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1Classical Field Theory.

Lagrange formalism for particles.

Classical mechanics.

  $L(q_n,{\dot{q}}_n)$ Lagrange function [assume no explicit time dependence], $q_n,n=1,\cdots ,N$ generalized coordinates.

$$S[q_n]={\int }_{t_1}^{t_2}L(q_n(t),{\dot{q}}_n(t))dt$$Action Functional of orbits $q_n(t)$ with fixed starting and end points $q_n(t_1,q_n(t_2)$.

  Equation of motion follows from the “action principle”. The orbit taken by the physical system is the one where the action is stationary: $$\delta S[q_n]=0.$$This implie the Euler-Lagrange equations $$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_n}-\frac{\partial L}{\partial q_n}=0.$$

  Transition to the Hamiltonian:

  Canonically conjugate momenta $$p_n\equiv \frac{\partial L}{\partial {\dot{q}}_n}.$$(1)Then $$H(q_n,p_n)\equiv \sum _n{p}_n{\dot{q}}_n-L(q_n,{\dot{q}}_n)$$where ${\dot{q}}_n={\dot{q}}_n(q_n,p_n)$ by solving (1) for ${\dot{q}}_n$.

  The canonical coordinates fulfill the equations $$\{q_n{p}_m{\}}_p={\delta }_{nm}\{q_n,q_m{\}}_p=\{p_n,p_m\}=0.$$$$\{\begin{array}{l}{\dot{q}}_n=\{q_n,H{\}}_p=\frac{\partial H}{\partial p_n}\\ {\dot{p}}_n=\{p_n,H{\}}_p=-\frac{\partial H}{\partial q_n}\end{array},$$where $\{A,B{\}}_p=\frac{\partial A}{\partial q_n}\frac{\partial B}{\partial p_n}-\frac{\partial A}{\partial p_n}\frac{\partial B}{\partial q_n}$.

Canonical quantization.

  $L(q_n,{\dot{q}}_n)$ given. Interpret $q_n,{\dot{q}}_n,p_n$ as operators on a Hilbert space.

  Impose the commutation relations $[\hslash \equiv 1]$, $$[q_n,p_m]=i{\delta }_{nm}[q_n,q_m]=[p_n,p_m]=0$$at a fixed time.

  The time evolution of the operators is $${\dot{q}}_n=\frac{1}{i}[q_n,H].$$Note: simply $\{,\}\to \frac{1}{i}[,]$.

Lagrange-formalism for fields.

  A field is a quantity defined at every point of space and time $(\vec{x},t)$ $$\varphi (\vec{x},t).$$Note: the concept of position has been relegated from a dynamical variable in classical mechanism to a mere label in field theory.

Example 1.1. E-M field.

$$\{\begin{array}{ll}\text{Electric field}:& \vec{E}(\vec{x},t)\\ \text{Magnetic field}:& \vec{B}(\vec{x},t)\end{array}:\text{spatial}3\text{-vectors},$$derive these two $3$-vectors from a single $4$-component filed $$A^{\mu }(\vec{x},t)=(\varphi ,\vec{A})[\mu =0,1,2,3],$$where $A^{\mu }$ is a vector in spacetime.

The electric and magnetic fields are $$\vec{E}=-\nabla \varphi -\frac{\partial \vec{A}}{\partial t}\text{and}\vec{B}=\nabla \times \vec{A}.$$

  The dynamics of the field is governed by a Lagrangian.

  In relativistic QFT one assumes that $L$ is an integral over Lagrangian density $$\begin{array}{rl}L& =\int d^3\vec{x}\mathcal{L}({\varphi }_n(x),\nabla {\varphi }_n(x),{\dot{\varphi }}_n(x))\\ & =\int d^3\vec{x}\mathcal{L}({\partial }_{\mu }{\varphi }_n(x),{\varphi }_n(x)).\end{array}$$It is difficult to construct non-local Lagrangians, which are compatible with relativistic causality.

  E-L equation: $$\begin{array}{rl}\delta S& =\delta \int d^{4}x\mathcal{L}({\varphi }_n,{\partial }_{\mu }{\varphi }_n)\\ & =\int d^{4}x[\frac{\partial \mathcal{L}}{\partial {\varphi }_n}\delta {\varphi }_n+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }_n)}\delta ({\partial }_{\mu }{\varphi }_n)]\end{array}$$Integration by parts and assume $\varphi (x)\to 0$ for $|\vec{x}|\to 0$, we get $${\partial }_{\mu }\left(\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }_n)}\right)-\frac{\partial \mathcal{L}}{\partial {\varphi }_n}=0.$$Then we define canonically conjugated fields: $${\pi }_n(x)\equiv \frac{\partial \mathcal{L}}{\partial ({\partial }_0{\varphi }_n(x))}.$$Hamilton density: $$H=\sum _n\int d^{3}x{\pi }_n(x){\partial }_0{\varphi }_n(x)-\int d^{3}x\mathcal{L}\equiv \int d^{3}x\mathcal{H}(x)$$$$\Rightarrow \mathcal{H}(x)=\sum _n{\pi }_n{\partial }_0{\varphi }_n-\mathcal{L}.$$

Suppose space were discrete with points $\vec{x}$ on a lattice, then there is a set ${\varphi }_{n,\vec{x}(t)}$ of canonical coordinates, and the previous discussion applie with substitutions

  Canonical quantization.

  Impose the equal-time commutation relations $$[{\varphi }_n(t,\vec{x}),{\pi }_m(t,\vec{y}]=i{\delta }_{nm}{\delta }^{(3)}(\vec{x}-\vec{y}),$$$$[{\varphi }_n(t,\vec{x}),{\varphi }_m(t,\vec{y})]=[{\pi }_n(t,\vec{x}),{\pi }_m(t,\vec{y})]=0.$$The E-L equation is equivalent to the Heisenberg equation $${\dot{\varphi }}_n(t,\vec{x})=\frac{1}{i}[{\varphi }_n(t,\vec{x}),H]$$for the field operator. The formal solution is $${\varphi }_n(t,\vec{x})=e^{iHt}{\varphi }_n(0,\vec{x})e^{-iHt},$$since $H$ does not explicitly depend on $t$.

Remark 1.2. If the commutation relations are imposed at one time, they are preserved at all times. $$\begin{array}{rl}[{\varphi }_n(t,\vec{x}),{\pi }_m(t,\vec{x})]& =[e^{iH(t-t_0)}{\varphi }_n(t_0,\vec{x})e^{-iH(t-t_0)},e^{iH(t-t_0)}{\pi }_m(t_0,\vec{x})e^{-iH(t-t_0)}]\\ & =e^{iH(t-t_0)}[{\varphi }_n(t_0,\vec{x}),{\pi }_m(t_0,\vec{x})]e^{-iH(t-t_0)}\\ & =i{\delta }_{nm}{\delta }^{(3)}(\vec{x}-\vec{y}).\end{array}$$

  Unless otherwise mentioned one uses the Heisenberg picture in QFT, where the time-dependence is in the operators, not the states.

Example 1.3. The Klein-Gorden Equation.

  Consider $L$ for a real scalar field $\varphi (\vec{x},t)$. $$\mathcal{L}=\frac{1}{2}{g}^{\mu \nu }{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -\frac{1}{2}{m}^2{\varphi }^2=\frac{1}{2}{\dot{\varphi }}^2-\frac{1}{2}(\vec{\nabla }\varphi {)}^2-\frac{1}{2}{m}^2{\varphi }^2,$$where we use the Minkowski space metric $g^{\mu \nu }=g_{\mu \nu }=\mathrm{diag}(+1,-1,-1,-1)$.

  We identify the kinetic energy of the field as $$T=\int d^{3}x\frac{1}{2}{\dot{\varphi }}^2,$$the potential energy as $$V=\int d^{3}x[\frac{1}{2}(\vec{\nabla }\varphi {)}^2+\frac{1}{2}{m}^2{\varphi }^2],$$where we call the first term by gradient energy and the second term by “true” potential energy.

  To determine the equation of motion (EOM) for $\varphi$, we compute $$\frac{\partial \mathcal{L}}{\partial \varphi }=-m^2\varphi \text{and}\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}={\partial }^{\mu }\varphi \equiv (\dot{\varphi },-\nabla \varphi ).$$The E-L euqation is then $$\ddot{\varphi }-{\nabla }^2\varphi +m^2\varphi =0\text{or}{\partial }_{\mu }{\partial }^{\mu }\varphi +m^2\varphi ,$$Klein-Gorden Equation.

  If we consider the Lagrangian with general potential $V(\varphi )$, $$\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -V(\varphi )\Rightarrow {\partial }_{\mu }{\partial }^{\mu }\varphi +\frac{\partial V}{\partial \varphi }=0.$$

One page is missing here.

Example 1.4. Maxwell’s Equation.

  From the Lagrangian $\mathcal{L}=-\frac{1}{2}({\partial }_{\mu }{A}_{\nu })({\partial }^{\mu }{A}^{\nu })+\frac{1}{2}({\partial }_{\mu }{A}^{\mu }{)}^2$.

  Note: $\mathcal{L}$ has no kinetic term ${\dot{A}}_0^2$ for $A_0$.

  Notice that $\frac{\partial ({\partial }_{\alpha }{A}_{\beta })}{\partial ({\partial }_{\mu }{A}_{\nu })}={\delta }_{\alpha }^{\mu }{\delta }_{\beta }^{\nu }$, we compute $$\begin{array}{rl}\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\nu })}& =\frac{\partial }{\partial ({\partial }_{\mu }{A}_{\nu })}[-\frac{1}{2}{\partial }_{\alpha }{A}_{\beta }{\partial }_{\rho }{A}_{\sigma }{g}^{\rho \alpha }{g}^{\sigma \beta }+\frac{1}{2}{\partial }_{\alpha }{A}_{\beta }{\partial }_{\rho }{A}_{\sigma }{g}^{\beta \alpha }{g}^{\sigma \rho }]\\ & =-\frac{1}{2}{\delta }_{\alpha }^{\mu }{\delta }_{\beta }^{\nu }{\partial }_{\rho }{A}_{\sigma }{g}^{\rho \alpha }{g}^{\sigma \beta }-\frac{1}{2}{\partial }_{\alpha }{A}_{\beta }{\delta }_{\rho }^{\mu }{\delta }_{\sigma }^{\nu }{g}^{\rho \alpha }{g}^{\sigma \beta }+\frac{1}{2}{\delta }_{\alpha }^{\mu }{\delta }_{\beta }^{\nu }{\partial }_{\rho }{A}_{\sigma }{g}^{\beta \alpha }{g}^{\sigma \rho }+\frac{1}{2}{\partial }_{\alpha }{A}_{\beta }{\delta }_{\rho }^{\mu }{\delta }_{\sigma }^{\nu }{g}^{\beta \alpha }{g}^{\sigma \rho }\\ & =-\frac{1}{2}{\partial }^{\mu }{A}^{\nu }-\frac{1}{2}{\partial }^{\mu }{A}^{\nu }+\frac{1}{2}{\partial }_{\rho }{A}^{\rho }{g}^{\mu \nu }+\frac{1}{2}{\partial }_{\alpha }{A}^{\alpha }{g}^{\mu \nu }\\ & =-{\partial }^{\mu }{A}^{\nu }+{\partial }_{\rho }{A}^{\rho }{g}^{\mu \nu }.\end{array}$$Then we have $$\begin{array}{rl}{\partial }_{\mu }[\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\nu })}]& =-{\partial }_{\mu }({\partial }^{\mu }{A}^{\nu })+{\partial }^{\nu }({\partial }_{\rho }{A}^{\rho })\\ & =-{\partial }_{\mu }[{\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }]\equiv -{\partial }_{\mu }{F}^{\mu \nu }\end{array}$$where the field strength is defined by $$F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }.$$Using $F^{\mu \nu }$, the Maxwell Lagrange is $$\mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }.$$

Noether’s theorem for fields

  The Noether theorem establishes a relation between symmetries and conserved quantities.

  There is no difference between classical and quantum field theory; $\phi (x)$ can be a classical field or quantum field operator. $$S[\phi ]=\int d^4\mathcal{L}(\phi ,{\partial }_{\mu }\phi ).$$A symmetry is a transformation of the fields, which leaves action invariant $${\phi }^{\prime }(x)=\phi (x)+\epsilon F(x),$$where $\epsilon$ is a constant and a global symmetry, $F$ depends on $\phi (x)$ and ${\partial }_{\mu }\phi (x)$. $$0=\delta S=\epsilon \int d^{4}x\frac{\delta S[\phi ]}{\delta \phi (x)}F(x)$$before the EOM are satisfied.

Example 1.5. Translation.

$${\phi }^{\prime }(x)=\phi (x+a)=\phi (x)+a^{\mu }{\partial }_{\mu }\phi (x)+\cdots ,$$then $a^{\mu }\sim \epsilon$ and ${\partial }_{\mu }\phi \sim F$.

Theorem 1.6. Noether theorem.

  For every continuous symmetry there is a conserved current $$j^{\mu }(x)=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi (x))}F(x)-{\kappa }^{\mu }(x)[{\partial }_{\mu }{j}^{\mu }(x)=0]$$and charge $$Q\equiv \int d^{3}x{j}^0(x)[\frac{dQ}{dt}=0].$$

Derivation.

  Since the action $S[\phi (x)]$ is invariant by assumption, $\mathcal{L}$ can only change by a total derivative, i.e. $$\delta S=\epsilon \int d^{4}x{\partial }_{\mu }{\kappa }^{\mu }(x)$$(2)for some ${\kappa }^{\mu }(x)$ (In particular ${\kappa }^{\mu }\equiv 0$ if $\mathcal{L}$ is invariant, not only $S[\phi ]$).

  One the other hand, $$\begin{array}{rl}\delta S& =\int d^{4}x\frac{\partial \mathcal{L}}{\partial \phi }\delta \phi +\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi )}\delta ({\partial }_{\mu }\phi )\\ & =\int d^{4}x\frac{\partial \mathcal{L}}{\partial \phi }\epsilon F(x)+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi )}\epsilon {\partial }_{\mu }F(x)\\ & =\int d^{4}x\epsilon F(x){\partial }_{\mu }[\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi )}]+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi )}\epsilon {\partial }_{\mu }F(x)\\ & =\int d^{4}x\epsilon {\partial }_{\mu }[\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\phi )}F].\end{array}$$(3)

  Taking (3) minus (1) gives $${\partial }_{\mu }{j}^{\mu }(x)=0$$for $j^{\mu }(x)$ as defined above. Also, $$\frac{dQ}{dt}=\int d^{3}x{\partial }^0{j}^0(x)=-\int d^{3}x\vec{\nabla }\cdot \vec{j}(x)=0,$$Assuming as usual that the fields vanish as $|\vec{x}|\to \infty$.

$\square$

Remark 1.7. The conservation of the current is only true for the fields satisfying the filed equations, because the E-L equation was used to obtain (3). In particular, in the quantum theory, this will not be true of every field configuration we sum over in the path integral, only the saddle-point configurations.

Translation and the Energy-Momentum Tensor.

  Consider the infinitesimal translation $x^{\nu }\to x^{\nu }-{\epsilon }^{\nu }$, then $$\begin{gathered} {\varphi }_n(x)\to {\varphi }_n(x)+{\epsilon }^{\nu }{\partial }_{\nu }{\varphi }_n(x) \\ \mathcal{L}(x)\to \mathcal{L}(x)+{\epsilon }^{\nu }{\partial }_{\nu }\mathcal{L}(x) \end{gathered}$$rewrite ${\epsilon }^{\nu }{\partial }_{\nu }\mathcal{L}(x)$ into ${\epsilon }^{\nu }{\partial }_{\mu }[{\delta }_{\nu }^{\mu }\mathcal{L}(x)]$.

  Then four conserved currents $(j^{\mu }{)}_{\nu }$, one for each of ${\epsilon }^{\nu }$, $\nu =0,1,2,3$, $$(j^{\mu }{)}_{\nu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }_n)}{\partial }_{\nu }{\varphi }_n-{\delta }_{\nu }^{\mu }\mathcal{L}(x)\equiv {T^{\mu }}_{\nu },$$where ${T^{\mu }}_{\nu }$ denote the Energy-Momentum tensor satisfying ${\partial }_{\mu }{T^{\mu }}_{\nu }=0$.

  Four conserved quantities are given by $$\text{the total energy of the field configuration}:E=\int d^{3}x{T}^{0,0},$$$$\text{the total momentum}:p^i=\int d^{3}x{T}^{0,i}.$$

Example 1.8. The scalar field theory: $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -g^{\mu \nu }\mathcal{L}$. Using EOM for $\varphi$, $({\partial }^2+m^2)\varphi =0$, then ${\partial }_{\mu }{T}^{\mu \nu }=0$.

  The conserved energy $E=\int d^{3}x\frac{1}{2}{\dot{\varphi }}^2+\frac{1}{2}(\vec{\nabla }\varphi {)}^2+\frac{1}{2}{m}^2{\varphi }^2$ and the total momentum $p^i=\int d^{3}x\dot{\varphi }{\partial }^i\varphi$.

Lorentz transformation

  Lorentz invariant is symmetry under rotations and boosts.

I ignore some introductions to math concepts.

  Among the most important symmetries of relativistic QFT are those which arise from the Lorentz transformations themselves.

  The spacetime inteval $d{s}^2=g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }$, where $g_{\mu \nu }=g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$.

  Any coordinate transformation $x^{\mu }\mapsto x^{\prime \mu }$ that satisfies $$g_{\mu \nu }d{x}^{\prime \mu }d{x}^{\prime }\nu =g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }\text{or}{g}_{\mu \nu }\frac{\partial x^{\prime \mu }}{\partial x^{\rho }}\frac{\partial x^{\prime \nu }}{\partial x^{\sigma }}=g_{\rho \sigma }$$is linear $$x^{\prime \mu }={{\Lambda }^{\mu }}_{\nu }{x}^{\nu }+a^{\nu }$$where $a^{\mu }$ are arbitrary constants and $\Lambda$ is a constant matrix satifying $$g_{\mu \nu }{{\Lambda }^{\mu }}_{\rho }{{\Lambda }^{\nu }}_{\sigma }=g_{\rho \sigma }\text{or}{\Lambda }^{T}G\Lambda =G$$(4)These transformations form a group.

  It’s easy to check the ransformation satisfies $$T(\overline{\Lambda },\overline{a})T(\Lambda ,a)=T(\overline{\Lambda }\Lambda ,\overline{\Lambda }a+\overline{a}).$$

  Taking the determinant of (4) gives $(\det \Lambda {)}^2=1$, so ${{\Lambda }^{\mu }}_{\nu }$ has an inverse $${({\Lambda }^{-1}{)}^{\rho }}_{\nu }={{\Lambda }_{\nu }}^{\rho }\equiv g_{\nu \mu }{g}^{\rho \sigma }{{\Lambda }^{\mu }}_{\sigma }$$

  The whole group of $T(\Lambda ,a)$ is Poincaré group.

  If $a^{\mu }=0$, $T(\Lambda ,0)$ is known as homogeneous Lorentz group.

Example 1.9. Some examples.

  A rotation by $\theta$ about the $x^3$-axis, $${{\Lambda }^{\mu }}_{\nu }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& \cos \theta & -\sin \theta & 0\\ 0& \sin \theta & \cos \theta & 0\\ 0& 0& 0& 1\end{array}\right).$$

  A boost by $v$ along the $x^1$-axis, $${{\Lambda }^{\mu }}_{\nu }=\left(\begin{array}{cccc}\gamma & -\gamma v& 0& 0\\ -\gamma v& \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$with $\gamma =\frac{1}{\sqrt{1-v^2}}$.

Remark 1.10. Poincaré group not only acts on spacetime coordinates, but also acts on the Hilbert space of a physical system, so it is not merely represented as $4\times 4$ matrices.

  $L$ has four connected components.

  Note that $\det \Lambda$ and $\mathrm{sgn}{{\Lambda }^0}_0$ are both continuous functions of ${{\Lambda }^{\mu }}_{\nu }$. Furthermore, $\det \Lambda =\pm 1$ and ${{\Lambda }^0}_0\ge 1$ or $\le -1$. Thus $\det \Lambda$ and $\mathrm{sgn}{{\Lambda }^0}_0$ must be constant on any one component. The four possibilities are

  The Lorentz transformation:

  Space inversion $I_s$: $$I_s=\mathrm{diag}(1,-1,-1,-1).$$

  Time inversion $I_t$: $$I_t=\mathrm{diag}(-1,1,1,1).$$

  Space-time inversion $I_{st}$: $$I_{st}=\mathrm{diag}(-1,-1,-1,-1)=-I.$$

  Clearly, $L_{-}^{\uparrow }\stackrel{I_s}{\to }{L}_{+}^{\uparrow }$, $L_{-}^{\downarrow }\stackrel{I_t}{\to }{L}_{+}^{\uparrow }$ and $L_{+}^{\downarrow }\stackrel{I_{st}}{\to }{L}_{+}^{\uparrow }$.

  The important subgroups of $L$:

    The orthochronous group $L^{\uparrow }=L_{+}^{\uparrow }\cup L_{-}^{\uparrow }$, ${{\Lambda }^0}_0\ge 1$.

    The proper Lorentz group $L_{+}=L_{+}^{\uparrow }\cup L_{+}^{\downarrow }$, $\det \Lambda =+1$.

    The orthochorous Lorentz group $L_0=L_{+}^{\uparrow }\cup L_{-}^{\downarrow }$, $\mathrm{sgn}{{\lambda }^0}_0\det \Lambda =1$.

    The proper orthochronous Lorentz group or the restricted Lorentz group $L_{+}^{\uparrow }$.

Classical field theory

  Scalar field: $\varphi (x)\stackrel{\Lambda }{\to }{\varphi }^{\prime }(x)=\varphi ({\Lambda }^{-1}x)$.

  Vector field: $A^{\mu }(x)\stackrel{\Lambda }{\to }{{\Lambda }^{\mu }}_{\nu }{A}^{\nu }({\Lambda }^{-1}x)$

Example 1.11. The Klein-Gorden field.

$${\partial }_{\mu }\varphi (x)\to {\partial }_{\mu }\varphi ({\Lambda }^{-1}x)={({\Lambda }^{-1}{)}^{\nu }}_{\mu }[{\partial }_{\nu }\varphi ]({\Lambda }^{-1}x)={{\Lambda }_{\mu }}^{\nu }({\partial }_{\nu }\varphi )({\Lambda }^{-1}x).$$$$\frac{\partial }{\partial x^{\mu }}=\frac{\partial }{\partial y^{\nu }}\frac{\partial y^{\nu }}{\partial x^{\mu }}=\frac{\partial }{\partial y^{\nu }}{({\Lambda }^{-1}{)}^{\nu }}_{\mu },y^{\nu }={({\Lambda }^{-1}{)}^{\nu }}_{\mu }{x}^{\mu }.$$

  The derivate term in the Lagrangian transform as $$\begin{array}{rl}{\mathcal{L}}_{deriv}(x)={\partial }_{\mu }\varphi (x){\partial }_{\nu }\varphi (x)g^{\mu \nu }& \to {{\Lambda }_{\mu }}^{\rho }{{\Lambda }_{\nu }}^{\sigma }[{\partial }_{\rho }\varphi ](y)[{\partial }_{\sigma }\varphi ](y)g^{\mu \nu }\\ & =g^{\rho \sigma }[{\partial }_{\rho }\varphi ](y)[{\partial }_{\sigma }\varphi ](y)\\ & ={\mathcal{L}}_{deriv}(y).\end{array}$$

  The potential terms transform in the same way ${\varphi }^2(x)\to {\varphi }^2(y)$, then $$\begin{array}{rl}S=\int d^{4}x\mathcal{L}(x)\to \int d^{4}x\mathcal{L}(y)& =\int d^{4}y\left|\frac{\partial x}{\partial y}\right|\mathcal{L}(y)\\ & =\int d^{4}y\mathcal{L}(y),\text{since}|\det \Lambda |=1.\end{array}$$

Quantum Lorentz Transformation and the Poincaré Algebra

For an infinitesimal inhomogeneous Lorentz transformation $${{\Lambda }^{\mu }}_{\nu }={{\delta }^{\mu }}_{\nu }+{{\omega }^{\mu }}_{\nu },a^{\mu }={\epsilon }^{\mu }.$$Then we have $$\begin{array}{rl}{g}_{\mu \nu }=& g_{\rho \sigma }({{\delta }^{\rho }}_{\mu }+{{\omega }^{\rho }}_{\mu })({{\delta }^{\sigma }}_{\nu }+{{\omega }^{\sigma }}_{\nu })\\ & =g_{\mu \nu }+{\omega }_{\mu \nu }+{\omega }_{\nu \mu }+o({\omega }^2).\end{array}$$Hence ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$ is antisymmetry, hence has $6$ independent components and $4$ components of ${\epsilon }^{\mu }$.

  An Poincaré transformation is described by $6+4=10$ parameters.

  The transformation $T(\Lambda ,a)$ induce a unitary linear transformation in the Hilbert space $$|\psi ⟩\to U(\Lambda ,a)|\psi ⟩.$$The operators $U$ satisfy $$U(\overline{\Lambda },\overline{a})U(\Lambda ,a)=U(\overline{\Lambda }\Lambda ,\overline{\Lambda }a+\overline{a}.$$

  Choose $U(1,0)=1$ the unit operator, then $$U(1+\omega ,\epsilon )=1+\frac{1}{2}i{\omega }_{\rho \sigma }{J}^{\rho \sigma }-i{\epsilon }_{\rho }{P}^{\rho }+\cdots$$