jasonc65
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The ground state wave functional for the photon theory is given as
\Psi_0[\tilde{a}] = \eta \exp \left(-\frac{1}{2} \int \frac{d^3k}{(2\pi)^3} \frac{(\vec{k}\times\tilde{a}(\vec{k}))\cdot(\vec{k}\times\tilde{a}(-\vec{k}))}{|\vec{k}|}\right)
where \tilde{a} is given as the Fourier transform of a, that is,
a_i(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\tilde{a}_i(\vec{k})e^{i\vec{k}\cdot\vec{x}}
Transforming back to a, the book now says that (10.81) is equivalent to
\Psi_0[a] = \eta \exp \left(-\frac{1}{(2\pi)^2} \int d^3x d^3y \frac{(\nabla\times\vec{a}(\vec{x}))\cdot(\nabla\times\vec{a}(\vec{y}))}{|\vec{x}-\vec{y}|^2}\right)
I've had to think about this for a long time, and I'm still not sure I understand it exactly.
\Psi_0[\tilde{a}] = \eta \exp \left(-\frac{1}{2} \int \frac{d^3k}{(2\pi)^3} \frac{(\vec{k}\times\tilde{a}(\vec{k}))\cdot(\vec{k}\times\tilde{a}(-\vec{k}))}{|\vec{k}|}\right)
(10.81)
where \tilde{a} is given as the Fourier transform of a, that is,
a_i(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\tilde{a}_i(\vec{k})e^{i\vec{k}\cdot\vec{x}}
(10.67)
Transforming back to a, the book now says that (10.81) is equivalent to
\Psi_0[a] = \eta \exp \left(-\frac{1}{(2\pi)^2} \int d^3x d^3y \frac{(\nabla\times\vec{a}(\vec{x}))\cdot(\nabla\times\vec{a}(\vec{y}))}{|\vec{x}-\vec{y}|^2}\right)
(10.83)
I've had to think about this for a long time, and I'm still not sure I understand it exactly.