# Sec2.1 Lorentz Invariance

\(\quad\)QFT is based on quantum mechanics, so we provide only the briefest version of summaries, in the generalized version of Dirac:

**Def(Ray):**

A \emph{ray} \(\mathscr{R}\) is a set of normalized vectors, i.e., \(\{\psi|\langle\psi,\psi\rangle=1\}\). Here \(\langle\cdot,\cdot\rangle\) is the inner product of Hilbert space, satisfying \(\forall\phi\psi\in H,\xi,\eta\in\mathbb{C}\),\par

1) \(\langle\phi,\psi\rangle=\langle\psi,\phi\rangle^{*}\);

2) \(\langle\phi,\xi_{1}\psi_{1}+\xi_{2}\psi_{2}\rangle=\xi_{1}\langle\phi,\psi_{1}\rangle+\xi_{2}\langle\phi,\psi_{2}\rangle\);

3) \(\langle\eta_{1}\phi_{1}+\eta_{2}\phi_{2},\psi\rangle=\eta_{1}^{*}\langle\phi_{1},\psi\rangle+\eta_{2}\langle\phi_{2},\psi\rangle\);

4) \(\langle\psi,\psi\rangle\geqslant0\quad\) and \(\langle\psi,\psi\rangle=0\Rightarrow\psi=0.\)

**Def:**

The \emph{self-adjoint} operator of one linear operator \(A\) of Hilbert space, denoted \(A^{\dagger}\), is defined as \(\forall\phi,\psi\in H, \langle\phi,A^{\dagger}\psi\rangle:=\langle A\phi,\psi\rangle\).

**Axiom(QM):**

1) Physical states are represented by rays in Hilbert space.

2) Observables are represented by Hermitian operators\footnote{More seriously, ’cause the domain of operators involve in some complicated problems, observables in quantum mechanics are essentially \emph{unbounded self-adjoint} operators in Functional Analysis.}, that is, \(A=A^{\dagger}\).

3) The probability of finding a state represented by \(\mathscr{R}\) in the mutually orthogonal states \(\mathscr{R}_{n}\) is \(P(\mathscr{R}\rightarrow\mathscr{R}_{n})=|\langle\psi,\psi_{n}\rangle|^{2}\).

**Def:**

A \emph{symmetry transformation} \(T\) is a change in the point of view that does not change the results of possible experiments. That is, observer \(\mathcal{O}\) sees a system in a state represented by ray \(\mathscr{R}_{i}\), and observer \(\mathcal{O}’\) looking at the same system will observe it in a different state, represented by ray \(\mathscr{R}’_{i}\). Then we always have

$$P(\mathscr{R}\rightarrow\mathscr{R}_{n})=P(\mathscr{R}’\rightarrow\mathscr{R}’_{n})$$

\(\quad\)Obviously the set of symmetry transformation \(T_{1}:\mathscr{R}\mapsto\mathscr{R}’\) form a group if we naturally define the trivial transformation \(\mathscr{R}\mapsto\mathscr{R}\) as its identity and the group product is defined as \(T_{2}T_{1}:\mathscr{R}\mapsto\mathscr{R}”\), where \(T_{2}:\mathscr{R}’\mapsto\mathscr{R}”\), and the inverse is \(T_{1}^{-1}:\mathscr{R}’\mapsto\mathscr{R}\).

**Theorem(Wigner):**

Any symmetry transformation \(T\) of rays can be represented as operators on Hilbert space, with \(U(T)\) either \emph{unitary and linear}:

$$\langle U\phi,U\psi\rangle=\langle\phi,\psi\rangle,\quad U(\eta\phi+\xi\psi)=\eta U\phi+\xi U\psi,$$

or \emph{antiunitary and antilinear}:

$$\langle U\phi,U\psi\rangle=\langle\phi,\psi\rangle^{*},\quad U(\eta\phi+\xi\psi)=\eta^{*} U\phi+\xi^{*} U\psi,$$

**Prooof:**

skipped.

\(\quad\)Since the set of symmetry transformation \(\{T\}\) is a group \(G\), we naturally find a map \(U:G\rightarrow GL(H), U=U(T)\), in the sense of Wigner theorem. But this map may not be \emph{homomorphism}. In fact, we can only get

$$U(T_{1})U(T_{2})=e^{\phi(T_{1},T_{2})}U(T_{1}T_{2})$$

from our former discussion. We call this non-homomorphism map as \emph{projective representation} of the symmetry transformation group.

In order to avoid tedious discussion about the phase of projective representation(we will return to this topic in the end of this chapter), we directly admitted that

**Assertion(non-projective representation):**

Any representations of symmetry transformation group with phases \(\phi\) can be canceled through group enlarging(without changing its physical implications). That is, we choose such a homomorphism map of \(U:G\rightarrow GL(H)\) that

$$U(T_{1})U(T_{2})=U(T_{1}T_{2}).$$

# Sec2.2 Lorentz Transformation

\(\quad\)By the \emph{Special equivalence principle} of Einstein, the unit distance holds in any inertia coordinates, i.e.,

\begin{equation}\label{2.2.1}\eta_{\mu\nu}\mathrm{d} x’^{\mu}\mathrm{d} x^{\nu}=\eta_{\mu\nu}\mathrm{d} x^{\mu}\mathrm{d} x^{\nu},\end{equation}

where the metric in our notation is

$$\eta_{\mu\nu}=\begin{pmatrix}+1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}.$$

Any transformations \(T:x^{\mu}\mapsto x’^{\mu}\) satisfying \eqref{2.2.1} have the linear form that

\begin{equation}\label{2.2.2}x’^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}+a^{\mu},\end{equation}

such that

\begin{equation}\label{2.2.3}\eta_{\mu\nu}\Lambda_{\rho}^{\mu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}.\end{equation}

Precisely denote the transformation as \(T(\Lambda,a)\), then from our former discussion, the set of \(T\) forms a group(the existence of inverse can be seen from \((\mathrm{det}\Lambda)^{2}=1)\), and its group product is

**Assertion:**

\begin{equation}\label{2.2.4}T(\bar{\Lambda},\bar{a})T(\Lambda,a)=T(\bar{\Lambda}\Lambda,\bar{\Lambda}a+\bar{a}).\end{equation}

**Proof:**

We firstly perform a transformation \(x^{\mu}\rightarrow x’^{\mu}\) as \eqref{2.2.2}, then continue to perform another transformation \(x’^{\mu}\rightarrow x”^{\mu}\), which gives

$$x”^{\mu}=\bar{\Lambda}_{\rho}^{\mu}\Lambda_{\nu}^{\rho}x^{\nu}+\left(\bar{\Lambda}_{\rho}^{\mu}a^{\rho}+\bar{a}^{\mu}\right).$$

And thus \(T(\Lambda,a)\) satisfy the claimed relation.

According to the discussion in the former section, transformation group \(\{T(\Lambda,a)\}\) induce a unitary representation \(U(T)\) acting on Hilbert space. Homomorphism transfer the product rule of transformation group \eqref{2.2.4} to \(GL(H)\), giving

\begin{equation}\label{2.2.5}U(\bar{\Lambda},\bar{a})U(\Lambda,a)=U(\bar{\Lambda}\Lambda,\bar{\Lambda}a+\bar{a}).\end{equation}

From \eqref{2.2.5}, we can easily write down the inverse map \(U^{-1}(\Lambda,a)\)

\begin{equation}\label{2.2.6}U^{-1}(\Lambda,a)=U(\Lambda^{-1},-\Lambda^{-1}a)\end{equation}

since \(U(\Lambda,a)U(\Lambda^{-1},-\Lambda^{-1}a)=U(1,0)=\mathbbold{1}\).

\(\quad\)Then we are to introduce an important symmetry group in physics.

**Def:**

The group of transformation \(T(\Lambda,a)\) is called \emph{inhomogeneous Lorentz group}, or \(\textit{Poincar}\acute{e} \textit{group}\).

\(\quad\)One crucial subgroup \(L\) of the \(\mathrm{Poincar\acute{e}}\) group’s is one that \(a^{\mu}=0\). Specifically,

$$L=\{T|T(\bar{\Lambda},0)T(\Lambda,0)=T(\bar{\Lambda}\Lambda,0).\}.$$

We call this \emph{homogeneous Lorentz group}, denoted as \(\mathrm{O}(1,3)\).

It is easy to show that group \(\mathrm{O}(1,1)\) has four distinct components of the form

$$\Lambda=\begin{pmatrix}\pm\cosh\theta&\mp\sinh\theta\\ \mp\sinh\theta&\pm\cosh\theta\end{pmatrix}.$$

Since \(\cosh\theta\) is always more than zero, these four components are disjoint with each other, indicating that group \(\mathrm{O}(1,1)\) is topologically not connected. Basing on our discussion in low-dimension conditions, most of physical textbooks directly claim as their please that so dose group \(\mathrm{O}(1,3)\). However, we should not randomly promote the disconnectedness of general Lorentz group because the topology structure of \(\mathrm{O}(1,n)\) is entirely different from the simple \(\mathrm{O}(1,1)\) case.

**Theorem(disconnectedness of \(\mathrm{O}(1,3)\))**

The Lorentz group \(L\) has four connected components. They are

\begin{align*}L_{+}^{\uparrow}&=\{\Lambda|\det\Lambda=1,\Lambda_{00}\geqslant1\},\\ L_{-}^{\uparrow}&=\{\Lambda|\det\Lambda=-1,\Lambda_{00}\geqslant1\},\\ L_{+}^{\downarrow}&=\{\Lambda|\det\Lambda=1,\Lambda_{00}\leqslant-1\},\\ L_{-}^{\downarrow}&=\{\Lambda|\det\Lambda=-1,\Lambda_{00}\leqslant-1\}.\end{align*}

Especially, component \(L_{+}^{\uparrow}\) is called the \emph{proper chronological Lorentz group}, or mathematically, \(\mathrm{SO}_{+}(1,3)\).

**Proof:**

Take the determinate of \eqref{2.2.3} immediately gives \(\det\Lambda=\pm1\). Also, let indices \(\rho=\sigma=0\), \eqref{2.2.3} gives \(1=\Lambda_{00}^{2}-\Lambda_{10}^{2}-\Lambda_{20}^{2}-\Lambda_{30}^{2}\). Thus, we have either \(\Lambda_{00}\geqslant1\) or \(\Lambda_{00}\leqslant-1\), which follows a disjoint union of open set:

$$L=L_{+}^{\uparrow}\bigcup L_{-}^{\uparrow}\bigcup L_{+}^{\downarrow}\bigcup L_{-}^{\downarrow}.$$

Moreover, since any Lorentz transformation in the three other set \(L_{-}^{\uparrow}, L_{+}^{\downarrow}, L_{-}^{\downarrow}\) can be written as product of discrete transformation(\(\mathscr{P}\equiv\mathrm{diag}\{-1,1,1,1\}\) or \(\mathscr{T}\equiv\mathrm{diag}\{1,-1,-1,-1\}\)) and one element of \(L_{+}^{\uparrow}\), thus it suffices to show the connectedness of \(L_{+}^{\uparrow}\). Let

$$H=\{x=(x^{0},\cdots,x^{3})^{T}\in\mathbb{R}^{4}|x^{\mu}x_{\mu}=1, x^{0}\geqslant1\},$$

certainly \((x^{0},\cdots,x^{3})^{T}\mapsto(x^{1},x^{2},x^{3})^{T}\) defines a diffeomorphism(note that \(x^{\mu}x_{\mu}\) confines \(x^{0}\)) of \(H\) with \(\mathbb{R}^{3}\). If \(v_{0}\in H\), we can complete \(v_{0}\) to an orthogonal normalized basis of \(\mathbb{R}^{4}\) through Gram-Schmidt procedure. Denote the super-vector(as an operator) as \(\mathcal{V}=(v_{0},\cdots,v_{3})\). Because by definition \(v_{0}^{0}>1\) and all basis are orthogonal to each other, implying that \(\det\mathcal{V}=1\), \(\mathcal{V}\) should belongs to \(L_{+}^{\uparrow}\). And for \(e_{0}=(1,0,0,0)^{T}\in H\),

$$(v_{0},\cdots,v_{3})e_{0}=\sum_{k=0}^{3}v_{i}e_{0}^{i}=v_{1},$$

thus the map \(\pi:L_{+}^{\uparrow}\rightarrow H\) given by \(\pi(\Lambda)=\Lambda e_{0}\) is onto(for \(v_{0}\) is arbitrary). It can be seen that the element of \(\pi^{-1}(e_{0})=\{\Lambda\in L_{+}^{\uparrow}|\Lambda e_{0}=e_{0}\}\) has the form of

$$\Lambda=\left(v_{0},v_{1},v_{2},v_{3}\right),$$

where \(v_{0}=(1,0,0,0)^{T}\). On the other hand, since the orthogonality demands \(\langle v_{0},v_{i}\rangle=0\Rightarrow v_{i}^{0}=0\) and \(\langle v_{i},v_{j}\rangle=0\), \(\Lambda\) must be an element of \(\mathrm{SO}(3)\) and therefore \(\pi^{-1}(e_{0})\cong\mathrm{SO}(3)\). It is also easy to see that \(\pi^{-1}(v^{0})=\mathcal{V}\pi^{-1}(e^{0})\).

Indeed, note that \(L_{+}^{\uparrow}\) is Lie subgroup of \(\mathrm{O}(1,3)\), \(\mathrm{SO}(3)\) acts on \(L_{+}^{\uparrow}\) to the right in such a way that \(\pi:L_{+}^{\uparrow}\rightarrow H\) is a principle fiber bundle over \(H\cong\mathbb{R}^{3}\) with group \(\mathrm{SO}(3)\). Any bundle over \(\mathbb{R}^{n}\) is trivial, and so

\(L_{+}^{\uparrow}\) is topologically \(\mathbb{R}^{3}\times\mathrm{SO}(3)\), which is obviously connected.

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