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## Main Question or Discussion Point

I am getting started with QFT and I'm having a hard time to understand the quantization procedure for the simples field: the scalar, massless and real Klein-Gordon field.

The approach I'm currently studying is that by Matthew Schwartz. In his QFT book he first solves the classical KG equation with Fourier transform and shows that the general solution is:

[tex]\phi(t,x)=\int \dfrac{d^3 \mathbf{p}}{(2\pi)^3}(a_{\mathbf{p}}e^{-ix^\mu p_\mu}+a_{\mathbf{p}}^\ast e^{i x^\mu p_\mu}).[/tex]

Here there is nothing fancy, it is the usual way to solve a DE by using Fourier transform and applying the additional conditions (in this case, that the field is real). In this setting, [itex]a_{\mathbf{p}}[/itex] are complex numbers.

After this, he

In Quantum Mechanics, in particular in the case of the SHO, it is pretty clear what we assume that we have and what do we derive. There we assume we have one state space of kets [itex]\mathcal{E}[/itex] on which there is one position observable [itex]X[/itex] together with its basis [itex]|x\rangle[/itex]. We also assume that there exists one momentum observable [itex]P[/itex] which acts as the generator of spatial translations, which in turn is equivalent to satisfy [itex][X,P]=i[/itex]. In that sense we know how [itex]P[/itex] acts on the position representation and we can relate both representations. All of this is assumed to be there, and this is not so hard to grasp.

Quantization then only means to pick the classical Hamiltonian [itex]H = p^2/2m + m\omega^2 x^2/2[/itex] and replace the classical dynamical variables by the operators, assumed to exist. This leads to the quantum Hamiltonian observable. We then try to factor the Hamitonian and we

In QFT everything is blurry. I mean, we have to find one

Considering all this there are some points to mention:

The approach I'm currently studying is that by Matthew Schwartz. In his QFT book he first solves the classical KG equation with Fourier transform and shows that the general solution is:

[tex]\phi(t,x)=\int \dfrac{d^3 \mathbf{p}}{(2\pi)^3}(a_{\mathbf{p}}e^{-ix^\mu p_\mu}+a_{\mathbf{p}}^\ast e^{i x^\mu p_\mu}).[/tex]

Here there is nothing fancy, it is the usual way to solve a DE by using Fourier transform and applying the additional conditions (in this case, that the field is real). In this setting, [itex]a_{\mathbf{p}}[/itex] are complex numbers.

After this, he

*simply*states the followingA little after that he says:Since the modes of an electromagnetic field have the same classical equations as a simple harmonic oscillator, we can quantize them in the same way. We introduce an annihilation operator [itex]a_{\mathbf{p}}[/itex] and its conjugate creation operator [itex]a_{\mathbf{p}}^\dagger[/itex] for each wavenumber [itex]\mathbf{p}[/itex] and integrate over them to get the Hamiltonian of the free theory: [tex] H_0 = \int \dfrac{d^3\mathbf{p}}{(2\pi)^3}\omega_{\mathbf{p}} \left(a_{\mathbf{p}}^\dagger a_{\mathbf{p}}+\dfrac{1}{2}\right)[/tex] with [itex] \omega_{\mathbf{p}} = |\mathbf{p}|[/itex]

Now I must confess I'm quite confused with this. It is completely unclear to me what do we actually assume that we have and what we are actually deriving and how.Now let us get a little more precise about what the Hamiltonian in Eq. (2.65) means. The natural generalization of

[tex][a,a^\dagger]=1[/tex]

are the equal-time commutation relations

[tex][a_{k},a_{p}^\dagger]=(2\pi)^3\delta(\mathbf{p}-\mathbf{k}).[/tex]

These [itex]a_{p}^\dagger[/itex] operators create particles with momentum [itex]p[/itex]:

[tex]a_{p}^\dagger |0\rangle = \dfrac{1}{\sqrt{2\omega_p}}|\mathbf{p}\rangle,[/tex]

where [itex]|\mathbf{p}\rangle[/itex] is a state with a single particle of momentum [itex]\mathbf{p}[/itex].

We then define quantum fields as integrals over creation and annihilation operators for each momentum:

[tex]\phi_0(\mathbf{x})=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3} \dfrac{1}{\sqrt{2\omega_p}}(a_p e^{ipx}+a_{p}^\dagger e^{-ipx}).[/tex]

In Quantum Mechanics, in particular in the case of the SHO, it is pretty clear what we assume that we have and what do we derive. There we assume we have one state space of kets [itex]\mathcal{E}[/itex] on which there is one position observable [itex]X[/itex] together with its basis [itex]|x\rangle[/itex]. We also assume that there exists one momentum observable [itex]P[/itex] which acts as the generator of spatial translations, which in turn is equivalent to satisfy [itex][X,P]=i[/itex]. In that sense we know how [itex]P[/itex] acts on the position representation and we can relate both representations. All of this is assumed to be there, and this is not so hard to grasp.

Quantization then only means to pick the classical Hamiltonian [itex]H = p^2/2m + m\omega^2 x^2/2[/itex] and replace the classical dynamical variables by the operators, assumed to exist. This leads to the quantum Hamiltonian observable. We then try to factor the Hamitonian and we

**derive in terms of [itex]X,P[/itex] which we have assumed to exist**, the ladder operators. Finally we prove several properties of these operators that allows us to find the spectrum of the Hamiltonian and the eigenstates.In QFT everything is blurry. I mean, we have to find one

**function that gives operators to events in spacetime**. This is not the same as assuming**one**operator [itex]X[/itex] exists. We must**find the dependency**[itex]\phi(x)[/itex] so that the field obeys a differential equation. It is not as in QM, where we suppose there exists the position and momentum operator satisfying the canonical commutation relations. Here we must find a specific functional form obeying a DE.Considering all this there are some points to mention:

- It is not clear how does one find out what is [itex]\phi(x)[/itex] in terms of the ladder operators. I mean, there is a huge jump from the classical solution to the quantum one. I don't see how this is obvious, actually it seems pretty not obvious.
- One usually writes down [itex]\phi(x)[/itex] in terms of ladder operators. But neither [itex]\phi(x)[/itex] neither those ladder operators have been defined. In truth this is in sharp contrast to the SHO: there we clearly have [itex]X,P[/itex] and their representations as starting point from which we derive the rest. Here it seems we have no starting point.
- The space is also not clear. In QM as I said we assume there is one Hilbert space spanned by some representation physically relevant based on the observable's algebra. Here it is not even clear what Hilbert space spanned by what basis do we have.