Classical equivalent of scalar free field in QFT

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SUMMARY

The discussion focuses on identifying a classical system that replicates the Lagrangian density of a free scalar field in Quantum Field Theory (QFT), represented as $$\mathcal{L} = \frac{1}{2}\left(\partial\phi\right)^2 - \frac{1}{2}m^2\phi^2$$. The proposed classical analogue is the transversal motion of an elastic string, with a Lagrangian of $$\mathcal{L} = \frac{1}{2}\left[\left(\partial_t\phi\right)^2 - \left(\partial_x\phi\right)^2\right]$$. The missing mass term $$\frac{1}{2}m^2\phi^2$$ can be attributed to a local elastic force acting on the string. The classical Lagrangian serves critical roles in QFT, including its application in the Feynman path integral formalism and the derivation of quantized free fields.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with Quantum Field Theory (QFT)
  • Knowledge of classical field theory
  • Basic principles of elastic forces and wave equations
NEXT STEPS
  • Explore the Feynman path integral formalism in QFT
  • Study the process of canonical quantization in field theory
  • Investigate the role of classical solutions in quantizing fields
  • Analyze the relationship between classical and quantum fields through Poisson and commutation brackets
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in theoretical physics, quantum mechanics, and classical mechanics, as well as students seeking to understand the connections between classical systems and quantum field theories.

eoghan
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Which classical system has the same lagrangian of a free scalar field in QFT?
Hi there,

In QFT, a free scalar field can be represented by the lagrangian density
$$\mathcal{L} = \frac{1}{2}\left(\partial\phi\right)^2 - \frac{1}{2}m^2\phi^2$$

I would like to find a classical system that has the same lagrangian. If we consider the transversal motion of an elastic string, the lagrangian would be
$$\mathcal{L} = \frac{1}{2}\left[\left(\partial_t\phi\right)^2 - \left(\partial_x\phi\right)^2\right]$$
This is very similar to the lagrangian of the free field, except that there is a missing term ##\frac{1}{2}m^2\phi^2##. Which classical force can give rise to this term? I was thinking of a local elastic force, that pulls each element of the string downwards. So basically it is as if the string was standing horizontal and each of its points are connected to the ground by vertical springs.

Is this correct?
 
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eoghan said:
Summary:: Which classical system has the same lagrangian of a free scalar field in QFT?

Hi there,

In QFT, a free scalar field can be represented by the lagrangian density
$$\mathcal{L} = \frac{1}{2}\left(\partial\phi\right)^2 - \frac{1}{2}m^2\phi^2$$

I would like to find a classical system that has the same lagrangian. If we consider the transversal motion of an elastic string, the lagrangian would be
$$\mathcal{L} = \frac{1}{2}\left[\left(\partial_t\phi\right)^2 - \left(\partial_x\phi\right)^2\right]$$
This is very similar to the lagrangian of the free field, except that there is a missing term ##\frac{1}{2}m^2\phi^2##. Which classical force can give rise to this term? I was thinking of a local elastic force, that pulls each element of the string downwards. So basically it is as if the string was standing horizontal and each of its points are connected to the ground by vertical springs.

Is this correct?

Just to make things rigorous: the quoted Lagrangian density you gave is from classical field theory. QFT uses the so-called "quantum fields", whose multiplication (and multiplication of space-time derivatives) is normally ill-defined.

So the classical Lagrangian is useful in QFT in two ways.
1. For the Feynman path integral formalism.
2. Its classical solutions of the EOM lead to quantized free fields. Putting classical Poisson brackets to quantum brackets (commutators of operators in a Fock space) is the key ingredient of quantization.
 
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3. To figure out the canonical field momenta for "canonical quantization", leading to the operator formalism.
 
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Thank you very much for all your answers :)
 

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