin{pmatrix}\psi^\dagger &-\psi\end{pmatrix}\begin{pmatrix}\psi\\ \psi^\dagger\end{pmatrix}$$ $${\cal H}=-\frac{\dot\iota}{2}(\psi^\dagger\partial_1\psi^\dagger-\psi\partial_1\psi)+\frac{m}{2}(\psi^\dagger\psi-\psi\psi^\dagger)$$ Using anticommutation relation of fermion we get, $${\cal H}=-\frac{\dot\iota}{2}(\psi^\dagger\partial_1\psi^\dagger-\psi\partial_1\psi)+m\psi^\dagger\psi$$ Now to answer your question

1. We have shown relation in Hamiltonian used by cond. mat. phy. people and Field Theorist people. Majorana fermion Hamiltonian used in cond. mat. phy. can be constructed from Majorana fermion Lagrangian used by field theory.
2. I have shown Legendre transformation from $$\cal L\mapsto H$$, I am sure you can do the reverse.

Problems in your question:

1. Lagrangian which you have written in Equation (2) is completely wrong.

The problem you need to work on

1. That $$\frac{\dot\iota}{2}$$ factor of Legendre Transformation and $$\frac{v}{2}$$ factor in Hamiltonian of cond. mat. phy.