# Given a QFT Hamiltonian, is there a unique Lagrangian?

Consider a QFT in one spatial dimension specified by the following Hamiltonian density:

$\mathcal{H} = -i \phi^\dagger \frac{\partial}{\partial x} \phi + V(\phi^\dagger,\phi)$

where $\phi$ is a scalar field which may or may not be relativistic.

It seems to me that for any constant $a$, we can get the above Hamiltonian density by doing the usual Legendre transform on the following Lagrangian density:

$\mathcal{L} = a i \phi^\dagger \frac{\partial}{\partial t} \phi + i \phi^\dagger \frac{\partial}{\partial x} \phi - V(\phi^\dagger,\phi)$

$\bf{Question}$. If I want to do calculations using the path integral formalism, what is the correct Lagrangian to use? Or do they all lead to the same physical results? (If so, is one particular Lagrangian more convenient?)

$\bf{Comment}$. I understand that the Lagrangian is never unique, even classically. However, the above non-uniqueness seems more substantial than just the addition of a total derivative.

• Is your Lagrangian Lorentz-invariant? – Robin Ekman Feb 7 '15 at 17:35
• @RobinEkman: Not OP, but this question has entered my mind. What does one do if the Lagrangian obtained from Legendre transform is not Lorentz-invariant? – Ryan Unger Feb 7 '15 at 20:17
• @RobinEkman The examples I have in mind are non-relativistic, but it would be nice to understand the answer to this question in both the non-relativistic and relativistic cases. – marlow Feb 7 '15 at 22:26
• @RobinEkman I wonder if you are hinting at the following possible answer to the question: ''All the Lagrangians whose Legendre transforms yield the correct Hamiltonian will give the same physical answers, but in a relativistic setting it is convenient to take the Lagrangian to be Lorentz-invariant." – marlow Feb 7 '15 at 22:29

## 1 Answer

I) Before we get to quantization and path integrals there are problems already at the classical level. The Legendre transformation is not well-defined without knowledge of the CCR. For instance if the CCRs for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ is zero, this would mean that OP's Hamiltonian density ${\cal H}$ is a pure potential term without kinetic terms. The Legendre transformation to the Lagrangian formulation would then become singular.

II) Here is a non-trivial example. Let us instead assume that the CCR for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ reads

$$\tag{1} [\hat{\phi}(x,t), \hat{\phi}^{\dagger}(y,t)] ~=~\hbar{\bf 1}~ \delta(x-y),$$

and other CCRs vanish. Equivalently in terms of Poisson brackets

$$\tag{2} \{\phi(x,t), \phi^{\ast}(y,t)\}~=~-i\delta(x-y).$$

We can expand the complex scalar field

$$\tag{3} \phi~=~(\phi^1+i\phi^2)/\sqrt{2}$$

in two real component fields $\phi^a$ , $a=1,2$. Then the CCR (2) becomes

$$\tag{4} \{\phi^1(x,t), \phi^{2}(y,t)\}~=~\delta(x-y).$$

Conclusion: We can identify $\phi^2$ as the momentum for $\phi^1$.

Next recall that OP's Hamiltonian density is (up to a total $x$-derivative)

$$\tag{5} {\cal H} ~=~\frac{i}{2}\left(\phi\partial_x\phi^{\ast}-\phi^{\ast}\partial_x\phi\right) + V(\phi,\phi^{\ast}) ~=~\frac{1}{2}\left(\phi^1\partial_x\phi^2-\phi^2\partial_x\phi^1\right) + V(\phi^1,\phi^2).$$

The corresponding Lagrangian density is then (up to a total $t$-derivative)

$$\tag{6} {\cal L}~=~\phi^2\dot{\phi}^1-{\cal H}~\sim~ i\phi^*\dot{\phi}-{\cal H}.$$

[Here the $\sim$ symbol means equality modulo total derivative terms.] This Legendre transformation (5)-(6) is explained in detail in this Phys.SE post. Note that the Lagrangian density (6) unconventionally depends on the momentum variable $\phi^2$. Nevertheless, the corresponding action $S=\int \! dt~dx~{\cal L}$ leads to the correct equations of motion, and serves as a starting point for the path integral formulation.

• Very interesting, thanks. I wonder what to do in the case that $\phi$ is a fermionic field? – marlow Feb 9 '15 at 3:52