# Quantum field theory 1, lecture 09

### 5 Relativistic scalar fields and O(N)-models

In the next chapter we discuss a first model with Lorentz symmetry. Lorentz symmetry is a key ingredient for elementary particle physics. We may focus on a simple model with a complex scalar. This is employed in order to understand how Lorentz symmetry is tightly connected with the existence of antiparticles or antimatter. We also will discuss the important concept of spontaneous symmetry breaking.

Examples for scalar fields. Neutral relativistic scalar fields are the neutral pion $$\pi ^0$$ in QCD, or the inflaton or cosmon. In this case a scalar field is a real function $$\chi (\vec{x},t).$$ In principle, its expectation value can be measured, similar to the electric or magnetic field. Complex scalar fields are the charged pions $$\pi ^\pm$$ and the kaons $$K^\pm$$, represented by a complex scalar field $$\chi (\vec{x},t).$$ An important complex field is the Higgs-doublet, represented by a two-component complex scalar field $$\varphi _a(t)$$ with $$a=1,2$$. In particle physics, its expectation value is responsible for the spontaneous breaking of the electroweak gauge symmetry, and the resulting masses of the W- and Z- bosons, quarks and charged leptons.

#### 5.1 Lorentz invariant action.

Action. To formulate the action we first need the fields which are now fields in Minkowski space $$\chi (x)$$ where $$x=(t, \vec x)$$. We consider local actions of the form \begin{equation*} S= \int _x \mathscr{L}(x),\quad \quad \quad \int _x = \int \; dt\;d^3x. \end{equation*} A typical form of the action is in an expansion in derivatives \begin{equation*} \mathscr{L}(x) = \mathscr{L}_{\text{kin}}+iV + \ldots \end{equation*} The action will reflect the symmetries of the model. One important symmetry is Lorentz symmetry.

Kinetic term. The kinetic term $$\mathscr{L}_{\text{kin}}$$ involves derivatives of fields. For non-relativistic free atoms we have found \begin{equation*} \mathscr{L}_{\text{kin}} = \chi ^{*}(x)\partial _t \chi (x) + \frac{i}{2M} \partial _j \chi ^{*}(x) \partial _j \chi (x),\quad \partial _j =\frac{\partial }{\partial x^j}. \end{equation*} The two space derivatives are needed for rotation symmetry. Lorentz-symmetry needs again two derivatives, \begin{equation*} \mathscr{L}_{\text{kin}} = i\partial ^\mu \chi ^{*}(x)\partial _\mu \chi (x), \end{equation*} with \begin{equation*} \partial _\mu =\left (\frac{\partial }{\partial t}, \frac{\partial }{\partial x^j}\right ) = (\partial _0, \partial _j), \end{equation*} and \begin{equation*} \partial ^\mu = \eta ^{\mu \nu } \partial _\nu , \quad \quad \quad \eta _{\mu \nu } = \eta ^{\mu \nu } = \begin{pmatrix} -1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}. \end{equation*} Derivatives of scalars $$\partial _{\mu } \chi$$ are covariant four-vectors. The scalar product of a covariant and a contravariant four-vector is invariant under Lorentz transformations. In momentum space the kinetic term involves the invariant squared momentum \begin{equation*} \mathscr{L}_{\text{kin}} = i q^2 \chi ^{*}(q)\chi (q) \;,\; q^2= q^{\mu }q_{\mu }= \eta ^{\mu \nu } q_{\mu } q_{\nu }. \end{equation*} We conclude that relativistic theories of scalars involves two time derivatives. It is a direct consequence of Lorentz symmetry that the number of time- derivatives equals the number of space- derivatives. The kinetic term can be formulated for real fields in the same way.

From complex to real fields. Writing a complex field as two real fields \begin{equation*} \chi = \frac{1}{\sqrt{2}}(\chi _1+i\chi _2) \end{equation*} one has \begin{equation*} \mathscr{L}_{\text{kin}} = \frac{i}{2} \sum _{a=1}^N \partial ^\mu \chi _a (x) \partial _\mu \chi _a(x). \end{equation*} Here $$N=1$$ for a real scalar, $$N=2$$ for a complex scalar and $$N=4$$ for the Higgs doublet.

Potential. The potential $$V$$ involves no derivatives. It is a function of the fields and we write \begin{equation*} V(x) = V(\chi (x)) = V(\chi ). \end{equation*} Internal symmetries yield further restrictions. Charge conservation corresponds for complex field to the symmetry \begin{equation*} \chi \to e^{i\alpha } \chi . \end{equation*} The potential can only depend on \begin{equation*} \rho = \chi ^{*}\chi = \frac{1}{2}(\chi ^{2}_{1} + \chi ^{2}_2). \end{equation*} For the Higgs doublet, the symmetry is $$\text{SU}(2)$$ such that \begin{equation*} \rho = \chi ^{\dagger }\chi = \frac{1}{2}\sum _{a=1}^4 \chi _{a}^{2}. \end{equation*}

Often one can expand \begin{equation*} V(\rho ) = \mu ^2 \rho + \frac{1}{2} \lambda \rho ^2 + \ldots \end{equation*} For potentials depending only on $$\rho$$, \begin{equation*} \mathscr{L}_{\text{kin}}+V(\rho ) \end{equation*} the action has $$O(N)$$ symmetry. Performing analytic continuation for a description of thermal equilibrium we recover the $$O(N)$$- symmetric models discussed in lecture 2.3. The classical statistical equilibrium and the quantum statistical equilibrium at vanishing temperature differ only by an additional dimension for the second, given by euclidean time. For the classical statistical setting the temperature enters as a parameter in the action, while for quantum statistics it appears in the periodic boundary condition.

#### 5.2 Lorentz invariance and antiparticles.

We next want to show that antiparticles are a natural consequence of Lorentz symmetry.

Two fields with one time-derivative. In the following we concentrate on a single complex scalar field. We want to see how the Lorentz invariant action actually describes two degrees of freedom, namely a charged scalar particle and its antiparticle with opposite charge. Both charged pions $$\pi ^{-}$$ and $$\pi ^{+}$$ are described by the same field.

In order to see this we recall that a differential equation with two derivatives is equivalent to two differential equations with one derivative. In other words, one complex field with two time-derivatives is equivalent to two complex fields with one time derivative. We will use this in order to rewrite the action in terms of two fields with only one time derivative. In this form we can make direct contact to the action for non-relativistic bosons that we have discussed previously.

Let us start with a free complex relativistic scalar field \begin{equation*} \mathscr{L} = i(\partial ^\mu \chi ^{*}\partial _\mu \chi + M^2 \chi ^{*}\chi ). \end{equation*} The potential $$V$$ describes only the mass $$M$$ of the scalar particle. In momentum space, $$\partial _t = \partial _0 = -\partial ^{0},$$ one has \begin{equation*} \mathscr{L}_p = -i\partial _t \chi ^{*} \partial _t \chi +i (\vec p^2 + M^2) \chi ^{*} \chi , \end{equation*} and the partition function is \begin{equation*} Z = \int \;D \chi \; e^{-\int dt \int _{\vec{p}} \mathscr{L}_p}, \end{equation*} where $$\int _{\vec p} = \int \frac{d^3p }{(2\pi )^3}$$.

We treat every $$\vec{p}$$ mode separately. In order to switch to a formulation with two complex fields and only one time derivative we insert a unit factor \begin{equation*} \int D\pi \; \text{exp} \{-i (\partial _t \chi ^{*}-\pi ^{*})(\partial _t \chi -\pi )\} = \text{const} \;"="\; 1, \end{equation*} where $$\pi (x)$$ is a field. This factor yields only a constant which is independent of $$\chi$$, as can be seen by a simple shift of the integration variable, $$\pi '= \pi -\partial _{t}\chi$$. Since multiplicative constants in $$Z$$ do not matter, we can write the partition function equivalently as \begin{equation*} \begin{split} Z &= \int D\chi D\pi ~ \text{exp}\left [-\int _t\left \{ -i\partial _t \chi ^{*}\partial _t\chi + i(\vec p^2 + M^2)\chi ^{*}\chi \right .\right .\\ & \qquad \qquad \qquad \qquad \qquad \quad + i\partial _t\chi ^{*} \partial _t \chi - i\partial _t\chi ^{*} \pi -i \pi ^{*}\partial _t\chi + i\pi ^{*}\pi \big \}\bigg ]. \end{split} \end{equation*} This eliminates the term with two derivatives. What remains are two complex fields $$\chi$$ and $$\pi$$ with one time derivative, \begin{equation*} Z= \int D\chi \;D\pi \; e^{-\int _t \mathscr{L}}, \end{equation*} where, after doing a partial integration, \begin{equation*} \mathscr{L} = i\chi ^{*}\partial _t \pi - i\pi ^{*}\partial _t \chi + i(\vec p^2 +M^2)\chi ^{*}\chi + i\pi ^{*}\pi . \end{equation*}

At this stage we have the wanted number of fields and only first time derivatives. The time derivative term mixes the fields $$\pi$$ and $$\chi$$. We want to diagonalise this term by suitable variable transformations, such that the independent degrees of freedom are clearly visible. For this purpose we perform the variable transformation \begin{equation*} \begin{split} &\chi (t) = \frac{1}{\sqrt{2}}(\vec p^2+ M^2)^{-\frac{1}{4}}(\varphi _1(t) + \varphi _2(-t)),\\ &\pi (t) = -\frac{i}{\sqrt{2}}(\vec p^2+M^2)^{\frac{1}{4}}(\varphi _1(t) - \varphi _2(-t)). \end{split} \end{equation*} This yields \begin{equation*} \begin{split} (\vec p^2+M^2)\chi ^{*}(t) \chi (t) &= \frac{1}{2}(\vec p^2+M^2)^{\frac{1}{2}} \left [\varphi ^{*}_1(t) \varphi _{1}(t)+\varphi ^{*}_{2}(-t) \varphi _{2}(-t) \right .\\ &\qquad \qquad \qquad \qquad \qquad \left . +\varphi ^{*}_{1}(t) \varphi _{2}(-t)+\varphi ^{*}_{2}(t) \varphi _{1}(t)\right ], \end{split} \end{equation*} Similarly, \begin{equation*} \begin{split} \pi ^*(t) \pi (t) &= \frac{1}{2}(\vec p^2+M^2)^{\frac{1}{2}}\left [\varphi ^{*}_1(t)\varphi _1(t) + \varphi ^{*}_2(-t) \varphi _2(-t) \right .\\ &\qquad \qquad \qquad \qquad \qquad \left . -\varphi ^{*}_1(t)\varphi _2(-t) - \varphi ^{*}_2(-t) \varphi _1(t)\right ], \end{split} \end{equation*} Summing both expressions gives \begin{equation*} \begin{split} i\left ((\vec p^2+M^2)\chi ^{*}\chi +\pi ^{*}\pi \right ) &= i(\vec p^2+M^2)^{\frac{1}{2}}\left [\varphi ^{*}_{1}(t)\varphi _{1}(t) +\varphi ^{*}_{2}(-t)\varphi _{2}(-t)\right ], \end{split} \end{equation*} and the mixed term involving time derivatives is \begin{equation*} \begin{split} i\left (\chi ^{*}\partial _{t}\pi - \pi ^{*}\partial _{t}\chi \right ) &=\frac{1}{2}\left \{(\varphi _1^{*}(t) + \varphi _2^{*}(-t))\partial _t(\varphi _1(t) -\varphi _2(-t)) \right .\\ &\qquad \qquad \qquad +\left .(\varphi _1^{*}(t) - \varphi _2^{*}(-t)) \partial _t(\varphi _1(t) + \varphi _2(-t))\right \}\\ &= \varphi _1^{*}(t) \partial _t \varphi _1(t) - \varphi _2^{*}(-t)\partial _t \varphi _2(-t). \end{split} \end{equation*} Under the $$t$$-integral one can replace $$-\varphi ^{*}_2(-t) \partial _t \varphi _2(-t) \to \varphi _2^{*}(t) \partial _t \varphi _2(t)$$.

Taking the terms together we find the action for two particles with dispersion relation $$E=\omega _{M} =\sqrt{\vec p^2+ M^2}$$, \begin{equation*} S= \int dt \left \{\varphi ^{*}_{1}\partial _{t} \varphi _{1} + \varphi ^{*}_{2} \partial _{t} \varphi _{2} - i\sqrt{\vec p^2 + M^2}(\varphi ^{*}_{1} \varphi _{1} + \varphi ^{*}_{2}\varphi _{2})\right \} \end{equation*} where $$\varphi _i = \varphi _i(t)$$. This has precisely the same form that we have encountered before for non-relativistic bosons as phonons. The only particularity is the form of the dispersion relation which reflects the relativistic relation between energy and momentum. The action is block-diagonal, and the two complex fields $$\phi _1$$ and $$\phi _2$$ describe two particles.

Antiparticles. The field $$\chi$$ with two time-derivatives describes a pair of fields $$\varphi _1, \varphi _2$$ with one time-derivative. One field is the antiparticle of the other. We want to show that the antiparticle has the opposite charge of the particle. For this purpose we couple the complex field $$\chi$$ to an ”external” electromagnetic field. The different field equations for $$\varphi _1$$ and $$\varphi _2$$ will then reveal their opposite charges.

The coupling to the electromagnetic potential $$A_{\mu }$$ is dictated by the principle of gauge invariance. This requires to replace every derivative $$\partial _{\mu }$$ by a covariant derivative $$D_{\mu }$$ according to \begin{equation*} \partial _\mu \to D_\mu \chi = (\partial _\mu -i e A_\mu )\chi . \end{equation*} We want to consider the particular field configuration $$A_i = 0$$ and constant electric potential $$A_0$$. In this case one only modifies the time derivative $$\partial _t \to \partial _t -i e A_0$$. Employing this modification also in the inserted unit factor one obtains for $$\mathscr{L}$$ an additional term \begin{equation*} \Delta \mathscr{L} = e A_0\left [\chi ^{*}(t) \pi (t) - \pi ^{*}(t)\chi (t)\right ]. \end{equation*} We express this addition in terms of the fields $$\varphi _1$$ and $$\varphi _2$$ \begin{equation*} \begin{split} \Delta \mathscr{L} &= e A_0 \left [-\frac{i}{2}(\varphi ^{*}_{1}(t)+\varphi ^{*}_{2}(-t))(\varphi _1(t)-\varphi _2(-t)) -\frac{i}{2}(\varphi ^{*}_1(t) -\varphi ^{*}_2(t))(\varphi _1(t)+\varphi _2(-t))\right ]\\ &= -ieA_0 (\varphi ^{*}_{1}(t) \varphi _1(t) -\varphi ^{*}_2(-t) \varphi _2(-t)). \end{split} \end{equation*} As a consequence, the time derivative part in the action becomes \begin{equation*} S= \int dt~ \left \{\varphi ^{*}_1 (\partial _t -ieA_0)\varphi _1 + \varphi ^{*}_2(\partial _t + ieA_0) \varphi _2 + \ldots \right \} \end{equation*} We conclude that $$\varphi _1$$ and $$\varphi _2$$ have opposite electric charge. An electric field, given by the gradient of $$A_0$$, will accelerate the two particles in opposite directions. The two fields show the characteristic properties of a pair of particle and antiparticle. They have the same mass, but opposite charge.

We have performed the insertion of unity and variable transformations merely in order to demonstrate the appearance of antiparticles in a simple way. In practise, one does not use this variable transformation. The reason is that the ”insertion of unity” for the introduction of the field $$\pi$$ is not compatible with the Lorentz symmetry - time is singled out. Since we have not changed the functional integral, the Lorentz symmetry still governs the dynamics if $$A_0=0$$. The presence of this symmetry is hidden for the action formulated in terms of $$\varphi _1$$ and $$\varphi _2$$. Since Lorentz invariance is such an important symmetry for particle physics one wants to work with an action for which this symmetry is manifest.

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