QFT: expanding the propagator in terms of Minkowski modes

Question

I'm trying to understand the usual "Fourier transform" of the free scalar propagator $ G(x,y) = \int \frac{d^{4}k}{(2\pi)^{4}} \frac{1}{\omega_{\mathbf{k}}^{2} + k^{2}} e^{i k \cdot (x - y)}$. I'd like to understand this as an expansion in terms of Minkowski modes.

The propagator is defined as the solution to the equation $[\square + m^{2}] G(x,y) = \delta^{(4)}(x-y)$. The Minkowski modes $\{ u_{\mathbf{k}}, u_{\mathbf{k}}^{\ast} \}_{\mathbf{k} \in \mathbb{R}^{3}}$ are solutions to the equation $[ \square + m^{2} ]u_{\mathbf{k}} = 0$, which are given by: $$ u_{\mathbf{k}}( x ) = u_{\mathbf{k}}( x^{0}, \mathbf{x} ) = \frac{ 1 }{ \sqrt{ 2 \omega_{\mathbf{k}} (2\pi)^{3} } } \exp\left( i \mathbf{k} \cdot \mathbf{x} - i \omega_{\mathbf{k}} x^{0} \right) $$

These solutions constitute a complete linearly independent set. They are also orthonormal with respect to the Klein-Gordon inner product, which is given by: $$ \langle \phi_{1}, \phi_{2} \rangle = i \int d^{3}\mathbf{x}\ \left[ \phi_{1}^{\ast}(x) \frac{\partial \phi_{2}}{\partial x^{0}} - \frac{\partial \phi^{\ast}_{1}}{\partial x^{0}} \phi_{2}(x) \right] $$

We integrate the above over a hypersurface $\Sigma$ of constant $x^{0}$. They are orthonormal such that: \begin{eqnarray*} \langle u_{\mathbf{k}}, u_{\mathbf{p}} \rangle = \delta^{(3)}(\mathbf{k} - \mathbf{p}) \\ \langle u_{\mathbf{k}}, u^{\ast}_{\mathbf{p}} \rangle = 0 \\ \langle u^{\ast}_{\mathbf{k}}, u^{\ast}_{\mathbf{p}} \rangle = - \delta^{(3)}(\mathbf{k} - \mathbf{p}) \end{eqnarray*}

Supposedly, the Minkowski modes are a complete set in the sense that: $$ \sum_{\mathbf{k}} u^{\ast}_{\mathbf{k}}(x) u_{\mathbf{k}}(y) = \delta^{(4)}(x-y) $$

I've been told that if I think of the solutions $u$ in terms of an eigenvalue problem $\square u_{\mathbf{k}} = \lambda_{\mathbf{k}} u_{\mathbf{k}}$, then I can use the above to write the propagator as: $$ G(x,y) \ = \ \sum_{\mathbf{k}} \frac{ u^{\ast}_{\mathbf{k}}(x) u_{\mathbf{k}}(y) }{\lambda_{\mathbf{k}}} \ = \ \int \frac{d^{3}\mathbf{k}}{(2\pi)^{3}} \frac{e^{i \mathbf{k} \cdot (\mathbf{x} - \mathbf{y})}}{\omega_{\mathbf{k}}^{2} - \mathbf{k}^{2}} $$

My Questions:

$\mathbf{1.}$ The Klein-Gordon inner product is over the space of $L^{2}(\mathbb{R}^{4})$? (The space of square-integrable functions, with variables in $\mathbb{R}^{4}$?)

$\mathbf{2.}$ How do can I understand the completeness relation $\sum_{\mathbf{k}} u^{\ast}_{\mathbf{k}}(x) u_{\mathbf{k}}(y) = \delta^{(4)}(x-y)$ with reference to the Klein-Gordon inner product? I'm a little confused how I can say this.

$\mathbf{3.}$ Why does $G(x,y) \ = \ \sum_{\mathbf{k}} \frac{ u^{\ast}_{\mathbf{k}}(x) u_{\mathbf{k}}(y) }{\lambda_{\mathbf{k}}} $? I simply don't understand this bit, and is the main reason I am posting.

Thank you in advance!

Answer 1

Consider a general function expanded out in terms of the eigenbasis:

$$ f(x) = \sum_{k} a_{k} u_{k}(x) $$

The action of $\Box$ on this is just given by:

$$ \Box f(x) = \sum_{k} \lambda_{k} a_{k} u_{k}(x) $$

where $\lambda_{k} = -k_{\mu}k^{\mu}$, which can be found by applying the d'Alembertian to $A\exp{(i k_{\mu}x^{\mu})}$.

If $f(x) = G(x;y)$ then:

$$ \sum_{k} \lambda_{k} a_{k} u_{k}(x) = \delta(x-y) $$

From the completeness relation you gave, it follows that:

$$ a_{k} = \frac{u^{*}_{k}(y)}{\lambda_{k}} $$

Which gives the desired result:

$$ f(x) = G(x;y) = \sum_{k} \frac{u^{*}_{k}(y)u_{k}(x)}{\lambda_{k}} $$

Update When you substitute the definition of $u_{k}(x)$ into the sum above and convert the sum to an integral over $\vec{k}$ you get:

$$ G(x;y) = \int \frac{d^{3}k}{(2\pi)^{3}} \frac{1}{2\omega_{\vec{k}}} \frac{1}{\omega_{\vec{k}}^{2} - \vec{k}^{2}} e^{i k_{\mu}(x^{\mu} - y^{\mu})} $$

Now, as I understand it, the bare classical propagator is given by

$$ G(x;y) = \int \frac{d^{4}k}{(2\pi)^{4}} \frac{1}{m^{2} - k_{\mu}k^{\mu}} e^{i k_{\mu}(x^{\mu} - y^{\mu})} $$

Separating this integral out into time and space components gives:

$$ G(x;y) = \int\frac{d^{3}k}{(2\pi)^{3}} e^{i \vec{k}(\vec{x} - \vec{y})} \int\frac{d\omega}{2\pi}\frac{1}{m^{2} - \vec{k}^2 + \omega^2} e^{i\omega(x^{0} - y^{0})} $$

or, since $-\omega_{k}^2 = m^2 - \vec{k}^2$,

$$ G(x;y) = \int\frac{d^{3}k}{(2\pi)^{3}} e^{i \vec{k}(\vec{x} - \vec{y})} \int\frac{d\omega}{2\pi}\frac{1}{\omega^2-\omega_{k}^2} e^{i\omega(x^{0} - y^{0})} $$

Which has poles at $\omega = \pm\omega_{k}$. Now, you can solve this by a complex contour integral and you pick up a residue of $\frac{2\pi i}{2\omega_{k}}$, which gives you:

$$ G(x;y) = \int\frac{d^{3}k}{(2\pi)^{3}} \frac{i}{2\omega_{k}} e^{i k_{\mu}(x^{\mu} - y^{\mu})} $$

This is reason the quantum case has an extra $-i$ (to cancel out the $i$ from the residue), and then this gives the Feynman propagator. However, in this case you still don't have a $\frac{1}{\omega_{\vec{k}}^{2} - \vec{k}^{2}}$ factor so I really don't know where this is coming from. Either you're missing a normalisation fact in your completeness relation, or the normalisation of the basis states is wrong.