Klein-Gordon Field Quantization
This article is one of Quantum Field Thoery.
Quantum field theory views the position and momentum of the field in each event in spacetime as a wave, and when all points are combined, it again represents a particle.
There is a saying in Buddhist scriptures:
色卽是空空卽是色
This means that substance is empty, and empty is substance.
2. The Klein-Gordon Field
2.1 The Necessity of the Field Viewpoint
Let's see single particle quantum mechanics. Consider the amplitude for a free particle to propagate from $\mathbf{x}_0$ to $\mathbf{x}$: $$U(t)=\langle \mathbf{x}|e^{-iHt}|\mathbf{x}_0\rangle.$$ In nonrelativistic quantm mechanics we have $E=\mathbf{p}^2/2m$, so $$U(t)= \langle \mathbf{x}|e^{-i(\mathbf{p}^2/2m)t}|\mathbf{x}_0\rangle\\ =\int \frac{d^3p}{(2\pi)^3}\langle \mathbf{p}|e^{-iHt}|\mathbf{p}_0\rangle\langle \mathbf{p}|\mathbf{x}_0\rangle\\ =\frac{1}{(2\pi)^3}\int d^3p\, e^{-i(\mathbf{p}^2/2m)t}|\mathbf{p}\rangle\langle \mathbf{p}|\mathbf{x}_0\rangle \\ =\frac{1}{(2\pi)^3}\int d^3p\, e^{-i(\mathbf{p}^2/2m)t}\cdot e^{i\mathbf{p}\cdot (\mathbf{x}-\mathbf{x}_0)}\\ =\left( \frac{m}{2\pi it}\right)^{3/2}e^{im(\mathbf{x}-\mathbf{x}_0)^2/2t}.$$ This expression is nonzero for all $x$ and $t$, indicating that a particle can propagate between any two points in an arbitrarily short time. In a relativistic theory, this conclusion would signal a violation of causality. One might hope that using the relativistivc expression $E=\sqrt{p^2+m^2}$ would help, but it does not.
2.2 Elements of Classical Field Theory
Lagrangian Field Theory
The funcdamental quantity of classical mechanics is the action, $S$, the time integral of the Lagrangian, $L$. In a local field theory the Lagrangian can be written as the spactial integral of a Lagrangian density, denoted by $\mathcal{L}$, which is a function of one or more fields $\phi(x)$ and their derivatives $\partial_\mu\phi$. Thus we have $$\begin{align}S=\int Ldt=\int \mathcal{L}(\phi,\partial_\mu \phi)d^4x.\end{align}$$ Since this is a book on field theory, we will refer to $\mathcal{L}$ simply as the Lagrangian.
The principle of least action states that when a system evolves from one given configuration to another between times $t_1$ and $t_2$, it does so along the "path in configuration space for which $S$ is an extremum (normally miminum). We can write this condition as $$\begin{align}0=\delta S\nonumber\\ =\int d^4x\left\{ \frac{\partial\mathcal{L}}{\partial\phi}\delta \phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)\right\}\nonumber\\ =\int d^4x\left\{ \frac{\partial\mathcal{L}}{\partial\phi}\delta\phi-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi +\partial_\mu \left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\right)\right\}.\end{align}$$ The last term can be turned into a surface integral over the boundary of the four-dimensional spacetime region of integration. Since the initial and final field configurations are assumed given, $\delta\phi$ is zero at the temporal beginning and end of this region. If we restrict our consideration to deformations $\delta\phi$ that vanish on the spatial boundary of the region as well, then the surface term is zero. Factoring out the $\delta\phi$ from the first two terms, we note that, since the integral must vanish for arbitrary $\delta\phi$, the quantity that multiplies $\delta\phi$ must vanish at all points. Thus we arrive at the Euler-Lagrane equation of motion for a field, $$\begin{align}\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial\mathcal{L}}{\partial\phi}=0.\end{align}$$ If the Lagrangian contins more than one field, there is one such equation for each.
Hamiltonian Field Theory
Recall that for a discrete system one can define a conjugate momentum $p\equiv \partial L./\partial\dot{q}$ (where $\dot{q}=\partial q/\partial t$) for each dynamical variable $q$. The Hamiltonian is then $H\equiv \sum p\dot{q}-L$. The generalization to a continuous system is best understood by pretending that the spatical poitns $\mathbf{x}$ are discretely spaced. We can define $$p(\mathbf{x})\equiv \frac{\partial L}{\partial \dot{\phi}(\mathbf{x})}=\frac{\partial}{\partial \dot{\phi}(\mathbf{x})}\int \mathcal{L}(\phi(\mathbf{y},\dot{\mathbf{y}}))d^3y\\ \sim \frac{\partial}{\partial\dot{\phi}(\mathbf{x})}\sum_\mathbf{y} \mathcal{L}(\phi(\mathbf{y}),\dot{\phi}(\mathbf{y}))d^3y =\pi(\mathbf{x})d^3x,$$ where $$\begin{align}\pi(\mathbf{x})\equiv\frac{\partial\mathcal{L}}{\partial\dot{\phi}(\mathbf{x})}\end{align}$$ us called the momentum density conjugate to $\phi(\mathbf{x})$. Thus the Hamiltonian can be written $$H=\sum_\mathbf{x}p(\mathbf{x})\dot{\phi}(\mathbf{x})-L.$$ Passing to the continuum, this becomes $$\begin{align}H=\int d^3x\, [\pi(\mathbf{x})\dot{\phi}(\mathbf{x})-\mathcal{L}]\equiv \int d^3\mathcal{H}.\end{align}$$
As a simple example, consider the theory of a single field $\phi(x)$, governed by the Lagrangian $$\begin{align}\mathcal{L}=\frac{1}{2}\dot{\phi}^2-\frac{1}{2}(\nabla\phi)^2-\frac{1}{2}m^2\phi^2\nonumber\\ =\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2.\end{align}$$ For now we take $\phi$ to be a real-valued field. From this Lagrangian the usual procedure gives the equation of motion $$\begin{align}\left( \frac{\partial^2}{\partial t^2}-\nabla^2+m^2\right)\phi=0\quad \mbox{or}\quad (\partial^\mu\partial_\mu+m^2)\phi=0,\end{align}$$ which is the well-known Klein-Gordon equation. Noting that the canocial momentum density conjugate to $\phi(x)$ is $\pi(x)=\dot{\phi}(x)$, we can also construct the Hamiltonian: $$\begin{align}H=\int d^3x\mathcal{H}=\int d^3x[\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+\frac{1}{2}m^2\phi^2].\end{align}$$
Noether's Theorem
This theorem concerns continuous transformations on the fields $\phi$, which in infinitesimal form can be written $$\begin{align}\phi(x)\rightarrow \phi'(x)=\phi(x)+\alpha\Delta\phi(x),\end{align}$$ where $\alpha$ is an infinitesimal parameter and $\Delta \phi$ is some deformation of the field configuration. We call this transformation a symmetry if it leaves the equation of motion invariant. This is insured if the action is invariant under (9). More generally, we can allow the action to change by a surface term, since the presence of such a term would not affect our derivation of the Euler-Lagrange equations of motion (3). The Lagrangian, therefore, must be invariant under (9) up to a 4-divergence: $$\begin{align}\mathcal{L}(x)\rightarrow \mathcal{L}(x)+\alpha\partial_\mu \mathcal{J}^\mu(x),\end{align}$$ for some $\mathcal{L}^\mu$. Let us compare this expectation for $\Delta\mathcal{L}$ to the result obtained by varying the fields: $$\begin{align}\alpha\Delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}(\alpha\Delta\phi)+\left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\partial_\mu(\alpha\Delta\phi)\nonumber\\ =\alpha\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\right)+\alpha\left[ \frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right]\Delta\phi.\end{align}$$ The second term vanishes by the Euler-Lagrange equation (3). We set the remaining term equal to $\alpha\partial_\mu\mathcal{J}^\mu$ and find $$\begin{align}\partial_\mu j^\mu(x)=0,\quad \mbox{for}\ j^\mu(x)=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi -\mathcal{J}^\mu.\end{align}$$ (If the symmetry involves more than one field, the first term of this expression for $j^\mu(x)$ should be replaced by a sum of such terms, one for each field.) This result states that the current $j^\mu(x)$ is conserved. For each continuous symmetry of $\mathcal{L}$, we have such a conservation law.
The conservation law can also be expressed by saying that the charge $$\begin{align}Q\equiv \int_{all\ space}j^0d^3x\end{align}$$ is a constant in time. Note, however, th at the formulation of field theory in terms of a local Lagrangian density leads directly to the local form of the conservation law, Eq. (2.12).
The easiest example of such a conservation law arises from a Lagrangian with only a kinetic term: $\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2$. The transformation $\phi\rightarrow \phi+\alpha$, where $\alpha$ is a constant, leaves $\mathcal{L}$ unchanged, so we conclude that the current $j^\mu=\partial^\mu\phi$ is conserved. As a less trivial example, consider the Lagrangian $$\begin{align}\mathcal{L}=|\partial_\mu\phi|^2-m^2|\phi|^2,\end{align}$$ where $\phi$ is now a complex-valued field. You can easily show that the equation of motion for this Lagrangian is again the Klein-Gordon equation, (7). This Lagrangian is invariant under the transforamtion $\phi\rightarrow e^{i\alpha}\phi$; for an infinitesimal transformation we have $$\begin{align}\alpha\Delta\phi=i\alpha\phi;\quad \alpha\Delta\phi^*=-i\alpha\phi^*.\end{align}$$ (We treat $\phi$ and $\phi^*$ as independent fields. Alternatively, we could work with the real and imaginary parts of $\phi$.) It is now a simple matter to show that the conserved Noether current is $$\begin{align}j^\mu=i[(\partial^\mu\phi^*)\phi-\phi^*(\partial^\mu\phi)].\end{align}$$ (The overall constant has been chosen arbitrarily.) You can check directly that the divergence of this current vanishes by using the Klein-Gordon equation. Later we will add terms to this Lagrangian that couple $\phi$ to an electromagnetic field. We will then interpret $j^\mu$ as the electromagnetic current density carried by the field, and the spatial integral of $j^0$ as its electric charge.
Noether's theorem can also be applied to spacetime tranformations such as translations and rotations. We can describe the infinitesimal translation $$x^\mu\rightarrow x^\mu-a^\mu$$ alternatively as a tansformation of the field configuration $$\phi(x)\rightarrow \phi(x+a)=\phi(x)+a^\mu\partial_\mu\phi(x).$$ The Lagrangian is also a scalar, so it must transfrom in the same way: $$\mathcal{L}\rightarrow \mathcal{L}+a^\mu\partial_\mu\mathcal{L}=\mathcal{L}+a^\nu\partial_\mu(\delta^\mu_\nu\mathcal{L}).$$ Comparing this equation to (20), we see that we now have a nonzero $\mathcal{L}^\mu$. Taking this into account, we can apply the theorem to obtain four separately conserved currents: $$\begin{align} T^\mu_\nu\equiv \frac{\partial\mathcal{L}}{\partial (\partial_\mu\phi)}\partial_\nu\phi-\mathcal{L}\delta^\mu_\nu.\end{align}$$ This is precisely the stress-energy tensor, also called the energy-momentum tensor, of the field $\phi$. The conserved charge associated with time translations is the Hamiltonian: $$\begin{align}H=\int T^{00}\, d^3x=\int \mathcal{H}\, d^3x.\end{align}$$ By computing this quantity for the Klein-Gordon field, one can recover the result (8). The conserved charge associated with spatial translations are $$\begin{align}P^i=\int T^{0i}\, d^3x=-\int \pi\partial_i\phi\, d^3x,\end{align}$$ and we naturally interpret this as the (physical) momentum carried by the field (not ot be confused with the canonical momentum).
2.3 The Klein-Gordon Field as Harmonic Oscillators
To quantizae the theory, we follow the same procedure as for any other dynamical system: We promote $\phi$ and $\pi$ to operators, and impose suitable commutation relations. Recall that for a discrete system of one or more particles the commutation relations are $$[q_i,p_i]=i\delta_{ij};\\ [q_i,q_j]=[p_i,p_j]=0.$$ For a continuous system the generalization is quite natural; since $\pi(\mathbf{x})$ is the momentum density, we get a Dirac delta function instead of a Kronecker delta: $$\begin{align}[\phi(\mathbf{x}),\pi(\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y});\nonumber\\ [\phi(\mathbf{x}),\phi(\mathbf{y})]=[\pi(\mathbf{x}),\pi(\mathbf{y})]=0.\end{align}$$ ( For now we work in the Schrodinger picture where $\phi$ and $\pi$ do not depend on time. When we switch to the Heisenberg pitcurre in the next section, these "equal time" commutation relations will still hold provided that both operators are considered at the same time.)
The Hamiltonian, being a functrion of $\phi$ and $\pi$, also becomes an opeartor. Our next task is to find the spectrum from the Hamiltonian. Since there is no obvious way to do this, let us seek guidance by writing the Klein-Gordon equation in Fourier space. If we expand the classical Klien-Gordon field as $$\phi(\mathbf{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\mathbf{p}\cdot \mathbf{x}}\phi(\mathbf{p},t)$$ (with $\phi^*(\mathbf{p})=\phi(-\mathbf{p})$ so that $\phi(\mathbf{x})$ is real), the Klein-Gordon equation (7) becomes $$\begin{align}\left[ \frac{\partial^2}{\partial t^2}+(|\mathbf{p}|^2+m^2)\right] \phi(\mathbf{p},t)=0.\end{align}$$ This is the same as the equation of motion for a simple harmonic oscillator with frequency $$\begin{align}\omega_\mathbf{p}=\sqrt{|\mathbf{p}|^2+m^2}.\end{align}$$
The simple harmonic oscillator is a system whose spectrum we already know how to find. Let us briefly recall how it is done. We write the Hamiltonian as $$H_{SHO}=\frac{1}{2}p^2+\frac{1}{2}\omega^2\phi^2.$$ To find the eigenvalues of $H_{SHO}$, we write $\phi$ and $p$ in terms of ladder operators: $$\begin{align}\phi=\frac{1}{\sqrt{2\omega}}(a+a^\dagger);\quad p=-i\sqrt{\frac{\omega}{2}}(a-a^\dagger).\end{align}$$ The canonical commutation relation $[\phi,p]=i$ is equivalent to $$\begin{align}[a,a^\dagger]=1.\end{align}$$ The Hamiltonian can now be rewritten $$H_{SHO}=\omega(a^\dagger a+\frac{1}{2}).$$ The state $|0\rangle$ such that $a|0\rangle=0$ is an eigenstate of $H$ with eigenvalue $\frac{1}{2}\omega$, the zero-point energy. Furthermore, the commutators $$[H_{SHO},a^\dagger]=\omega a^\dagger,\quad [H_{SHO},a]=-\omega a$$ make it easy to verify that the states $$|n\rangle \equiv (a^\dagger)^n|0\rangle$$ are eigenstates of $H_{SHO}$ with eigenvalues $(n+\frac{1}{2})\omega$. These states exhaust the spectrum.
We can find the spectrum of the Klein-Gordon Hamiltonian using the same trick, but now each Fourier mode of the field is treated as an independent oscillator with its own $a$ and $a^\dagger$. In analogy with (23) we write $$\begin{align}\phi(\mathbf{x})=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\mathbf{p}}}\left(a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}}+a^\dagger_\mathbf{p}e^{-i\mathbf{p}\cdot\mathbf{x}}\right);\\ \pi(\mathbf{x})= \int \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_\mathbf{p}}{2}}\left(a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}}-a^\dagger_\mathbf{p}e^{-i\mathbf{p}\cdot\mathbf{x}}\right).\end{align}$$ The inverse expressions for $a_\mathbf{p}$ and $a^\dagger_\mathbf{p}$ in terms of $\phi$ and $\pi$ are easy to derive but rarely needed. In the calculations below we will find it useful to rearrange (25) and (26) as follows: $$\begin{align}\phi(\mathbf{x})=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\mathbf{p}}}(a_\mathbf{p}+a^\dagger_{-\mathbf{p}})e^{i\mathbf{p}\cdot \mathbf{x}};\\ \pi(\mathbf{x})\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_\mathbf{p}}{2}}(a_\mathbf{p}-a^\dagger_{-\mathbf{p}})e^{i\mathbf{p}\cdot\mathbf{x}}.\end{align}$$ The commutation relation (24) becomes $$\begin{align}[a_\mathbf{p},a^\dagger_{\mathbf{p}'}]=(2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{p}'),\end{align}$$ from which you can verify that the commutator of $\phi$ and $\pi$ works out correctly: $$\begin{align}[\phi(\mathbf{x}),\pi(\mathbf{x}')]=\int \frac{d^3p\, d^3p'}{(2\pi)^6}\frac{-i}{2}\sqrt{\frac{\omega_{\mathbf{p}'}}{\omega_\mathbf{p}}}\left( [a^\dagger_{-\mathbf{p}},a_{\mathbf{p}'}]-[a_\mathbf{p},a^\dagger_{-\mathbf{p}'}]\right) e^{i(\mathbf{p}\cdot \mathbf{x}+\mathbf{p}'\cdot\mathbf{x}')}\nonumber\\ =i\delta^{(3)}(\mathbf{x}-\mathbf{x}').\end{align}$$
We are not ready to express the Hamiltonian in terms of ladder operators. Starting from its expression (8) in terms of $\phi$ and $\pi$, we have $$\begin{align}H=\int d^3x\int\frac{d^3p\, d^3p'}{(2\pi)^6}e^{i(\mathbf{p}+\mathbf{p}')\cdot\mathbf{x}}\left\{ -\frac{\sqrt{\omega_\mathbf{p}\omega_{\mathbf{p}'}}}{4}(a_\mathbf{p}-a^\dagger_{-\mathbf{p}})(a_{\mathbf{p}'}-a^\dagger_{-\mathbf{p}'})\nonumber\\ +\frac{-\mathbf{p}\cdot\mathbf{p}'+m^2}{4\sqrt{\omega_\mathbf{p}\omega_{\mathbf{p}'}}}(a_\mathbf{p}+a^\dagger_{-\mathbf{p}})(a_{\mathbf{p}'}+a^\dagger_{-\mathbf{p}'})\right\}\nonumber\\ =\int\frac{d^3p}{(2\pi)^3}\omega_\mathbf{p}\left( a^\dagger_\mathbf{p}a_\mathbf{p}+\frac{1}{2}[a_\mathbf{p},a^\dagger_\mathbf{p}]\right).\end{align}$$ The second term is proportional to $\delta(0)$, an infinite $c$-number. It is simple the sum over all modes of the zero-point energies $\omega_\mathbf{p}/2$, so its presence is completely expected, if somewhat disturbing. This infinite energy shift cannot be detected experimentally, since experiments measure only energy differences from the ground state of $H$. We will therefore ignore this infinite constant term in all of our calculations.
Using this expression for the Hamiltonian in t erms of $a_\mathbf{p}$ and $a^\dagger_\mathbf{p}$, it is easy to evaluate the commutators $$\begin{align}[H,a^\dagger_\mathbf{p}]=\omega_\mathbf{p}a^\dagger_\mathbf{p};\quad [H,a_\mathbf{p}]=-\omega_\mathbf{p}a_\mathbf{p}.\end{align}$$ We can now write down the spectrum of the theory, just as for the harmonic oscillator. The state $|0\rangle$ such that $a_\mathbf{p}|0\rangle=0$ for all $\mathbf{p}$ is ground state or vacuum, and has $E=0$ after the frop the infinite constant in (31). All other energy eigenstates can be built by acting on $|0\rangle$ with creation operators. in general, the state $a^\dagger_\mathbf{p}a^\dagger_\mathbf{q}\cdots|0\rangle$ is an eigenstate of $H$ with energy $\omega_\mathbf{p}+\omega_\mathbf{q}+\cdots$. These states exhaust the spectrum.
Having found the spectrum of the Hamiltonian, let us try to interpret its eigenstates. From (19) and a calculation similar to (31) we can write down the total momemtum operator, $$\begin{align}\mathbf{P}=-\int d^3x\ \pi(\mathbf{x})\nabla\phi(\mathbf{x})=\int \frac{d^3p}{(2\pi)^3}\mathbf{p}a^\dagger_\mathbf{p}a_\mathbf{p}.\end{align}$$ So the operator $a^\dagger_\mathbf{p}$ creates momentum $\mathbf{p}$ and energy $\omega_\mathbf{p}=\sqrt{|\mathbf{p}^2+m^2}$. Similarly, the state $a^\dagger_\mathbf{p}a^\dagger_\mathbf{q}\cdots|0\rangle$ has momentum $\mathbf{p}+\mathbf{q}+\cdots$. It is quite naturla to call these excitations particles, since they are discrete entities that have the proper relativistic energy-momentum relations. From now on we will refer to $\omega_\mathbf{p}$ as $E_\mathbf{p}$ (or simply $E$), since it really is the energy of a particle. Note, by the way, that the energy is always positive: $E_\mathbf{p}=+\sqrt{|\mathbf{p}|^2+m^2}$.
We naturally choose to normalize the vacuum state so that $\langle 0|0\rangle=1$. The one-particle states $|\mathbf{p}\rangle \propto a^\dagger_\mathbf{p}|0\rangle$ will also appear quite often, and it is worthwhile to adopt a convention for their normalization. The simplest normalization $\langle \mathbf{p}|\mathbf{q}\rangle=(2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q})$ is not Lorentz invariant, as we can demonstrate by considering the effect of a boost in the 3-direction. Under such a boost we have $p'_3=\gamma(p_3+\beta E)$, $E'=\gamma(E+\beta p_3)$. Using the delta function identity $$\begin{align}\delta(f(x)-f(x_0))=\frac{1}{|f'(x_0)|}\delta(x-x_0),\end{align}$$ we can compute $$\delta^{(3)}(\mathbf{p}-\mathbf{q})=\delta^{(3)}(\mathbf{p}'-\mathbf{q}')\cdot \frac{dp'_3}{dp_3} =\delta^{(3)}((\mathbf{p}'-\mathbf{q}')\gamma \left( 1+\beta\frac{dE}{dp_3}\right) =\delta^{(3)}(\mathbf{p}'-\mathbf{q}')\frac{\gamma}{E}(E+\beta p_3)=\delta^{(3)}(\mathbf{p}'-\mathbf{q}')\frac{E'}{E}.$$ The problem is that the volumes are not invariant under boosts; a box whose volume is $V$ in its rest frame has volume $V/\gamma$ in a boosted frame, due to Lorentz contraction. But from the above calculation, we see that the quantity $E_\mathbf{p}\delta^{(3)}(\mathbf{p}-\mathbf{q})$ is Lorentz invariant. We therefore define $$\begin{align}|\mathbf{p}\rangle=\sqrt{2E_\mathbf{p}}a^\dagger_\mathbf{p}|0\rangle,\end{align}$$ so that $$\begin{align}\langle \mathbf{p}|\mathbf{q}\rangle=2E_\mathbf{p}(2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}).\end{align}$$
On the Hilbert space of quantum states, a Lorentz transformation $\Lambda$ will be implemented as some unitary operator $U(\Lambda)$. Our normalization condition (35) then implies that $$\begin{align}U(\Lambda)|\mathbf{p}\rangle=|\Lambda\mathbf{p}\rangle.\end{align}$$ If we prefer to think of this transformation as acting on the operator $a^\dagger_\mathbf{p}$, we can also write $$\begin{align}U(\Lambda)a^\dagger_\mathbf{p}U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_\mathbf{p}}}a^\dagger_{\Lambda\mathbf{p}}.\end{align}$$
With this normalization we must divide by $2E_\mathbf{p}$ in other places. For example, the completeness relation for the one-particle states is $$\begin{align}(\mathbf{1})_{1-particle}=\int\frac{d^3p}{(2\pi)^3}|\mathbf{p}\rangle\frac{1}{2E_\mathbf{p}}\langle \mathbf{p}|,\end{align}$$ where the operator on the left is simply the identity within the subspace of one-particle states, and zero in the rest of the Hilbert space. Integrals of this form will occur quite often; in fact, the integral $$\begin{align}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}=\int\frac{d^4p}{(2\pi)^4}(2\pi)\left. \delta(p^2-m^2)\right|_{p^0>0}\end{align}$$ is a Lorentz-invariant 3-momentum integral, in the sense that if $f(p)$ is Lorentz-invariant, so is $\int d^3p\, f(p)/(2E_\mathbf{p})$. The integration can be thought of as being over the $p^0>0$ branch of the hyperboloid $p^2=m^2$ in 4-momentum space (see Fig. 2.2).
Finally let us consider the interpretation of the state $\phi(\mathbf{x})|0\rangle$. From the expansion (25) we see that $$\begin{align}\phi(\mathbf{x})|0\rangle=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}|\mathbf{p}\rangle\end{align}$$ is a linear superposition of single-particle states that have well-defined momentum. Except for the factor $1/2E_\mathbf{p}$, this is the same as the familiar nonrelativistic expression for the eigenstate of position $|\mathbf{x}\rangle$; in fact the extra factor is nearly constant for small interpretation, and claim that the operator $\phi(\mathbf{x})$, acting on the vacuum, creates a particle at position $\mathbf{x}$. This interpretation is further confirmed when we compute $$\begin{align}\langle 0|\phi(\mathbf{x})|\mathbf{p}\rangle=\langle 0|\int \frac{d^3p'}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}'}}}\left( a_{\mathbf{p}'}e^{i\mathbf{p}'\cdot\mathbf{x}}+a^\dagger_{\mathbf{p}'}e^{-i\mathbf{p}'\cdot\mathbf{x}}\right)\sqrt{2E_\mathbf{p}}a^\dagger_\mathbf{p}|0\rangle\end{align}$$ We can interpret this as the position-space representation of the single-particle wavefunction of the state $|\mathbf{p}\rangle$, just at in nonrelativistic quantum mechanics $\langle \mathbf{x}|\mathbf{p}\rangle\propto e^{i\mathbf{p}\cdot\mathbf{x}}$ is the wavefunction of the state $|\mathbf{p}\rangle$.
2.4 The Klein-Gordon Field in Space-Time
In the Heisenberg pricture, we make the oeprators $\phi$ and $\pi$ time-dependent in the useal way: $$\begin{align}\phi(x)=\phi(\mathbf{x},t)=e^{iHt}\phi(\mathbf{x})e^{-iHt},\end{align}$$ and similarly for $\pi(x)=\pi(\mathbf{x},t)$. The Heisenberg equation of motion, $$\begin{align}i\frac{\partial}{\partial t}\mathcal{O}=[\mathcal{O},H],\end{align}$$ allows us to compute the time dependence of $\phi$ and $\pi$: $$i\frac{\partial}{\partial t}\phi(\mathbf{x},t)=\left[ \phi(\mathbf{x},t),\int d^3x'\left\{ \frac{1}{2}\pi^2(\mathbf{x}',t)+\frac{1}{2}(\nabla\phi(\mathbf{x}',t))^2+\frac{1}{2}m^2\phi^2(\mathbf{x}',t)\right\}\right]\\ =\int d^3x'\left( i\delta^{(3)}(\mathbf{x}-\mathbf{x}')\pi(\mathbf{x}',t)\right)\\ =i\pi(\mathbf{x},t);$$ $$i\frac{\partial}{\partial t}\pi(\mathbf{x},t)=\left[ \pi(\mathbf{x},t),\int d^3x'\left\{ \frac{1}{2}\pi^2(\mathbf{x}',t)+\frac{1}{2}\phi(\mathbf{x}',t)(-\nabla^2+m^2)\phi(\mathbf{x}',t)\right\}\right]\\ =\int d^3x'\left( -i\delta^{(3)}(\mathbf{x}-\mathbf{x}')(-\nabla^2+m^2)\phi(\mathbf{x}',t)\right) \\ =-i(-\nabla^2+m^2)\phi(\mathbf{x},t).$$ Combining the two results gives $$\begin{align} \frac{\partial^2}{\partial t^2}\phi=(\nabla^2-m^2)\phi,\end{align}$$ which is just the Klein-Gordon equation.
We can better undertand the time dependence of $\phi(x)$ and $\pi(x)$ by writing them in terms of creation and annihilation operators. First note that $$Ha_\mathbf{p}=a_\mathbf{p}(H-E_\mathbf{p}),$$ and hence $$H^na_\mathbf{p}=a_\mathbf{p}(H-E_\mathbf{p})^n,$$ for any $n$. A similar relation (with $-$ replaced by $+$) holds for $a^\dagger_\mathbf{p}$. Thus we have derived the identities $$\begin{align}e^{iHt}a_\mathbf{p}e^{-iHt}=a_\mathbf{p}e^{-iE_\mathbf{p}t},\quad e^{iHt}a^\dagger_\mathbf{p}e^{-iHt}=a^\dagger_\mathbf{p}e^{iE_\mathbf{p}t},\end{align}$$ which we can use on expression (25) for $\phi(\mathbf{x})$ to find the desired expression for the Heisenberg operator $\phi(x)$, according to (43). The result if $$\begin{align}\phi(\mathbf{x},t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}\left. \left( a_\mathbf{p}e^{-ip\cdot x}+a^\dagger_\mathbf{p}e^{ip\cdot x}\right)\right|_{p^0=E_\mathbf{p}};\nonumber\\ \pi(\mathbf{x},t)=\frac{\partial}{\partial t}\phi(\mathbf{x},t).\end{align}$$
It is worth mentioning that we can perform the same manipulations with $\mathbf{P}$ instead of $H$ to relate $\phi(\mathbf{x})$ to $\phi(0)$. In analogy with (46), one can show $$\begin{align}e^{-i\mathbf{P}\cdot \mathbf{x}}a_\mathbf{p}e^{i\mathbf{P}\cdot \mathbf{x}}=a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}},\quad e^{-i\mathbf{P}\cdot\mathbf{x}}a^\dagger_\mathbf{p}e^{i\mathbf{P}\cdot\mathbf{x}}=a^\dagger_\mathbf{p}e^{-i\mathbf{p}\cdot\mathbf{x}},\end{align}$$ and therefore $$\begin{align}\phi(x)=e^{i(Ht-\mathbf{P}\cdot\mathbf{x})}\phi(0)e^{-i(Ht-\mathbf{P}\cdot\mathbf{x})}=e^{iP\cdot x}\phi(0)e^{-iP\cdot x},\end{align}$$ where $P^\mu=(H,\mathbf{P})$.
Equation (47) makes explicit the dual particle and wave interpretations of the quautum field $\phi(x)$. On the one hand, $\phi(x)$ is written as a Hilbert space operator, which creates and destroys the aprticles that rae the quanta of field excitation. On the other hand, $\phi(x)$ is written as a linear combination of solutions ($e^{ip\cdot x}$ and $e^{-ip\cdot x}$) of the Klein-Gordon equation. Both signs of the time dependence in the xponential appear: We find both $e^{-ip^0t}$ and $e^{+ip^0t}$, although $p^0$ is always positive. If these were single-particle wavefunctions, they would correspond to states to positive and negative energy; let us refer to them more generally as positive- and negative-frequanecy modes. The connection between the particel creation operators and the waveforms displayed here is always valid for free quantum fields: A positive-frequency solution of the field equation has as its coefficient the operator that destroys a particle in that single-particle wavefunction. A negative-frequency solution of the field equation, being the Hermitian conjugate of a positive-frequency solution, has as its coefficient the operator that creates a particle in that positive-energy single-particle wavefunction. In this way, the fact that relativistic wave equations have both positive- and negative-frequency solutions is reconciled with the requirement that a sensible quantum theory contain only positive excitation energies.
Causality
In our present formalism, still working in the Heisenberg picture, the amplitude for a particle to propagate from $y$ to $x$ is $\langle 0|\phi(x)\phi(y)|0\rangle$. We will call this quantity $D(x-y)$. Each operator $\phi$ is a sum of $a$ and $a^\dagger$ operators, but only the term $\langle0|a_\mathbf{p}a^\dagger_\mathbf{q}|0\rangle=(2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q})$ survives in this expression. It is easy to check that we are left with $$\begin{align}D(x-y)=\langle 0|\phi(x)\phi(y)|0\rangle=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}e^{-ip\cdot (x-y)}.\end{align}$$ We have already argued in (40) that integrals of this form are Lorentz invariant. Let us now evaluate this integral for some particular values of $x-y$.
First consider the case where the difference $x-y$ is purely in the time-direction: $x^0-y^0=t$, $\mathbf{x}-\mathbf{y}=0$. Then we have $$\begin{align}D(x-y)=\frac{4\pi}{(2\pi)^3}\int^\infty_0dp\frac{p^2}{2\sqrt{p^2+m^2}}e^{-\sqrt{p^2+m^2}t}\nonumber\\ =\frac{1}{4\pi^2}\int^\infty_mdE\sqrt{E^2-m^2}e^{-iEt}\sim_{t\rightarrow \infty}e^{-imt}.\end{align}$$
Next consider the case where $x-y$ is purely spatial: $x^0-y^0=0$, $\mathbf{x}-\mathbf{y}=\mathbf{r}$. The amplitude is then $$D(x-y)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{r}}\\ =\frac{2\pi}{(2\pi)^3}\int^\infty_0dp\frac{p^2}{2E_\mathbf{p}}\frac{e^{ipr}-e^{-ipr}}{ipr}\nonumber\\ =\frac{-i}{2(2\pi)^2r}\int^\infty_{-\infty}dp\frac{pe^{ipr}}{\sqrt{p^2+m^2}}.$$ The integrand, considered as a complex function of $p$, has branch cuts on the imaginary axis starting at $\pm im$ (see Fig. 2.3). To evaluate the integral we push the contour up to warp around the upper branch cut. Defining $\rho=-ip$, we obtain $$\begin{align}\frac{1}{4\pi^2r}\int^\infty_mdp\frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}\sim_{r\rightarrow \infty}e^{-mr}.\end{align}$$ So again we find that outside the light-cone, the propagation amplitude is exponentially vanishing but nonzero.
To really discuss causality, however, we should ask not whether particles can propagate over spacelike intervals, but whether a measurement performed at one point can affect a measurement at another point whose separation from the first is spacelike. The simplest thing we could try to measure is the field $\phi(x)$, so we should compute the commutator $[\phi(x),\phi(y)]$; if this commutator vanishes, one measurment cannot affect the other. In fact, if the commuator vanishes for $(x-y)^2<0$, causality is preserved quite generally, since commutators involving any function of $\phi(x)$, including $\pi(x)=\partial\phi/\partial t$, would also have to vanish. Of course we know from Eq. (20) that the commutator vanishes for $x^0=y^0$; now let's do the more general computation: $$\begin{align}[\phi(x),\phi(y)]=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}} \int \frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{q}}}\nonumber\\ \times \left[ (a_\mathbf{p}e^{-ip\cdot x}+a^\dagger_\mathbf{p}e^{ip\cdot x}),(a_\mathbf{p}e^{-ip\cdot x}+a^\dagger_\mathbf{p}e^{ip\cdot x})\right]\nonumber\\ =\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}(e^{-ip\cdot (x-y)}-e^{ip\cdot (x-y)})\nonumber\\ =D(x-y)-D(y-x).\end{align}$$ When $(x-y)^2<0$, we can perform a Lorentz tarnsformation on the second term, taking $(x-y)\rightarrow -(x-y)$, as shown in Fig. 2.4. The two terms are therefore equal and cancel to give zero; causality is preserved. Note that if $(x-y)^2>0$ there is no continuous Lorentz transformation that takes $(x-y)\rightarrow -(x-y)$. In this case, by Eq. (51), the amplitude is nonzero, roughly $(e^{-imt}-e^{imt})$ for the special case $\mathbf{x}-\mathbf{y}=0$. Thus we conclude that no measurement in the Klein-Gordon theory can affect another measurement outside the light-cone.
The Klein-Gordon Propagator
Let us study the commutator $[\phi(x),\phi(y)]$ a little further. Since it is a $c$-number, we can write $[\phi(x),\phi(y)]=\langle 0|[\phi(x),\phi(y)]|0\rangle$. This can be rewritten as a four-dimensional integral as follows, assuming for now that $x^0>y^0$: $$\begin{align}\langle0|[\phi(x),\phi(y)]|0\rangle=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}(e^{-ip\cdot (x-y)}-e^{ip\cdot (x-y)})\nonumber\\ =\int \frac{d^3p}{(2\pi)^3}\left\{ \frac{1}{2E_\mathbf{p}}\left. e^{-ip\cdot (x-y)}\right|_{p^0=E_\mathbf{p}}+\frac{1}{-2E_\mathbf{p}}\left. e^{-ip\cdot (x-y)}\right|_{p^0=-E_\mathbf{p}}\right\}\nonumber\\ =_{x^0>y^0}\int \frac{d^3p}{(2\pi)^3}\int \frac{dp^0}{2\pi i}\frac{-1}{p^2-m^2}e^{-ip\cdot (x-y)}.\end{align}$$ In the last step the $p^0$ integral is to be performed along the following contour:
For $x^0>y^0$ we can close the contour below, picking up both poles to obtain the previous line of (54). For $x^0<y^0$ we may close the contour above, giving zero. Thus the last line of (54), together with the prescription for going around the poles, is an expression for what we will call $$\begin{align}D_R(x-y)\equiv \theta(x^0-y^0)\langle 0|[\phi(x),\phi(y)]|0\rangle.\end{align}$$
To understand this quantity better, let's do another computation: $$\begin{align}(\partial^2+m^2)D_R(x-y)=(\partial^2\theta(x^0-y^0))\langle 0|[\phi(x),\phi(y)]|0\rangle\nonumber\\ +2(\partial_\mu \theta(x^0-y^0))(\partial^\mu \langle 0|[\phi(x),\phi(y)]|0\rangle )+\theta(x^0-y^0)(\partial^2+m^2)\langle 0[\phi(x),\phi(y)]|0\rangle\nonumber\\ =-\delta(x^0-y^0)\langle 0|[\pi(x),\phi(y)]|0\rangle +2\delta(x^0-y^0)\langle 0|[\pi(x),\phi(y)]|0\rangle+0\nonumber\\ =-i\delta^{(4)}(x-y).\end{align}$$ This says that $D_R(x-y)$ is a Green's function of the Klein-Gordon operator. Since it vanishes for $x^0<y^0$, it is retarded Green's function.
If we had not already derived expression (54), we could find it by Fourier transformation. Writing $$\begin{align}D_R(x-y)=\int \frac{d^4p}{(2\pi)^4}e^{-ip\cdot (x-y)}\tilde{D}_R(p),\end{align}$$ we obtain an algebraic expression for $\tilde{D}_R(p)$: $$(-p^2+m^2)\tilde{D}_R(p)=-i.$$ Thus we immediately arrive at the result $$\begin{align}D_R(x-y)=\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{-ip\cdot (x-y)}.\end{align}$$
The $p^0$-integral of (58) can be evaluated accoring to four different contours, of which that used in (54) is only one. In Chapter 4 we will find that a different pole prescription,
is extremely useful; it is called the Feynman prescription. A convenient way to remember it is to write $$\begin{align}D_F(x-y)\equiv \int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}e^{-ip\cdot (x-y)},\end{align}$$ since the poles are then at $p^0=\pm (E_\mathbf{p}-i\epsilon)$, displaced properly above and below the real axis. When $x^0>y^0$ we can perform the $p^0$ integral by closing the contour below, obtaining exactly the propagation amplitude $D(x-y)$ (50). When $x^0<y^0$ we close the contour above, obtaining the same expression but with $x$ and $y$ interchanged. Thus we have $$\begin{align}D_F(x-y)=\begin{cases}D(x-y)&\mbox{for } x^0>y^0\\ D(y-x)\quad \mbox{for } x^0<y^0\end{cases}\nonumber\\ =\theta(x^0-y^0)\langle 0|\phi(x)\phi(y)|0\rangle +\theta(y^0-x^0)\langle 0|\phi(y)\phi(x)|0\rangle\nonumber\\ \equiv \langle 0|T\phi(x)\phi(y)|0\rangle.\end{align}$$
Particle Creation by a Classical Source
There is one type of interaction, however, that we are already equipped to handle. Consider a Klein-Gordon field coupled to an external, classical source field $j(x)$. That is, consider the field equation $$\begin{align}(\partial^2+m^2)\phi(x)=j(x),\end{align}$$ where $j(x)$ is some fixed, known function of space and time is nonzero only for a finite time interval. If we start in the vacuum state, what will be find after $j(x)$ has been truned on and off again?
The field equation (61) follows from the Lagrangian $$\begin{align}\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2+j(x)\phi(x).\end{align}$$ But it $j(x)$ is turned on for only a finite time, it is easiest to solve the problem using the field equation directly. Before $j(x)$ is turned on, $\phi(x)$ has the form $$\phi_0(x)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}(a_\mathbf{p}e^{-ip\cdot x}+a^\dagger_\mathbf{p}e^{ip\cdot x}).$$ If there were no source, this would be the solution for all time. With a source, the solution of the equation of motion can be constructed using the retarded Green's function: $$\begin{align}\phi(x)=\phi_0(x)+i\int d^4y\, D_R(x-y)j(y)\nonumber\\ =\phi_0(x)+i\int d^4y\, \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}\theta(x^0-y^0)(e^{-ip\cdot (x-y)}-e^{ip\cdot (x-y)})j(y).\end{align}$$ If we wait until all of $j$ is in the past, the theta function equals 1 in the whole domain of integration. Then $\phi(x)$ involves only the Fourier transform of $j$, $$\tilde{j}(p)=\int d^4y\, e^{ip\cdot y}j(y),$$ evaluated at 4-momenta $p$ such that $p^2=m^2$. It is natural to group the positive-frequency terms together with $a_\mathbf{p}$ and the negative-frequency terms with $a^\dagger_\mathbf{p}$; this yields the expression $$\begin{align}\phi(x)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}\left\{ \left( a_\mathbf{p}+\frac{i}{\sqrt{2E_\mathbf{p}}}\tilde{j}(p)\right)e^{-ip\cdot x}+\mbox{h.c.}\right\}.\end{align}$$
You can now guess the form of the Hamiltonian after $j(x)$ has acted: Just replace $a_\mathbf{p}$ with ($a_\mathbf{p}+i\tilde{j}(p)/\sqrt{2E_\mathbf{p}}$) to obtain $$H=\int \frac{d^3p}{(2\pi)^3}E_\mathbf{p}\left( a^\dagger_\mathbf{p}-\frac{i}{\sqrt{2E_\mathbf{p}}}\tilde{j}^*(p)\right) \left( a_\mathbf{p}+\frac{i}{\sqrt{2E_\mathbf{p}}}\tilde{j}(p)\right).$$ The energy of the system after the source has been turned off is $$\begin{align}\langle 0|H|0\rangle =\int \frac{d^3p}{(2\pi)^3}\frac{1}{2}|\tilde{j}(p)|^2,\end{align}$$ where $|0\rangle$ still denotes the ground state of the free theory. We can interpret these results in terms of particles by identifying $|\tilde{j}(p)|^2/2E_\mathbf{p}$ as the probability density for creating a particle in the mode $p$. Then the total number of particles produced is $$\begin{align}\int dN=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}|\tilde{j}(p)|^2.\end{align}$$ Only those Fourier components of $j(x)$ that are in resonance with on-mass-shell (i.e., $p^2=m^2$) Klein-Gordon waves are effective at creating particles.
Problems
2.1
Classical electromagnetism (with no sources) follows from the action $$S=\int d^4x\left( -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right),\quad \mbox{where } F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$
(a) Derive Maxwell's equations as the Euler-Lagrange equations of this action, treating the components $A_\mu(x)$ as the dynamical variables. Write the equations in standard from by identifying $E^i=-F^{0i}$ and $\epsilon^{ijk}B^k=-F^{ij}$.
(b) Construct the energy-momentum tensor for this theory. Note that the usual procedure does not result in a symmetric tensor. To remedy that, we can add to $T^{\mu\nu}$ a term of the form $\partial_\lambda K^{\lambda\mu\nu}$, where $K^{\lambda\mu\nu}$ is antisymmetric in its first two indices. Such an object is automatically divergenceless, so $$\hat{T}^{\mu\nu}=T^{\mu\nu}+\partial_\lambda K^{\lambda\mu\nu}$$ is an equally good energy-momentum tensor with the same globally conserved energy and momentum. Show that this construction, with $$K^{\lambda\mu\nu}=F^{\mu\lambda}A^\nu,$$ leads to an energy-momentum tensor $\hat{T}$ that is symmetric and yields the standard formulae for the electromagnetic energy and momentum densities $$\mathcal{E}=\frac{1}{2}(E^2+B^2);\quad \mathbf{S}=\mathbf{E}\times \mathbf{B}.$$
Sol)
(a) The Lagrangian density can be written as $$\mathcal{L} = \frac{1}{2} \left(\partial_\mu A_\nu\right)^2 - \frac{1}{2}\partial_\mu A_\nu \partial^\nu A^\mu.$$ Substitute into Eular-Lagrange equation $$\frac{\partial\mathcal{L}}{\partial A_\mu} = \partial_\nu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\nu A_\mu)}\right),$$ the LHS is zero, while the derivative in the bracket of RHS becomes $$\begin{align*}\frac{\partial\mathcal{L}}{\partial(\partial_\nu A_\mu)} & = -\frac{1}{2} F^{\alpha\beta} \left(\delta^\nu_\alpha \delta^\mu_\beta - \delta^\nu_\beta \delta^\mu_\alpha\right)\\ & = -F^{\nu\mu}.\end{align*}$$ So the equation of motion is $$\partial_\nu F^{\nu\mu} = 0.$$ For $\mu = 0$, it yields: $$\partial_\nu F^{\nu 0} = \partial_i E^i = 0,$$ which is nothing but $\nabla\cdot\vec{E} = 0$. For $\mu = i$, it becomes $$\begin{align*}0 = \partial_\nu F^{\nu i} & = \partial_0 F^{0i} + \partial_j F^{ji}\\ & = -\frac{\partial E^i}{\partial t} + \epsilon_{ijk}\partial_j B^k.\end{align*}$$ Written in vector form, this is just $$\nabla\times\vec{B} = \frac{\partial \vec{E}}{\partial t}.$$
(b) Consider an infinitesimal translation $x^\mu \to x^\mu + \epsilon^\mu$, the Neother current is $$T^{\mu\nu} = \frac{\partial\mathcal{L}}{(\partial_\mu A_\lambda)} \partial^\nu A_\lambda - \mathcal{L} g^{\mu\nu}.$$ Substitute lagrangian into above, we get $$T^{\mu\nu} = -F^{\mu\lambda} \partial^\nu A_\lambda + \frac{1}{4} (F_{\alpha\beta})^2 g^{\mu\nu}$$ so that $$\begin{align*} \widehat{T}^{\mu\nu} & = T^{\mu\nu} + \partial_\lambda F^{\mu\lambda} A^\nu + F^{\mu\lambda} \partial_\lambda A^\nu \\ & = F^{\mu\lambda} F_\lambda{}^\nu + \frac{1}{4}(F_{\alpha\beta})^2 g^{\mu\nu}, \end{align*}$$ where we have used maxwell eq. in the second step. Now the energy and momentum density can be read as $$\mathcal{E} = \widehat{T}^{00} = F^{0i}F_i{}^0 + \frac{1}{2}(B^2 - E^2) = \frac{1}{2} (B^2 + E^2),$$ $$S^i = \widehat{T}^{i0} = F^{0j}F_j{}^i = E^j B^k \epsilon_{jki}.$$ That is $$\vec{S} = \vec{E} \times \vec{B}.$$
2.2 The complex scalar field.
Consider the field theory of a complex-valued scalar field obeying the Klein-Gordon equation. The action of this theory is $$S=\int d^4x(\partial_\mu\phi^*\partial^\mu\phi-m^2\phi^*\phi).$$ It is easiest to analyze this theory by considering $\phi(x)$ and $\phi^*(x)$, rather than the real and imaginary parts of $\phi(x)$, as the basic dynamical variables.
(a) Find the conjugate momenta to $\phi(x)$ and $\phi^*(x)$ and the canonical commutation relations. Show that the Hamiltonian is $$H=\int d^3x(\pi^*\pi+\nabla\phi^*\cdot\nabla\phi+m^2\phi^*\phi).$$ Compute the Heisenberg equation of motion for $\phi(x)$ and show that it is indeed the Klein-Gordon equation.
(b) Diagonalize $H$ by introducing creation and annihilation operators. Show that the theory contains two sets of particles of mass $m$.
(c) Rewrite the conserved charge $$Q=\int d^3x\frac{i}{2}(\phi^*\pi^*-\pi\phi)$$ in terms of creation and annihilation operators, and evaluate the charge of the particle of each type.
(d) Consider the case of two complex Klein-Gordon fields with the same mass. Label the fields as $\phi_a(x)$, where $a=1,2$. Show that there are now four conserved charges, one given by the generalization of part (c), and the other three given by $$Q^i=\int d^3x\frac{i}{2}(\phi^*_a(\sigma^i)_{ab}\pi^*_b-\pi_a(\sigma^i)_{ab}\phi_b),$$ where $\sigma^i$ are the Pauli sigma matrices. Show that these charges have the commutation relations of angular momentum ($SU(2)$). Generalize these results to the case of $n$ identical complex scalar fields.
Sol)
(a) The conjugate momenta are $$\begin{align} \pi & = \frac{\partial\mathcal{L}}{\partial\dot{\phi}} = \dot{\phi}^\ast, \\ \pi^\ast & = \frac{\partial\mathcal{L}}{\partial\dot{\phi}^\ast} = \dot{\phi}. \end{align}$$ $\pi(x)$ and $\phi(x)$ should satisfy the canonical commutation relations: $$[\phi(t, \vec{x}), \phi(t, \vec{y})] = [\phi(t, \vec{x}), \phi^\ast(t, \vec{y})] = 0;$$ $$[\pi(t, \vec{x}), \pi(t, \vec{y})] = [\pi(t, \vec{x}), \pi^\ast(t, \vec{y})] = 0;$$ $$[\pi(t, \vec{x}), \phi^\ast(t, \vec{y})] = [\pi^\ast(t, \vec{x}), \phi(t, \vec{y})] = 0;$$ $$[\phi(t, \vec{x}), \pi(t, \vec{y})] = [\phi^\ast(t, \vec{x}), \pi^\ast(t, \vec{y})] = i\delta^3(\vec{x} - \vec{y}).$$ The Hamiltonian is $$\begin{align} H & = \int d^3x (\pi\dot{\phi} + \pi^\ast\dot{\phi}^*) - L \nonumber\\ & = \int d^3x \left(\pi^*\pi + \nabla\phi^* \cdot \nabla\phi + m^2 \phi^*\phi\right). \end{align}$$ Heisenburg equation for $\phi(x)$ is simply $$i\dot{\phi} = [\phi, H] = i\pi^*.$$ While for $\pi(x)$ is $$ \begin{align} i\dot{\pi}(x) = [\pi(x), H(t)] & = \left[\pi(x), \int d^3y (\partial_i \phi^*(y) \partial_i\phi(y) + m^2 \phi^*(y)\phi(y))\right]\nonumber\\ & = \left[\pi(x), \int d^3y \phi(y)(-\nabla^2 + m^2)\phi^\ast(y)\right]\nonumber\\ & = i(\nabla^2 - m^2) \phi^*(x), \end{align}$$ where $x^0 = y^0 = t$, and we have integrated by parts at second step. Combining (69) and (70) we obtain the equation of motion $$\ddot{\phi} = \dot{\pi}^\ast = (\nabla^2 - m^2)\phi,$$ which is nothing but the Klein-Gordon equation $(\Box + m^2)\phi = 0$.
(b) $\phi(x)$ can be expanded in terms of plain waves as $$\phi(x) = \sum_k \left(a_k \varphi_k + b^\dagger_k \varphi^*_k \right),$$ where $a_k$ and $b_k$ are annihilation operators, and the plain wave function $\varphi_k(x)$ is $$\varphi_k(t, \vec{x}) = \frac{1}{\sqrt{2V\omega_k}}e^{-i\omega_kt + i\vec{k}\cdot\vec{x}},$$ where $\omega_k = \sqrt{k^2 + m^2}$. Define inner product $\langle\cdot,\cdot\rangle$ as follow $$\langle f, g \rangle \equiv i \int d^3x (f^*\dot{g} - \dot{f}^* g).$$ So the inner products of the plain wave functions are $$\begin{align*} \langle \varphi_k, \varphi_{k'} \rangle & = \delta_{kk'};\\ \langle \varphi^*_k, \varphi^*_{k'} \rangle & = -\delta_{kk'};\\ \langle \varphi_k, \varphi^*_{k'} \rangle & = 0. \end{align*}$$ Notice that the second slot of the inner product is linear, while the first slot is antilinear. Then the operators $a_k$ and $b_k$ can be expressed as $$\begin{align} a_k & = \langle \varphi_k, \phi \rangle = \int d^3x \, \varphi^*_k \left(\omega_k\phi + i\pi^*\right);\\ b_k & = -\langle \varphi^*_k, \phi \rangle^\dagger = \int d^3x \, \varphi^*_k \left(\omega_k\phi^* + i\pi\right). \end{align}$$ Using CCR, we get the commutation relations for operators $a_k$ and $b_k$: $$[a_k,a_{k'}] = [b_k, b_{k'}] = 0;$$ $$[a_k,b_{k'}] = [a_k, b_{k'}^\dagger] = 0;$$ $$[a_k, a_{k'}^\dagger] = [b_k, b_{k'}^\dagger] = \delta_{kk'}.$$ We get the Hamiltonian written in terms of creation and annihilation operators $$\begin{align} H & = \sum_{k, k'} \int d^4x \, \left[ \left(\omega_k\omega_{k'} + \vec{k}\cdot\vec{k}'\right) \left(a_k^\dagger\varphi_k^* - a_k \varphi_k\right) \left(a_{k'}\varphi_{k'} - a_{k'}^\dagger\varphi_{k'}^*\right)\right.\nonumber\\ & \qquad \left. + m^2 \left(a_k^\dagger\varphi_k^* + a_k\varphi_k\right) \left(a_{k'}\varphi_{k'} + a_{k'}^\dagger\varphi_{k'}^*\right) \right]\\ & = \sum_{k, k'}\left(\omega_k\omega_{k'} + \vec{k}\cdot\vec{k}' + m^2\right) \int d^3x \, \left(a^\dagger_k a_{k'} \varphi^*_k \varphi_{k'} + b_k b^\dagger_{k'} \varphi_k \varphi^*_{k'}\right)\nonumber\\ & = \sum_k \omega_k \left(a^\dagger_k a_k + b^\dagger_k b_k + 1\right). \end{align}$$ So this theory contains two sets of particles with mass \textit{m}, created by $a^\dagger_k$ and $b^\dagger_k$, respectively. Notice that the energy of ground state is infinity, but it's irrelavent, since we only concern energy differences.
(c) We get $$\begin{align} Q & = \frac{i}{2} \left[\sum_{k, k'} (-i\omega_{k'}) \int d^3x \, \left(a^\dagger_k\varphi^\ast_k + b_k\varphi_k\right) \left(a_{k'} \varphi_{k'} - b^\dagger_{k'}\varphi^*_{k'}\right) - h.c.\right]\nonumber\\ & = \frac{i}{2} \left[-\frac{i}{2} \sum_k \left(a^\dagger_k a_k - b_k b^\dagger_k\right) - h.c.\right]\nonumber\\ & = \frac{1}{2} \sum_k \left(a^\dagger_k a_k - b^\dagger_k b_k\right),\end{align}$$ where we have ignored an infinite constant in the last step. So the two sets of particles created by $a^\dagger_k$ and $b^\dagger_k$ have charges $\frac{1}{2}$ and $-\frac{1}{2}$, respectively.
(d) We first consider the general case of \textit{n} fields, the case of two fields can be obtained by setting $n = 2$. Let the \textit{n} fields to be $\phi_a(x)$, where $a = 1, 2, \cdots, n$, define $$\phi = \begin{pmatrix} \phi_1\\ \phi_2\\ \vdots\\ \phi_n \end{pmatrix}, \qquad \pi = \left(\pi_1, \pi_2, \cdots, \pi_n\right).$$ The Lagrangian can be written as $$\mathcal{L} = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi.$$ It is invariant under global transformation $$\phi \to A \phi,$$ where $A \in U(n)$ is an $n \times n$ matrix. Since $U(n) = SU(n) \otimes U(1)$, the infinitesimal transformation can be expressed as $$\phi \to (I - i\alpha - i t^i \theta^i)\phi,$$ where $\alpha$ and $\theta^i$ are infinitesimal parameters, correspond to $U(1)$ and $SU(n)$ respectively, matrices $t^i$ are generators of $SU(n)$, $i = 1, 2, \cdots, (n^2 - 1)$. By Neother's theorem, $\alpha$ yields one conserved charge: $$\begin{align*} q & = \frac{1}{\alpha}\int d^3x \, \left(\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)}(-i\alpha\phi) + (i\alpha\phi^\dagger)\frac{\partial\mathcal{L}}{\partial(\partial_0\phi^\dagger)}\right)\\ & = i\int d^3x \, \left(\phi^\dagger \pi^\dagger - \pi\phi\right). \end{align*}$$ The $SU(n)$ transformation, on the other hand, yields $n^2 - 1$ charges: $$\begin{align*} Q^i & = \frac{1}{\theta^i}\int d^3x \, \left(\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)} (-it^i\phi) + (it^i \phi^\dagger)\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)} \right)\\ & = i\int d^3x \, \left(\phi^\dagger t^i \pi^\dagger - \pi t^i \phi\right). \end{align*}$$ There are totally $n^2$ charges. The commutator between them is $$\begin{align*} [Q^i, Q^j] & = -\int d^3x\int d^3y (t^i)_{ab}(t^j)_{cd} \left[ \phi^*_a(x)\pi^*_b(x) - \pi_a(x)\phi_b(x), \phi^*_c(y)\pi^*_d(y) - \pi_c(y)\phi_d(y) \right]\\ & = -\int d^3x\int d^3y (t^i)_{ab}(t^j)_{cd}\big( \left[\phi^*_a(x)\pi^*_b(x), \phi^*_c(y)\pi^*_d(y)\right]\\ & \qquad + \left[\pi_a(x)\phi_b(x), \pi_c(y)\phi_d(y)\right] \big)\\ & = -i\int d^3x (t^i)_{ab}(t^j)_{cd}\big( \phi^*_c\pi^*_b\delta_{ad} - \phi^*_a\pi^*_d\delta_{bc} + \pi_a\phi_d\delta_{bc} - \pi_c\phi_b\delta_{ad} \big)\\ & = i\int d^3x \left(\phi^\dagger[t^i, t^j]\pi^\dagger - \pi[t^i, t^j]\phi\right)\\ & = -f^{ijk} \int d^3x \left(\phi^\dagger t^k\pi^\dagger - \pi t^k\phi\right)\\ & = if^{ijk}Q^k, \end{align*}$$ where $f^{ijk}$ are the structure constants, i.e. $[t^i, t^j] = if^{ijk}t^k$. In the case $n = 2$, there are 4 charges, and $t^i = \sigma^i / 2$.
2.3
Evaluate the function $$\langle 0|\phi(x)\phi(y)|0\rangle=D(x-y)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}e^{-ip\cdot (x-y)},$$ for $(x-y)$ spacelike so that $(x-y)^2=-r^2$, explicitly in terms of Bessel functions.
Sol)
$$\begin{align*} D(x - y) & = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}} e^{i\vec{p}\cdot(\vec{x} - \vec{y})}\\ & = \frac{1}{4\pi^2} \int_0^{+\infty} \frac{p^2}{2E_{\mathbf{p}}} d p \int_{-1}^1 d (\cos\theta) e^{ipr\cos\theta}\\ & = \frac{1}{8\pi^2ir} \int_0^{+\infty} \frac{p}{E_{\mathbf{p}}} \left(e^{ipr} - e^{-ipr}\right)\\ & = \frac{1}{8\pi^2ir} \int_{-\infty}^{+\infty} \frac{p}{\sqrt{p^2 + m^2}} e^{ipr} d p\\ & = \frac{1}{8\pi^2ir} \left[\int_{+im}^{+i\infty} \frac{p}{i\sqrt{-p^2 - m^2}} e^{ipr} d p + \int_{+i\infty}^{im} \frac{p}{-i\sqrt{-p^2 - m^2}} e^{ipr} d p \right]\\ & = \frac{1}{4\pi^2r} \int_m^{+\infty} \frac{x}{\sqrt{x^2 - m^2}} e^{-rx} d x\\ & = \frac{1}{4\pi^2r} \int_0^{+\infty} e^{-r\sqrt{p^2 + m^2}} d p\\& = \frac{m}{4\pi^2r} \int_0^{+\infty} e^{-mr\cosh t} \cosh t d t\\ & = \frac{m}{4\pi^2r} K_1(mr),\end{align*}$$ $$K_\alpha(x) = \int_0^{+\infty} e^{-x\cosh t} \cosh{\alpha t} d t, \quad \mathrm{Re}(x) > 0$$ is the modified Bessel function.
Reference
Peskin