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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?

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marked as duplicate by ACuriousMind, Community Feb 19 at 14:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Possible duplicates: physics.stackexchange.com/q/78508/2451 and links therein. – Qmechanic Feb 19 at 6:57
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The premise of the question is false, there is no general preference for the Lagrangian. The canonical quantization procedure for instance uses the Hamiltonian, and it is still the first thing most do in QFT. It's just a bit inconvenient to use the Hamiltonian in a relativistic theory because Lorentz invariance is not manifest when doing so, but both variants have the advantages in different situations. – ACuriousMind Feb 19 at 14:24
up vote 8 down vote accepted

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.

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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.

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Symmetries are also prominently used in non-relativistic QM. But there we use the in-variance of the Hamiltonian under symmetry operations rather than using Lagrangian formalism. Like here. The answer is most probably mathematical simplicity. The question is why does that arise in QFT and not in QM ? – biryani Feb 19 at 6:58
    
QFT is a many body theory and the complexity is great when one considers ground states and creation and annihilation operators of all possible fields from the table of elementary particles. Simplicity is important, and also there is a sense of "beauty", when the mathematics is "clean". – anna v Feb 19 at 7:12

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.

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Could you explain a bit more. It would be nice if the answer was self contained to some extent. If I had the necessary background to read Weinberg I wouldn't ask this question. – biryani Feb 19 at 6:41
    
@biryani In fact, I'm not sure what kind of detail you want to know. I'm not suggesting you to read the whole book. You can just read the introductory part of ch. 9. I'm sure that anyone who has some sense of QFT can read it. – Wen Chern Feb 19 at 7:05
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The Hamiltonian is $H = p\dot q - L$ where $p = \partial L / \partial \dot{q}$, right? But $\dot{q} = \partial q / \partial t$, so to pass to the Hamiltonian formalism you have to choose a time-like direction. However, you can write the invariant expression $\dot q = u^\mu \partial_\mu q $ and say that what you get is the Hamiltonian according to an observer with 4-velocity $u^\mu$. I think to understand this, you should recall that $H$ is the time component of a 4-vector, so observers in relative motion will not agree on $H$. More plainly, $H$ generates time evolution and observers disagree – Robin Ekman Feb 19 at 7:35
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on what is time and what is space. But if $L$ is Lorentz invariant, no observer is privileged and there's no problem. Note that I didn't invoke anything quantum - this is entirely about relativity and is equally valid in a classical field theory. It also applies to Hamiltonian formalisms in general relativity. (It's "worse" then, because in special relativity at least inertial frames are privileged. In general relativity, you have no such help.) – Robin Ekman Feb 19 at 7:41
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Besides $\dot{q}$ there is another source for non-invariance of the Hamiltonian. The mentioned formula uses not a Lagrange density $\mathcal{L}$ or action $S$ which are scalars but a Lagrange function $L=\int d^3x \mathcal{L}$, $S=\int dt L$. Everything in the Hamiltonian formalism is about living on a consant time slice which breaks Lorentz invariance manifestly. – OON Feb 19 at 9:39

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.

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protected by Qmechanic Feb 19 at 7:27

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