Added quantization of scalar field

Author: Songtech-0912

content/quantum-field-theory/chapter-2.md

@@ -33,6 +33,174 @@ To go from the quantum harmonic oscillator to quantum field theory, we need to m
 - The Hamiltonian must sum over an infinite number of harmonic oscillators, so the sum over oscillators becomes an integral
 - The states $|0\rangle, |1\rangle, |2\rangle, \dots |n\rangle$ are now the states of the _quantum field_ $\hat \phi$, where the field has now become an operator with $|0\rangle, |1\rangle, |2\rangle, \dots |n\rangle$ being its eigenstates
 
+### Sidenote on notation
+
+In quantum field theory, there are a billion different conventions for almost *anything*. This leads to some pretty bizarre results:
+
+- Momentum ($p$) becomes the same thing as the wavevector ($k$), because in natural units where $\hbar = 1$ (like we're using), $p = \hbar k$ becomes $p = k$
+- For photons, angular frequency $\omega$ and energy $E$ become equivalent, because in natural units where $c = 1$ (like we're using), $E = \hbar \omega$ becomes $E = \omega$
+- Einstein's relativistic energy-momentum relation of $E^2 = (pc)^2 + (mc^2)^2$ becomes simply $E^2 = p^2 + m^2$, so we get $E = \sqrt{p^2 + m^2}$. You may also see this denoted as $E_p = \sqrt{p^2 + m^2}$, such as in Peskin and Schroeder's _An Introduction To Quantum Field Theory_.
+
+This means that every text often has its own notational system, especially if a certain text doesn't use natural units. Here, we will **always use natural units** unless otherwise denoted.
+
+### Equations of motion of a free scalar field
+
+The "simplest" quantum field theory is the theory of a free scalar field. We should note that "simple" is a rather relative term; approaching even this relatively simple quantum field theory can be an intimidating task, so it's important that we get the basics down.
+
+Since the free scalar field theory is after all a _field theory_, we start with its field Lagrangian. In this case, the field Lagrangian of a free scalar field is given by:
+
+{% math() %}
+\mathscr{L} = \partial_\mu \phi \partial^\mu \phi - m^2 \phi^2
+{% end %}
+
+Using the Euler-Lagrange equations, we can find the equations of motion of the field $\phi$. For the free scalar field, the Euler-Lagrange equations take the form:
+
+{% math() %}
+\dfrac{\partial \mathscr{L}}{\partial \phi} - \partial_\beta \left(\dfrac{\partial \mathscr{L}}{\partial(\partial_\beta \phi)}\right) = 0
+{% end %}
+
+Let's now compute the partial derivatives. Let's start with the derivative with respect to $\phi$, which is relatively simple to find:
+
+{% math() %}
+\dfrac{\partial \mathscr{L}}{\partial \phi} = -2m^2 \phi
+{% end %}
+
+The derivative with respect to $\partial_\beta \phi$ is not as easy to find, so we'll show the steps below:
+
+{% math() %}
+\begin{align*}
+\dfrac{\partial \mathscr{L}}{\partial(\partial_\beta \phi)}
+&= \dfrac{\partial}{\partial(\partial_\beta \phi)} \bigg(\partial_\mu \phi \partial^\mu \phi - m^2 \phi^2\bigg) \\
+&= \dfrac{\partial}{\partial(\partial_\beta \phi)} \bigg(\partial_\mu \phi \partial^\mu \phi\bigg) - \cancel{\dfrac{\partial}{\partial(\partial_\beta \phi)}\bigg(m^2 \phi^2\bigg)}^0 \\
+&= \dfrac{\partial}{\partial(\partial_\beta \phi)} \underbrace{\bigg(\partial_\beta \phi \partial^\beta \phi\bigg)}_\text{relabel dummy indices} \\
+&= \dfrac{\partial}{\partial(\partial_\beta \phi)} 
+\bigg(\partial_\beta \phi\, \underbrace{\eta^{\alpha \beta}\partial_\alpha \phi}_\text{lower indices}\bigg) \\
+&= \dfrac{\partial}{\partial(\partial_\beta \phi)} 
+\bigg(\partial_\beta \phi\, \eta^{\alpha \beta} \underbrace{\delta^\beta{}_\alpha \partial_\beta}_\text{relabel}  \phi\bigg) \\
+&= \eta^{\alpha \beta} \delta^\beta{}_\alpha\dfrac{\partial}{\partial(\partial_\beta \phi)} 
+\bigg(\partial_\beta \phi\, \partial_\beta \phi\bigg)
+\end{align*}
+{% end %}
+
+Let's recap on the steps we've done so far. First, since the $m^2 \phi^2$ term doesn't depend on the derivatives of $\phi$, its partial derivative is zero. This leaves us with just the $\partial_\mu \phi \partial^\mu \phi$ term left. Since $\mu$ is repeated over the upper and lower indices, we know that it is a **dummy index** so we can relabel it whatever we want - here we chose $\mu \to \beta$ because we're differentiating with respect to $\beta$, because again, since it's a dummy index, $\partial_\mu \phi \partial^\mu \phi = \partial_\beta \phi \partial^\beta \phi$.
+
+Finally, we needed to get the $\partial^\beta \phi$ term as a lower index. To do this, we used the Minkowski metric $\eta^{\alpha \beta}$ to lower the index, and then used the Kronecker delta {% inlmath() %}\delta^\beta{}_\alpha{% end %} to relabel. The end result we got was {% inlmath() %}\partial^\beta \phi = \eta^{\alpha \beta} \delta^\beta{}_\alpha \partial_\beta \phi{% end %}, from which we could pull out the (constant) factor of {% inlmath() %}\eta^{\alpha \beta} \delta^\beta{}_\alpha{% end %}. This gives us everything inside the derivative _purely_ in terms of {% inlmath() %}\partial_\beta \phi{% end %}, at which point we can _finally_ differentiate:
+
+{% math() %}
+\begin{align*}
+\dfrac{\partial \mathscr{L}}{\partial (\partial_\beta \phi)} &= \eta^{\alpha \beta} \delta^\beta{}_\alpha\dfrac{\partial}{\partial(\partial_\beta \phi)} 
+\bigg(\partial_\beta \phi\, \partial_\beta \phi\bigg) \\
+&= \eta^{\alpha \beta} \delta^\beta{}_\alpha (2\partial_\beta \phi) \\
+&= \underbrace{2\eta^{\alpha \beta}\delta^\beta_\alpha \partial_\beta \phi}_\text{now contract indices} \\
+&= 2 \partial^\beta \phi
+\end{align*}
+{% end %}
+
+Here, we took the derivative in the standard way (it is just taking the derivative of $(\partial_\beta \phi \partial_\beta \phi)^2$ which yields $2\partial_\beta \phi$). Then, we contracted the indices with the Minkowski metric and Kronecker delta terms, giving us $\eta^{\alpha \beta}\delta^\beta_\alpha \partial_\beta \phi = \partial^\beta \phi$. So, substituting into the Euler-Lagrange equations, we have:
+
+{% math() %}
+\begin{gather*}
+-2m^2 \phi - \partial_\beta(2\partial^\beta \phi) = 0 \\
+\Rightarrow \partial_\beta \partial^\beta \phi + m^2 \phi = 0
+\end{gather*}
+{% end %}
+
+This is our **equation of motion**, and it is called the **Klein-Gordon equation**, although it is usually written with the dummy index $\mu$ instead of $\beta$, in which it takes the form:
+
+{% math() %}
+\partial_\mu \partial^\mu \phi + m^2 \phi = 0
+{% end %}
+
+We may solve the Klein-Gordon equation with a [Fourier decomposition](https://ese-msc.github.io/preinduction/edsml/primer/notebooks/c_mathematics/differential_equations/11_pde_fourier.html), which gives the general solution:
+
+{% math() %}
+\phi(x^\mu) = \int \dfrac{d^3 p}{(2\pi)^32E} \left[a(p) e^{i\mathbf{p} \cdot \mathbf{x}} + a^*(p) e^{-i\mathbf{p} \cdot \mathbf{x}}\right] 
+{% end %}
+
+Where $E = \sqrt{p^2 + m^2}$ comes from the energy-momentum relation, as we discussed earlier. We may promote this to a field if we make the identifications of $a \to \hat a$, $a^* \to a^\dagger$, which are the _creation and annihilation operators_ which generalize the raising and lowering operators we previously saw. Thus, our _quantum_ field $\hat \phi$ is now an operator:
+
+{% math() %}
+\hat \phi(x^\mu) = \int \dfrac{d^3 p}{(2\pi)^32E} \left[\hat a(p) e^{i\mathbf{p} \cdot \mathbf{x}} + \hat a^\dagger(p) e^{-i\mathbf{p} \cdot \mathbf{x}}\right] 
+{% end %}
+
+For mathematical convenience we write $\hat a(p)$ as $\hat a_p$ and $\hat a^\dagger(p)$ as $\hat a^\dagger_p$, and we write $\hat \phi(x^\mu)$ as just $\hat \phi$. Thus our field operator becomes:
+
+{% math() %}
+\hat \phi = \int \dfrac{d^3 p}{(2\pi)^3 2E} \left[\hat a_p e^{i\mathbf{p} \cdot \mathbf{x}} + \hat a^\dagger_p e^{-i\mathbf{p} \cdot \mathbf{x}}\right] 
+{% end %}
+
+Where we _define_ our raising and lowering operators such that:
+
+{% math() %}
+\begin{align*}
+\hat a_p |n\rangle &= \sqrt{n}\,|n - 1\rangle \\
+\hat a_p^\dagger|n\rangle &= \sqrt{n + 1}\,|n + 1\rangle
+\end{align*}
+{% end %}
+
+Which is just like the quantum harmonic oscillator. We note that for the vacuum state $|0\rangle$, we have:
+
+{% math() %}
+\hat a_p|0\rangle = 0, \quad \hat a_p^\dagger|0\rangle = |1\rangle
+{% end %}
+
+The raising operator $\hat a_p$ raises the field to a higher-energy state, whereas the lowering operator $\hat a_p^\dagger$ acting on a state drops the field to a lower-energy state - in the case of the vacuum state, we end up with just zero. Let us now take an example of using the field operator $\hat \phi$. For instance, what happens if we act it on the ground state? Well, if we do the calculation, we find that:
+
+{% math() %}
+\begin{align*}
+\hat \phi|0\rangle &= \int \dfrac{d^3 p}{(2\pi)^32E} \left[\hat a_p e^{i\mathbf{p} \cdot \mathbf{x}} + \hat a^\dagger_p e^{-i\mathbf{p} \cdot \mathbf{x}}\right] |0\rangle \\
+% step 1
+&= \int \dfrac{d^3 p}{(2\pi)^32E} \bigg[\underbrace{\hat a_p|0\rangle}_\text{distribute} e^{i\mathbf{p} \cdot \mathbf{x}} + \underbrace{\hat a^\dagger_p|0\rangle}_\text{distribute} e^{-i\mathbf{p} \cdot \mathbf{x}}\bigg] \\
+% step 2
+&= \int \dfrac{d^3 p}{(2\pi)^32E} \bigg[\underbrace{\cancel{\hat a_p|0}\rangle^0}_{\text{since } \hat a_p|0\rangle = 0 } e^{i\mathbf{p} \cdot \mathbf{x}} + \underbrace{|1\rangle e^{-i\mathbf{p} \cdot \mathbf{x}}}_{\text{since } \hat a_p^\dagger|0\rangle = |1\rangle}\bigg] \\
+&= \int \dfrac{d^3 p}{(2\pi)^32E}|1\rangle \, e^{-i\mathbf{p} \cdot \mathbf{x}} \\
+&= \int d^3 p\dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^32E}|1\rangle
+\end{align*}
+{% end %}
+
+So we've found that the field operator acting on the vacuum state $|0\rangle$ produces a **superposition** of the **first excited state** $|1\rangle$. The physical interpretation is that the field acting on $|0\rangle$ has _created_ a particle!
+
+Let's break down what this means. An integral is just an infinite sum, so the integral over $p$ gives a superposition of the first excited state $|1\rangle$. It is a _superposition_ because when the field acts on the vacuum state $|0\rangle$ to produce the first excited state $|1\rangle$ (which we interpret as the field creating a particle), the resultant particle may have a range of different momenta, so we must sum over all possible momenta. This is a profound result: a quantum field can _create_ particles and (we'll soon see) _annihilate_ particles by acting on a state to produce another state.
+
+Let's show our result even more explicitly by finding the _expectation value_ of the field for a single-particle excited state. We *define* a **single-particle state** with momentum $p$ (and thus energy $E = \sqrt{p^2 + m^2}$) as $|p\rangle$ where $|p\rangle$ is given by:
+
+{% math() %}
+|p\rangle = 2E\, \hat a_p^\dagger |0\rangle = 2E\,|1\rangle
+{% end %}
+
+If our physical interpretation is correct, then the single-particle state $|p\rangle$ represents a particle created by the field - and we'll show that this is true. Before we begin though, we also need to *define* our states to satisfy the following orthogonality relation:
+
+{% math() %}
+\langle p |p'\rangle = 2E (2\pi)^3 \delta^3(p-p')
+{% end %}
+
+Where $\delta^3$ is the three-dimensional Dirac delta function, which obeys $\displaystyle \int d^3 p\, \delta^3(p) = 1$, a fact that will be _extremely helpful_ later.
+
+> **Why this particular orthogonality relation?** The answer is that however we choose to define orthogonality between two states, any scaling constants are **a matter of personal preference**. Here, we just chose the constant factor $(2\pi)^3$ because it is convenient for our later calculations, but it doesn't really matter. After all, we cannot measure quantum states (directly) anyways, so it really **does not matter** how we define them. As long as our definitions are **consistent** and our states are **normalized** (such that you can never have a probability greater than one!), we may choose whatever normalization constants we want.
+
+To calculate the expectation value of the single-particle state $\langle 0 |\hat \phi| p\rangle$, we just "flip" our previous result for $\hat |0\rangle$ around to get:
+
+{% math() %}
+\langle 0| \hat \phi = \langle 1|\int d^3 p\dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^32E}
+{% end %}
+
+Then, substituting in $|p\rangle = 2E\, \hat a_p^\dagger |0\rangle = 2E\,|1\rangle$, we have:
+
+{% math() %}
+\begin{align*}
+\langle 0 |\hat \phi |p\rangle &= \langle 1|\int d^3 p\dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^32E}|p\rangle \\
+&= \langle 1|\int d^3 p\dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^32E} \underbrace{2E\,|1\rangle}_{|p\rangle = 2E\,|1\rangle} \\
+&= \int d^3 p \dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^3} \underbrace{\langle 1|1\rangle}_\text{orthogonal} \\
+&= \int d^3 p \dfrac{e^{-i\mathbf{p} \cdot \mathbf{x}}}{(2\pi)^3}(2\pi)^3 \delta^3(p - p') \\
+&= \int d^3 p \underbrace{\left[e^{-i\mathbf{p} \cdot \mathbf{x}}\delta^3(p - p')\right]}_\text{Dirac delta integrates to 1} \\
+&= e^{-i\mathbf{p} \cdot \mathbf{x}}
+\end{align*}
+{% end %}
+
+So the expectation value of the field for the single-particle state is just $e^{-i\mathbf{p} \cdot \mathbf{x}}$, _exactly_ the same as the wavefunction of a single particle with momentum $\mathbf{p}$. Physically-speaking, we have thus shown that our quantum field $\hat \phi$ can _create_ particles, and this is why we often call $\hat a^\dagger_p$ the **creation operator**. Our field $\hat \phi$ can also _annihilate_ particles, and this is why we often call $\hat a_p$ the **annihilation operator**.
+
+This means that, unlike non-relativistic quantum mechanics, where the particle (such as the electron) is the fundamental entity, in quantum field theory, the *field* is the fundamental entity, out of which particles arise and disappear - which is why quantum field theory is the basic theoretical framework for **particle physics**. Indeed, the free scalar field theory is a theory that can describe particles called [pions](https://en.wikipedia.org/wiki/Pion) (also called pi mesons), although not much else. We'll soon see more "interesting" (but also more complicated) theories that describe particles that are much more well-known, like photons, electrons, and quarks.
+
 ### Scattering and propagators
 
 Relativistic QFT is most often used in the calculation of **scattering cross-sections** that describes what happens when elementary particles interact. To understand how this procedure works, let us consider a QFT with Hamiltonian $\hat H$. This Hamiltonian can be divided into two parts: the free Hamiltonian $\hat H_0$, and the _interaction_ Hamiltonian $\hat H_i$. Thus we have: