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In class we derived the field for the non-homogeneous Klein-Gordon equation

\begin{equation}\label{eqn:nonhomoKGhamiltonian:20}

\begin{aligned}

\phi(x)

&= \int \frac{d^3 p}{(2\pi)^3} \inv{\sqrt{2 \omega_\Bp}}

\evalbar{

\lr{

e^{-i p \cdot x} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }

+

e^{i p \cdot x} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }

}

}

{

p^0 = \omega_\Bp

} \\

&= \int \frac{d^3 p}{(2\pi)^3} \inv{\sqrt{2 \omega_\Bp}}

\lr{

e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }

+

e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }

}.

\end{aligned}

\end{equation}

This means that we have

\begin{equation}\label{eqn:nonhomoKGhamiltonian:40}

\begin{aligned}

\pi = \dot{\phi}

&= \int \frac{d^3 p}{(2\pi)^3} \frac{i \omega_\Bp}{\sqrt{2 \omega_\Bp}}

\lr{

– e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }

+

e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }

} \\

(\spacegrad \phi)_k =

&= \int \frac{d^3 p}{(2\pi)^3} \frac{i p_k}{\sqrt{2 \omega_\Bp}}

\lr{

e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }

–

e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }

},

\end{aligned}

\end{equation}

and could plug these into the Hamiltonian

\begin{equation}\label{eqn:nonhomoKGhamiltonian:60}

H = \int d^3 p \lr{ \inv{2} \pi^2 + \inv{2} \lr{ \spacegrad \phi}^2 + \frac{m^2}{2} \phi^2 },

\end{equation}

to find \( H \) in terms of \( \tilde{j} \) and \( a_\Bp^\dagger, a_\Bp \). The result was mentioned in class, and it was left as an exercise to verify.

There’s an easy way and a dumb way to do this exercise. I did it the dumb way, and then after suffering through two long pages, where the equations were so long that I had to write on the paper sideways, I realized the way I should have done it.

The easy way is to observe that we’ve already done exactly this for the case \( \tilde{j} = 0 \), which had the answer

\begin{equation}\label{eqn:nonhomoKGhamiltonian:80}

H = \inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger }.

\end{equation}

To handle this more general case, all we have to do is apply a transformation

\begin{equation}\label{eqn:nonhomoKGhamiltonian:100}

a_\Bp \rightarrow

a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}},

\end{equation}

to \ref{eqn:nonhomoKGhamiltonian:80}, which gives

\begin{equation}\label{eqn:nonhomoKGhamiltonian:120}

\begin{aligned}

H

&=

\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }^\dagger\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} } +\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }^\dagger } \\

&=

\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} } \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} } +\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }

}.

\end{aligned}

\end{equation}

Like the \( \tilde{j} = 0 \) case, we can use normal ordering. This is easily seen by direct expansion:

\begin{equation}\label{eqn:nonhomoKGhamiltonian:140}

\begin{aligned}

\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} } \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }

&=

a_\Bp^\dagger a_\Bp

– \frac{i \tilde{j}^\conj(p) a_\Bp}{\sqrt{2 \omega_\Bp}}

+ \frac{ a_\Bp^\dagger i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}

+ \frac{\Abs{j}^2}{2 \omega_\Bp} \\

\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }

&=

a_\Bp^\dagger a_\Bp

+ \frac{i \tilde{j}^\conj(p) a_\Bp^\dagger}{\sqrt{2 \omega_\Bp}}

– \frac{ a_\Bp i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}

+ \frac{\Abs{j}^2}{2 \omega_\Bp}.

\end{aligned}

\end{equation}

Because \( \tilde{j} \) is just a complex valued function, it commutes with \( a_\Bp, a_\Bp^\dagger \), and these are equal up to the normal ordering, allowing us to write

\begin{equation}\label{eqn:nonhomoKGhamiltonian:160}

:H: =

\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}} \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} },

\end{equation}

which is the result mentioned in class.