Quantization of Scalar Fields: Creation and Annihilation Operators

1. Why Quantum Field Theory?

Your background in physics has equipped you with special relativity and quantum mechanics. The most direct attempt to merge these, known as relativistic quantum mechanics, involves promoting the relativistic energy-momentum relation $E^2=p^2{c}^2+m^2{c}^4$ to a quantum wave equation, the Klein-Gordon equation:
$(\square +m^2)\psi =0$
Interpreting $\psi$ as the wavefunction for a single relativistic particle, however, leads to several conceptual problems.

As the video explains, these issues signal that the single-particle interpretation is flawed. QFT resolves this by a radical change in perspective. Instead of quantizing particles, we quantize the classical field itself. In this new picture, particles are not fundamental but emerge as the quantized excitations of the field, much like photons are the quanta of the electromagnetic field. This approach elegantly incorporates the possibility of particle creation and annihilation from the outset.

2. The Recipe: Canonical Quantization

The procedure we will use is called canonical quantization. It's a general recipe for turning a classical field theory into a quantum one. For you, with your experience in Hamiltonian mechanics, the logic will be very familiar. It's a direct extension of how one quantizes a point particle or a simple harmonic oscillator.

Let's briefly recall the simple harmonic oscillator (SHO) as our guiding example.
The Hamiltonian is $H=\frac{1}{2}{p}^2+\frac{1}{2}{\omega }^2{q}^2$. Canonical quantization proceeds by:

1. Promoting the position $q$ and momentum $p$ to Hermitian operators, $\hat{q}$ and $\hat{p}$.
2. Imposing the canonical commutation relation: $[\hat{q},\hat{p}]=i$ (using units where $\hslash =1$).
3. Defining annihilation and creation operators:
   $a=\sqrt{\frac{\omega }{2}}\left(\hat{q}+\frac{i}{\omega }\hat{p}\right),a^{†}=\sqrt{\frac{\omega }{2}}\left(\hat{q}-\frac{i}{\omega }\hat{p}\right)$
   These operators satisfy $[a,a^{†}]=1$.
4. Rewriting the Hamiltonian in terms of these operators: $H=\omega (a^{†}a+\frac{1}{2})$.
5. Constructing the Hilbert space (a Fock space):
   • The ground state or vacuum, $|0⟩$, is defined by $a|0⟩=0$.
   • Excited states are created by repeatedly applying $a^{†}$: $|n⟩\propto (a^{†}{)}^n|0⟩$. The operator $N=a^{†}a$ is the number operator, as $N|n⟩=n|n⟩$.

The strategy for a field is exactly the same, but instead of one degree of freedom $q$, we have an infinite number: the value of the field $\varphi (x)$ at each point in space.

3. Quantizing the Real Scalar Field

Let's apply this recipe to the simplest relativistic field theory: the free, real scalar field $\varphi (x)$.

Step 1: Classical Lagrangian and Hamiltonian

The theory is defined by the Lagrangian density:
$\mathcal{L}=\frac{1}{2}{\eta }^{\mu \nu }({\partial }_{\mu }\varphi )({\partial }_{\nu }\varphi )-\frac{1}{2}{m}^2{\varphi }^2=\frac{1}{2}{\dot{\varphi }}^2-\frac{1}{2}(\nabla \varphi {)}^2-\frac{1}{2}{m}^2{\varphi }^2$
The canonical momentum density conjugate to $\varphi$ is:
$\pi (x)=\frac{\partial \mathcal{L}}{\partial \dot{\varphi }(x)}=\dot{\varphi }(x)$
The Hamiltonian density is $\mathcal{H}=\pi \dot{\varphi }-\mathcal{L}$, and the total Hamiltonian is its integral over space:
$H=\int d^{3}x\mathcal{H}=\int d^{3}x\left[\frac{1}{2}{\pi }^2+\frac{1}{2}(\nabla \varphi {)}^2+\frac{1}{2}{m}^2{\varphi }^2\right]$
This Hamiltonian represents the total energy of the field configuration.

Step 2: Impose Commutation Relations

Now we quantize. We promote the field $\varphi (t,\mathbf{x})$ and its momentum $\pi (t,\mathbf{x})$ to operators. Since we are working in the Schrödinger picture, these operators are defined at a fixed time, say $t=0$, and we denote them $\varphi (\mathbf{x})$ and $\pi (\mathbf{x})$. We impose the equal-time commutation relations (ETCRs):
$[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$
$[\varphi (\mathbf{x}),\varphi (\mathbf{y})]=0,[\pi (\mathbf{x}),\pi (\mathbf{y})]=0$
The Dirac delta function ${\delta }^{(3)}(\mathbf{x}-\mathbf{y})$ is the continuum analogue of the Kronecker delta ${\delta }_{ij}$; it enforces that the field and its momentum at different spatial points are independent degrees of freedom.

Step 3: From Fields to Oscillators

The Hamiltonian is a complicated object involving spatial derivatives that couple the field at neighboring points. The key to simplifying it is to decompose the field into its Fourier modes. This diagonalization is analogous to finding the normal modes of a system of coupled oscillators.

We expand the operators $\varphi (\mathbf{x})$ and $\pi (\mathbf{x})$ in terms of new operators, $a_{\mathbf{p}}$ and $a_{\mathbf{p}}^{†}$, which will become our creation and annihilation operators. The expansion is a Fourier transform:
$\varphi (\mathbf{x})=\int \frac{d^{3}p}{(2\pi {)}^3}\frac{1}{\sqrt{2{\omega }_{\mathbf{p}}}}\left(a_{\mathbf{p}}{e}^{i\mathbf{p}\cdot \mathbf{x}}+a_{\mathbf{p}}^{†}{e}^{-i\mathbf{p}\cdot \mathbf{x}}\right)$
$\pi (\mathbf{x})=\int \frac{d^{3}p}{(2\pi {)}^3}(-i)\sqrt{\frac{{\omega }_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}{e}^{i\mathbf{p}\cdot \mathbf{x}}-a_{\mathbf{p}}^{†}{e}^{-i\mathbf{p}\cdot \mathbf{x}}\right)$
Here, ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ is the relativistic energy for a particle with momentum $\mathbf{p}$. The normalization factors like $1/\sqrt{2{\omega }_{\mathbf{p}}}$ are chosen carefully to make the final results simple and Lorentz invariant.

The crucial step is to demand that these operators $a_{\mathbf{p}}$ and $a_{\mathbf{p}}^{†}$ satisfy the commutation relations of harmonic oscillator ladder operators for each momentum mode $\mathbf{p}$:
$[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$
$[a_{\mathbf{p}},a_{\mathbf{q}}]=0,[a_{\mathbf{p}}^{†},a_{\mathbf{q}}^{†}]=0$
One can show that if the $a_{\mathbf{p}}$ operators satisfy these relations, then the $\varphi (\mathbf{x})$ and $\pi (\mathbf{y})$ operators defined by the expansion above will automatically satisfy the correct ETCRs.

Step 4: The Hamiltonian and Particle Interpretation

The real magic happens when we substitute these expansions back into the Hamiltonian operator $H$. After a significant but straightforward calculation, the Hamiltonian simplifies dramatically.

The result of this calculation is:
$H=\int \frac{d^{3}p}{(2\pi {)}^3}{\omega }_{\mathbf{p}}\left(a_{\mathbf{p}}^{†}{a}_{\mathbf{p}}+\frac{1}{2}[a_{\mathbf{p}},a_{\mathbf{p}}^{†}]\right)$
Using the commutation relation for $a_{\mathbf{p}}$ and $a_{\mathbf{p}}^{†}$, the second term becomes $\frac{1}{2}(2\pi {)}^3{\delta }^{(3)}(0)$. This is an infinite constant, the sum of the zero-point energies $\frac{1}{2}{\omega }_{\mathbf{p}}$ of all the oscillators. For now, we will discard this infinite energy by a procedure called normal ordering, which means we simply define the Hamiltonian to be:
$H=\int \frac{d^{3}p}{(2\pi {)}^3}{\omega }_{\mathbf{p}}{a}_{\mathbf{p}}^{†}{a}_{\mathbf{p}}$
This reveals the central result: the quantum scalar field is mathematically equivalent to an infinite collection of independent simple harmonic oscillators, one for each momentum mode $\mathbf{p}$.

4. The Fock Space of Particles

Now we can build the state space and understand what the particles are.

1. The Vacuum: We define the ground state of the theory, the vacuum $|0⟩$, as the state with no excitations. It is annihilated by all annihilation operators:
   $a_{\mathbf{p}}|0⟩=0\text{for all }\mathbf{p}$
   This state has zero energy (after we discard the infinite zero-point energy).

2. Single-Particle States: We create an excitation by acting on the vacuum with a creation operator. Let's define the state $|\mathbf{p}⟩\equiv a_{\mathbf{p}}^{†}|0⟩$. What is this state? We can find out by acting on it with the Hamiltonian and total momentum operators. The total momentum operator can be shown to be $\mathbf{P}=\int \frac{d^{3}p}{(2\pi {)}^3}\mathbf{p}{a}_{\mathbf{p}}^{†}{a}_{\mathbf{p}}$.

   • Energy: $H|\mathbf{p}⟩={\omega }_{\mathbf{p}}|\mathbf{p}⟩$
   • Momentum: $\mathbf{P}|\mathbf{p}⟩=\mathbf{p}|\mathbf{p}⟩$

   The state $|\mathbf{p}⟩$ is an eigenstate of energy ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and momentum $\mathbf{p}$. These are precisely the properties of a single relativistic particle of mass $m$. We have found the particles!

3. Multi-Particle States: We can create multiple particles by acting with multiple creation operators. For example, $|{\mathbf{p}}_1,{\mathbf{p}}_2⟩=a_{{\mathbf{p}}_1}^{†}{a}_{{\mathbf{p}}_2}^{†}|0⟩$. Because $[a_{{\mathbf{p}}_1}^{†},a_{{\mathbf{p}}_2}^{†}]=0$, the state is symmetric under the exchange of the two particles: $|{\mathbf{p}}_1,{\mathbf{p}}_2⟩=|{\mathbf{p}}_2,{\mathbf{p}}_1⟩$. This means the quanta of the scalar field are bosons, as expected for a spin-0 particle.

The full Hilbert space, constructed by all possible applications of creation operators on the vacuum, is called the Fock space.

Conclusion

Today we have taken the first crucial step into quantum field theory. We have seen how the formal procedure of canonical quantization, when applied to a classical scalar field, resolves the problems of relativistic quantum mechanics and gives rise to a consistent theory of multiple, identical, non-interacting bosons.

Key Takeaways:

• Canonical quantization is a recipe to create a quantum theory from a classical Hamiltonian system by promoting fields and their conjugate momenta to operators and imposing equal-time commutation relations.
• For the free scalar field, this procedure reveals that the theory is equivalent to an infinite collection of simple harmonic oscillators, one for each momentum mode.
• Creation ($a_{\mathbf{p}}^{†}$) and annihilation ($a_{\mathbf{p}}$) operators act on these oscillator modes. They add or remove quanta of excitation from the field.
• These quanta are interpreted as particles. A state $a_{\mathbf{p}}^{†}|0⟩$ has energy ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and momentum $\mathbf{p}$, corresponding to a relativistic particle of mass $m$.
• The Hilbert space of the theory is a Fock space, built by applying creation operators to a ground state called the vacuum.

We have successfully quantized a field in flat, static Minkowski spacetime. However, our ultimate goal is to understand quantum fields in the expanding, curved spacetime of our universe. What happens to our notion of particles and the vacuum when spacetime itself is dynamic? This is a deep and fascinating question.

Preview of the Next Lesson:

Our definition of particles was tied to the modes $e^{-ikx}$ that are "positive frequency" with respect to the standard time coordinate $t$. In a general curved spacetime, there is no unique, globally preferred time coordinate. Different observers may naturally choose different time coordinates, leading them to disagree on what constitutes a "positive frequency" mode. This leads to a startling conclusion: the very concept of a particle, and the definition of the vacuum state, becomes observer-dependent. Our next lesson will explore this profound idea by tackling the learning outcome: Explain the ambiguity in defining particles in a general curved spacetime due to the lack of a global timelike Killing vector.