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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.