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When considering a real scalar field with Lagrangian $$\mathcal{L} = - \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) - \frac{1}{2}m^2\phi^2$$ the equation of motion is the Klein-Gordon equation $(\Box_{x} - m^2) \phi(x)=0$.

In texts on QFTs in curved spacetimes the quantization of the field $\phi$ is performed by summing over Minkowski modes $\{ u_{\mathbf{k}} \}_{\mathbf{k} \in \mathbb{R}^3}$ (the positive frequency modes) and $\{ u^{\ast}_{\mathbf{k}} \}_{\mathbf{k} \in \mathbb{R}^3}$ (the negative frequency modes) which are plane-wave solutions to the KG equation labelled by the three-momentum $\mathbf{k} \in \mathbb{R}^3$: $$ u_{\mathbf{k}}(x) \ = \ u_{\mathbf{k}}(x^0,\mathbf{x}) \ = \ \left( (2\pi)^3 2 \sqrt{ |\mathbf{k}|^2 + m^2 } \right)^{-\frac{1}{2}} e^{ - i \sqrt{ |\mathbf{k}|^2 + m^2 }\ x^0 + i \mathbf{k} \cdot \mathbf{x} } $$ The Minkowski modes have normalizations $\left( (2\pi^3) 2 \sqrt{ |\mathbf{k}|^2 + m^2 } \right)^{-\frac{1}{2}}$ because they are normalized with respect to the Klein-Gordon inner product, defined for any complex-valued functions $f,g$ as $$ \langle f,g\rangle = i \int_{\Sigma} d^{3}\mathbf{x} \ \left[ f^{\ast}(x) \frac{\partial g}{\partial x^0} - \frac{\partial f^{\ast}}{\partial x^0} g(x) \right] $$

Where $\Sigma$ is a 3D hypersurface of constant time $x^0$ (because the function being integrated is a conserved current, it follows that the value of $\langle f,g\rangle$ is independent of the choice of $\Sigma$ used to integrate it). The Minkowski modes are normalized such that: $$ \langle u_{\mathbf{k}}, u_{\mathbf{p}} \rangle = \delta(\mathbf{k} - \mathbf{p}) \\ \langle u_{\mathbf{k}}, u^{\ast}_{\mathbf{p}} \rangle = 0 \\ \langle u^{\ast}_{\mathbf{k}}, u^{\ast}_{\mathbf{p}} \rangle = - \delta(\mathbf{k} - \mathbf{p}) $$

At this point, the texts will often say the Minkowski modes are complete, which is why we then are able to expand the scalar field $\phi$ as $\phi(x) = \sum u_{\mathbf{k}}(x) a_{\mathbf{k}} + u^{\ast}_{\mathbf{k}}(x) a^{\dagger}_{\mathbf{k}}$, and then quantize $a_{\mathbf{k}}$ and $a_{\mathbf{k}}^{\dagger}$, and so on.

My Question: What does it precisely mean for the Minkowski modes to be complete here?

The texts all seem to glaze over this point. I want to say that there should be a completeness relation $\sum_{\mathbf{k}} u^{\ast}_{\mathbf{k}}(x)u_{\mathbf{k}}(y)$ which is proportional to either $\delta^{(3)}(\mathbf{x} - \mathbf{y})$ or maybe $\delta^{(4)}(x - y)$ but this doesn't seem to be true. I am not even sure what is the vector space which is complete here.

EDIT 1: I am working in ordinary rectangular Minkowski coordinates (flat space) with metric $\eta^{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$. I am interested in how to construct the modes in this simplest case (the curved spacetime texts then generalize this procedure to arbitrary manifolds)

EDIT 2: I am guessing that the way to understand the completeness here is something along the lines as you would in QM. If $\{ |n\rangle \}_{n=1}^{N}$ is a complete set of states for some $N$-dimensional Hilbert space $\mathcal{H}$, then we have $\sum_{n=1}^N |n\rangle\langle n| = \mathbb{I}_{N\times N}$, which allows for the expansion of an arbitrary state $|\psi\rangle$ as $|\psi\rangle = \sum_{n=1}^N \langle n|\psi\rangle |n\rangle $. The expansion done on the field $\phi(x)$ is exactly $\phi(x) = \int d^3\mathbf{k}\ \big[ u_{\mathbf{k}}(x) a_{\mathbf{k}} + u^{\ast}_{\mathbf{k}}(x) a_{\mathbf{k}}^{\dagger} \big]$, where $a_{\mathbf{k}}=\langle u_{\mathbf{k}}, \phi\rangle$ and $a^{\dagger}_{\mathbf{k}}=\langle u^{\ast}_{\mathbf{k}}, \phi\rangle$.

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    $\begingroup$ I've improved the equations formatting. When using Dirac's notation I suggest that to render $\langle,\rangle$ use respectively \langle and \rangle instead of the less than < and greater than > signs. $\endgroup$ – user1620696 Jul 13 '18 at 20:52
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The modes are a complete basis of the vector space of solutions to the Klein-Gordon equations, with the $u_\mathbf{k}$ modes having positive eigenenergies while their conjugates have negative ones, reflecting relativity's quadratic dispersion relation. A general solution can be written as the sum of two integrals over $\mathbf{k}$, one per energy sign. The modes' $\mathbf{k}$-dependent coefficients in the integrands are Bogoliubov coefficients.

In a conformally flat spacetime of dynamic size, these coefficients evolve in an interesting way: an initially positive-energy wave eventually gets a negative-energy component. Birrell and Davies discusses its role in cosmological particle creation.

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  • $\begingroup$ So what does the modes being complete mean here? $\endgroup$ – Greg.Paul Jul 14 '18 at 20:24
  • $\begingroup$ @Greg.Paul See my first sentence. A set of objects in a vector space is a complete basis of that space if it spans them (in this case, in the Hilbert sense that allows infinitely many to have non-zero coefficients). The Klein-Gordon equation's solutions are a vector space. $\endgroup$ – J.G. Jul 14 '18 at 21:02
  • $\begingroup$ I see, thanks for that. But let's say I found a set of mode functions $\{ M_{k}(x) \}_{k}$ ($k$ being a collection of arbitrary mode labels) which solve the KG equation in arbitrary coordinates (but still in flat Minkowski space). Let's say I would like to check that the set $\{ M_{k}(x) \}_{k}$ is a complete set - what would I have to check? $\endgroup$ – Greg.Paul Jul 14 '18 at 22:03
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    $\begingroup$ They are "complete" in that they span the $L^2({\mathbb R}^3)$ Hilbert space that consists of squre integrable functions (non necessarily solutions of the KG equation) at time $x_0$. If you can expand $\delta^3(x-y)$, you can expand anything. It is this fixed-time completeness that allows the $[a_{\bf k},a_{{\bf k}'}^\dagger]$ commutator to give the $[\phi, \dot\phi]$ commutator. Such completeness is property of the eigenfunctions of any self-adjoint operator. $\endgroup$ – mike stone Jul 15 '18 at 15:54
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    $\begingroup$ I ran out of space. If you can expand arbitrary Cauchy data at time $x_0$, then the resulting solutions of KG can be expanded in terms of the time-evolved eiegnmodes. It is in this sense that the ${\mathbb R}^3$ completeness of the $u_{\bf k}(x,x_0)$ at $x_0$ allows the expansion of any solution of KG at all times. $\endgroup$ – mike stone Jul 15 '18 at 16:04
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I don't think you are using curved space in your equations, so I will assume flat Minkowski space-time.

The completeness is the relation that is needed to start from the presumed mode expansion, and from it to show that $$ [\phi(x), \partial_t \phi(y)]_{x_0=y_0}= \delta^3(x-y). $$ It basically arises from the completeness of the plane wave eignefunctions of $-\nabla^2$. That is, from $$ \int_{{\mathbb R}^3}\frac{d^3k}{(2\pi)^3} e^{ik\cdot (x-y)} = \delta^3(x-y). $$ It's been a while since I did the algebra, but I expect that the normalisation factor that arises from the time derivative in the K-G inner product compensates for the time derivative of the field in the commutator.

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  • $\begingroup$ Oops I edited my question; yes I am interested in flat space. Working backwards from your second expression I can write $$\int d^{3}\mathbf{k}\ -i \frac{\partial}{\partial x^0} \left[ u^{\ast}_{\mathbf{k}}(x) u_{\mathbf{k}}(y) + u_{\mathbf{k}}(x) u^{\ast}_{\mathbf{k}}(y) \right] \bigg|_{x^0=y^0} = \delta^{(3)}(\mathbf{x} - \mathbf{y})$$ but I don't understand why this is done. How does this constitute a completeness of the Minkowski modes? $\endgroup$ – Greg.Paul Jul 13 '18 at 19:23

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