Does the Dirac Hamiltonian uniquely define the Dirac Lagrangian?

Question

Example Hamiltonian: the linearised graphene Hamiltonian

In condensed matter, we typically write down Hamiltonians instead of Lagrangians. An example is given by the Hamiltonian for graphene. When we linearise the theory about the Dirac points of the model, we obtain a pair of Dirac Hamiltonians, see Eq. (18) of this review for example. Let's focus on one of these Hamiltonians, it is given by

$$ H = \int dx \psi^\dagger (-i\alpha^i \partial_i + \beta m) \psi \equiv \int dx \mathcal{H}, $$

where I have set the $v_F = 1$ and $\alpha^i$, $\beta$ are the usual Dirac alpha and beta matrices used for expressing the Dirac equation in the form of a Schrodinger equation.

What is the Lagrangian of this Hamiltonian?

Now I ask, what is the Lagrangian of this model? Well, I could write the obivous Lagrangian as

$$ \mathcal{L} = i \psi^\dagger \partial_t \psi - \mathcal{H} = \bar{\psi}( i \gamma^\mu \partial_\mu-m) \psi$$

where I have combined the space and time derivatives into a single object $\partial_\mu = (\partial_t , \partial_i)$ and defined the objects $\gamma^0 = \beta$, $\gamma^i = \beta \alpha^i$ and $\bar{\psi} = \psi^\dagger \gamma^0$ which brings the action into its standard covariant form. This is nothing but the Dirac Lagrangian and it returns the Hamiltonian after a Legendre transformation as expected.

However, what if I decided to insert some arbitrary matrix $M$ into the time derivative part of my Lagrangian such as

$$ \mathcal{L}'= i \psi^\dagger M \partial_t \psi -\mathcal{H}$$

where I could let $M$ be horribly space-dependent if I wish. This Lagrangian $\mathcal{L}'$ would yield a different equation of motion to the Dirac equation, however it would still yield the same Hamiltonian. Using the definition of the Legendre transformation, we have

$$ \mathcal{H}' = \frac{\partial \mathcal{L'}}{\partial (\partial_t \psi)} \partial_t \psi - \mathcal{L}'= i\psi^\dagger M \partial_t \psi-\mathcal{L}'=\mathcal{H}$$

My question

It appears that the Hamiltonian can be derived from both of the Lagrangians $\mathcal{L}$ and $\mathcal{L}'$, however the equations of motion of these two Lagrangians differ. The equations of motion are

$$ \mathcal{L} \quad \Rightarrow \quad i \gamma^\mu \partial_\mu \psi = 0 $$ $$ \mathcal{L}' \quad \Rightarrow \quad i \gamma^0 M \partial_t \psi + i \gamma^i \partial_i \psi = 0$$

This is puzzling, because when I quantise the theory, I have to construct the mode expansion from the equations of motion. This will diagonalise the Hamiltonian. But it appears that this process would yield diferent results depending on whether I let $M \neq \mathbb{I}$ or not. For this reason my question is the following: if the Lagrangian corresponding to a Hamiltonian is not unique, which Lagrangian do I choose?

Answer 1

Putting the $M$ in changes the commutation relation of $\psi$ and $\psi^\dagger$. The conjugate $\Pi$ field is now $\Pi=\psi^\dagger M$. You will therefore have $\{\psi^a,(\psi^b)^\dagger\}= (M^{-1})^{ab}\delta^3(x-x')$ and $H$ may look the same, but it does not have the same hamlitonian equations of motion.

Answer 2

1. An expression of the form $$ L = \text{stuff} - H$$ does not actually make sense without further qualification. The Lagrangian is a function on the tangent bundle, the Hamiltonian is a function on the cotangent bundle, these cannot be equal, nor "differ" only by $\text{stuff}$, just like you cannot meaningfully take the difference of a function $f:X\to\mathbb{R}$ and a function $g:Y\to\mathbb{R}$ when $X\neq Y$ - the expression $f = c+ g$ just doesn't mean anything. See this answer of mine for a longer discussion of where the Lagrangian and Hamiltonian live and how to talk about them while avoiding this confusion.

2. So, if we restore the proper arguments what you really want to say is that you have a Lagrangian like $$ L(q,\dot{q}) = q\dot{q} - H(q,p_L(q,\dot{q}))$$ where $p_L(q,\dot{q}) = \frac{\partial L}{\partial \dot{q}}(q,\dot{q})$ is the canonical momentum for $H(q,p)$ obtained as the Legendre transform of $L$. You know want to write something like $$ L'(q,\dot{q}) = qM\dot{q} - H(q,p_L(q,\dot{q}))$$ and claim that this $L'$ still has $H$ as its Hamiltonian. But this is not true - note that the $p_L(q,\dot{q})$ is still associated with $L$ and not $L'$, and this equation would need to read $$ L'(q,\dot{q}) = qM\dot{q} - H(q,p_{L'}(q,\dot{q}))$$ where $p_{L'}(q,\dot{q}) = \frac{\partial L'}{\partial \dot{q}}(q,\dot{q})$ for this claim to be true, but $p_{L'}\neq p_L$ for $M\neq \mathbf{1}$, so this is not true.

Answer 3

If you include the matrix $M$ in your Lagrangian then your new canonical momentum will be $$p_\psi=i\psi^\dagger M $$ which means you will have to replace $\psi^\dagger$ in your Hamiltonina with $-ip_\psi M^{-1}$. More importantly when you quantize this system $\psi$ and $p_\psi$ will obey the equal-time anticommunication relations, not $\psi$ and $\psi^\dagger$. So really these are different systems.