Quantization

B.2 Quantization

In this section, we briefly sketch canonical quantization, a process of turning a classical field theory into a QFT. It is based on the Hamiltonian formalism, in close analogy to the quantization of classical mechanics $\to$ QM. The result makes manifest the connection between quantum fields and their associated particles.

An alternative quantization approach not discussed here is based on the path integral formulation (see Section B.1.1.). As with most alternative mathematical prescriptions of the same physics, it provides useful insight into the theory and can simplify certain calculations. Further detail can be found, for example, in Peskin and Schroeder [81] Chapter 9.

B.2.1 Canonical quantization

The process of quantizing a classical system in QM can be summarized as (1) promoting the canonical coordinates to quantum operators, and (2) imposing the canonical Poisson bracket relations as quantum commutator relations:

\begin{matrix} \begin{aligned} x \to \hat{x}, & p \to \hat{p}, \\ {x, p} = 1 & \to [\hat{x}, \hat{p}] = i ℏ . \end{aligned} \end{matrix}

(B.2.1)

Canonical quantization of a field theory is done analogously, with fields becoming operator-valued and obeying their own canonical commutation relations based on Eq. B.1.15. For our free scalar field theory (Eq. 3.1.4), this means promoting the integration constants in the classical solution (Eq. B.1.9 and B.1.20) to operators:

\begin{matrix} \begin{aligned} ϕ (x, t) & = \int \frac{d^{3} p}{{(2 π)}^{3}} \frac{1}{\sqrt{2 ω_{p}}} (â_{p} e^{ip \cdot x} + â_{p}^{†} e^{- ip \cdot x}), \\ π (x, t) & = - i \int \frac{d^{3} p}{{(2 π)}^{3}} \sqrt{\frac{ω_{p}}{2}} (â_{p} e^{ip \cdot x} - â_{p}^{†} e^{- ip \cdot x}), \end{aligned} \end{matrix}

(B.2.2)

where again $p \cdot x = p_{μ} x^{μ}$ is the 4D spacetime inner product and $p_{μ} = (ω_{p} = \sqrt{{| p |}^{2} + m^{2}}, p)$ . Recall that the integration constants $a (p)$ and $a^{*} (p)$ arose from a SHO equation for each momentum $p$ (Eq. B.1.8); thus, quantized, we expect them to correspond to the raising ( $â^{†}$ ) and lowering ( $â$ ) operators of a quantum harmonic oscillator (QHO), again one for each momentum mode $p$ . We can check this by deriving their commutation relations. Indeed, imposing the canonical commutation relationships:

[ϕ (x, t), ϕ (y, t)] = [π (x, t), π (y, t)] = 0, [ϕ (x, t), π (y, t)] = i δ^{3} (x - y),

(B.2.3)

reproduces (continuous versions of) the raising and lowering operator commutation relationships for a QHO:

[â_{p}, â_{q}] = [â_{p}^{†}, â_{q}^{†}] = 0, [â_{p}, â_{q}^{†}] = {(2 π)}^{3} δ^{3} (p - q) .

(B.2.4)

Next, we look at the commutators with the Hamiltonian and the resulting Hilbert space.

B.2.2 The Hamiltonian and the vacuum catastrophe

The quantized Hamiltonian, from Eq. B.1.21, can be found to be:

\begin{matrix} \begin{aligned} H & = \int d^{3} x (\frac{1}{2} π^{2} + \frac{1}{2} {(\nabla ϕ)}^{2} + \frac{1}{2} m^{2} ϕ^{2}) \\ = \int \frac{d^{3} p}{{(2 π)}^{3}} ω_{p} [â_{p}^{†} â_{p} + \frac{1}{2} {(2 π)}^{3} δ^{3} (0)] . \end{aligned} \end{matrix}

(B.2.5)

This looks a lot like the Hamiltonian for a QHO, $H = ω (a^{†} a + \frac{1}{2})$ , for each momenta, but with an unwieldy delta function. This latter term is called the zero-point energy and represents the energy of the vacuum state. It is infinite, and, indeed, is one of the many infinities that have to be dealt with in QFT. In this case, since it is a constant energy term, it does not affect the dynamics of the system and can simply be ignored / subtracted for our purposes.² However, the vacuum energy density does affect Einstein’s equations of general relativity, and the disagreement between the large zero-point energy we expect from QFT and the small observed value is known as the cosmological constant problem (or, more dramatically, the vacuum catastrophe) [429, 430].

An infinity of harmonic oscillators

Subtracting away the zero-point energy gives us

H = \int \frac{d^{3} p}{{(2 π)}^{3}} ω_{p} â_{p}^{†} â_{p},

(B.2.6)

whose commutators with the raising and lowering operators are:

[H, â_{p}^{†}] = ω_{p} â_{p}^{†}, [H, â_{p}] = - ω_{p} â_{p},

(B.2.7)

just as for a QHO. This tells us that given an eigenstate of $H$ , $| E ⟩$ , with eigenvalue $E$ , $â_{p}^{†} | E ⟩$ and $â_{p} | E ⟩$ are also eigenstates with eigenvalues $E + ω_{p}$ and $E - ω_{p}$ , respectively:

\begin{matrix} \begin{aligned} H â_{p}^{†} | E ⟩ & = (â_{p}^{†} H + ω_{p} â_{p}^{†}) | E ⟩ = (E + ω_{p}) â_{p}^{†} | E ⟩, \\ H â_{p} | E ⟩ & = (â_{p} H - ω_{p} â_{p}) | E ⟩ = (E - ω_{p}) â_{p} | E ⟩ . \end{aligned} \end{matrix}

(B.2.8)

B.2.3 Particles

To understand these states further, we can also quantize the total momentum of the field density we found from Noether’s theorem (Eq. 3.1.14):³

P = \int d^{3} x π \nabla ϕ = \int \frac{d^{3} p}{{(2 π)}^{3}} p â_{p}^{†} â_{p} .

(B.2.9)

Acting with $P$ on $| p ⟩$ gives us:

\begin{matrix} \begin{aligned} P | p ⟩ & = \int \frac{d^{3} k}{{(2 π)}^{3}} k â_{k}^{†} â_{k} â_{p}^{†} | 0 ⟩ \\ = \int \frac{d^{3} k}{{(2 π)}^{3}} k â_{k}^{†} (â_{p}^{†} â_{k} - {(2 π)}^{3} δ^{3} (k - p)) | 0 ⟩ \\ = p | p ⟩ . \end{aligned} \end{matrix}

(B.2.10)

Thus the states $| p ⟩$ are eigenstates of $P$ as well, with eigenvalues $p$ . Putting this together, we have a Hilbert space spanned by the states $| p ⟩$ , which each have momentum $p$ and energy $ω_{p} = | \sqrt{p^{2} + m^{2}} |$ , i.e. the relativistic energy-momentum relation for a free particle. Thus, we see $| p ⟩$ exactly corresponds to the momentum eigenstate for a single particle of mass $m$ and momentum $p$ ! One can similarly quantize the total angular momentum of the field $J$ and show that $J | p = 0 ⟩ = 0$ , i.e. the particle has spin 0.

This is one of the miracles of QFT: what from quantizing a free, relativistic field looked bizarrely like an infinite series of QHOs, actually gives the intuitive physical result of discrete particle states. The Fock space hence is the space spanned by different numbers of discrete particles per each continuous momentum mode $p$ . The number of particles $n$ in a particular state of the Fock space is given by the number operator $N$ , essentially the Hamiltonian density divided by $ω_{p}$ :

N = \int \frac{d^{3} p}{{(2 π)}^{3}} â_{p}^{†} â_{p} \Rightarrow N | p_{1}, \dots, p_{n} ⟩ = n | p_{1}, \dots, p_{n} ⟩ .

(B.2.11)

Note that the number operator $N$ commutes with the Hamiltonian $H$ , $[N, H] = 0$ , which means particle number is conserved; however, this will not be the case for interacting theories in the next section.

Normalization of states and wavepackets

Note that we cannot simply choose the normalization of momentum eigenstates as $⟨ q | p ⟩ = δ^{3} (q - p)$ , as in nonrelativistic QM, because the delta function alone is not Lorentz-invariant. Instead, we choose the normalization in Eq. 3.1.21, which is a Lorentz scalar.

Like in QM, however, these momentum eigenstates are not normalized to 1: $⟨ p | p ⟩ = 2 E_{p} δ^{3} (0)$ , so they are not exactly physical one-particle states. Physical particles must exist in the form of a wavepacket:

| φ ⟩ = \int d^{3} p φ (p) | p ⟩,

(B.2.12)

with some spread in momenta $φ (p)$ . However, as long as this variation is smaller than the resolution of our detector (as we will assume), for all practical purposes and calculations we can continue to treat particles as momentum eigenstates. This assumption is further motivated in Peskin and Schroeder [81] Chapter 4.5.

B.2.4 The complex scalar field and antiparticles

The complex scalar field Lagrangian from Eq. 3.1.15 has the EOMs:

\begin{matrix} \begin{aligned} (\partial_{μ} \partial^{μ} + m^{2}) ψ & = 0, \\ (\partial_{μ} \partial^{μ} + m^{2}) ψ^{*} & = 0, \end{aligned} \end{matrix}

(B.2.13)

with solutions:

\begin{matrix} \begin{aligned} ψ (x) & = \int \frac{d^{3} p}{{(2 π)}^{3}} \frac{1}{\sqrt{2 ω_{p}}} (b (p) e^{ip \cdot x} + c^{*} (p) e^{- ip \cdot x}) . \end{aligned} \end{matrix}

(B.2.14)

Note that because the field is complex, the coefficients $b$ and $c^{*}$ need not be complex conjugates of each other as for a real field. This field can be quantized analogously to above:

\begin{matrix} \begin{aligned} ψ (x, t) & = \int \frac{d^{3} p}{{(2 π)}^{3}} \frac{1}{\sqrt{2 ω_{p}}} ({\hat{b}}_{p} e^{ip \cdot x} + ĉ_{p}^{†} e^{- ip \cdot x}), \\ ψ^{†} (x, t) & = \int \frac{d^{3} p}{{(2 π)}^{3}} \frac{1}{\sqrt{2 ω_{p}}} ({\hat{b}}_{p}^{†} e^{- ip \cdot x} + ĉ_{p} e^{ip \cdot x}), \end{aligned} \end{matrix}

(B.2.15)

where we now have two sets of creation and annihilation operators, ${{\hat{b}}^{†}, \hat{b}}$ and ${ĉ^{†}, ĉ}$ . One can check each pair individually satisfies the canonical commutation relations from Eq. B.2.4, and mutually commutes with each other. Thus, they are interpreted as corresponding to two different particles, with the same mass $m$ and spin 0, but, as we saw, with opposite charges under the $U (1)$ internal symmetry. Such pairs are considered particles and antiparticles.

Finally, let us revisit and quantize the conserved charge associated with the $U (1)$ symmetry (Eq. 3.1.16):

Q = \int d^{3} x i (ψ^{*} \partial^{0} ψ - ψ \partial^{0} ψ^{*}) \to \int \frac{d^{3} p}{{(2 π)}^{3}} ({\hat{b}}_{p}^{†} {\hat{b}}_{p} - ĉ_{p}^{†} ĉ_{p}) = N_{b} - N_{c} .

(B.2.16)

This is saying the difference in the number of particles and antiparticles is conserved, which for a single charged particle-antiparticle pair, is equivalent to charge conservation. This will be more significant for interacting theories, in which $N_{b}$ and $N_{c}$ are not individually conserved but as long as the interactions retain the $U (1)$ symmetry, $Q$ is.

Negative energy states?

Note that the full, time-dependent formula for the field $ψ (x, t)$ contains both the $e^{- ip \cdot x} \propto e^{- iEt}$ and $e^{ip \cdot x} \propto e^{iEt}$ terms. As single-particle plane-wave solutions to the nonrelativistic Schrödinger equation, these would correspond to positive and negative energy states, the latter of which does not make physical sense.⁴ Our solution is to refer to these states instead as positive- and negative-frequency modes, which, as we saw, are always associated to operators that create and destroy positive-energy (anti)particles, respectively.

²Equivalently, we can consider the normal-ordered Hamiltonian (see, e.g., Tong QFT [74] Chapter 2.3). An alternative way of resolving the infinity is to introduce an ultra-violet cut-off scale $Λ$ in the integral over momenta.

³Technically, we show here the normal-ordered momentum.

⁴This is related to the problem Dirac faced in developing his relativistic quantum theory of the electron, except we are dealing with bosons instead of fermions, so we cannot rely on the fermionic Dirac sea “solution”.