- #1
LAHLH
- 409
- 1
Hi,
I'm just trying to convince myself that the field in the interaction picture (IP) [tex] \phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}[/tex] satisfies the Klein Gordan equation: [tex] (\tfrac{\partial^2}{\partial t^2}-\nabla^2+m^2)\phi_I(x,t)=0 [/tex].
I have so far worked out that the time derivative is:
[tex] \frac{\partial\phi_I}{\partial t}=-i[\phi_I,H_0] [/tex]
and the second time deriv is:
[tex] \frac{\partial^2\phi_I}{\partial^2 t}=[H_0, [\phi_I,H_0]] [/tex].
I'm now trying to work out the commutator in these expressions. My [tex] H_0=\int d^3x \left(\tfrac{1}{2}\Pi(x,0)^2+\tfrac{1}{2}(\nabla\phi(x,0))^2+\tfrac{1}{2}m^2\phi^2(x,0)\right) [/tex]. This can also be expressed as [tex] H_0=\int \tfrac{d^3\vec{k}\omega}{(2\pi)^3} a^{\dag}(\vec{k})a(\vec{k}) [/tex]
So considering [tex][\phi_I,H_0]=e^{iH_0t}\phi(x,0)e^{-iH_0t}H_0-H_0e^{iH_0t}\phi(x,0)e^{-iH_0t}=e^{iH_0t}[\phi(x,0), H_0]e^{-iH_0t} [/tex] I see I need to evalate the commutator in the Schroedinger picture:
[tex][\phi(x,0), H_0]=\phi(x,0)H_0-H_0\phi(x,0) [/tex]
I'm having issues doing this however, would it be better to work with H and the field expressed in terms of creation and annihilation or is it easier just to stay in the phi Pi representation?
I'm just trying to convince myself that the field in the interaction picture (IP) [tex] \phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}[/tex] satisfies the Klein Gordan equation: [tex] (\tfrac{\partial^2}{\partial t^2}-\nabla^2+m^2)\phi_I(x,t)=0 [/tex].
I have so far worked out that the time derivative is:
[tex] \frac{\partial\phi_I}{\partial t}=-i[\phi_I,H_0] [/tex]
and the second time deriv is:
[tex] \frac{\partial^2\phi_I}{\partial^2 t}=[H_0, [\phi_I,H_0]] [/tex].
I'm now trying to work out the commutator in these expressions. My [tex] H_0=\int d^3x \left(\tfrac{1}{2}\Pi(x,0)^2+\tfrac{1}{2}(\nabla\phi(x,0))^2+\tfrac{1}{2}m^2\phi^2(x,0)\right) [/tex]. This can also be expressed as [tex] H_0=\int \tfrac{d^3\vec{k}\omega}{(2\pi)^3} a^{\dag}(\vec{k})a(\vec{k}) [/tex]
So considering [tex][\phi_I,H_0]=e^{iH_0t}\phi(x,0)e^{-iH_0t}H_0-H_0e^{iH_0t}\phi(x,0)e^{-iH_0t}=e^{iH_0t}[\phi(x,0), H_0]e^{-iH_0t} [/tex] I see I need to evalate the commutator in the Schroedinger picture:
[tex][\phi(x,0), H_0]=\phi(x,0)H_0-H_0\phi(x,0) [/tex]
I'm having issues doing this however, would it be better to work with H and the field expressed in terms of creation and annihilation or is it easier just to stay in the phi Pi representation?