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QFT: expanding the propagator in terms of Minkowski modes

Physics Asked on February 3, 2021

I’m trying to understand the usual "Fourier transform" of the free scalar propagator $ G(x,y) = int frac{d^{4}k}{(2pi)^{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:
$$
u_{mathbf{k}}( x ) = u_{mathbf{k}}( x^{0}, mathbf{x} ) = frac{ 1 }{ sqrt{ 2 omega_{mathbf{k}} (2pi)^{3} } } expleft( i mathbf{k} cdot mathbf{x} – i omega_{mathbf{k}} x^{0} right)
$$

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:
$$
G(x,y) = sum_{mathbf{k}} frac{ u^{ast}_{mathbf{k}}(x) u_{mathbf{k}}(y) }{lambda_{mathbf{k}}} = int frac{d^{3}mathbf{k}}{(2pi)^{3}} frac{e^{i mathbf{k} cdot (mathbf{x} – mathbf{y})}}{omega_{mathbf{k}}^{2} – mathbf{k}^{2}}
$$

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!

One Answer

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 $Aexp{(i k_{mu}x^{mu})}$.

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:

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

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

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

Separating this integral out into time and space components gives:

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

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

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

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

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

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.

Answered by gautampk on February 3, 2021

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