hamiltonian formalism - Quantization of a free field: Klein-Gordon case

I am a beginner and reading this course text on QFT.

The author first introduces the KG equation:

$$\partial_\mu\partial^{\mu}\phi+m^2\phi=0$$

[with Minkowski signature $(+,-,-,-)$]. Then the Fourier transform is used to obtain:

$$\phi(\mathbf{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\mathbf{p}\cdot\mathbf{x}}\phi(\mathbf{p},t)$$

My first question is related perhaps to notation choice or a typo. Should we use the same function used for the filed to write the Fourier transform? Or we should put $\Phi(\mathbf{p}, t)$ instead of $\phi(\mathbf{p},t)$?

Now if we apply the Fourier transform to the KG equation we have:

$$(\frac{\partial^2}{\partial t^2}+ p^2+m^2)\phi(\mathbf{p},t)=0$$ which is the equation of an oscillator vibrating at frequency $\sqrt{(p^2+m^2)}$.

( I am still confused here because we need to say that the Fourier transform of the field $\Phi(\mathbf{p},t)$ is an oscillator).

My second question is that why we use $\mathbf{p}$ here, we could label it something else. Below we call it a 3-momentum. This is also confusing for me. We introduce the conjugate momentum of the field by $\pi(\mathbf{x})$. But a 3-momentum does not make sense at this stage because we have not yet discussed particles and can use any other label.The theory simply does not need the notion of particles. I see that later on we call excitations of the field as particles with energies $\sqrt{(p^2+m^2)}$.

If we want to quantize this oscillator we recall from the quantum mechanical Hamiltonian formalism that the generalized coordinates can be given in terms of creation and annihilation operators as:

$$q=\frac{1}{\sqrt{2\omega}}(a+a^{\dagger})$$

The author now gives the equation of the field as a linear sum of an infinite number of creation and annhilation operators indexed by the 3-momentum $\mathbf{p}$:

$$\phi(\mathbf{x})=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}[a_{\mathbf{p}}e^{i\mathbf{p}.\mathbf{x}}+a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}.\mathbf{x}}]$$

My third question is that why we have $e^{-i\mathbf{p}.\mathbf{x}}$. Shouldn't it be $e^{i\mathbf{p}.\mathbf{x}}$ instead.

My fourth and fifth questions. Isn't it true that we just need to replace $\Phi$ with $q$ in the second equation above? And where is $t$?

Answer

My first question is related perhaps to notation choice or a typo. Should we use the same function used for the filed to write the Fourier transform? Or we should put Φ(p,t) instead of ϕ(p,t)?

This is a standard awful physicist convention. Clearly the functional forms of the two "phis" are different, but the symbol being used is the same. This really is an awful abuse of notation, it relies on the convention that we "know" that "p" is a letter conventionally used for momenta and "x" is a letter conventionally used for position. In truth it is better to write (note the tilde over the second phi):

$$\phi(\mathbf{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\mathbf{p}.\mathbf{x}} \tilde \phi(\mathbf{p},t)$$ and, indeed, you will see this notation sometimes.

My second question is that why we use p here, we could label it something else. Below we call it a 3-momentum.

This is in anticipation of its subsequent interpretation as a momentum. There's no reason this interpretation should be clear until you have further studied the implications of the equation.

My third question is that why we have $e^{-i\mathbf{p}.\mathbf{x}}$. Shouldn't it be $e^{i\mathbf{p}.\mathbf{x}}$ instead.

Since $\omega$ only depends on the magnitude of the momentum you can change variables from $\vec p$ to $-\vec p$ if you want. This just will change your definition of $a^\dagger_{\vec p}$. This is the conventional definition.

[My fourth and fifth questions...] Isn't it true that we just need to replace $\Phi$ with $q$ in the second equation above? And where is $t$?

You have to extend the idea of "q" to a momentum-dependent quantity. And, as discussed above, there is a conventional way to do this.

The variable t does not appear in the equation you gave because, as this point, you are only Fourier transforming with respect to position. To anticipate/understand what is going on think about operators in the Schrodinger picture. They have no explicit time dependence. The time dependence will come in later as you move to the Heisenburg picture and sandwich your operators between $e^{-\hat H t}$ and $e^{i\hat H t}$.