The Formalism

The language of particle physics is the Quantum Field Theory. In this formalism to each fundamental excitation corresponds a quantum field

$$\chi_{\alpha}(x)$$ (13)

where $\alpha$ indicates the necessary group of indeces to characterize the field. It is also possible to easily add eventual symmetries introducing invariant Lagrangians (or Actions) in front of a given symmetry operations. That is to say, functions which do not change when the fields are subjected to the operation of a $G$ group of symmetry transformations. In this case, the field $\chi_{\alpha}(x)$ is a member of an irreducible multiplet, that is to say, under the operation $a$ of the $G$ group

$$\chi_{\alpha}(x) \rightarrow \chi'_{\alpha}(x) = {\cal D}_{\alpha\beta}(a) \chi_{\beta}(x)$$ (14)

where ${\cal D}_{\alpha}(a)$ is the corresponding representation of $G$. So, if

$$\chi_{\alpha}(x) \stackrel{a}{\rightarrow}\chi'_{\alpha}(x)\... ...\chi_{\alpha}(x) \stackrel{a''} {\rightarrow}\chi''_{\alpha}$$ (15)

$$a'' = a . a'$$ (16)

then

$${\cal D}_{\alpha\beta}(a') {\cal D}_{\beta\gamma}(a) = {\cal D}_{\alpha\gamma}(a'')$$ (17)

These operations imply, seen from the Hilbert space, the existence of a unitary operator $U(a)$ for the operation $a$, so it satisfies

$$U(a) U(a') = U(a'')$$ (18)

and acting on the field verifies the relation

$$U^{-1}(a) \chi_{\alpha}(x) U(a) = \chi'_{\alpha}(x) = {\cal D}_{\alpha\beta}(a) \chi_{\beta}(x)$$ (19)

which indicates that

If we are dealing with continuous transformations, the study can be restricted to the infinitesimal transformations of the form

$$U(\delta a) = 1 + i {\delta} a_{j} G_{j}$$ (20)

where $G_{j}$ are the generators of the group algebra. As unitary groups are concerned, $G_{j}$ are hermitic operators. The Lie algebra of the group is defined by

$$[G_{i},G_{j}]= i C_{ijk} G_{k}$$ (21)

with $C_{ijk}$ the structure constants of the group. The corresponding representations $g_{j}$ of the generators $G_{j}$ are defined by

$${\cal D}_{\alpha\beta}(\delta a) = \delta_{\alpha\beta} + i {\delta}a_{j} (g_{j})_{\alpha\beta}$$ (22)

with the obvious property

$$[g_{i},g_{j}]= i C_{ijk} g_{k}$$ (23)

Being $\chi_{\alpha}(x)$ an irreducible tensorial operator, one can immediately write the relation

$$[G_{j},\chi_{\alpha}(x)]= - (g_{i})_{\alpha\beta} \chi_{\beta}(x)$$ (24)

A theory is invariant under a group $G$ if after the symmetry transformation it is verified that

$$S = \int d^{4}x {\cal L}(\chi_{\alpha}, \partial_{\mu}\ch... ...{4}x {\cal L}(\chi'_{\alpha}, \partial_{\mu}\chi'_{\alpha})$$ (25)

that is to say

$$\delta S = 0$$ (26)

The invariance leads to the presence, via Noether theorem, of a conserved current. As the field variation is written

$$\delta \chi_{\alpha} = \chi'_{\alpha} - \chi_{\alpha} = i {\delta}a_{j} (g_{j})_{\alpha\beta} \chi_{\beta}$$ (27)

it results

$$j_{j}^{\mu}(x) = \frac{\partial{\cal L}}{\partial\partial_{... ...a}(x)} \frac{1}{i} (g_{j})_{\alpha\beta} \chi_{\beta}(x)$$ (28)

with the property

$$\partial j_{j}^{\mu}(x) = 0$$ (29)

which also implies that the generator (the charge)

$$G_{j} = \int d^{3}x j_{j}^{0}(x)$$ (30)

is independent of the time, that is to say

$$[H,G_{j}]= 0$$ (31)

which is other way of expressing the invariance of the theory under the group $G$. Let us study now the states. A state of one particle is written: $\mid p;\alpha>$ with $p^{2} = - m^{2}$.

If and only if the vacuum is $G$ invariant, that is, if and only if

$$U(a) \mid0> = \mid0>$$ (32)

that is

$$G_{j} \mid0> = 0$$ (33)

then the states of a particle associated with the field $\chi_{\alpha}(x)$ are analogously transformed to $\chi_{\alpha}(x)$ Let us remember the reduction scheme LSZ

$$\mid p;\alpha> = \lim_{x^{0}\rightarrow\pm\infty}\int d^{3}x... ...tackrel{\leftrightarrow}{\partial_{0}} \chi_{\alpha}(x)\mid 0>$$ (34)

and then

$$U^{-1}(a) \mid p;\alpha> = \lim_{x^{0}\rightarrow\pm\infty}\int d^{3}x e^{ipx} \frac{1}{i}\stackrel{\leftrightarrow}{\partial_{0}} U^{-1}(a)\chi_{\alpha}(x) \mid 0>$$ (35)

$$= \lim_{x^{0}\rightarrow\pm\infty}\int d^{3}x e^{ipx} \frac{1}{i}\stackrel{\leftrightarrow}{\partial_{0}} U^{-1}(a)\chi_{\alpha}(x)U(a) \mid 0>$$ (36)

$$= \lim_{x^{0}\rightarrow\pm\infty}\int d^{3}x e^{ipx} \frac{1}{i... ...el{\leftrightarrow}{\partial_{0}} {\cal D}_{\alpha\beta}(a)\chi_{\beta} \mid 0>$$ (37)

that is to say

$$U^{-1}(a) \mid p;\alpha> = {\cal D}_{\alpha\beta}(a) \mid p;\beta>$$ (38)

This property implies that all the states of the multiplete $\mid p;\alpha>$ have the same mass: If $p^{\mu} \equiv (m_{\alpha},0)$, the rest state $\mid p;\alpha>_{rest}$ is defined through

$$H \mid p;\alpha>_{rest} = m_{\alpha} \mid p;\alpha>_{rest}$$ (39)

On the other hand, due to the invariance under $G$ it is possible to write

$$[H,U^{-1}(a)]= 0$$ (40)

then

$$0 = [H,U^{-1}] \mid p;\alpha>_{rest}$$ (41)

$$= {\cal D}(a)_{\alpha\beta} (m_{\beta} - m_{\alpha}) \mid p;\alpha>_{rest}$$ (42)

which taking into account the arbitrariness of ${\cal D}(a)$ shows that $m_{\beta} = m_{\alpha}$

We conclude that the particles associated with fields which irreducibly transform under a group $G$ have the same mass. This is the Wigner-Weyl realization of the symmetry. As an example we remember the isospin symmetry which implies $m_{p} = m_{n}$ when electromagnetism is absent.

What happens when the vaccum is not $G$ invariant, that is to say when

$$U^{-1}(a) \mid p;\alpha> \neq 0$$ (43)

Then the fundamental state is not unique: the fundamental state is degenerate. As a consequence the symmetry in this case does not imply multipletes of particles of the same mass. That is

$$U^{-1}(a) \mid p;\alpha> \neq {\cal D}_{\alpha\beta}(a) \mid p;\beta>$$ (44)

even if

$$[H,G_{j}]= 0$$ (45)

In this case we speak of a Nambu-Goldstone realization of the symmetry and the following theorem is verified: If a theory is $G$ invariant and the vaccum state (fundamental) is not $G$ invariant then there appear null mass excitations called Goldstone bosons. Let us see why. The relation

$$[G_{j},\chi_{\alpha}(x)]= - (g_{j})_{\alpha\beta} \chi_{\beta}(x)$$ (46)

leads to

$$<0\mid [G_{j},\chi_{\alpha}(x)] \mid 0> = - (g_{j})_{\alpha\beta} <0\mid \chi_{\beta}(x) \mid 0>$$ (47)

and as the realization of the symmetry is $\grave{a}\; la$ Nambu-Goldstone

$$< 0\mid \chi_{\beta} \mid 0> \neq 0$$ (48)

Besides, we can explicitly write

$$I = - (g_{j})_{\alpha\beta}<0\mid \chi_{\beta}(x)\mid 0>$$

$$= \int d^{3}x <0\mid j_{j}^{0}(y) \chi_{\alpha}(x) - \chi_{\alpha}(x) j_{j}^{0}(y) \mid 0>$$

$$= R$$ (49)

Introducing now intermediate states $\mid n>$ and the traslation

$$j_{i}^{0}(y) = e^{-ipy} j_{i}^{0}(0) e^{ipy}$$ (50)

results

$$I = \sum_{n} \int d^{3}y\{e^{ip_{n}y} <0\mid j_{j}^{0}\mid n> <n\mid \chi_{\alpha}(x)\mid 0> -$$

$$e^{-ip_{n}y}<0\mid \chi_{\alpha}(x)\mid n><n\mid j_{j}^{0}\mid 0>\}$$ (51)

$$= \sum_{n} (2 \pi)^{3} \delta^{3}(\vec{p}_{n})\{e^{-ip_{n}^{0}y^{0}}<0\mid j_{j}^{0}(0)\mid n><n\mid \chi_{\alpha}(x)\mid 0> -$$

$$e^{ip_{n}^{0}y^{0}}<0\mid \chi_{\alpha}(x)\mid n><n\mid j_{j}^{0}(0)\mid 0>\}$$ (52)

On the other hand

$$\partial_{\mu} j_{j}^{\mu} = 0 \Rightarrow \frac{\partial}{\... ...{3}y j_{j}^{0}(x) + \int d^{3}y \vec{\nabla}\vec{ j_{j}} = 0$$ (53)

and then

$$\frac{\partial}{\partial y^{0}}\int d^{3}y<0\mid [ j_{j}^{0}(y),\chi_{\alpha}(x)]\mid 0> = -\int d^{3}y <0\mid [\vec{\nabla}\vec{j_{j}},\chi_{\alpha}(x)]\mid 0>$$

$$= -\int_{S}d\vec{\sigma}<0\mid [\vec{j}_{j},\chi_{\alpha}]\mid 0>$$

$$= \frac{\partial R}{\partial y^{0}} = 0$$ (54)

If we now observe in I the terms $\mid n>$ with $m_{n}\neq 0$ we see that they lead to factors

$$e^{\pm i m_{n}y^{0}}$$ (55)

which imply an explicit dependence in $y^{0}$ not allowed and therefore it must be

$$<n\mid j_{j}^{0} \mid 0> \equiv 0$$ (56)

Besides, the state $\mid n> = \mid 0>$ makes explicitly $I=0$. We conclude then that there must be some state $\mid \tilde{n}>$ so that

$$<0\mid j_{j}^{0} \mid \tilde{n}> \neq 0$$ (57)

independent from $y^{0}$ with the property

$$m_{\tilde{n}} \equiv 0$$ (58)

which is precisely the state that is called Goldstone boson. The previous development shows us that for each generator $G_{j}^{NA}$ that does not annihilate the vaccum there appears a Goldstone boson. In other words, the action of each $G_{j}^{NA}$ on $\mid 0>$ makes appear one state $\mid p;j>_{G}$ with $m^{2}=0$. Summarizing a bit more formally: If a system presents a continuous symmetry related to a group $G$ of $n$ parameters and the solutions exhibit, on the contrary, a symmetry associated to a subgroup of $m<n$ parameters, then there appear in the theory $(n-m)$ scalar (or pseudoscalar) partcles without mass: my bosons, as Goldstone would say.