I Classical equivalent of scalar free field in QFT

eoghan
Messages
201
Reaction score
7
TL;DR 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?
 
  • Like
Likes Delta2
Physics news on Phys.org
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.
 
  • Like
Likes vanhees71
3. To figure out the canonical field momenta for "canonical quantization", leading to the operator formalism.
 
  • Like
Likes eoghan
Thank you very much for all your answers :)
 
I am not sure if this belongs in the biology section, but it appears more of a quantum physics question. Mike Wiest, Associate Professor of Neuroscience at Wellesley College in the US. In 2024 he published the results of an experiment on anaesthesia which purported to point to a role of quantum processes in consciousness; here is a popular exposition: https://neurosciencenews.com/quantum-process-consciousness-27624/ As my expertise in neuroscience doesn't reach up to an ant's ear...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
I am reading WHAT IS A QUANTUM FIELD THEORY?" A First Introduction for Mathematicians. The author states (2.4 Finite versus Continuous Models) that the use of continuity causes the infinities in QFT: 'Mathematicians are trained to think of physical space as R3. But our continuous model of physical space as R3 is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum...
Back
Top