Why the minima of potential of classical Lagrangian called ''vacuum expectation''?

In summary, the minima of the potential of classical Lagrangian is referred to as the "vacuum expectation value of Phi (field function)." This is because the lowest energy state of a system can be defined as the vacuum, and the Lagrangian gives the energies of the system. The expectation value of the field operator at the vacuum state depends on the properties of the vacuum, which can be shifted by a constant to make it zero. This is known as spontaneous symmetry breaking, where the choice of vacuum state breaks the symmetry of the problem.
  • #1
ndung200790
519
0
Please teach me this:
Why the minima of potential of classical Lagrangian is called the ''vacuum expectation value of Phi(field function)''.Is it really a vacuum expectation value of field operator at the vacuum states(at this state,the potential part of classical Lagrangian equals zero)?
Thankyu very much in advanced.
 
Physics news on Phys.org
  • #2


Well, the answer lies in the fact that you can always define your vacuum as the state with the lowest energy, and since your Lagrangian gives you the energies of your system, the lowest energy state lies in the minimum of the Lagrangian.
 
  • #3


It seem to me that the expectation value of operator of field at the vacuum states depends on the properties of the vacuum states.In some case the one point(one leg) correlation function does not equal to zero.Then the expectation value of operator of field does not equal zero.So that the spontaneously breaking of symmetry has the origination from the properties of vacuum states.Is that correct?
 
  • #4


Well, the properties of your vacuum state depend on your definition of the Lagrangian. In the case of your vacuum state not being zero, you can simply shift your field by some constant to make it zero. This amounts to a redefinition of your Lagrangian.
Spontaneous symmetry breaking also depends (obviously) on the definition of your Lagrangian. The most prominent example would be the [tex]\varphi^4[/tex]-potential, which admits a U(1)-symmetry. One has now continuously many possible vacuum states, and by chosing one you break the symmetry of your problem while the Lagrangian itself remains symmetric. The degree of freedom of chosing one of those vacuum states amounts to the existence of the so-called Nambu-Goldstone boson.

Edit: By chosing a vacuum state I mean taking it as a starting point for your perturbative expansion.
 
Last edited:

1. What is a potential in classical Lagrangian?

A potential in classical Lagrangian is a mathematical function that describes the forces acting on a particle in a given system. It is used to calculate the energy of the system in terms of the positions and velocities of the particles.

2. Why are the minima of potential in classical Lagrangian called "vacuum expectation"?

The minima of potential in classical Lagrangian are called "vacuum expectation" because they represent the lowest energy state of the system, or the state in which there are no forces acting on the particles. This is similar to the concept of vacuum in quantum field theory, where it represents the lowest energy state of a field.

3. How are the minima of potential in classical Lagrangian related to quantum mechanics?

The minima of potential in classical Lagrangian are related to quantum mechanics through the concept of vacuum expectation. In quantum mechanics, the vacuum expectation value is the average value of a quantum field in its lowest energy state. This value is related to the minima of potential in classical Lagrangian, as they both represent the lowest energy state of their respective systems.

4. Can the minima of potential in classical Lagrangian change?

Yes, the minima of potential in classical Lagrangian can change depending on the forces acting on the particles in the system. If the forces change, the energy of the system will also change, resulting in a shift in the minima of potential.

5. How are the minima of potential in classical Lagrangian used in physics?

The minima of potential in classical Lagrangian are used in physics to understand the behavior and dynamics of physical systems. They help to determine the equilibrium points of a system, as well as the stability of these points. They are also used to calculate the energy of a system and how it changes over time as the particles move.

Similar threads

  • Quantum Physics
Replies
1
Views
734
Replies
3
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
2K
  • Quantum Physics
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
742
  • Quantum Physics
Replies
5
Views
2K
Replies
3
Views
745
  • Quantum Physics
Replies
6
Views
4K
Replies
16
Views
2K
Back
Top