Why does Lagrangian in QFT only include first order derivative of field?

In summary, the Lagrangian in QFT does not include high order derivatives of field because they are irrelevant at low energies. This is why the theories are not renormalizable. Higher derivatives in the Lagrangian will also result in higher derivatives in the equations of motion. However, classical situations where higher derivatives are required can be related to the nonrenormalizability of higher-derivative actions.
  • #1
ndung200790
519
0
Please teach me this:
Why the Lagrangian in QFT does not include high order derivative of field?Is it correct the reason being all fields obey the only Dirac and Klein-Gordon equations?
Thank you very much for your kind helping.
 
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  • #2
One explanation I've heard but don't fully understand is that the contribution from higher-order terms would be negligible at low energies anyway, and can therefore be ignored. These terms would also make the theories not renormalizable.
 
  • #4
ndung200790 said:
Why the Lagrangian in QFT does not include high order derivative of field?
A renormalizable QFT cannot have higher derivatives. Effective (nonrenormalizable) theories can.
 
  • #5
Also, classically speaking, higher derivatives in the Lagrangian will result in higher derivatives in the equations of motion. In particular, if an equation of motion has more than two time derivatives, it becomes difficult to make physical sense out of it (because it seems to require too much initial data to define a unique solution).
 
  • #6
Ben Niehoff said:
Also, classically speaking, higher derivatives in the Lagrangian will result in higher derivatives in the equations of motion. In particular, if an equation of motion has more than two time derivatives, it becomes difficult to make physical sense out of it (because it seems to require too much initial data to define a unique solution).

Not really. A k-th order differential equation requires as initial conditions the field and the first k-1 derivatives
at the initial time. The case k=2 is nothing special here.
 
  • #7
A. Neumaier said:
Not really. A k-th order differential equation requires as initial conditions the field and the first k-1 derivatives
at the initial time. The case k=2 is nothing special here.

Yes, I'm aware of that. However, our vast experience with classical mechanics indicates that k=2 is indeed special. For example, for a classical system to be a Hamiltonian flow on phase space requires k=2.

The one case I know of in which k>2, the radiation reaction on a moving, charged source, has issues with causality and runaway solutions, precisely because of the third time derivative that appears in the equation of motion.

These are classical situations, but I think deep down they have some relationship to the nonrenormalizability of higher-derivative actions. There is something fundamentally quirky about a physical system that requires information about higher time derivatives.
 
  • #8
Ben Niehoff said:
Yes, I'm aware of that. However, our vast experience with classical mechanics indicates that k=2 is indeed special. For example, for a classical system to be a Hamiltonian flow on phase space requires k=2.

It depends very much on the variables one is using. The Hamiltonian formulation has in fact k=1, and not always can a Hamiltonian system be rewritten in second order form.

Of course, Hamiltonian flow is very special, but it has nothing to do with the number of derivatives.
 
  • #9
Ben Niehoff said:
Also, classically speaking, higher derivatives in the Lagrangian will result in higher derivatives in the equations of motion. In particular, if an equation of motion has more than two time derivatives, it becomes difficult to make physical sense out of it (because it seems to require too much initial data to define a unique solution).

A. Neumaier said:
Not really. A k-th order differential equation requires as initial conditions the field and the first k-1 derivatives
at the initial time. The case k=2 is nothing special here.

Ben Niehoff said:
Yes, I'm aware of that. However, our vast experience with classical mechanics indicates that k=2 is indeed special. For example, for a classical system to be a Hamiltonian flow on phase space requires k=2.

The one case I know of in which k>2, the radiation reaction on a moving, charged source, has issues with causality and runaway solutions, precisely because of the third time derivative that appears in the equation of motion.

These are classical situations, but I think deep down they have some relationship to the nonrenormalizability of higher-derivative actions. There is something fundamentally quirky about a physical system that requires information about higher time derivatives.

Dear Ben Niehoff,

I am afraid I am somewhat skeptical about your arguments for the following reason. For example, the Dirac equation is generally equivalent to a 4th order PDE for just one component (furthermore, this component can be made real by a gauge transform). Source:
http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (an article in Journ. Math. Phys.) Or another example: the equations of scalar electrodynamics (the Klein-Gordon-Maxwell electrodynamics) are generally equivalent (after algebraic elimination of the matter field) to higher order equations for electromagnetic field. Source: http://www.akhmeteli.org/akh-prepr-ws-ijqi2.pdf (an article in Int. Journ. Quantum Inf.). A somewhat cleaner proof can be found in my recent preprints.
 
  • #10
Hmm, ok, maybe I'm wrong.
 
  • #11
Hollowood, http://arxiv.org/abs/0909.0859

"... we should allow all possible operators consistent with spacetime symmetries. In the case of a scalar field, all powers of the field and its derivatives ... (p12)"

"...the principle of universality that allows us to formulate our theories in terms of simple actions. All we need do is include the relevant couplings: all the irrelevant couplings can be taken to vanish. ... it is sufficient to write the simple action ... (p21)"
 

1. Why does Lagrangian in QFT only include first order derivative of field?

One reason for this is that the Lagrangian, which is a function used to describe the dynamics of a system, is derived from the principle of least action. This principle states that a system will take the path of least action, which is the path that minimizes the difference between the initial and final states. In QFT, this means that the Lagrangian must only depend on the first order derivatives of the field, as this is the minimum necessary to describe the dynamics of the system.

2. Are higher order derivatives of the field not important in QFT?

Higher order derivatives of the field can be important in certain situations, but they are not included in the Lagrangian for a few reasons. Firstly, including higher order derivatives would result in equations of motion that are higher order, making them more difficult to solve. Additionally, many physical systems can be accurately described using only the first order derivatives. Finally, there is a principle in physics known as Occam's razor, which states that the simplest explanation is often the best. In this case, including only the first order derivatives of the field is the simplest and most elegant way to describe the system.

3. Can the Lagrangian in QFT include higher order derivatives if needed?

Yes, the Lagrangian in QFT can include higher order derivatives if needed. However, this is typically only done in situations where it is necessary, such as in certain quantum field theories or when studying systems with higher order dynamics. In most cases, the first order derivative is sufficient to accurately describe the dynamics of the system.

4. What is the advantage of using the Lagrangian in QFT?

The use of the Lagrangian in QFT has several advantages. Firstly, it allows for a more elegant and concise description of the system's dynamics compared to other methods. Additionally, the Lagrangian formalism is closely related to the Hamiltonian formalism, which is useful for studying the energy and momentum of a system. Finally, the use of the Lagrangian allows for the application of the principle of least action, which is a powerful tool for understanding the behavior of physical systems.

5. Are there any limitations to using the Lagrangian in QFT?

While the Lagrangian is a powerful tool in QFT, it does have some limitations. One limitation is that it may not be applicable to all systems, particularly those with highly nonlinear dynamics. Additionally, the Lagrangian formalism may not be the most efficient or convenient method for solving certain problems. In these cases, other methods, such as the Hamiltonian formalism, may be more useful. However, overall, the Lagrangian is a valuable and widely used tool in QFT that has greatly contributed to our understanding of the fundamental laws of nature.

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