- #1
maverick280857
- 1,789
- 4
Hi everyone
The Hamiltonian of the Klein Gordon field can be written as
[tex]H = \frac{1}{2}\int d^{3}E_p \left[a^{\dagger}(p)a(p) + a(p)a^{\dagger}(p)\right][/tex]
and we have
[tex][H,a(p')] = -E_{p'}a(p')[/tex]
[tex][H,a(p')] = +E_{p'}a^{\dagger}(p')[/tex]
The book I'm reading states that
What does this mean?
Thanks.
The Hamiltonian of the Klein Gordon field can be written as
[tex]H = \frac{1}{2}\int d^{3}E_p \left[a^{\dagger}(p)a(p) + a(p)a^{\dagger}(p)\right][/tex]
and we have
[tex][H,a(p')] = -E_{p'}a(p')[/tex]
[tex][H,a(p')] = +E_{p'}a^{\dagger}(p')[/tex]
The book I'm reading states that
What was the positive energy component of the classical field now annihilates the quantum, and the negative energy component now creates the quantum.
What does this mean?
Thanks.