Why is the Lagrangian approach preferred over the Hamiltonian approach in QFT?

Question

Going from non-relativistic quantum mechanics(QM) to QFT there is a marked change in the approach used. QM almost exclusively uses Hamiltonains. Lagrangian based methods like the path-integrals are seldom used. In QFT a Lagrangian based approach seems to be more widespread, although Hamiltonians can also be used. Why is this so? Is the reason historical or is it a matter of " this method works better for this situation"? Or does it have anything to do with the relativistic nature of QFT?

Answer 1

The assumption in the question is not entirely accurate since the canoncial formulation of QFT does use the Hamiltonian.

In QFT the Hamiltonian now has field operators, with eigenstates now being functionals. In ordinary wave mechanics the wave functions are real valued functions that are functions of position and time. In QFT, the "wave functions" become wave functionals: the probability density takes in time and wave functions (rather than position) as independent variables. Imagine you have the Hamiltonian of a single harmonic oscillator: $H_{i}=P^{2}/2m+m\omega Q^{2}/2$. Now add several harmonic oscillators: $H=\sum H_{i}+ V_{int}$. Thus, now you have many momentum and position operators in your configuration space. Fields have an infinite number of "harmonic oscillators", and hence an infinite number of degrees of freedom in the configuration space. Thus, instead of having many position operators, we have an uncountably infinite number of them that need a label besides $i$ that we used above: now we use $x$ as a parameter of the field (very different than as an operator), and our wave function becomes a wave functional that takes in fields parameterized by position and time.

Here is an excellent reference that describes this process: https://physics.ucsd.edu/students/courses/fall2015/physics200a/Hamiltonian%20Formulations%20for%20Continuua-RFS.pdf

Now dealing with the functional Schrodinger equation is no fun: http://arxiv.org/abs/hep-th/9306161

Thus, that is one reason: having to use the functional schrodinger equation is difficult.

A second reason: manifest Lorentz invariance of the Lagrangian since time and space are treated equally, as opposed to the Hamiltonian which singles out an extra time derivative as special.

A third reason: Although the canonical formulation of QFT does use the Hamiltonian, when we impose commutation relations of the fields in the free Hamiltonian we get a much simpler Hamiltonian as a function of creation and annihilation operators (bypassing the difficulty of using the functional schrodinger equation), but this is not as elegant as conceptually summing over all field configurations, which is what we think actually happens in QM.

That's what came to the top of my head for the moment.

Answer 2

In particle physics, where quantum mechanical formalisms are a necessity, symmetries and conservation laws are prominent in the data. The Lagrangian is connected with Noether's theorem, which cleanly gives the conserved quantities from it, and the symmetries that emerge from the data: SU(3)xSU(2)xU(1).

This question is relevant a kind of Noether's theorem for the Hamitonian

I believe that "mathematical simplicity" is the answer.

Answer 3

To add to an answer by @SalehHamdan (with which I totally agree).

Another reason why path integrals dominate the field of QFT calculation methods comes from the fact that we are interested primarily in Green's functions, which are spacetime dependent and have a clear meaning in terms of the path integral formalism.

On contrast, in QM we aren't interested in Green's functions. Take, for example, the harmonic oscillator. The corresponding Green's function depends on the two instances of time and thus provides no spatial information. It cannot be interpreted as a probability amplitude of a particle exchange between two spacetime points since it does not depend on spatial coordinates!

Thus, in first quantization, we are interested in another kind of quantities (such as matrix elements of the evolution operator). And these are precisely the kind of problems which can be solved conveniently in the canonical quantization formalism.

Answer 4

The Lagrangian formalism makes the Lorentz invariance of the theory more transparent. The Hamiltonian formalism, though is essentially covariant, breaks the Lorentz invariance formally. For more detail, you may read S. Weinberg, The Quantum Theory of Fields (1995), ch. 7,9.