Chapter 3 Free quantum Klein-Gordon field

This chapter studies the canonical quantization of the free real Klein-Gordon field from Examples 2.1 and 2.13. It will also be illustrated in which sense a QFT has an associated theory of particles.

3.1 Field operator and dynamics

Recall that, in quantum mechanics, canonical quantization amounts to replacing the canonical coordinates \(q_a\) and \(p^a\) of the Hamiltonian formalism by operators that satisfy the canonical commutation relations (CCRs) from (1.8). It is important to emphasize that these commutation relations are formulated at a fixed instance of time \(t_0\in \bbR \), so when being more pedantic with notation one should write \(q_a := q_a(t_0)\) and \(p^a := p^a(t_0)\). We will mostly choose \(t_0=0\), which simplifies our formulas as it avoids expressions of the form \(t-t_0\).

The analog of the CCRs for a real Klein-Gordon field in the Hamiltonian formalism from Example 2.13 is given by considering the field \(\Phi (\mathbf {x}):= \Phi (t_0,\mathbf {x})\) and its canonical momentum \(\Pi (\mathbf {x}) := \Pi (t_0,\mathbf {x})\) at a fixed instance of time \(t_0\in \bbR \), which we suppress from the notation, and demand the commutation relations

where \(\delta (\mathbf {x}-\mathbf {y})\) denotes the Dirac delta function. Because all operators are considered at the same value of time \(t_0\in \bbR \), but importantly not necessarily at the same positions in space \(\mathbf {x},\mathbf {y}\in \bbR ^{d-1}\), one often calls (3.1) the equal-time commutation relations. Comparison between (3.1) and (1.8) shows that, in the context of field theory, the index \(a=1,\dots ,N\) on \(q_a\) and \(p^a\) that counts the degrees of freedom is generalized to a continuous index \(\mathbf {x}\in \bbR ^{d-1}\) that corresponds to the different points in space. This is why people often say that QFT deals with infinitely many degrees of freedom, at least one for each point \(\mathbf {x}\in \bbR ^{d-1}\) in space. Note that the equal-time commutation relations are local in space, by which one means that any two operators corresponding to different points \(\mathbf {x}, \mathbf {y}\in \bbR ^{d-1}\) commute with each other. (Informally speaking, the Dirac delta function \(\delta (\mathbf {x}-\mathbf {y})\) is \(0\) for \(\mathbf {x}\neq \mathbf {y}\).)

The (too naive, as explained below) Hamiltonian operator for the free real Klein-Gordon field is given by promoting the classical Hamiltonian (2.13) to an operator

\begin{flalign} \label {eqn:KGHamiltonianoperator} H = \int _{\bbR ^{d-1}} \frac {1}{2} \,\Big (\Pi ^2 + (\nabla \Phi )^2 + m^2\,\Phi ^2\Big )\,\dd \mathbf {x}\quad , \end{flalign} where it is important to emphasize that \(\Pi \) and \(\Phi \) are now operators in contrast to the classical fields in (2.13). Due to the spatial derivatives \(\nabla \) on \(\Phi \), the Hamiltonian mixes between the field operators \(\Phi (\mathbf {x})\) corresponding to different points \(\mathbf {x}\in \bbR ^{d-1}\) in space. (Recall that the derivative of a function \(f(x)\) is defined by \(f^\prime (x) = \lim _{\epsilon \to 0}\, (f(x+\epsilon )-f(x))/\epsilon \), hence it depends on \(f\) at the different, albeit infinitesimally close, points \(x\) and \(x+\epsilon \).) Such interplay complicates finding solutions of Heisenberg’s equation for the time evolution of operators, so thinking ahead we should do something about it. There is a simple way to get rid of this mixing coming from the spatial derivatives \(\nabla \), which is given by going from position space \(\mathbf {x}\in \bbR ^{d-1}\) to Fourier space \(\mathbf {k}\in \bbR ^{d-1}\) via a Fourier transform: Indeed, from the Fourier transformation formulas in Section 1.4, one easily sees that a derivative \(\nabla f(\mathbf {x})\) in position space becomes a multiplication \(\ii \,\mathbf {k} \,\widetilde {f}(\mathbf {k})\) in Fourier space. Writing the operators

\begin{flalign} \label {eqn:FourierKG} \Phi (\mathbf {x}) = \int _{\bbR ^{d-1}} \widetilde {\Phi }(\mathbf {k})\, e^{\ii \,\mathbf {k}\,\mathbf {x}} \,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad ,\quad \Pi (\mathbf {x}) = \int _{\bbR ^{d-1}} \widetilde {\Pi }(\mathbf {k})\, e^{\ii \,\mathbf {k}\,\mathbf {x}} \,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}} \end{flalign} in Fourier space, one finds for the Hamiltonian operator (3.2)

\begin{flalign} \nn H&= \int _{\bbR ^{d-1}} \frac {1}{2}\bigg ( \widetilde {\Pi }(\mathbf {k})\, \widetilde {\Pi }(-\mathbf {k})+ \big (\mathbf {k}^2 + m^2\big )~ \widetilde {\Phi }(\mathbf {k})\, \widetilde {\Phi }(-\mathbf {k}) \bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ &= \int _{\bbR ^{d-1}} \frac {1}{2}\bigg ( \widetilde {\Pi }(\mathbf {k})\, \widetilde {\Pi }(-\mathbf {k})+ \omega _{\mathbf {k}}^2~\widetilde {\Phi }(\mathbf {k})\, \widetilde {\Phi }(-\mathbf {k}) \bigg )~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \label {eqn:KGHamiltonianoperatorFourier} \end{flalign} where in the second line we have identified the relativistic energy

To carry out this calculation, which you should definitely do, one has to use the identity

\begin{flalign} \int _{\bbR ^{d-1}} e^{\ii \,(\mathbf {k}+\mathbf {q})\,\mathbf {x}}\,\dd \mathbf {x} = (2\pi )^{d-1}\,\delta (\mathbf {k}+\mathbf {q}) \end{flalign} for the \(d-1\)-dimensional Dirac delta function. The Fourier transforms of the equal-time commutation relations (3.1) read as

\begin{flalign} \label {eqn:CCRrealKGFourier} \big [\widetilde {\Phi }(\mathbf {k}),\widetilde {\Phi }(\mathbf {q})\big ] = 0 = \big [\widetilde {\Pi }(\mathbf {k}),\widetilde {\Pi }(\mathbf {q})\big ] \quad ,\qquad \big [\widetilde {\Phi }(\mathbf {k}),\widetilde {\Pi }(\mathbf {q})\big ] = \ii \,(2\pi )^{d-1}\, \delta (\mathbf {k}+\mathbf {q}) \quad , \end{flalign} as one can easily check using the formulas from Section 1.4.

A further simplification can be achieved by introducing, in analogy to the quantum harmonic oscillator from quantum mechanics, annihilation operators \(a(\mathbf {k})\) and their adjoints the creation operators \(a^\dagger (\mathbf {k}):= (a(\mathbf {k}))^\dagger \), for all \(\mathbf {k}\in \bbR ^{d-1}\). Because of the nonstandard Hermiticity conditions \((\widetilde {\Phi }(\mathbf {k}))^\dagger = \widetilde {\Phi }(-\mathbf {k})\) and \((\widetilde {\Pi }(\mathbf {k}))^\dagger = \widetilde {\Pi }(-\mathbf {k})\) in Fourier space, which follow from the usual Hermiticity conditions \((\Phi (\mathbf {x}))^\dagger =\Phi (\mathbf {x})\) and \((\Pi (\mathbf {x}))^\dagger =\Pi (\mathbf {x})\) in position space and the fact that taking adjoints changes the sign in the complex exponentials in (3.3), the annihilation and creation operators are defined slightly differently than in quantum mechanics in order to ensure that they are adjoints of each other. In short, the expressions

\begin{flalign} \label {eqn:PhiPiviaAnCre} \widetilde {\Phi }(\mathbf {k}) = \frac {1}{\sqrt {2\omega _{\mathbf {k}}}} \,\Big (a(\mathbf {k}) + a^\dagger (-\mathbf {k})\Big )\quad ,\qquad \widetilde {\Pi }(\mathbf {k}) = -\ii \,\sqrt {\frac {\omega _{\mathbf {k}}}{2}} \,\Big (a(\mathbf {k}) -a^\dagger (-\mathbf {k})\Big ) \end{flalign} are compatible with the required Hermiticity conditions (note the \(-\mathbf {k}\) in the argument of \(a^\dagger \)), and when inverted give

\begin{flalign} a(\mathbf {k}) = \sqrt {\frac {\omega _{\mathbf {k}}}{2}}~\widetilde {\Phi }(\mathbf {k}) + \frac {\ii }{\sqrt {2\omega _{\mathbf {k}}}} ~\widetilde {\Pi }(\mathbf {k}) \quad ,\qquad a^\dagger (\mathbf {k}) = \sqrt {\frac {\omega _{\mathbf {k}}}{2}}~\widetilde {\Phi }(-\mathbf {k}) - \frac {\ii }{\sqrt {2\omega _{\mathbf {k}}}} ~\widetilde {\Pi }(-\mathbf {k}) \quad . \end{flalign} From (3.7) one then obtains the commutation relations

for the annihilation and creation operators. Furthermore, inserting (3.8) into the Hamiltonian operator (3.4), one finds after a calculation (that you should do!) the expression

\begin{flalign} H = \int _{\bbR ^{d-1}} \frac {\omega _{\mathbf {k}}}{2}~\Big (a^\dagger (\mathbf {k})\,a(\mathbf {k}) + a(\mathbf {k})\,a^\dagger (\mathbf {k})\Big )~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}} \end{flalign} in terms of the annihilation and creation operators.

This form of the Hamiltonian shows that the free quantum Klein-Gordon field in Fourier space is simply an infinite family of quantum harmonic oscillators, one for each \(\mathbf {k}\in \bbR ^{d-1}\). However, this infinite number of degrees of freedom bites back! If we want to rewrite the Hamiltonian in the standard way such that the \(a^\dagger \)’s are to the left of the \(a\)’s, we find using the commutation relations (3.10) that

\begin{flalign} \label {eqn:Hamiltoniandivergent} H = \int _{\bbR ^{d-1}} \omega _{\mathbf {k}}~a^\dagger (\mathbf {k})\,a(\mathbf {k}) ~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}} +(2\pi )^{d-1}\,\delta (\mathbf {0})\,\int _{\bbR ^{d-1}} \frac {\omega _{\mathbf {k}}}{2}~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \end{flalign} which is a nonsensible expression for two very different reasons: First, the Dirac delta function \((2\pi )^{d-1}\, \delta (\mathbf {0}) = \int _{\bbR ^{d-1}} \dd \mathbf {x} \) at zero is mathematically ill-defined. Note that this divergence results from the infinite volume of position space, i.e. large distances, which is why it is called an infrared divergence. Second, the integral \(\int _{\bbR ^{d-1}} \frac {\omega _{\mathbf {k}}}{2}~\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\) diverges because \(\omega _{\mathbf {k}}\) grows as \(\vert \mathbf {k}\vert \) for large \(\mathbf {k}\). Note that this divergence results from large momenta, i.e. short distances in position space, which is why it is called an ultraviolet divergence. The way how to deal with such types of divergences is to establish a better operator ordering.

Let us now derive the Heisenberg picture field operator \(\Phi (x) := \Phi (t,\mathbf {x})\), which is an operator on spacetime \(x=(t,\mathbf {x})\in \bbR ^d\) (in contrast to only space \(\mathbf {x}\in \bbR ^{d-1}\)) that is determined by solving Heisenberg’s equation

\begin{flalign} \label {eqn:KGHeisenbergeqauation} \frac {\partial }{\partial t} \Phi (x) = \ii \,\big [\noor {H},\Phi (x)\big ]\quad , \end{flalign} subject to the initial condition \(\Phi (0,\mathbf {x}) = \Phi (\mathbf {x})\). (We choose the initial time \(t_0=0\) to be zero in order to avoid cluttering our formulas with a lot of \(t_0\)’s. Of course, you are free to take any initial time \(t_0\in \bbR \) that you like.) Recall that the solution of this equation is given exponentiation

\begin{flalign} \Phi (x) = e^{\ii \,t\, [\noor {H},-]}\,\Phi (\mathbf {x}) = e^{\ii \,t \,\noor {H}}\,\Phi (\mathbf {x})\,e^{-\ii \, t \,\noor {H}}\quad . \end{flalign} Combining (3.3) and (3.8), we can write the initial condition as

\begin{flalign} \nn \Phi (\mathbf {x}) &= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~\bigg ( a(\mathbf {k}) \,e^{\ii \,\mathbf {k}\,\mathbf {x}}+ a^\dagger (-\mathbf {k})\, e^{\ii \,\mathbf {k}\,\mathbf {x}}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ &= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ \bigg ( a(\mathbf {k}) \,e^{\ii \,\mathbf {k}\,\mathbf {x}}+ a^\dagger (\mathbf {k})\, e^{-\ii \,\mathbf {k}\,\mathbf {x}}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad , \end{flalign} where in the second line we have transformed the second term under the integral via \(\mathbf {k}\mapsto -\mathbf {k}\). Using further that \([\noor {H},a(\mathbf {k})]=-\omega _{\mathbf {k}}\,a(\mathbf {k})\) and that \([\noor {H},a^\dagger (\mathbf {k})]=\omega _{\mathbf {k}}\,a^\dagger (\mathbf {k})\), which can be derived from (3.13) and (3.10), we can compute

\begin{flalign} \nn \Phi (x) &= e^{\ii \,t\,[\noor {H},-]}\,\Phi (\mathbf {x}) = \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ \bigg ( e^{\ii \, t\, [\noor {H},-]}\,a(\mathbf {k}) \,e^{\ii \,\mathbf {k}\,\mathbf {x}}+ e^{\ii \,t\,[\noor {H},-]}\,a^\dagger (\mathbf {k})\, e^{-\ii \,\mathbf {k}\, \mathbf {x}}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ &= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ \bigg ( e^{-\ii \,\omega _{\mathbf {k}}\,t}\,a(\mathbf {k}) \,e^{\ii \,\mathbf {k}\,\mathbf {x}}+ e^{\ii \,\omega _{\mathbf {k}}\,t}\,a^\dagger (\mathbf {k})\, e^{-\ii \,\mathbf {k}\,\mathbf {x}}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{flalign} Introducing the relativistic Fourier momentum

\begin{flalign} \label {eqn:relativisticFouriermomentum} k :=\begin{pmatrix} \omega _{\mathbf {k}} \\ \mathbf {\mathbf {k}} \end {pmatrix} \in \bbR ^d \end{flalign} that, as a consequence of (3.5), satisfies the relativistic energy-momentum relation (also called on-shell condition in the context of QFT)

we can write the Heisenberg picture field operator in a very compact form

where the exponents involve the Minkowski inner product \(k\,x :=\eta _{\mu \nu } \,k^\mu \,x^\nu \). Last but not least, we note that this field operator satisfies the Klein-Gordon equation

\begin{flalign} \label {eqn:KGquantumEOM} \nn (-\partial ^2 + m^2)\Phi (x)&= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ \bigg (a(\mathbf {k}) \, (-\partial ^2 + m^2)e^{\ii \,k\,x}+ a^\dagger (\mathbf {k})\, (-\partial ^2 + m^2)e^{-\ii \,k\,x}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ \nn &= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ \bigg (a(\mathbf {k}) \, (k^2 + m^2)e^{\ii \,k\,x}+ a^\dagger (\mathbf {k})\, (k^2 + m^2)e^{-\ii \,k\,x}\bigg )\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\\ &=0\quad , \end{flalign} where in the last step we have used the on-shell condition (3.19). Hence, the quantum dynamics of the free Klein-Gordon field is governed by the same equation that arises as the Euler-Lagrange equation for the classical free Klein-Gordon field. It is easy to check that the Heisenberg picture operator corresponding to the canonical momentum is given by the time derivative \(\Pi (x) = \dot {\Phi }(x)\) of (3.20), hence it contains no new information.

We conclude this section by studying the commutation relation \([\Phi (x),\Phi (y)]\) between two Heisenberg picture field operators. In contrast to the equal-time commutation relations in (3.1), this now involves two arbitrary spacetime points \(x,y\in \bbR ^d\) that do not necessarily have the same time coordinates. With a short calculation using (3.20) and (3.10), one obtains

Observe that the right-hand side is just a complex number, which is sometimes called the commutator function and denoted by

\begin{flalign} \Delta (x-y):= \big [\Phi (x),\Phi (y)\big ]\quad . \end{flalign} We can rewrite this as an integral over all \(k=(k^0,\mathbf {k})\in \bbR ^d\) (not subject to the on-shell condition (3.19))

\begin{flalign} \label {eqn:KGcommutatorfunction2} \Delta (x-y) = \int _{\bbR ^{d}} \delta (k^2 + m^2)\,\,\Theta (k^0) ~\Big (e^{\ii \,k\,(x-y)} - e^{-\ii \,k\,(x-y)} \Big )\,\frac {\dd k}{(2\pi )^{d-1}}\quad , \end{flalign} where

\begin{flalign} \Theta (k^0) = \begin{cases} 1 & ,~\text {for }k^0> 0\quad ,\\ \frac {1}{2} & ,~\text {for }k^0=0\quad ,\\ 0 & ,~\text {for }k^0< 0\quad ,\\ \end {cases} \end{flalign} denotes the Heaviside step function. Indeed, using \(k^2+m^2 = -(k^0)^2 + \mathbf {k}^2 + m^2 = -(k^0)^2 +\omega _{\mathbf {k}}^2\) and standard properties of the Dirac delta function, one finds that

\begin{flalign} \nn \delta (k^2 + m^2)\,\Theta (k^0) &= \delta \Big ((\omega _\mathbf {k}+ k^0)\, (\omega _\mathbf {k}- k^0)\Big )\, \Theta ( k^0)\\ &= \frac {1}{\omega _\mathbf {k}+ k^0}\, \delta (\omega _\mathbf {k}- k^0)\, \Theta (k^0)\quad , \label {eqn:masshelltrick} \end{flalign} from which one shows that (3.24) agrees with (3.22) by carrying out the integral over \(k^0\). The advantage of the expression (3.24) is that it is manifestly invariant under proper and orthochronous Poincaré transformations, see Section 1.4.

With these preparations, we can show that the commutator function \(\Delta (x-y)=\big [\Phi (x),\Phi (y)\big ]=0\) vanishes whenever the points \(x,y\in \bbR ^d\) are spacelike separated, i.e. \((x-y)^2 >0\). It is a standard exercise in special relativity to show that for each pair of spacelike separated points there exists a proper and orthochronous Poincaré transformation such that the transformed difference \(x^\prime -y^\prime = (0,\mathbf {z})\) has a zero time component. (In words, for each pair of spacelike separated points there exists a choice of coordinates such that the associated events happen at the same time.) Since the commutator function (3.24), and hence also (3.22), is invariant under such Poincaré transformations, we can transform to the new coordinates and find that

\begin{flalign} \label {eqn:KGcausality} \Delta (x-y) = \Delta ^\prime (x^\prime -y^\prime ) = \int _{\bbR ^{d-1}} \frac {1}{2\,\omega _{\mathbf {k}^\prime }}\,\Big (e^{\ii \,\mathbf {k}^\prime \,\mathbf {z}} - e^{-\ii \,\mathbf {k}^\prime \,\mathbf {z}} \Big ) \,\frac {\dd \mathbf {k}^\prime }{(2\pi )^{d-1}} =0\quad ,~~\text {for }(x-y)^2>0\quad , \end{flalign} because the integrand is an odd function in \(\mathbf {k}^\prime \) and hence it integrates to zero. This result can be interpreted as a statement about the causality of the quantum Klein-Gordon field. Indeed, the main feature of commuting operators in quantum theory is that they are independent in the sense that they do not influence or alter the measurement of each other, which in our context means that the field operators \(\Phi (x)\) and \(\Phi (y)\) associated with spacelike separated points are independent of each other.