# Proving Interaction picture field satisfies KG eqn

## Main Question or Discussion Point

Hi,

I'm just trying to convince myself that the field in the interaction picture (IP) $$\phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}$$ satisfies the Klein Gordan equation: $$(\tfrac{\partial^2}{\partial t^2}-\nabla^2+m^2)\phi_I(x,t)=0$$.

I have so far worked out that the time derivative is:

$$\frac{\partial\phi_I}{\partial t}=-i[\phi_I,H_0]$$

and the second time deriv is:

$$\frac{\partial^2\phi_I}{\partial^2 t}=[H_0, [\phi_I,H_0]]$$.

I'm now trying to work out the commutator in these expressions. My $$H_0=\int d^3x \left(\tfrac{1}{2}\Pi(x,0)^2+\tfrac{1}{2}(\nabla\phi(x,0))^2+\tfrac{1}{2}m^2\phi^2(x,0)\right)$$. This can also be expressed as $$H_0=\int \tfrac{d^3\vec{k}\omega}{(2\pi)^3} a^{\dag}(\vec{k})a(\vec{k})$$

So considering $$[\phi_I,H_0]=e^{iH_0t}\phi(x,0)e^{-iH_0t}H_0-H_0e^{iH_0t}\phi(x,0)e^{-iH_0t}=e^{iH_0t}[\phi(x,0), H_0]e^{-iH_0t}$$ I see I need to evalate the commutator in the Schroedinger picture:

$$[\phi(x,0), H_0]=\phi(x,0)H_0-H_0\phi(x,0)$$

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?

Related Quantum Physics News on Phys.org
There must be an easier way to show the simple fact that the interaction field obeys the Klein Gordon eqn. I just did a billion pages of algebra, working out commutators in terms of the fields, only to find I'd prob made a mistake.

LAHLH,

why don't you use the explicit form of the quantum field? For example, in the scalar case it is

$$\phi(\mathbf{x}, t) = \int d\mathbf{p} \left( e^{-i \mathbf{px}+i\omega_p t} a_p + e^{i \mathbf{px}-i\omega_p t} a^{\dag}_p \right)$$

and it is very easy to show that KG equation is satisfied.

Eugene.

I think the field you quoted is in the Heisenburg picture, I'm trying to show that the field $$\phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}$$ satisifes the KG eqn. My general Hamiltonian is $$H=H_0+H_1$$. This is problem 9.5 in Srendicki to make things clearer: http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

Cheers

I think the field you quoted is in the Heisenburg picture, I'm trying to show that the field $$\phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}$$ satisifes the KG eqn. My general Hamiltonian is $$H=H_0+H_1$$. This is problem 9.5 in Srendicki to make things clearer: http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

Cheers
The field that I wrote evolves in time with the help of the non-interacting Hamiltonian. This is exactly what you need. See eq. (4.5) in Srednicki.

Eugene.

samalkhaiat
$$\phi_I(x,t)=e^{iH_0t}\phi_{S}(x)e^{-iH_0t}$$

Expand this using the Baker-Hausdroff identity, you find

$$\phi_{I}(x,t) = \phi_{S}(x) +i[tH_{0},\phi_{S}] + \frac{i^{2}}{2!}[tH_{0},[tH_{0},\phi_{S}]] + ... \ \ \ (1)$$

In the Schrodinger picture, we expand the field operator $\phi_{S}(x)$ in terms of some base functions $u_{k}(x)$, as

$$\phi_{S}(x) = \sum_{k} \left( a_{k} u_{k}(x) + a^{\dagger}_{k} u^{*}_{k}(x) \right) \ \ \ (2)$$

Putting eq(2) in eq(1), the problem get reduced to calculating the commutators of $a_{k}$ and $a^{\dagger}_{k}$ with the free Hamiltonian, which I will take to be

$$H_{0} = \sum_{k} \omega_{k} a^{\dagger}_{k} a_{k}$$

Using the canonical algebra

$$[a_{n},a_{k}] = [a^{\dagger}_{n},a^{\dagger}_{k}] = 0$$

$$[a_{n},a^{\dagger}_{k}] = \delta_{nk}\ ,$$

we find

$$[H_{0},a_{k}] = \omega_{k}a_{k} \ ,$$
$$[H_{0},[H_{0},a_{k}]] = \omega^{2}_{k} a_{k} \ , ....$$

and similar results for $a^{\dagger}_{k}$. If you now insert the whole lot back into eq(1), you find

$$\phi_{I}(x,t) = \sum_{k} \left( a_{k}e^{-i\omega_{k}t} u_{k}(x) + a^{\dagger}_{k}e^{+i\omega_{k}t} u^{*}(x) \right)$$

It is now easy to show that the field $\phi_{I}$ satisfies the K-G equation when

$$\omega_{k} = + \sqrt{ k^{2} + m^{2}} , \ u_{k}(x) = \exp(ip.x)$$

regards

sam

Last edited:
Thanks to you both. I think I've proved this in my own (rather complicated and perhaps uneccessary) way in the end before I came back to this thread.
$$\phi_I(x,t)=e^{iH_0t}\phi(x,0)e^{-iH_0t}$$
From which it follows that:

$$\frac{\partial^2\phi_I}{\partial^2 t}=-[ [\phi_I,H_0], H_0]$$

So 1) Calculate $$[\phi_I,H_0]$$. Expanding: $$[\phi_I,H_0]=e^{iH_0t} \phi(x,0)e^{-iH_0t}H_0-H_0e^{iH_0t}\phi(x,0)e^{-iH_0t}=e^{iH_0t}[\phi(x,0), H_0]e^{-iH_0t}$$ (eqn 1)

So what is$$[\phi(x,0), H_0]$$? : Remembering that $$H_0=\int d^3x \left(\tfrac{1}{2}\Pi(x,0)^2+\tfrac{1}{2}(\nabla\phi(x,0))^2+\tfrac{1}{2}m^2\phi^2(x,0)\right)$$
Then
$$[\phi(x,0), H_0]=\tfrac{1}{2} \int d^3y [\phi(x,0), \Pi^2(y,0)]$$
$$[\phi(x,0), H_0]=\tfrac{1}{2} \int d^3y [\phi(x,0), \Pi(y,0)] \Pi(y,0)+ \Pi(y,0) [\phi(x,0), \Pi(y,0)]$$
Using equal time commuation relations now
$$[\phi(x,0), H_0]=\tfrac{1}{2} \int d^3y i\delta^3(x-y)\Pi(y,0)+ \Pi(y,0) i\delta^3(x-y)$$
$$[\phi(x,0), H_0]=i\Pi(x,0)$$

Plugging this back into eqn 1:

$$[\phi_I,H_0]=e^{iH_0t}[\phi(x,0), H_0]e^{-iH_0t} =i e^{iH_0t}\Pi(x,0)e^{-iH_0t}=i\Pi_I(x,t)$$

Step 2) Plug this result back into $$\frac{\partial^2\phi_I}{\partial^2 t}=-[ [\phi_I,H_0], H_0] =-i[ \Pi_I(x,t), H_0]$$

So now what is $$[ \Pi_I(x,t), H_0] =e^{iH_0t} \Pi(x,0)e^{-iH_0t}H_0-H_0e^{iH_0t}\Pi(x,0)e^{-iH_0t}=e^{iH_0t}[\Pi(x,0), H_0]e^{-iH_0t}$$?

Now considering $$[\Pi(x,0), H_0]$$:

$$[ \Pi(x,0), H_0]= \tfrac{1}{2} \int d^3y [\Pi(x,0), \nabla^{i}\phi(y,0)\nabla^{i}\phi(y,0)+m^2\phi(y,0)\phi(y,0)]$$

$$[ \Pi(x,0), H_0]= \tfrac{1}{2} \int d^3y \left( [\Pi(x,0), \nabla^{i}\phi(y,0)]\nabla^{i}\phi(y,0)+\nabla^{i}\phi(y,0) [\Pi(x,0), \nabla^{i}\phi(y,0)]+m^2([\Pi(x,0)\phi(y,0)]\phi(y,0)+\phi(y,0)[\Pi(x,0)\phi(y,0)])\right)$$

Since the dels are acting only on y we can bring out of the commutator completely as follows

$$[ \Pi(x,0), H_0]= \tfrac{1}{2} \int d^3y \left( \nabla^{i}_{y}[\Pi(x,0), \phi(y,0)]\nabla^{i}\phi(y,0)+\nabla^{i}\phi(y,0) \nabla^{i}_{y}[\Pi(x,0), \phi(y,0)]+m^2([\Pi(x,0)\phi(y,0)]\phi(y,0)+\phi(y,0)[\Pi(x,0)\phi(y,0)])\right)$$

Finally using the canonical equal time commutation relations:

$$[ \Pi(x,0), H_0]= -i \int d^3y \left( \nabla^{i}_{y}\delta^3(x-y)\nabla^{i}\phi(y,0)+m^2\delta^3(x-y)\phi(y,0)\right)$$
$$[ \Pi(x,0), H_0]= -i \left( -\nabla^{2}\phi(x,0)+m^2\phi(x,0)\right)$$

Thus:

$$[ \Pi_I(x,t), H_0] =-ie^{iH_0t}\left(\nabla^{2}\phi(x,0)+m^2\phi(x,0)\right)e^{-iH_0t} =-i\left(\nabla^{2}+m^2\right)\phi_I(x,t)$$

This finally.....gives:

$$\frac{\partial^2\phi_I}{\partial^2 t}=-[ [\phi_I,H_0], H_0] =-i[ \Pi_I(x,t), H_0]=-\left(-\nabla^{2}+m^2\right)\phi_I(x,t)=\left(\nabla^{2}-m^2\right)\phi_I(x,t)$$

So we have recovered the KG equation:

$$\frac{\partial^2\phi_I}{\partial^2 t}=\left(\nabla^{2}-m^2\right)\phi_I(x,t)$$

1) $$[\phi(x,0), H_0]=\tfrac{1}{2} \int d^3y [\phi(x,0), \Pi^2(y,0)]$$

This is true because \phi(x,0) obviously commutes with the mass term $$m^2\phi^2(x,0)$$, but it is also true that $$\int d^3y [\phi(x,0), \nabla^{i}\phi(y,0)\nabla^{i}\phi(y,0)]=0$$, as I will now show:

$$\int d^3y [\phi(x,0), \nabla^{i}\phi(y,0)\nabla^{i}\phi(y,0)]=\int d^3y [\phi(x,0), \nabla^{i}\phi(y,0)]\nabla^{i}\phi(y,0)+\nabla^{i}\phi(y,0)[\phi(x,0), \nabla^{i}\phi(y,0)]$$

Since the nabla's are acting on y's we can pull them outside the commutators:

$$\int d^3y [\phi(x,0), \nabla^{i}\phi(y,0)\nabla^{i}\phi(y,0)]=\int d^3y \nabla^{i}_{y}[\phi(x,0), \phi(y,0)]\nabla^{i}\phi(y,0)+\nabla^{i}\phi(y,0)\nabla^{i}_{y}[\phi(x,0), \phi(y,0)]$$

This is obviously zero since $$[\phi(x,0), \phi(y,0)]=0$$. So we have the result that:

$$[\phi(x,0), (\nabla^{i}\phi(y,0))^2]=0$$

2) $$[ \Pi(x,0), H_0]= -i \int d^3y \left( \nabla^{i}_{y}\delta^3(x-y)\nabla^{i}\phi(y,0)+m^2\delta^3(x-y)\phi(y,0)\right)$$

Concentrating on the first term:

$$\int d^3y \left( \nabla^{i}_{y}\delta^3(x-y)\nabla^{i}\phi(y,0) \right)$$

I use integration by parts here to shift the nabla off the delta:

$$\left[\delta^3(x-y)\nabla^{i}_{y}\phi(y,0)\right]- \int d^3y\delta^3(x-y)\nabla^{2}_{y}\phi(y,0)$$

The first term dies by compact support and we're left with$$-\int d^3y\delta^3(x-y)\nabla^{2}_{y}\phi(y,0)=-\nabla^{2}\phi(x,0)$$

This is the origin of the -ve term in my KG equation on the del squared.

I feel like maybe this was a waste of time now, if one can just use the expansion sam posted, alas............................

I'd still be grateful if anyone could comment if my proof is actually correct. The only niggle I have with it is this part:

$$e^{iH_0t}\left(\nabla^{2}\phi(x,0)+m^2\phi(x,0)\right)e^{-iH_0t} =-\left(\nabla^{2}+m^2\right)\phi_I(x,t)$$

Am I allowed to pass $$e^{iH_0t}$$ through the $$\nabla^{2}$$ to turn my $$\phi(x,0)$$ into $$\phi_I(x,t)$$ as I require? Since $$e^{iH_0t}$$ is obviously an operator and I can't see an obvious reason that is should commute with the operator $$\nabla^{2}$$?

I feel like maybe this was a waste of time now, if one can just use the expansion sam posted, alas............................
The calculation is much much easier if both $$H_0$$ and $$\phi(x,t)$$ are expanded in the momentum space as samalkhaiat showed you.

I'd still be grateful if anyone could comment if my proof is actually correct. The only niggle I have with it is this part:

$$e^{iH_0t}\left(\nabla^{2}\phi(x,0)+m^2\phi(x,0)\right)e^{-iH_0t} =-\left(\nabla^{2}+m^2\right)\phi_I(x,t)$$

Am I allowed to pass $$e^{iH_0t}$$ through the $$\nabla^{2}$$ to turn my $$\phi(x,0)$$ into $$\phi_I(x,t)$$ as I require? Since $$e^{iH_0t}$$ is obviously an operator and I can't see an obvious reason that is should commute with the operator $$\nabla^{2}$$?
$$H_0$$ can be regarded as differentiation on t, so it commutes with $$\nabla^{2}$$, which differentiates on x.

Eugene.

$$H_0$$ can be regarded as differentiation on t, so it commutes with $$\nabla^{2}$$, which differentiates on x.
Eugene.
Thanks again Eugene. Could you elaborate a little on this part for me. Is this just because $$\dot{A}=[A,H]$$ ? or something similar?

Thanks again Eugene. Could you elaborate a little on this part for me. Is this just because $$\dot{A}=[A,H]$$ ? or something similar?
Yes, the field derivative on t is given by the field commutator with $$H_0$$

$$\frac{\partial\phi}{\partial t}=-i[\phi,H_0]$$

$$H_0=\frac{1}{(2\pi)^3 } \int d^3\vec{k}\omega_k a^{\dag}(\vec{k})a(\vec{k})$$

The field derivative on $$\vec{x}$$ is given by the field commutator with the total momentum operator $$\vec{P}$$

$$\frac{\partial\phi}{\partial \vec{x}}=-i[\phi, \vec{P}]$$

$$\vec{P} = \frac{1}{(2\pi)^3 } \int d^3\vec{k} \vec{k} a^{\dag}(\vec{k})a(\vec{k})$$

One can show easily that the two operators commute $$[H_0, \vec{P}] = 0$$.

Eugene.

Thanks for the help