<reponame>jacopok/notes
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\section{Symmetries and conservation laws}

\marginpar{Tuesday\\ 2020-3-17, \\ compiled \\ \today}

Our aim is to describe the fundamental constituents of matter with a Quantum Field Theory. 
The method used to derive the equations of motion is a variational principle: we will find a Lagrangian density for various particles, and then apply the variational principle to find their equations of motion. 

A guiding principle on the description of these fundamental particles is based on using their symmetries. 
We have Nöether's theorem in Quantum Field Theory: from these symmetries we are able to find conserved quantities. 

These symmetries are described with groups, since we can compose their application; the theory describing groups is very rich. 
For this lecture we will base ourselves on Peskin's chapter 2 \cite[]{peskinConceptsElementaryParticle2019}.

A group \(G\) is a set of elements endowed with an operation. The set of elements can be either discrete or continuous. Examples of discrete transformations are the parity transformation \(P\): \(P \vec{x} = -\vec{x}\) and the time swap \(T\): \(T x^{\mu } = (-x^{0}, \vec{x})\). Continuous symmetries, on the other hand, are parametrized by one or more continuous-valued parameters. 

We distinguish: 
\begin{enumerate}
  \item \textbf{spacetime} symmetries: groups which transform our coordinate system for spacetime, such as Lorentz and Poincaré transformations;
  \item \textbf{internal} symmetries: groups which transform a certain field, or a certain property of our quantum system.
\end{enumerate}

For our set to be a group, we need to be able to define an operation --- we will usually call it multiplication --- between the elements of the group, such that if \(a, b \in G\) then \(ab \in G\). Also, we must have 
\begin{enumerate}
  \item associativity: \((ab)c = a(bc)\);
  \item existence of the identity \(\mathbb{1}\), such that \(\mathbb{1} a = a \mathbb{1} = a\);
  \item existence of inverses: there exists \(a^{-1}\) such that \(a a^{-1} = a^{-1} a = \mathbb{1}\). 
\end{enumerate}

What is of interest to us is the association of the group with a transformation which is a symmetry: this is called a \emph{representation}, which associates to each \(g \in G\) a unitary operator \(U_g\) acting on the quantum states. We ask that this representation should preserve the group structure, that is to say, \(U_{gh} = U_g U_h\) and \(U_{g^{-1}} = U_{g}^{-1}\).

We call a transformation a symmetry if, after performing the transformation, the dynamics of the system do not change. 

For a quantum mechanical system, we are interested in the observables: these are described by operators, whose eigenvalues are the observations we make, and which in the Heisenberg picture evolve like 
%
\begin{align}
- i \hbar \dv{}{t} O(t) = [H, O(t)]
\,;
\end{align}
%
if an operator commutes with the Hamiltonian, \([H,O]=0\), then the operator's expectation value on any state is constant --- which is to say, the operator is constant. 

If we perform a transformation in the form 
%
\begin{align}
\ket{\psi } \rightarrow \ket{\psi'} = U \ket{\psi }
\,,
\end{align}
%
then the operators will change by 
%
\begin{align}
O \rightarrow O^{\prime } = U ^\dag O U 
\,.
\end{align}

Note that whether we have \(U ^\dag O U\) or \(U O U ^\dag\) does not matter, since we ask observables \(O\) to be Hermitian, so \(O = O ^\dag\). 

We know that these transformations must always be unitary, because the conservation of probability implies that we must have \(\braket{\psi }{\psi } = \const\): so, 
%
\begin{align}
U ^\dag U = \mathbb{1}
\,.
\end{align}

This can be also stated as \(U ^\dag = U^{-1}\). 

So, the function associating a unitary operator \(U\) to an element \(g\) of the group is called its \emph{unitary representation}. 

A transformation \(G\) is a symmetry if \(\forall a \in G\) we have 
%
\begin{align}
[U(a), H]  =0
\,,
\end{align}
%
that is, the unitary representation of the group element always commutes with the Hamiltonian.

If we have a state \(\ket{\psi }\) with energy \(H \ket{\psi } = E \ket{\psi }\), then the transformation commuting with the Hamiltonian means that \(\ket{\psi'} = U \ket{\psi }\) has the same energy: 
%
\begin{align}
H (U \ket{\psi }) = HU \ket{\psi } \overset{[H, U] = 0}{=} UH \ket{\psi } =
 U E \ket{\psi } = E \qty(u \ket{\psi })
\,,
\end{align}
%
so the eigenvalue of \(U \ket{\psi }\) is the same as that of \(\ket{\psi }\).

Now, we can move to an example, taken from Peskin \cite[eq.\ 2.38 onward]{peskinConceptsElementaryParticle2019}. 
Consider the discrete group \(\mathbb{Z}_{2}\), which only has the elements \(1\) and \(-1\), with the same multiplication rules as those we would have if these elements were integers. 
So, the group is closed with respect to multiplication.
It can be easily checked that this is indeed a group based on our definition. 

In order for this to be of interest to us, we can consider a quantum mechanical system and find a unitary representation acting on its Hilbert space. 

Let us suppose we have a QM system with a basis made of two states \(\ket{\pi^{+}}\) and \(\ket{\pi^{-}}\). Let us define the \emph{charge conjugation} operator \(C\), by: 
%
\begin{align}
C \ket{\pi^{+}} = \ket{\pi^{-}} 
\qquad \text{and} \qquad
C \ket{\pi^{-}} = \ket{\pi^{+}} 
\,.
\end{align}

So, we can find a unitary representation of \(\mathbb{Z}_{2}\) in this system: we need to define \(U(1)\) and \(U(-1)\). We define 
%
\begin{subequations}
\begin{align}
U(1) = \mathbb{1} =\left[\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array}\right] 
\qquad \text{and} \qquad
U(-1) = \sigma_{x} = \left[\begin{array}{cc}
0 & 1 \\ 
1 & 0
\end{array}\right]
\,,
\end{align}
\end{subequations}
%
where the matrices are to be interpreted as acting on vectors expressed to the basis \(\qty{\ket{\pi_{+}}, \ket{\pi_{-}}}\).

So, we can say that our unitary representation looks like 
%
\begin{align}
\mathbb{Z}_{2} \rightarrow \qty{\mathbb{1}, C}
\,.
\end{align}

Now, if \([C, H] =0\) (and \(\mathbb{1}\) commutes with \(H\), which is always the case) then we say that ``\(H\) has the symmetry \(\mathbb{Z}_{2}\)'': this implies that the energies of the two \(\pi_{\pm}\) particles are equal.

The interesting question to determine will be whether this is actually the case for our given group. 

Groups can be subdivided into abelian and non-abelian ones.
A group is abelian if for every \(a, b\) in \(G\) we have \(ab = ba\), or equivalently, \([a, b] =0\). 
It is not if this is not the case, that is, there exist \(a, b\) such that \(ab \neq ba\).

The condition on the elements directly translates to a condition on the matrices of the unitary representation. 
If we have commuting matrices, we can simultaneously diagonalize them: for example, in the case of \(\mathbb{Z}_{2}\) we can go to a basis in which 
%
\begin{subequations}
\begin{align}
C = \left[\begin{array}{cc}
1 & 0 \\ 
0 & -1
\end{array}\right]
\,,
\end{align}
\end{subequations}
%
specifically the states on which this matrix will act will need to be 
%
\begin{align}
\ket{\pi_{1}} = \frac{\ket{\pi^{+}} + \ket{\pi^{-}}}{\sqrt{2}} 
\qquad \text{and} \qquad
\ket{\pi_{2}} = \frac{\ket{\pi^{+}} - \ket{\pi^{-}}}{\sqrt{2}}
\,,
\end{align}
%
since then \(C \ket{\pi_1 } = \ket{\pi_1 }\) (we write \(C = +1\)) and \(C \ket{\pi_{2}} = - \ket{\pi_2 }\) (we write \(C = -1\)). 
We will often use this notation, confusing operator and eigenvalue.

In the case of nonabelian groups it is not in general possible to diagonalize all the matrices; we can however do a change of basis and write the matrices as a block matrix with the smallest possible blocks:
%
\begin{subequations}
\begin{align}
U_{R} = \left[\begin{array}{ccc}
U_1  & 0 & 0 \\ 
0 & U_2  & 0 \\ 
0 & 0 & \dots
\end{array}\right] 
\,,
\end{align}
\end{subequations}
%
where the matrices \(U_i\) are called the \textbf{irreducible unitary representations of \(G\)}.
The dimension of the matrices \(U_i\) tells us the dimension of these irreducible unitary representations.

\todo[inline]{Add more details on irreps --- maybe not here? They can be found in professor Rigolin's intro to groups.}

Do note that some elements of a nonabelian group can commute: for example, in the rotation group we have 
%
\begin{align}
[J^{i}, J^{j}] = \epsilon^{ijk} J_{k}
\,,
\end{align}
%
so if we take \(i = j\), that is, we consider rotations along the same axis, they will commute since then the Kronecker symbol is equal to zero. 

\subsection{Continuous transformations: space translations}

An element of the group can be written as 
%
\begin{align}
U(a) = e^{ - i a P }
\,,
\end{align}
%
where the operator \(P\), whose eigenvalue is the momentum, is called the generator of the transformation. 

If we consider a plane wave we can clearly see how this action works: if we start from
%
\begin{align}
\braket{x}{p} = e^{i p x}
\,,
\end{align}
%
we can apply the operator \(U(a)\) to \(\ket{p}\), which will yield \(e^{-ipa}\) (since eigenvectors of an operator are also eigenvectors of its exponential): so we find
%
\begin{align}
\bra{x} U(a) \ket{p} = e^{i p (x-a)}
\,,
\end{align}
%
which means that by acting with this operator we have effectively performed a translation with displacement \(a\). 

If our system is invariant under translations, then Nöether's theorem tells us that the momentum is conserved. 

In order to be a physical observable \(P\) needs to be Hermitian: \(P = P ^\dag\). 

So, the adjoint of the transformation \(U(a)\) is 
%
\begin{align}
U ^\dag (a) =
\sum _{n} \qty(\frac{(-iaP)^{n}}{n!}) ^\dag
= \sum _{n} \frac{(iaP ^\dag)^{n}}{n!}
=
e^{i a P ^\dag} = e^{iaP} = U^{-1}(a)
\,,
\end{align}
%
which confirms the fact that the transformation is unitary. 

Let us suppose that the momentum operator \(P\) commutes with the Hamiltonian: \([P,H] = 0\). Then, 
%
\begin{align}
[U(a), H] = 0 
\,,
\end{align}
%
that is, the Hamiltonian is translation-invariant. 

All this is to say that a constant of motion \(O\) corresponds to an operator \(O\) which commutes with the Hamiltonian. 
This is formalized by Nöther's theorem, which establishes the equivalence between symmetries and conservation laws:
%
\begin{align}
[O, H] = 0 
\iff 
[U_O, H] = 0
\,.
\end{align}

As an example, take the group \(G\) of 3D rotations. 
They depend on a continuous parameter \(\vec{\alpha}\), just like translations depended on the parameter \(a\). 

The rotation is written as 
%
\begin{align}
U(\vec{\alpha}) = e^{-i \vec{\alpha} \cdot \vec{J}}
\,,
\end{align}
%
where the components of the angular momentum have the following commutation relations: 
%
\begin{align}
[J^{i}, J^{j}] = i \epsilon^{ijk} J^{k}
\,.
\end{align}

We will be able to compose the representations of rotations: 
%
\begin{align}
U(\vec{\beta}) U(\vec{\alpha}) = U(\vec{\gamma})
\,.
\end{align}

This space of 3D rotations is called SO(3), since every rotation corresponds to a 3x3 matrix which is a rotation matrix --- it is orthogonal and has determinant 1.

Now, we seek \textbf{representations} of these rotations: so, we choose the dimension \(d\) of a quantum-mechanical vector and describe how it changes upon the action of the unitary matrices found by exponentiating certain \(d\)-dimensional generators \(J^{i}\), which must have the algebra discussed above. 

If we look for 1D representations of the generators \(J^{i}\) the only option we find is \(J^{i} = 0\), which means that we are not actually performing a rotation. This is because scalars commute with each other.
Which states transform this way? These are scalar states, with spin 0. 

For 2D representations, we have 
%
\begin{align}
J^{i} =\frac{1}{2} \sigma^{i} 
\,,
\end{align}
%
where the \(\sigma^{i}\) are the Pauli matrices.

We can also find 3D representations, which look like 
%
\begin{subequations}
\begin{align}
J^{1}= \left[\begin{array}{ccc}
0 & 0 & 0 \\ 
0 & 0 & -i \\ 
0 & i & 0
\end{array}\right] 
\qquad 
J^{2}= \left[\begin{array}{ccc}
0 & 0 & i \\ 
0 & 0 & 0 \\ 
-i & 0 & 0
\end{array}\right] 
\qquad 
J^{1}= \left[\begin{array}{ccc}
0 & -i & 0 \\ 
i & 0 & 0 \\ 
0 & 0 & 0
\end{array}\right] 
\,
\end{align}
\end{subequations}
%
and represent a spin 1 particle. In general, spin \(s\) corresponds to a \(2s+1\)-dimensional representation.

A rotation in 2D, represented by an element of SO(2), corresponds to a phase shift, so we can say that it is equivalent to an element of U(1).
This then allows us to see that SO(2) is abelian. 

In general, we write for a unitary \(n \times n\) representation
%
\begin{align}
U(n) \rightarrow e^{-i \alpha^{n} t^{a}}
\,,
\end{align}
%
where the generators \(t^{a}\) are Hermitian matrices corresponding to Hermitian operators. 
In particular, conventionally we say that one of these is the identity: \(t^{0} = \mathbb{1}\) (which must always be included in the group, lest we lose closure).

So, we omit it and say that we have \(n^2-1\) generators for the SU\((n)\) group.
We shall see that each of these generators corresponds to a particle, and for the weak interaction we will have \(2^2-1 = 3\) particles, while for the strong one we will have \(3^2-1=8\). 

In general, if \(t^{a}\) are the generators of an abstract Lie group, we can describe the algebra of the group by 
%
\begin{align}
\qty[t^{a}, t^{b}] = i f^{abc} t^{c}
\,,
\end{align}
%
so, the commutator is decomposed into a linear combination of the generators, whose coefficients \(f^{abc}\) are called the \textbf{structure constants} of the group.

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