Expressing the Klein Gordon Hamiltonian in terms of ladder operators

In summary, the conversation is about expressing the terms of the Hamiltonian in terms of ladder operators a(p) and [itex]a^{\dagger}(p). The first term is calculated as \int d^{3}x \frac{E_{p}}{2}\left[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)-a(p)a(-p)-a^{\dagger}(p)a^{\dagger}(-p)\right], while the book states that it should be \int d^{3}x \frac{E_p}{2}\left[-a(p)a(-p)e^{-2iE_{p}t} + a(p)a^
  • #1
maverick280857
1,789
4
Hi everyone

I'm trying to express each term of the Hamiltonian

[itex]H = \int d^{3}x \frac{1}{2}\left[\Pi^2 + (\nabla \Phi)^2 + m^2\Phi^2\right][/tex]

in terms of the ladder operators [itex]a(p)[/itex] and [itex]a^{\dagger}(p)[/itex].

This is what I get for the first term

[tex]\int d^{3}x \frac{E_{p}}{2}\left[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)-a(p)a(-p)-a^{\dagger}(p)a^{\dagger}(-p)\right][/tex]

whereas the book I'm reading from says

[tex]\int d^{3}x \frac{E_p}{2}\left[-a(p)a(-p)e^{-2iE_{p}t} + a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)-a^{\dagger}(p)a^{\dagger}(-p)e^{-2iE_{p}t}\right][/tex]

Is this because the time dependence must be explicitly accounted for? It so happens that the explicit time dependence goes away through the other two terms...but is my own computation correct?

Thanks.

(PS -- This is not homework.)
 
Last edited:
Physics news on Phys.org
  • #2
Ok I think I get it. I have to convert from the Schrodinger to the Heisenberg picture. Is that correct?
 
  • #3
You can do it in either picture. The H-picture is more commonly used in field theory, and then you get those time-dep phases.
 

1. What is the Klein Gordon Hamiltonian?

The Klein Gordon Hamiltonian is a mathematical expression used in quantum field theory to describe the dynamics of a free scalar field. It is a sum of the kinetic and potential energies of the field.

2. What are ladder operators?

Ladder operators are mathematical operators used in quantum mechanics to describe the raising and lowering of energy levels in a quantum system. They are used to represent the creation and annihilation of particles in quantum field theory.

3. How are the Klein Gordon Hamiltonian and ladder operators related?

The Klein Gordon Hamiltonian can be expressed in terms of ladder operators by using the ladder operators to represent the energy levels of the free scalar field. This allows for a more concise and elegant formulation of the Hamiltonian.

4. Why is it useful to express the Klein Gordon Hamiltonian in terms of ladder operators?

Expressing the Klein Gordon Hamiltonian in terms of ladder operators allows for a more intuitive and physically meaningful representation of the dynamics of a free scalar field. It also simplifies calculations and makes it easier to study the behavior of the field.

5. Are there any limitations to expressing the Klein Gordon Hamiltonian in terms of ladder operators?

While expressing the Hamiltonian in terms of ladder operators is useful and widely used in quantum field theory, it is not always possible to do so for more complex systems. In these cases, other mathematical techniques may need to be used to describe the dynamics of the system.

Similar threads

Replies
3
Views
278
  • Quantum Physics
Replies
3
Views
834
Replies
24
Views
1K
  • Quantum Physics
Replies
4
Views
804
Replies
9
Views
480
Replies
1
Views
555
  • Quantum Physics
Replies
0
Views
667
Replies
8
Views
1K
  • Quantum Physics
Replies
1
Views
611
  • Quantum Physics
Replies
4
Views
1K
Back
Top