Advanced Chemistry

October 3, 2026

The Quark Model

The quark model is a theory in particle physics that explains hadrons, such as protons and neutrons, as being composed of fundamental particles called quarks.

In 1961, Murray Gell-Mann developed the Eightfold Way and found that the classification of the eight lightest spin-½ baryons could be explained using the SU(3) symmetry group. Since a basis vector that transforms according to the fundamental representation of SU(3) has three components, he proposed that the eight spin-½ baryons are composed of three fundamental particles called quarks:

$u=\begin{pmatrix}1\\0\\0\end{pmatrix}\;\;\;\;d=\begin{pmatrix}0\\1\\0\end{pmatrix}\;\;\;\;s=\begin{pmatrix}0\\0\\1\end{pmatrix}$

where the labels $u$ , $d$ and $s$ will become clear shortly.

Gell-Mann further suggested that the properties of isospin and hypercharge that characterise the baryons also apply to the three quarks. Given that the third component of the isospin operator, which is defined by convention as $\hat{I}_z=\frac{\hat{\lambda}_3}{2}$ where $\hat{\lambda}_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$ is the third Gell-Mann matrix, acts on a basis quark eigenstate to give an eigenvalue corresponding to the third component of the quark’s isospin, we have

$\hat{I}_zu=\frac{1}{2}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=+\frac{1}{2}u$

$\hat{I}_zd=\frac{1}{2}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=-\frac{1}{2}d$

$\hat{I}_zs=\frac{1}{2}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}0\\0\\1\end{pmatrix}=0s$

Therefore, $u$ denotes the “up” quark, referring to its positive isospin projection of $+\frac{1}{2}$ , while $d$ denotes the “down” quark, with an isospin projection of $-\frac{1}{2}$ . The terminology is analogous to the familiar “spin up” and “spin down” notation, although these quantum numbers refer to isospin rather than spin angular momentum.

Similarly, the hypercharge operator is defined as

$\hat{Y}=\frac{\hat{\lambda}_8}{\sqrt{3}}$

where $\hat{\lambda}_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$ is the eighth Gell-Mann matrix.

Hence,

$\hat{Y}u=\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=+\frac{1}{3}u$

$\hat{Y}d=\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=+\frac{1}{3}d$

$\hat{Y}s=\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\begin{pmatrix}0\\0\\1\end{pmatrix}=-\frac{2}{3}s$

Question

Why are the eighth and third Gell-Mann matrices used to define the hypercharge and the third component of the isospin operators?

Answer

In quantum mechanics, two observables can have simultaneous definite values only if their corresponding operators commute. The third and eighth Gell-Mann matrices are the only two diagonal and mutually commuting generators of SU(3). Therefore, they are used to define two independent quantum numbers, conventionally identified with the third component of isospin and hypercharge.

To make these quantum numbers consistent with the baryon quantum numbers, each quark is assigned a baryon number of $B=+\frac{1}{3}$ . Using the hypercharge relation $Y=B+S$ and the Gell-Mann–Nishijima formula $Q=I_3+\frac{Y}{2}$ gives $S=0$ and $Q=+\frac{2}{3}$ for the $u$ quark, $S=0$ and $Q=-\frac{1}{3}$ for the $d$ quark, and $S=-1$ and $Q=-\frac{1}{3}$ for the $s$ quark (see table below). Hence, the third quark, with a non-zero strange quantum number, is labelled $s$ , for “strange”. The up, down and strange properties of quarks are collectively known as flavours.

The definition of flavours allows Gell-Mann to assign combinations of the three quarks to the spin-½ baryons such that the additive quantum numbers of the constituent quarks, given by the eigenvalues of

$(\hat{O}_1\otimes\hat{I}\otimes\hat{I}+\hat{I}\otimes\hat{O}_2\otimes\hat{I}+\hat{I}\otimes\hat{I}\otimes\hat{O}_3)q_1\otimes q_2\otimes q_3\equiv (\hat{O}_1+\hat{O}_2+\hat{O}_3)q_1q_2q_3$

where $\hat{I}$ is the identity operator, reproduce the corresponding quantum numbers of the baryon. For example, the proton is composed of two $u$ quarks and one $d$ quark, $p=uud$ , where $Q=+1$ , $I_3=+\frac{1}{2}$ , $B=1$ , $S=0$ and $Y=1$ . The quark compositions of the remaining baryons are listed in the table below.

Question

Explain why $(\hat{O}_1\otimes\hat{I}\otimes\hat{I}+\hat{I}\otimes\hat{O}_2\otimes\hat{I}+\hat{I}\otimes\hat{I}\otimes\hat{O}_3)q_1\otimes q_2\otimes q_3$ is equivalent to $(\hat{O}_1+\hat{O}_2+\hat{O}_3)q_1q_2q_3$ ?

Answer

When the Kronecker-product operators act on the composite state, each operator acts only on its corresponding substate, while the identity operators leave the other substates unchanged:

$(\hat{O}_1\otimes\hat{I}\otimes\hat{I})q_1\otimes q_2\otimes q_3=\hat{O}_1q_1\otimes\hat{I}q_2\otimes\hat{I}q_3=\varepsilon_1(q_1\otimes q_2\otimes q_3)$

$(\hat{I}\otimes\hat{O}_2\otimes\hat{I})q_1\otimes q_2\otimes q_3=\hat{I}q_1\otimes\hat{O}_2q_2\otimes\hat{I}q_3=\varepsilon_2(q_1\otimes q_2\otimes q_3)$

$(\hat{I}\otimes\hat{I}\otimes\hat{O}_3)q_1\otimes q_2\otimes q_3=\hat{I}q_1\otimes\hat{I}q_2\otimes\hat{O}_3q_3=\varepsilon_3(q_1\otimes q_2\otimes q_3)$

Adding the three above equations gives:

$(\hat{O}_1\otimes\hat{I}\otimes\hat{I}+\hat{I}\otimes\hat{O}_2\otimes\hat{I}+\hat{I}\otimes\hat{I}\otimes\hat{O}_3)q_1\otimes q_2\otimes q_3=(\varepsilon_1+\varepsilon_2+\varepsilon_3)(q_1\otimes q_2\otimes q_3)$

which in shorthand notation is $(\hat{O}_1+\hat{O}_2+\hat{O}_3)q_1 q_2 q_3=(\varepsilon_1+\varepsilon_2+\varepsilon_3)q_1 q_2 q_3$ .

How about the nine spin-0 mesons? It would be mathematically impossible to construct these mesons, which have baryon number of $B=0$ , from three quarks (see above table). But if mesons can be arranged alongside baryons according to the Eightfold Way, then they should also, at least theoretically, be composed of quarks. Gell-Mann resolved this apparent contradiction by proposing that, unlike baryons, mesons are composed of a quark and an antiquark.

To generate the quantum numbers corresponding to antiquarks, we need to determine the antiquark operators. Consider the the quark electric charge operator $\hat{Q}=\hat{I}_3+\frac{\hat{Y}}{2}=\frac{\hat{\lambda}_3}{2}+\frac{\hat{\lambda}_8}{2\sqrt{3}}$ acting on the flavour vector $\begin{pmatrix}u\\d\\s\end{pmatrix}$ :

$\hat{Q}\begin{pmatrix}u\\d\\s\end{pmatrix}=\left\[\frac{1}{2}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}+\frac{1}{6}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\right\]\begin{pmatrix}u\\d\\s\end{pmatrix}=\begin{pmatrix}+\frac{2}{3}&0&0\\0&-\frac{1}{3}&0\\0&0&-\frac{1}{3}\end{pmatrix}\begin{pmatrix}u\\d\\s\end{pmatrix}$

The diagonal entries of the 3×3 matrix are the electric-charge eigenvalues corresponding to the $u$ , $d$ and $s$ quarks respectively. For antiquarks, which are antiparticles, their quantum numbers have the opposite sign to those of quarks (see table below). So, we require

$\hat{Q}\begin{pmatrix}\bar{u}\\\bar{d}\\\bar{s}\end{pmatrix}=\begin{pmatrix}-\frac{2}{3}&0&0\\0&+\frac{1}{3}&0\\0&0&+\frac{1}{3}\end{pmatrix}\begin{pmatrix}\bar{u}\\\bar{d}\\\bar{s}\end{pmatrix}$

which is only possible if $\hat{\lambda}_{3,anti}=-\hat{\lambda}_3=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&0\end{pmatrix}$ and $\hat{\lambda}_{8,anti}=-\hat{\lambda}_8=\frac{1}{\sqrt{3}}\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&+2\end{pmatrix}$ .

Additionally, the generators ( $\frac{\hat{\lambda}_{i,anti}}{2}$ ) defining the antiquark representation of SU(3) must satisfy the commutation relation $[\lambda_a,\lambda_b]=2i\sum_{c=1}^8f_{abc}\lambda_c$ , where $f_{abc}$ are real coefficients known as the structure constants. However, this relation breaks down if we simply substitute $\lambda_i=-\lambda_{i,anti}$ into it.

Taking the complex conjugate of the commutation relation gives $[\lambda_a^*,\lambda_b^*]=-2i\sum_{c=1}^8f_{abc}\lambda_c^*$ , or equivalently,

$[-\lambda_a^*,-\lambda_b^*]=2i\sum_{c=1}^8f_{abc}(-\lambda_c^*)$

Therefore, the antiquark generators must be chosen as $\hat{\lambda}_{i,anti}=-\hat{\lambda}_i^*$ . To see how this is used to derive antiquark operators, we refer to the quark raising operator

$\hat{I}_+=\hat{I}_1+i\hat{I}_2=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$

where $\hat{I}_1=\frac{\hat{\lambda}_1}{2}=\frac{1}{2}\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}$ , $\hat{I}_2=\frac{\hat{\lambda}_2}{2}=\frac{1}{2}\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}$ , and $\lambda_1$ and $\lambda_2$ are the first two Gell-Mann matrices.

If the quark basis states, which span a three-dimensional complex vector space called the 3 representation of SU(3), are written as

$u=\begin{pmatrix}1\\0\\0\end{pmatrix}\;\;\;\;d=\begin{pmatrix}0\\1\\0\end{pmatrix}\;\;\;\;s=\begin{pmatrix}0\\0\\1\end{pmatrix}$

then

$\hat{I}_+d=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}=u$

Similarly, $\hat{I}_+u=0$ and $\hat{I}_+s=0$ .

Question

Show that $\hat{I}_-u=d$ .

Answer

$\hat{I}_-u=(\hat{I}_1-i\hat{I}_2)u=\frac{1}{2}\left\[\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}-\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix}\right\]\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}=d$

The corresponding antiquark raising operator is

$\hat{I}_{+,anti}=\hat{I}_{1,anti}+i\hat{I}_{2,anti}=-(\hat{I}_1)^*+i[-(\hat{I}_2)^*]=-\hat{I}_1+i\hat{I}_2=\begin{pmatrix}0&0&0\\-1&0&0\\0&0&0\end{pmatrix}$

$\hat{I}_{+,anti}$ acting on the basic vector $\begin{pmatrix}1\\0\\0\end{pmatrix}$ gives

$\hat{I}_{+,anti}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0&0&0\\-1&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\-1\\0\end{pmatrix}$

Thus, if the antiquark basis states, which span the complex conjugate vector space of the 3 representation of SU(3), are written as

$\bar{u}=\begin{pmatrix}1\\0\\0\end{pmatrix}\;\;\;\;\bar{d}=\begin{pmatrix}0\\1\\0\end{pmatrix}\;\;\;\;\bar{s}=\begin{pmatrix}0\\0\\1\end{pmatrix}$

then $\hat{I}_{+,anti}\bar{u}=-\bar{d}$ and $\hat{I}_{+,anti}\bar{s}=0$ .

Similarly, $\hat{I}_{+,anti}\bar{d}=0$ , and

$\hat{I}_{-,anti}=\hat{I}_{1,anti}-i\hat{I}_{2,anti}=\begin{pmatrix}0&-1&0\\0&0&0\\0&0&0\end{pmatrix}$

Thus, $\hat{I}_{-,anti}\bar{d}=-\bar{u}$ .

Using $\hat{O}_{total}=\hat{O}_1\otimes \hat{I}+\hat{I}\otimes\hat{O}_2$ , where

	$\hat{O}_{i,quark}$	$\hat{O}_{i,anti}$
$\hat{I}_3$	$\frac{1}{2}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$	$\frac{1}{2}\begin{pmatrix}-1&0&0\\0&1&0\\0&0&0\end{pmatrix}$
$\hat{Y}$	$\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$	$\frac{1}{3}\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&2\end{pmatrix}$
$\hat{Q}$	$\begin{pmatrix}+\frac{2}{3}&0&0\\0&-\frac{1}{3}&0\\0&0&-\frac{1}{3}\end{pmatrix}$	$\begin{pmatrix}-\frac{2}{3}&0&0\\0&+\frac{1}{3}&0\\0&0&+\frac{1}{3}\end{pmatrix}$

we can determine the flavour states of the nine spin-0 mesons by requiring the additive quantum numbers of their constituent quarks to match those of the mesons, which had already been determined experimentally before the Eightfold Way was developed.

For example, the three $\pi$ mesons ( $I=1$ ) form an isospin triplet $\{\pi^+,\pi^0,\pi^-\}$ . The $\pi^+$ , corresponding to $\vert I=1,I_3=+1\rangle$ , is composed of an up quark and a down antiquark, giving the state: $\pi^+=\vert u\bar{d}\rangle$ because

$I_3=+1$

$B=\frac{1}{3}+\left$-\frac{1}{3}\right$=0$

$Y=\frac{1}{3}+\left$-\frac{1}{3}\right$=0$

$S=Y-B=0$

$Q=I_3+\frac{Y}{2}=+1$

This leads to the flavour states of the nine spin-0 mesons shown in the table below.

Interestingly, $\pi^0$ , $\eta$ and $\eta'$ are constructed from linear combinations of quark-antiquark pairs rather than from a single pair. To understand this, we begin by expressing the quark-antiquark pairs as composite quantum states:

$\vert i\bar{j}\rangle=\vert i\rangle\otimes\vert\bar{j}\rangle=\begin{pmatrix}u\\d\\s\end{pmatrix}\otimes\begin{pmatrix}\bar{u}\\\bar{d}\\\bar{s}\end{pmatrix}=\begin{pmatrix}u\bar{u}\\u\bar{d}\\u\bar{s}\\d\bar{u}\\d\bar{d}\\d\bar{s}\\s\bar{u}\\s\bar{d}\\s\bar{s}\end{pmatrix}$

The resulting 9×1 vector is characterised by the nine basis vectors $e_1=\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}$ , $e_2=\begin{pmatrix}0\\1\\\vdots\\0\end{pmatrix}$ , …, $e_9=\begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix}$ which span a nine-dimensional vector space. It can also be expressed as the matrix

$M=\begin{pmatrix}u\bar{u}&u\bar{d}&u\bar{s}\\d\bar{u}&d\bar{d}&d\bar{s}\\s\bar{u}&s\bar{d}&s\bar{s}\end{pmatrix}=\begin{pmatrix}u\\d\\s\end{pmatrix}\otimes\begin{pmatrix}\bar{u}&\bar{d}&\bar{s}\end{pmatrix}$

where $E_{11}=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$ , $E_{12}=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$ , …, $E_{33}=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$ are nine elementary matrices that equivalently span the nine-dimensional vector space.

Next, any 3×3 matrix can be decomposed into a traceless part and a part proportional to the identity: $M=T+\frac{1}{3}Tr(M)I$ , where $T$ is traceless. So, we can write

$\begin{pmatrix}u\bar{u}&u\bar{d}&u\bar{s}\\d\bar{u}&d\bar{d}&d\bar{s}\\s\bar{u}&s\bar{d}&s\bar{s}\end{pmatrix}=\begin{pmatrix}\frac{2u\bar{u}-d\bar{d}-s\bar{s}}{3}&u\bar{d}&u\bar{s}\\d\bar{u}&\frac{2d\bar{d}-u\bar{u}-s\bar{s}}{3}&d\bar{s}\\s\bar{u}&s\bar{d}&\frac{2s\bar{s}-u\bar{u}-d\bar{d}}{3}\end{pmatrix}+\frac{1}{3}(u\bar{u}+d\bar{d}+s\bar{s})\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

Each entry of the traceless matrix contributes to a degree of freedom. The six off-diagonal elements are unconstrained and give rise to six independent states ( $\pi^+,\pi^-,K^+,K^0,K^-,\bar{K}^0$ ). However, the three diagonal elements are restricted by the traceless condition, removing one degree of freedom and resulting in two independent linear combinations. Together, these eight independent states form an octet.

On the other hand, the trace part has only one degree of freedom, given by the scalar multiplier $\frac{1}{3}(u\bar{u}+d\bar{d}+s\bar{s})$ , and hence corresponds to one independent state (a singlet). The normalised singlet state $\frac{1}{\sqrt{3}}(u\bar{u}+d\bar{d}+s\bar{s})$ cannot describe the $\pi^0$ , which is part of the pion isospin triplet $\{\pi^+,\pi^0,\pi^-\}$ , with isospin states $\vert I=1,I_3=+1\rangle$ , $\vert I=1,I_3=0\rangle$ and $\vert I=1,I_3=-1\rangle$ . This is because $\hat{I}^2\left\[\frac{1}{\sqrt{3}}(u\bar{u}+d\bar{d}+s\bar{s})\right\]=0$ rather than the required $I=1$ .

Question

Show that $\hat{I}^2\left\[\frac{1}{\sqrt{3}}(u\bar{u}+d\bar{d}+s\bar{s})\right\]=0$ .

Answer

With reference to eq116, $\hat{I}^2=\hat{I}_-\hat{I}_++\hat{I}_3^2+\hat{I}_3$ in units where $\hbar=1$ . Since $\hat{I}_3s=0$ , $\hat{I}_3\bar{s}=0$ , $\hat{I}_+s=0$ and $\hat{I}_{+,anti}\bar{s}=0$ , we have $\hat{I}_3(s\bar{s})=0$ , $\hat{I}_3^2(s\bar{s})=0$ and $\hat{I}_-\hat{I}_+(s\bar{s})=0$ . For the up and down quark–antiquark pairs, $\hat{I}_3(u\bar{u})=0$ , $\hat{I}_3^2(u\bar{u})=0$ , $\hat{I}_3(d\bar{d})=0$ and $\hat{I}_3^2(d\bar{d})=0$ . Using the previously derived identities of $\hat{I}_+d=u$ , $\hat{I}_+u=0$ , $\hat{I}_{+,anti}\bar{d}=0$ and $\hat{I}_{+,anti}\bar{u}=-=\bar{d}$ , we obtain

$\hat{I}_{+}(u\bar{u})=\bar{u}\hat{I}_+u+u\hat{I}_{+,anti}\bar{u}=-u\bar{d}$

$\hat{I}_{+}(d\bar{d})=\bar{d}\hat{I}_+d+d\hat{I}_{+,anti}\bar{d}=u\bar{d}$

Thus, $\hat{I}_{+}(u\bar{u}+d\bar{d})=0$ , which implies that $\hat{I}_-\hat{I}_{+}(u\bar{u}+d\bar{d})=0$ . Therefore, $\hat{I}^2\left\[\frac{1}{\sqrt{3}}(u\bar{u}+d\bar{d}+s\bar{s})\right\]=0$ .

To determine the flavour state $\pi^0=\vert I=1,I_3=0\rangle$ , we let the isospin analogue of eq147

$\hat{I}_-\vert I,I_3\rangle=\sqrt{I(I+1)-I_3(I_3-1)}\vert I,I_3-1\rangle$

act on $\pi^+=\vert u\bar{d}\rangle=\vert I=1,I_3=+1\rangle$ to yield $\hat{I}_-\vert I,I_3\rangle=\sqrt{2}\vert I,I_3-1\rangle$ , or equivalently, $\vert I=1,I_3=0\rangle=\frac{1}{\sqrt{2}}\hat{I}_-\vert u\bar{d}\rangle$ . Using the identities $\hat{I}_{-,anti}\bar{d}=-\bar{u}$ and $\hat{I}_{-}u=d$ results in

$\vert I=1,I_3=0\rangle=\frac{1}{\sqrt{2}}(d\bar{d}-u\bar{u})=-\frac{1}{\sqrt{2}}(u\bar{u}-d\bar{d})$

$\frac{1}{\sqrt{2}}(d\bar{d}-u\bar{u})$ and $\frac{1}{\sqrt{2}}(u\bar{u}-d\bar{d})$ differ only by an overall phase of -1 and therefore represent the same physical state. The choice of which overall phase to assign to $\pi^0$ is a convention. Here, we choose

$\frac{1}{\sqrt{2}}(u\bar{u}-d\bar{d})$

This leaves $\frac{1}{\sqrt{3}}(u\bar{u}+d\bar{d}+s\bar{s})$ and the remaining diagonal linear combination, which is orthonormal to every state, $\frac{1}{\sqrt{6}}(u\bar{u}+d\bar{d}-2s\bar{s})$ , unassigned. Since both have isospin $I=0$ , they are identified as $\eta_1$ and $\eta_8$ , where the subscripts 1 and 8 denote the singlet and octet representations. The physical $\eta'$ and $\eta$ mesons are, in general, linear combinations of $\eta_1$ and $\eta_8$ . However, as an approximation, the $\eta'$ and $\eta$ can be identified with the pure $\eta_1$ and $\eta_8$ states respectively.

Question

How is the orthonormality of the nine spin-0 meson states verified?

Answer

The orthonormality of the nine spin-0 meson states is verified by taking the inner product of any two states, treating each distinct quark-antiquark pair as a basis vector. For example, $u\bar{u}$ and $u\bar{d}$ are orthonormal to each other because

$u\bar{u}\cdot u\bar{d}=\begin{pmatrix}1&0&0&\cdots\end{pmatrix}\begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix}=0$

$u\bar{u}\cdot u\bar{u}=\begin{pmatrix}1&0&0&\cdots\end{pmatrix}\begin{pmatrix}1\\0\\0\\\vdots\end{pmatrix}=1$

$u\bar{d}\cdot u\bar{d}=\begin{pmatrix}0&1&0&\cdots\end{pmatrix}\begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix}=1$

Thus, for any two of the nine distinct quark–antiquark basis states, $\langle i\vert j\rangle=\delta_{ij}$ .

Hence, Gell-Mann predicted the existence of quarks and antiquarks from a group-theoretical analysis of the then-known light baryons and mesons, which could be arranged according to the Eightfold Way. Definitive evidence for the internal structure of hadrons came subsequently from deep inelastic scattering experiments conducted in the late 1960s and early 1970s at the SLAC National Accelerator Laboratory. For example, when high-energy electrons were fired at protons, they were scattered at large angles rather than passing through as they would if the proton consisted of a uniform cloud of positive charge. Although quarks and antiquarks cannot be isolated individually due to colour confinement, these experiments demonstrated that protons contain point-like, electrically charged constituents.

As more hadrons were discovered in the years that followed, it became increasingly clear that the original $u$ , $d$ and $s$ quarks, and their corresponding antiquarks, might not be sufficient to account for the growing spectrum of particles. Physicists therefore began to ask whether additional quark flavours might exist, particularly heavier ones that would give rise to new families of hadrons. This possibility soon acquired a strong theoretical motivation from the emerging theory of the weak interaction. In this framework, the $u$ and $d$ quarks formed a weak-interaction doublet, suggesting that the strange quark should likewise have a corresponding partner. This led to the prediction of a fourth quark, the charm quark, whose existence would eventually be confirmed experimentally. The discovery of charm, followed in time by the bottom and top quarks, extended the quark model far beyond its original three-flavour picture and revealed a much richer structure underlying the hadron spectrum.

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October 3, 2026October 16, 2026

The Eightfold Way

The Eightfold Way is a classification scheme proposed by Murray Gell-Mann and Yuval Ne’eman that organises hadrons into groups based on their shared properties and underlying symmetries.

The scheme arranges particles discovered before 1970, into geometric patterns according to properties such as electric charge and strangeness. Specifically, the eight lightest spin-½ baryons, including the proton, neutron, and several strange particles such as the $\Lambda$ , $\Sigma$ and $\Xi$ , can be organised into a baryon octet on a strangeness-charge diagram, with the electric charge represented along a sloping axis (see diagram above). Similarly, nine spin-0 mesons, including the pions, kaons and etas mesons, form an octet together with an additional singlet state, producing a similar pattern on a strangeness-charge diagram.

These patterns revealed that seemingly different particles were related by an underlying symmetry and suggested that they were not fundamental particles themselves, but rather combinations of more basic constituents. Specifically, Gell-Mann found that the eightfold way classification of the baryons and mesons could be explained using the SU(3) symmetry group. This idea ultimately led to the quark model, in which baryons are composed of three quarks and mesons are composed of a quark and an antiquark.

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October 3, 2026October 3, 2026

Quantum Numbers and Conservation Laws of Hadrons and Leptons

Quantum Numbers and Conservation Laws of Hadrons and Leptons refers to the properties used to classify hadrons and leptons and the rules governing which of these properties remain conserved during particle interactions.

Hadrons

The behaviour of hadrons in particle interactions is governed by several important conservation laws, which are established through observations of particle interactions. For example, since a proton is heavier than the combined masses of a positron and a neutral pion, nothing based solely on the principle of energy conservation prevents a proton from decaying as

$p\rightarrow e^++\pi^0$

Yet, despite extensive searches, scientists have never observed a proton decay. To account for this apparent prohibition, a quantum number known as baryon number, $B$ , is introduced. Every baryon is assigned a value of $B=+1$ , every antibaryon a value of $B=-1$ , and particles that are neither baryons nor antibaryons a value of $B=0$ . The proposed decay would therefore change the total baryon number from $B=+1$ to $B=0$ , and is consequently forbidden by baryon-number conservation.

In contrast, proton-proton fusion provides a classic example of a reaction that satisfies baryon-number conservation. It is the first step of the fusion process that powers the Sun:

$p+p\rightarrow d+e^++\nu_e$

Here, the initial state has a total baryon number of $B=+2$ , since each proton has $B=+1$ . The deuteron ( $d$ ) also has $B=+2$ , as it consists of one proton and one neutron, while the positron and electron neutrino have $B=0$ . Thus, the total baryon number is conserved in the reaction.

Another quantum number that governs how particles interact is the isospin $I$ . When protons collide with protons, $p+p\rightarrow p+p$ , the particles experience both the repulsive Coulomb force and the strong nuclear force. In proton-neutron collisions, $p+n\rightarrow p+n$ , the interaction is dominated by the strong nuclear force because the neutron has no net electric charge.

In the early 1930s, scattering experiments comparing proton–proton and proton–neutron interactions measured how the number of particles scattered varied with scattering angle. The proton and neutron have very similar masses and, consequently, very similar kinematic properties in nuclear collisions. After accounting for the Coulomb interaction in proton–proton scattering, the angular distributions of the scattered particles were found to be remarkably similar. Together, the similar masses and scattering behaviour suggested that protons and neutrons have closely related properties under the strong interaction, providing evidence that the strong nuclear force acts on them in almost the same way. This led Heisenberg to propose that the proton and neutron might not be fundamentally different kinds of particle, but rather two different states of the same underlying particle, which he called the nucleon.

To describe these two states, Heisenberg exploited the same mathematical framework used for angular momentum, in which the number of possible states is $2l+1$ . Since the nucleon $\vert nc\rangle=\begin{pmatrix}p\\n\end{pmatrix}$ has two states, its isospin is $I=\frac{1}{2}$ . The proton $\vert p\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and the neutron $\vert n\rangle=\begin{pmatrix}0\\1\end{pmatrix}$ then correspond to the two possible projections of the nucleon’s isospin, $I_3(p)=+\frac{1}{2}$ and $I_3(n)=-\frac{1}{2}$ (where the subscript 3 denotes the third axis). As $\vert p\rangle$ and $\vert n\rangle$ have the same mathematical form as the spin– $\frac{1}{2}$ states of an electron, they transform under the SU(2) group.

It follows that a component of the isospin operator acts on a basis eigenstate of a particle to give an eigenvalue corresponding to a component of the particle’s isospin. For the nucleon, the three components of the isospin operator are represented in the two-dimensional SU(2) isospin space by $\hat{I}_{k=1,2,3}=\frac{\hat{\sigma}_{k=x,y,z}}{2}$ , where $\hat{\sigma}_k$ denotes one of the three Pauli matrices. For example, using $\hat{I}_3=\hat{\sigma}_z$ , where $\hat{\sigma}_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ , we obtain

$\hat{I}_3\vert p\rangle=\frac{1}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=+\frac{1}{2}\vert p\rangle$

$\hat{I}_3\vert n\rangle=\frac{1}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=-\frac{1}{2}\vert n\rangle$

Since the Pauli matrices satisfy the commutation relations $[\hat{J}_i,\hat{J}_j]=i\varepsilon_{ijk}\hat{J}_k$ , where $\hat{J}_i=\frac{\hat{\sigma}_i}{2}$ and $\varepsilon_{ijk}$ is the Levi-Civita symbol, the isospin operators likewise satisfy:

$[\hat{I}_i,\hat{I}_j]=i\varepsilon_{ijk}\hat{I}_k$

We can also define, just as in quantum angular momentum, the isospin ladder operators $\hat{I}_{\pm}=\hat{I}_1\pm i\hat{I}_2$ , where $\hat{I_1}=\frac{\hat{\sigma}_x}{2}=\frac{1}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $\hat{I_2}=\frac{\hat{\sigma}_y}{2}=\frac{1}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix}$ . Explicitly,

$\hat{I}_{+}=\hat{I}_1+i\hat{I}_2=\begin{pmatrix}0&1\\0&0\end{pmatrix}\;\;\;\;\hat{I}_{-}=\hat{I}_1-i\hat{I}_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}$

Consequently,

$\hat{I}_{+}\vert n\rangle=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\vert p\rangle$

$\hat{I}_{-}\vert p\rangle=\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\vert n\rangle$

This is consistent with the ladder operators raising or lowering the eigenvalue of a basis state by one unit. Specifically,

$\hat{I}_3\hat{I}_{+}\vert n\rangle=\hat{I}_3\vert p\rangle=+\frac{1}{2}\vert p\rangle=+\frac{1}{2}\hat{I}_+\vert n\rangle$

$\hat{I}_3\hat{I}_{-}\vert p\rangle=\hat{I}_3\vert n\rangle=-\frac{1}{2}\vert n\rangle=-\frac{1}{2}\hat{I}_-\vert p\rangle$

In other words, the raising operator $\hat{I}_{+}$ transforms $\vert n\rangle$ into an eigenvector ( $\hat{I}_{+}\vert n\rangle$ ) of $\hat{I}_{3}$ with an eigenvalue that is one unit higher than $-\frac{1}{2}$ . Similarly, the lowering operator $\hat{I}_-$ transforms $\vert p\rangle$ into an eigenvector ( $\hat{I}_{-}\vert p\rangle$ ) of $\hat{I}_{3}$ with an eigenvalue that is one unit lower than $+\frac{1}{2}$ . These operators are important for understanding the quark model.

If the strong interaction cannot distinguish between the proton and neutron states, then changing a nucleon from its proton state to its neutron state does not change the strong-interaction physics. This approximate symmetry is called isospin symmetry, and the corresponding isospin quantum numbers are conserved in strong interactions. Just as the proton and neutron form an isospin doublet with $I=\frac{1}{2}$ , other hadrons with similar masses and related strong-interaction properties can form isospin multiplets. For example, the three pions $\{\pi^+,\pi^0,\pi^-\}$ form an isospin triplet with $I=1$ and $I_3(\pi^+)=+1$ , $I_3(\pi^0)=0$ , $I_3(\pi^-)=-1$ . Therefore, isospin conservation in strong interactions allows reactions such as:

$\pi^-+p\rightarrow \pi^0+n$

However, isospin is not conserved in weak interactions. Unlike the strong interaction, the weak interaction can distinguish between the proton and neutron states and can transform one into the other. For example, in beta decay, a neutron can decay into a proton through the weak interaction:

$n\rightarrow p+e^-+\tilde{\nu}_e$

In terms of isospin, the nucleon changes from $I_3=-\frac{1}{2}$ to $I_3=+\frac{1}{2}$ . This suggests that the proton and neutron may not be fundamental particles, but are instead composed of more elementary particles whose properties determine how they interact through the strong and weak interactions.

Question

Can decay occur through the strong and electromagnetic interactions, instead of the weak interaction?

Answer

Yes. For example, $\rho^0\rightarrow \pi^++\pi^-$ is a strong decay, while $\pi^0\rightarrow \gamma+\gamma$ is an electromagnetic decay.

The conservation of baryon number and isospin, however, is not sufficient to explain all the patterns observed in hadronic interactions. Consider the reaction

$\pi^-+p\rightarrow K^0+\Lambda$

At first glance, there is nothing unusual about this reaction: the total electric charge, baryon number and isospin are conserved. However, experiments revealed that the $K^0$ and $\Lambda$ exhibited unusual production and decay patterns. They could be produced rapidly through the strong interaction, yet their subsequent decays (e.g. $\Lambda\rightarrow p+\pi^-$ ) were much slower than typical strong decays. Since particle decays can occur through the strong interaction, this raised an important question: why could these particles be produced through the strong interaction but not undergo equally rapid strong decays?

To account for this unusual behaviour, Murray Gell-Mann and Kazuhiko Nishijima proposed that these particles carried an additional quantum number that was conserved in strong interactions but could be violated by the weak interaction. This new quantum number was called strangeness, denoted by $S$ .

The $K^0$ and $\Lambda$ were were assigned opposite strangeness quantum numbers, $S(K^0)=+1$ and $S(\Lambda)=-1$ , while the non-strange particles $\pi^-$ and $p$ were assigned $S(\pi^-)=0$ and $S(p)=0$ . Thus, strangeness is conserved in the strong-interaction reaction $\pi^-+p\rightarrow K^0+\Lambda$ . It follows that the decay $\Lambda\rightarrow p+\pi^-$ is forbidden by the strong interaction and instead proceed through the weak interaction because the strangeness changes from $S=-1$ to $S=0$ , with $\Delta S\neq 0$ , explaining its much longer lifetime.

The introduction of strangeness provides a useful way to distinguish particles that otherwise have similar properties, but it also suggests a deeper relationship between the quantum numbers used to classify hadrons. In particular, Gell-Mann and Nishijima noticed that hadrons could be systematically expressed in terms of their baryon number and strangeness. They introduced a new quantum number called hypercharge, denoted by $Y$ , which is defined as

$Y=B+S$

For the non-strange particles in the reaction $\pi^-+p\rightarrow K^0+\Lambda$ , the pion has $B=0$ and $S=0$ , giving $Y(\pi^-)=0$ , while the proton has $B=1$ and $S=0$ , giving $Y(p)=1$ . For the strange particles, the $K^0$ has $B=0$ and $S=+1$ , so $Y(K^0)=+1$ , whereas the $\Lambda$ has $B=1$ and $S=-1$ , giving $Y(\Lambda)=0$ . Thus, hypercharge is also conserved in this strong-interaction reaction.

Furthermore, Gell-Mann and Nishijima found that the electric charges of hadrons could be expressed in terms of hypercharge and the third component of isospin through the empirical relation known as the Gell-Mann–Nishijima formula:

$Q=I_3+\frac{Y}{2}$

where $Q$ is the electric charge in units of the elementary charge.

This relation provides a systematic connection between electric charge, isospin, baryon number and strangeness, and provides a unified framework for classifying the growing number of hadrons discovered in experiments. For example,

Leptons

The behaviour of leptons in particle interactions is also governed by important conservation laws. For example, consider the decay of a muon,

$\mu^-\rightarrow e^-+\gamma$

This decay is allowed by energy and electric-charge conservation. However, it has never been observed. To describe the observed patterns of lepton interactions, a quantum number known as lepton number, $L$ , is introduced. Every lepton is assigned a value of $L=+1$ , every antilepton a value of $L=-1$ , and particles that are neither leptons nor antileptons a value of $L=0$ . Thus, total lepton number is conserved in processes such as muon decay,

$\mu^-\rightarrow e^-+\tilde{\nu}_e+\nu_{\mu}$

However, the conservation of total lepton number alone does not explain why the decay $\mu^-\rightarrow e^-+\gamma$ is not observed. To account for this, the lepton number of each lepton family $L_e$ , $L_{\mu}$ and $L_{\tau}$ is required to be conserved in a reaction. The decay $\mu^-\rightarrow e^-+\gamma$ would therefore change the individual lepton-family numbers, with $L_{e}\neq 0$ and $L_{\mu}\neq 0$ , and is consequently forbidden.

Finally, leptons do not participate in the strong interaction, so they have zero baryon number and strangeness. The strong isospin and hadronic hypercharge used to classify hadrons are therefore not applicable to leptons in the same sense.

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Hadrons and Leptons

Hadrons are composite particles made up of quarks that participate in the strong interaction, whereas leptons are elementary particles that do not participate in the strong interaction. These two broad classes of subatomic particles therefore differ fundamentally in their internal structure and interactions.

Hadrons can be either fermions or bosons, depending on their total spin. They are broadly classified into two groups: baryons, which are composed of three quarks and have half-integer spin angular momentum quantum number, and mesons, which consist of a quark and an antiquark and have integer spin (see above diagram). Thus, baryons are fermionic hadrons, while mesons are bosonic hadrons. Baryons, such as protons and neutrons, are especially important because they form atomic nuclei and account for most of the mass of ordinary matter. Mesons, on the other hand, are generally unstable particles that decay rapidly into lighter particles. Although they do not make up ordinary matter, mesons play an important role in the study of the strong interaction and nuclear physics. For example, pions, the lightest mesons, contribute to the residual strong force that binds protons and neutrons within atomic nuclei.

The existence of a particle with approximately the pion’s properties was theoretically anticipated by Hideki Yukawa in 1935. At the time, scientists knew that protons and neutrons were held together by a strong, short-range nuclear force that became negligible beyond nuclear distances of roughly 10^-15 m. Yukawa proposed that this force could be explained in terms of the exchange of a particle, analogous to the way electromagnetic interactions can be described in terms of the exchange of photons.

In photon exchange, one particle emits a photon and the other absorbs it. The photon is called a virtual particle because it exists only as an intermediate part of the interaction and cannot be detected as a free photon. The word “exchange“ therefore describes the quantum-mechanical process connecting the interacting particles, rather than the literal transfer of an observable particle between them. Yukawa proposed that a similar process could account for the nuclear force, with a nucleon emitting a virtual particle that is subsequently absorbed by another nucleon:

$N_1\rightarrow N_1+\pi\;\;\;\;N_2+\pi\rightarrow N_2$

Using the energy-time uncertainty principle, he suggested that a virtual particle can be exchanged over a characteristic time determined approximately by $\Delta E\Delta t\sim\hbar$ . Taking the energy associated with the virtual particle to be approximately its rest energy, $\Delta E\sim mc^2$ , gives $\Delta t\sim \frac{\hbar}{mc^2}$ . If the virtual particle travels at approximately the speed of light during this time, the corresponding range of the interaction is $x\sim c\Delta t\sim \frac{\hbar}{mc^2}$ , or equivalently, $mc^2\sim \frac{\hbar c}{x}$ . Using $\hbar c\approx 197$ MeV fm, and $x\approx 1.4$ fm, $mc^2\approx 141$ MeV.

Yukawa therefore predicted a particle with a mass of approximately 140 MeV/c², roughly 275 times the electron mass. This prediction was remarkably close to the mass scale of the pion, which was subsequently identified as the lightest meson. The discovery and study of the pion therefore provided an important early connection between Yukawa’s theoretical description of the short-range nuclear force and the experimentally observed spectrum of hadrons.

Unlike hadrons, all leptons are fermions because they have half-integer spin. The charged leptons, including electrons, muons and taus, participate in the electromagnetic and weak interactions, while neutrinos interact primarily through the weak interaction. Of these particles, the electron plays a central role in ordinary matter, forming part of atoms and governing many of their chemical and electromagnetic properties. Muons and taus are much heavier and unstable, while neutrinos are electrically neutral and have extremely small masses.

The neutrino was predicted by Wolfgang Pauli before it was experimentally detected. In 1930, Pauli proposed the existence of an additional neutral particle to explain the continuous energy spectrum observed in nuclear beta decay, which appeared to violate the conservation of momentum and energy if the decay involved only the emitted electron and the recoiling nucleus. In a two-body decay of a stationary nucleus, conservation of momentum requires the daughter nucleus and electron to recoil back-to-back with fixed momenta, resulting in a fixed electron energy. However, experiments showed a continuous electron-energy spectrum. This suggested that another particle was carrying away some of the momentum and energy. Pauli therefore proposed the existence of a neutral particle, later identified in 1956 by Cowan and Reines as the electron antineutrino:

$n\rightarrow p+e^-+\tilde{\nu}_e$

Together, hadrons and leptons encompass the particles that make up much of the visible matter in the universe. Hadrons include the protons and neutrons found in atomic nuclei, while leptons such as electrons form the surrounding electron clouds. Understanding the distinction between these two classes, as well as their corresponding quantum numbers and conservation laws, is therefore essential to understanding the composition and behaviour of matter at the subatomic level.

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Fermions and Bosons

Fermions are subatomic particles with half-integer spin that obey the Pauli exclusion principle, while bosons have integer spin and can occupy the same quantum state.

Although both types of particles are essential to the structure of matter and the forces of nature, they behave in very different ways. Fermions include particles such as electrons, protons, neutrons and quarks. They have half-integer spin and obey Fermi–Dirac statistics, named after physicists Enrico Fermi and Paul Dirac. These statistics incorporate the Pauli exclusion principle, which states that two identical fermions cannot occupy the same quantum state. Such a restriction plays an important role in the structure of atoms and matter, helping to explain why electrons occupy different energy levels and why ordinary matter takes up space.

Bosons, on the other hand, include particles such as photons, gluons and the Higgs boson. They have integer spin and obey Bose–Einstein statistics, named after physicists Satyendra Nath Bose and Albert Einstein. Unlike fermions, identical bosons can occupy the same quantum state in any number. As a result, bosons can exhibit collective quantum behaviour, such as the coherent light produced by lasers and the formation of Bose–Einstein condensates at extremely low temperatures.

The distinction between fermions and bosons is closely connected to their intrinsic angular momentum, or spin. In quantum mechanics, the allowed values of the spin quantum number are restricted to integer or half-integer values: 0,1/2,1, 3/2, 2, …. Thus, particles can have either integer spin, such as 0 or 1, or half-integer spin, such as 1/2 or 3/2. The spin–statistics theorem further establishes a fundamental connection between these two classes of spin and quantum statistics, providing the basis for the two fundamental categories into which ordinary subatomic particles fall.

This connection also determines how the quantum states of identical particles behave under exchange. Bosons have wavefunctions that are symmetric under the exchange of two identical particles, whereas fermions have wavefunctions that are antisymmetric. The antisymmetry of fermionic states leads directly to the Pauli exclusion principle, while the symmetry of bosonic states allows identical bosons to occupy the same quantum state.

Together, fermions and bosons form the foundation of the Standard Model of particle physics. Fermions constitute matter, while bosons include particles that mediate fundamental interactions. Their contrasting quantum statistics are therefore central to understanding both the microscopic structure of matter and many of the remarkable collective phenomena observed in the quantum world.

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Particles and Antiparticles

Particles are constituents of matter and energy, while antiparticles are their corresponding counterparts with the same mass and spin but opposite values of certain physical properties, such as electric charge.

The concept of antiparticles emerged from the development of relativistic quantum mechanics. In special relativity, the energy–momentum relation for a particle of mass $m$ is $E^2=p^2c^2+m^2c^4$ , which gives two possible solutions:

$E=\pm\sqrt{p^2c^2+m^2c^4}$

For an ordinary particle, the positive-energy solution is physically intuitive: transitions to higher positive-energy states require the particle, such as an electron, to absorb energy from an external source. The negative-energy solution, however, posed a serious conceptual problem. Transitions to lower negative-energy states would instead be energetically favourable and could occur spontaneously, with the excess energy emitted as radiation. Moreover, the electron could continue transitioning to increasingly negative-energy states, releasing ever greater amounts of energy at each step. Since there would be no ground state for the electron to occupy, this process could, in principle, lead to an infinite and catastrophic release of radiation, making ordinary matter fundamentally unstable.

In 1930, the British physicist Paul Dirac sought to solve this problem by proposing that a vacuum is not truly empty. He imagined that all available negative-energy states in the vacuum were already occupied by electrons, forming what became known as the Dirac sea, while the positive-energy states were completely empty (see diagram below). Just as deep-sea fish experience the crushing, uniform and isotropic pressure of the surrounding water as their baseline environment, Dirac proposed that the sea of negative-energy electrons was similarly uniform, to the extent that we could regard it as the zero-energy state, $E_{DS}=0$ . Hence, any electron added to the vaccum could not simply fall into a negative-energy state because of the Pauli exclusion principle. It would instead occupy a positive-energy state, consistent with ordinary observations.

Furthermore, if one of the electrons in this sea, with energy $E_e=-E$ , where $E>0$ , were removed, the resulting hole would carry the opposite charge of the missing electron and have positive energy relative to the filled Dirac sea: $E_{DS}-E_e=+E$ . Such a hole would behave as a particle with the same mass and spin as an electron but with the opposite electric charge. Dirac therefore interpreted the hole as a positively charged counterpart of the electron: the anti-electron, or positron. This gives the negative-energy solutions of the relativistic energy–momentum relation a physical interpretation.

Question

If the baseline energy of the Dirac sea is defined as zero, shouldn’t all electrons in the sea have zero energy?

Answer

Although the sea as a whole is defined to have zero-energy, each electron in the sea does not lose its identity; that is, each sea electron still occupies a specific negative-energy state.

Imagine that your bank account has exactly negative one million dollars. You decide to redefine your financial baseline starting right now. You reset your mental accounting app so that this exact state is treated as your “zero point” (baseline = 0). Even though you call your current state “zero,” the actual debt making up that balance has not disappeared. If you remove a single one-dollar debt ( $E_e=-$1$ ) from that pile, your total debt decreases. Relative to your newly defined baseline, your account balance shifts towards the positive side: $0-(-$1)=+$1$ .

The hypothesis also suggested a mechanism for producing particle–antiparticle pairs. If sufficient energy were supplied to the vacuum, an electron could be promoted from a negative-energy state into a positive-energy state, leaving a hole behind in the Dirac sea and, hence, generating an electron and a positron simultaneously.

However, Dirac’s conjecture had a flaw: it implied an infinite amount of energy in the vacuum. The Dirac sea concept was eventually superseded by quantum field theory, in which the vacuum is permeated by quantum fields. According to the Heisenberg Uncertainty Principle, a quantum field in a vacuum cannot simultaneously have both an exactly zero field value and an exactly zero conjugate momentum. As a result, a vacuum field unavoidably exhibits tiny quantum fluctuations. These fluctuations are sometimes described heuristically as particles appearing and disappearing randomly for extremely short periods of time. However, real particles require stable excitations of a field and therefore a definite enery source to be produced. Their formation is triggered by physical processes that supply energy to the field, such as particle collisions in accelerators or the decay of an unstable nucleus.

Nevertheless, Dirac’s prediction was particularly significant because no such particle had been observed at the time. The existence of a positively charged particle with the same mass as the electron was confirmed in 1932 by the American physicist Carl Anderson. In his experiments, he exposed a cloud chamber, filled with supersaturated water vapour, to naturally occurring high-energy radiation from deep space, known as cosmic rays (see diagram below). As charged cosmic-ray particles passed through the chamber, they ionised the gas along their paths, leaving visible particle tracks that curved under the influence of a magnetic field.

Anderson observed tracks that curved in the direction expected for positively charged particles. However, their curvature and rate of energy loss indicated that these particles had a mass comparable to that of an electron, rather than that of the much heavier proton. He therefore concluded that he had observed a new particle with the same mass as an electron but with the opposite electric charge. This particle was subsequently identified as the positron, providing the first experimental evidence for antiparticles and confirming one of the most remarkable predictions arising from Dirac’s theory.

The electron is not the only particle with an antiparticle. Many particles have corresponding antiparticles with the same mass and spin but opposite additive quantum numbers. For example, the antiparticle of a quark is an antiquark, while the proton and neutron have corresponding antiparticles known as the antiproton and antineutron. Some particles, such as the photon, are their own antiparticles because they have no relevant distinguishing additive quantum numbers.

Particles and antiparticles interact in a distinctive process known as annihilation. When a particle encounters its corresponding antiparticle, they can be converted into other particles while conserving quantities such as energy, momentum and electric charge. For example, an electron and a positron can annihilate to produce photons. This process illustrates the equivalence of mass and energy described by Einstein’s relation $E=mc^2$ .

The reverse process, known as pair production, can also occur as predicted by Dirac. Under suitable conditions, sufficient energy can be converted into a particle and its corresponding antiparticle. Pair production and annihilation demonstrate the close relationship between matter and energy and are important processes in high-energy particle physics.

Together, particles and antiparticles form an essential part of the modern description of the physical universe. The prediction and subsequent discovery of the positron demonstrated that antimatter could arise naturally from the laws of relativistic quantum mechanics. Since then, the study of antiparticles has become an important part of particle physics, providing insight into the fundamental structure of matter and the interactions that govern the universe.

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Particle accelerator

A particle accelerator is a machine that uses electric and magnetic fields to propel charged particles, such as electrons or protons, to extremely high speeds and energies.

Particle accelerators are among the most important tools in modern physics. By accelerating particles and directing them into targets or into collisions with one another, scientists can study the fundamental building blocks of matter and the forces that govern them.

There are two major types of particle accelerators: linear accelerators (linacs) and circular accelerators. In a linear accelerator, particles travel along a straight path through a series of accelerating sections. Alternating electric fields are carefully timed so that each section gives the particles another push, increasing their energy as they move forward (see diagram below). Magnetic fields can also be used to focus and steer the beam, keeping the particles traveling along the desired path.

Circular accelerators work differently: particles travel repeatedly around a circular or nearly circular track, passing through accelerating structures each time they complete a circuit (refer to the storage ring in the diagram below). Powerful dipole magnets bend the particles’ paths and keep the beam focused. As relativistic particles travel around curves, they emit energy in the form of radiation, which is directed along beamlines and subsequently used as a radiation source for experiments. To replenish the kinetic energy lost through radiation, radio-frequency accelerating cavities (RF cavities), which contain electric fields oscillating at radio frequencies, are placed around the track to increase the particles’ energy with each successive pass.

A synchrotron is a specialised accelerator that combines the principles of linear and circular acceleration (see diagram above). Source particles initially pass through a linear accelerator before entering the circular accelerator known as a storage ring. The strength of the magnetic fields and the timing of the accelerating electric fields in the ring are synchronised with the particles as their energy increases. This allows particles to remain on the same circular path in the storage ring even as they approach extremely high energies. Additionally, undulators and wigglers (collectively known as insertion devices) are placed in the straight sections of the storage ring. They use an array of alternating magnetic poles (North-South-North-South) to force the travelling charged particles into a periodic “wiggling” path, causing them to emit intensely bright, highly focused synchrotron radiation. This radiation is particularly useful for studying the structure, chemical composition and physical properties of materials at the atomic and molecular levels.

The Large Hadron Collider (LHC) is a famous example of a synchrotron and the world’s largest and most powerful particle accelerator (see diagram below). Located at CERN near Geneva, Switzerland, it uses powerful superconducting magnets to guide two beams of protons around a 27-kilometre circular tunnel in opposite directions (see diagram below). When the beams reach their target energies, magnets steer them into collisions at specific points around the accelerator. Scientists study the particles produced in these collisions using large detectors, allowing them to investigate fundamental questions about matter and the forces of nature. One of the LHC’s most significant achievements was the discovery of the Higgs boson in 2012, providing experimental confirmation of a particle predicted by the Standard Model of particle physics.

A particularly striking example of what can happen in a particle accelerator is the collision between a proton and an antiproton. A proton is made up of three quarks, while an antiproton is made up of three antiquarks. When the two particles collide at sufficiently high energy, their matter and antimatter constituents can annihilate, converting their energy into other particles. This does not mean that all the energy simply disappears. Instead, according to Einstein’s famous equation $E=mc^2$ , the energy carried by the colliding particles can be transformed into the mass and kinetic energy of newly created particles. For example, when a proton and an antiproton collide at high energy, their energy can be converted into short-lived particles such as pions, kaons, or heavier particles, depending on the energy and the interaction that takes place (see diagram below for an example). These particles may then decay into other particles, like muons and neutrinos.

Typically, the collision occurs inside a chamber exposed to a uniform magnetic field and containing many layers of highly sensitive sensors. As the newly produced particles travel outwards from the collision point, they pass through these sensors and leave tiny signals at different locations. Computers collect these signals and use the pattern of sensor readings to reconstruct the trajectories of the particles.

For example, the diagram above shows the negatively and positively charged particles produced in the annihilation being deflected clockwise and anticlockwise respectively in a uniform magnetic field directed out of the page. By measuring the radii of curvature of their tracks, we can calculate their momenta and, hence, determine their masses. This allows physicists to reconstruct what happened during the collision and learn about the fundamental interactions between matter and antimatter.

Finally, such technologies are not limited to fundamental research. Particle accelerators are also used in medicine, industry, and scientific research. In cancer treatment, for example, accelerators can produce beams used in radiation therapy to destroy cancer cells. They can also help manufacture specialised materials, inspect industrial components, and produce radioactive isotopes used in medical diagnosis.

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Subatomic and elementary particles

Subatomic particles are particles smaller than atoms, while elementary particles are fundamental particles that are not known to be made of smaller constituents.

Although atoms were once considered the smallest indivisible units of matter, scientific discoveries have shown that atoms themselves have a complex internal structure. An atom consists of a small, dense nucleus containing protons and neutrons, surrounded by electrons. Further investigation revealed that protons and neutrons are themselves composed of even smaller particles called quarks. This discovery led to a deeper understanding of matter and gave rise to the modern field of particle physics.

Subatomic particles can broadly be divided into composite particles, meaning they are made of smaller constituents, and elementary particles, which are currently regarded as fundamental (see above diagram). Protons and neutrons, for example, are composite particles made of quarks held together by the strong nuclear force. Electrons, on the other hand, are considered elementary particles because no smaller internal structure has been observed within them. Other elementary particles include quarks, neutrinos, photons, gluons and the particles associated with the Higgs field. There are a total of 61 elementary particles and over 300 composite particles.

These particles differ greatly in their properties, including mass and electric charge. They are also associated with different spin quantum numbers, which determine whether they are fermions or bosons and, consequently, whether they follow Fermi–Dirac or Bose–Einstein statistics, respectively (see above diagram). Furthermore, their varied properties influence how they interact and behave under the four fundamental forces: gravitational, electromagnetic, strong nuclear, and weak nuclear. Therefore, subatomic particles are also commonly classified according to the fundamental forces with which they interact (see diagram below).

Additionally, subatomic particles can be distinguished based on both their internal structure and their interactions with the strong nuclear force. Hadrons are composite particles made up of quarks that participate in the strong interaction, whereas leptons are elementary particles that do not participate in the strong interaction.

In essence, the study of subatomic and elementary particles provides insight into the structure of matter. This is achieved using particle accelerators to produce short-lived particles, revealing how matter is constituted, how particles interact, and the fundamental laws governing the universe.

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SU(3) group

The SU(3) group, or special unitary group of degree 3, is defined by its fundamental representation of $3\times 3$ unitary matrices $U$ with determinant 1, with the generators $H_k$ of all its $n\times n$ representations satisfying the commutation relation $[H_a,H_b]=i\sum_{c=1}^8f_{abc}H_c$ .

An element of SU(3) can be expressed as

$U=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$

where $det(U)=1$ and each entry is a complex number.

Since each complex element contains two real variables, the matrix initially contains eighteen real variables. The unitarity condition $U^{\dagger}U=I$ imposes nine independent real constraints, while the condition $det(U)=1$ removes one further degree of freedom. Consequently, every element of SU(3) depends on eight independent real parameters. In other words, the matrices of SU(3) describe an eight-dimensional manifold.

Question

Show that $U^{\dagger}U=I$ imposes nine independent real constraints.

Answer

$U^{\dagger}U=I$ becomes

$U=\begin{pmatrix}a^*&d^*&g^*\\b^*&e^*&h^*\\c^*&f^*&i^*\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

Carrying out the multiplication gives three equations corresponding to diagonal entries

$\vert a\vert^2+\vert d\vert^2+\vert g\vert^2=1$

$\vert b\vert^2+\vert e\vert^2+\vert h\vert^2=1$

$\vert c\vert^2+\vert f\vert^2+\vert i\vert^2=1$

and six equations corresponding to the off-diagonal entries

$\begin{matrix}a^*b+d^*e+g^*h=0\\a^*c+d^*f+g^*i=0\\b^*a+e^*d+h^*g=0\\\vdots\end{matrix}$

Therefore, these nine simultaneous equations impose nine independent real constraints.

As with SU(2), every element of SU(3) can be generated by exponentiating a linear combination of the group’s generators. However, instead of the three 2×2 Pauli matrices, SU(3) requires the generators to be 3×3 Hermitian matrices, each with zero trace. Futhermore, the number of independent real parameters specifying a group element must equal the number of generator matrices (see this article for details), resulting in eight linearly independent 3×3 generator matrices with the following general form:

$\lambda_i=\begin{bmatrix}a&x-iy&u-iv\\x+iy&b&s-it\\u+iv&s+it&-(a+b)\end{bmatrix}$

Another important observation is that the SU(3) rotation operator $\hat{U}$ transforms the basis vectors $\phi_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\phi_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\phi_3=\begin{pmatrix}0\\0\\1\end{pmatrix}$ , which span the three-dimensional complex vector space, into linear combinations of the basis vectors. The simplest of such a transformation mixes a pair of basis vectors; that is, it transforms each vector in the pair into a linear combination of the two while leaving the third basis vector unchanged:

$\hat{U}\phi_1=a\phi_1+b\phi_2$

$\hat{U}\phi_2=c\phi_1+d\phi_2$

$\hat{U}\phi_3=\phi_3$

Consequently, it is natural to choose generators that mix the three pairs of basis vectors, $(\phi_1,\phi_2)$ , $(\phi_1,\phi_3)$ and $(\phi_2,\phi_3)$ . The most basic traceless Hermitian generator matrix that mixes $(\phi_1,\phi_2)$ is

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}$

To see how this symmetric generator matrix is related to an SU(3) transformation, we rewrite it as

$\lambda_1=\begin{pmatrix}\sigma_x&0\\0&0\end{pmatrix}$

where $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is one of three Pauli matrices.

The Taylor expansion of the rotation operator $\hat{U}=e^{-i\frac{\mu}{2}\lambda_1}$ is

$e^{-i\frac{\mu}{2}\lambda_1}=I+\left$-i\frac{\mu}{2}\lambda_1\right$+\frac{1}{2!}\left$-i\frac{\mu}{2}\lambda_1\right$^2+\cdots$

Since every power of $\lambda_1$ has the same block structure $\lambda_1^n=\begin{pmatrix}\sigma_x^n&0\\0&0\end{pmatrix}$ , every term in the Taylor series acts only on the first two components a basis vector, leaving the third component unchanged. Furthermore, the identity matrix is $I=\begin{pmatrix}I_{2\times 2}&0\\0&1\end{pmatrix}$ . Therefore, the SU(3) rotation operator becomes

$\hat{U}=\begin{pmatrix}e^{-i\frac{\mu}{2}\sigma_x}&0\\0&1\end{pmatrix}=e^{-i\frac{\mu}{2}\lambda_1}$

From eq129, where $n_y=n_z=0$ , we have the one-parameter subgroup of SU(3):

$\hat{U}\phi_1=\begin{pmatrix}cos\frac{\mu}{2}&-isin\frac{\mu}{2}&0\\-isin\frac{\mu}{2}&cos\frac{\mu}{2}&0\\0&0&1\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}cos\frac{\mu}{2}\\-isin\frac{\mu}{2}\\0\end{pmatrix}=cos\frac{\mu}{2}\phi_1-isin\frac{\mu}{2}\phi_2$

$\hat{U}\phi_2=-isin\frac{\mu}{2}\phi_1+cos\frac{\mu}{2}\phi_2$

$\hat{U}\phi_3=\phi_3$

demonstrating how $\lambda_1$ is related to an SU(3) transformation.

Question

What is the geometric interpretation of the transformation $\hat{U}(\mu)=\begin{pmatrix}cos\frac{\mu}{2}&-isin\frac{\mu}{2}&0\\-isin\frac{\mu}{2}&cos\frac{\mu}{2}&0\\0&0&1\end{pmatrix}$ ?

Answer

In general, an element of SU(3), which is an 8-dimensional manifold, can be described locally by eight independent real parameters:

$\hat{U}=e^{i\theta_1H_1+i\theta_2H_2+\cdots+i\theta_8H_8}$

where $H_k=\frac{\lambda_k}{2}$ .

If $\theta_2=\cdots=\theta_8=0$ and $\theta_1=\mu$ , then $\hat{U}=\hat{U}(\mu)$ . As $\mu$ varies, $\hat{U}(\mu)$ traces out a one-dimensional curve within the 8-dimensional manifold.

Another basic traceless Hermitian generator matrix that mixes $(\phi_1,\phi_2)$ is

$\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}$

which is the antisymmetric counterpart of $\lambda_1$ .

Notably, $\lambda_2$ contains the Pauli matrix $\sigma_y$ , in the same way that $\lambda_1$ contains $\sigma_x$ . In the early 1960s, Murray Gell-Mann introduced a particularly convenient set of SU(3) generators known as the Gell-Mann matrices. The first three generators contain the Pauli matrices in their upper left blocks, producing SU(3) transformations that mix $(\phi_1,\phi_2)$ :

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$

It follows that the corresponding symmetric and antisymmetric generators that mix $(\phi_1,\phi_3)$ and $(\phi_2,\phi_3)$ are

$\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\;\;\;\;\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}$

and

$\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\;\;\;\;\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}$

respectively.

All generators must be mutually orthogonal with respect to the Frobenius inner product $Tr(\lambda_i\lambda_j)=h\delta_{ij}$ , where $h$ is a scaling constant, to satisfy $U=e^{-\frac{i}{2}\bold{\mu}\cdot\bold{\lambda}}$ . The first seven generators are conveniently chosen to meet this criterion. The final matrix

$\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$

is then scaled to be orthogonal to the first seven such that $Tr(\lambda_i\lambda_j)=2\delta_{ij}$ . These eight linearly independent Hermitian generators are known as the Gell-Mann matrices:

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}$

$\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\;\;\;\;\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\;\;\;\;\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\;\;\;\;\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$

Notably, these matrices satisfy the commutation relation $[H_a,H_b]=i\sum_{c=1}^8f_{abc}H_c$ , where $H_k=\frac{\lambda_k}{2}$ , and $f_{abc}$ are the structure constants of SU(3), whose values are determined by the specific combination of indices $a,b,c$ (see this article for details).

The significance of the Gell-Mann matrices extends beyond providing a mathematical basis for SU(3). In particle physics, the same SU(3) symmetry appears as the flavour symmetry of quarks and the colour symmetry of the strong interaction.

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SU(2) group

The SU(2) group, or special unitary group of degree 2, is defined by its fundamental representation of $2\times 2$ unitary matrices $U$ with determinant 1, with the generators $J_k$ of all its $n\times n$ representations satisfying the commutation relation $[J_a,J_b]=i\varepsilon_{abc}J_c$ .

Question

Why is $det(U)=1$ and why is SU(2) called a special group?

Answer

In general, SO(n) is the subgroup of SU(n) formed by the elements of SU(n) whose matrices entries are real. Under this restriction, the unit-determinant condition is preserved, requiring $det(U)=1$ , consistent with the defining condition of SO(n). For example, When the SU(3) transformations are restricted to real matrices, the symmetry descends from SU(3) to SO(3).

Since the determinant of a complex matrix is a complex number, let $det(U)=z=re^{i\theta}$ . Then,

$det(\hat{U}^{\dagger})det(\hat{U})=z^*z=\vert z\vert^2=r^2$

From $U^{\dagger}U=I$ , we have $det(U^{\dagger}U)=det(I)=1$ . Using $det(AB)=det(A)det(B)$ (see Property 6 of this article for the proof),

$det(\hat{U}^{\dagger}\hat{U})=det(\hat{U}^{\dagger})det(\hat{U})=1$

Therefore, $r^2=1$ , which implies that $r=1$ because $r=\sqrt{x^2+y^2}$ is the modulus (distance from the origin) and is non-negative. It follows that

$det(U)=e^{i\theta}$

and SU(n) is the special case where $det(U)=1$ , corresponding to $\theta=2\pi k$ , where $k\in\mathbb{Z}$ .

To derive an element of an SU(2) fundamental representation, we begin with $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , where each entry is a complex number. Using the matrix identity $(u^{-1})_{qk}=\frac{C_{kq}}{det(U)}$ where $C_{kq}$ is the cofactor of $U_{kq}$ (see Property 11 of this article for the proof), we have $U^{-1}=\frac{1}{det(U)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ since $det(U)=1$ . Furthermore, $U^{-1}=U^{\dagger}$ (see this article for proof), which means $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}a^*&c^*\\b^*&d^*\end{pmatrix}$ . We can therefore express a general SU(2) matrix as: $U=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix}$ .

Although $U$ , which corresponds to a linear operator $\hat{U}$ , initially contains four real variables because $a=x_1+ix_2$ and $b=x_3+ix_4$ , the condition $det(U)=1$ gives $aa^*+bb^*=\vert a\vert^2+\vert b\vert^2=1$ , or equivalently,

$\vert x_1+ix_2\vert^2+\vert x_3+ix_4\vert^2=x_1^2+x_2^2+x_3^2+x_4^2=1$

This results in three independent variables (or three degrees of freedom), which means that every element of SU(2) can be specified by three independent real parameters. For example, an element representing a rotation operation can be parameterised by three angles $(\omega,\theta,\phi)$ , where $\omega$ is rotation angle, while $\theta$ and $\phi$ are the spherical coordinates specifying the orientation of the unit rotation axis $\bold{n}$ . This unit axis can also be expressed as $\boldsymbol{n}=n_x\boldsymbol{i}+n_y\boldsymbol{j}+n_z\boldsymbol{k}$ , which contains only two independent variables because $n_x^2+n_y^2+n_z^2=1$ .

Question

Explain why the two-dimensional complex vector space of SU(2) does not contradict the requirement of having three independent real variables in the matrices.

Answer

Consider rotations in a plane, which is a two-dimensional vector space. Although the vectors being rotated have two components, every rotation is completely specified by a single parameter: the angle of rotation. The dimension of the vector space describes the space on which the transformations act, whereas the number of parameters describes the independent ways of specifying a transformation. These are different concepts.

A similar idea appears in geometry. The surface of a sphere is embedded in three-dimensional space, yet only two parameters (for example, latitude and longitude) are needed to specify a point on the surface. Although the sphere is curved globally, each small region looks approximately like a flat plane. Such an object is called a manifold.

Likewise, the set of all SU(2) matrices, each with three degrees of freedom, forms a three-dimensional manifold. A general 2×2 complex matrix contains eight real parameters, but the conditions of unitarity and unit determinant constrain the allowed matrices so that only three independent real parameters remain. Thus, the two-dimensional complex vector space refers to the vectors on which SU(2) acts, while the three independent real parameters describe the manifold of allowed SU(2) transformations.

To illustrate the action of such SU(2) transformations, consider a rotation of the electron spin states $\alpha=\begin{pmatrix}1\\0\end{pmatrix}$ and $\beta=\begin{pmatrix}0\\1\end{pmatrix}$ , with the rotation operator (see this article for derivation) given by

$\hat{D}_n(\mu)=e^{-i\frac{\mu}{\hbar}(\bold{n}\cdot\hat{\bold{S})}}$

where $\hat{\bold{S}}=\hat{S}_x\bold{i}+\hat{S}_y\bold{j}+\hat{S}_z\bold{k}$ is the spin angular momentum operator.

Since $\hat{\bold{S}}=\hat{S}_x\bold{i}+\hat{S}_y\bold{j}+\hat{S}_z\bold{k}=\frac{\hbar}{2}\hat{\bold{\sigma}}$ (see this article for derivation), where $\hat{\bold{\sigma}}=\hat{\sigma}_x\bold{i}+\hat{\sigma}_y\bold{j}+\hat{\sigma}_z\bold{k}$ , with the Pauli matrices given by

$\hat{\sigma}_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\;\;\;\;\hat{\sigma}_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\;\;\;\;\hat{\sigma}_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$

the rotation operator becomes

$\hat{D}_n(\mu)=e^{-i\frac{\mu}{2}\hat{A}}$

where $\hat{A}=n_x\hat{\sigma}_x+n_y\hat{\sigma}_y+n_z\hat{\sigma}_z$ , with $\frac{\hat{\sigma}_k}{2}$ being the infinitesimal generators of a representation of the SU(2) group.

Notably, the Pauli matrices satisfy the commutation relations $[\hat{J}_a,\hat{J}_b]=i\varepsilon_{abc}\hat{J}_c$ , where $\hat{J}_k=\frac{\hat{\sigma}_k}{2}$ . Moreover, the number of independent real parameters specifying a group element must equal the number of generator matrices so that the dot product in the exponential is well defined (see this article for details).

Question

Show that the generators of SU(n) are traceless.

Answer

The transformation $\hat{U}=e^{-i\frac{\mu}{\hbar}(\bold{n}\cdot\hat{\bold{H})}}$ applies to all unitary groups, and every element of SU(n) can be written as $U=e^{-iH}$ , where $H$ is an $n\times n$ Hermitian matrix. Using the identity $det(e^H)=e^{Tr(H)}$ (see property 15 of this article for the proof),

$det(U)=det(e^{-H})=e^{-iTr(H)}$

Since every SU(n) matrix must satisfy $det(U)=1$ , we require $e^{-iTr(H)}=1$ , which implies that $Tr(H)=0$ .

Expanding the exponential as a Taylor series gives:

$e^{-i\frac{\mu}{2}A}=I+\left$-i\frac{\mu}{2}A\right$+\frac{1}{2!}\left$-i\frac{\mu}{2}A\right$^2+\cdots$

Using the Pauli matrix identity $A^2=(\bold{n}\cdot\hat{\bold{\sigma}})^2=I$ , we have $A^3=A$ , $A^4=I$ , $A^5=A$ and so on. Therefore,

$e^{-i\frac{\mu}{2}A}=\left\[1-\frac{1}{2!}\left$\frac{\mu}{2}\right$^2+\frac{1}{4!}\left$\frac{\mu}{2}\right$^4-\cdots\right\]I-i\left\[\frac{\mu}{2}-\frac{1}{3!}\left$\frac{\mu}{2}\right$^3+\frac{1}{5!}\left$\frac{\mu}{2}\right$^5-\cdots\right\]A+\cdots$

Since $cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ and $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ ,

$e^{-i\frac{\mu}{2}A}=cos\left$\frac{\mu}{2}\right$I-isin\left$\frac{\mu}{2}\right$A$

Thus, if $e^{-i\frac{\mu}{2}A}$ is Hermitian, $A$ must also be Hermitian.

Substituting $A=n_x\begin{pmatrix}0&1\\1&0\end{pmatrix}+n_y\begin{pmatrix}0&-i\\i&0\end{pmatrix}+n_z\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ into the above equation results in the general form of the matrices that form an irreducible representation of SU(2):

$\hat{D}_n(\mu)=\begin{bmatrix}cos\frac{\mu}{2}-in_zsin\frac{\mu}{2}&(-in_x-n_y)sin\frac{\mu}{2}\\(-in_x+n_y)sin\frac{\mu}{2}&cos\frac{\mu}{2}+in_zsin\frac{\mu}{2}\end{bmatrix}\;\;\;\;\;\;\;\;129$

with $\alpha=\begin{pmatrix}1\\0\end{pmatrix}$ and $\beta=\begin{pmatrix}0\\1\end{pmatrix}$ being the basis states that transform according to the representation.

It is important to recognise that the irreducible representation contains an infinite number of elements because the parameter $\mu$ is a continuous variable. In spherical coordinates, $n_x=sin\theta cos\phi$ , $n_y=sin\theta sin\phi$ and $n_z=cos\theta$ (see this article for details). Substituting these equations into $\hat{D}_n(\mu)$ and simplifying gives:

$\hat{D}_n(\mu)=\begin{pmatrix}cos\frac{\mu}{2}-icos\theta sin\frac{\mu}{2}&-isin\theta e^{-i\phi}sin\frac{\mu}{2}\\-isin\theta e^{i\phi}sin\frac{\mu}{2}&cos\frac{\mu}{2}+icos\theta sin\frac{\mu}{2}\end{pmatrix}$

$\hat{D}_n(\mu)$ satisfies the conditions $U=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix}$ and $det(\hat{D}_n)=1$ , which can be easily verified. If we let $\hat{D}_n(\mu)$ act on either $\alpha=\begin{pmatrix}1\\0\end{pmatrix}$ or $\beta=\begin{pmatrix}0\\1\end{pmatrix}$ , the resulting state is a linear combination of the two spin states. For example,

$\hat{D}_n(\mu)\alpha=\hat{D}_n(\mu)\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}cos\frac{\mu}{2}-icos\theta sin\frac{\mu}{2}\\-isin\theta e^{i\phi}sin\frac{\mu}{2}\end{pmatrix}=a\alpha+b\beta$

where $a=cos\frac{\mu}{2}-icos\theta sin\frac{\mu}{2}$ and $b=-isin\theta e^{i\phi} sin\frac{\mu}{2}$ .

This is expected because the rotated basis vector is another vector in the two-dimensional complex vector space, and every vector in this space can be expressed as a linear combination of the basis vectors. In quantum mechanics, $\alpha$ and $\beta$ are eigenstates of $\hat{S}_z$ . After the rotation, the transformed state is generally no longer an eigenstate of $\hat{S}_z$ , since the spin has been rotated to point along a different direction in space. Consequently, a measurement of the spin component along the $z$ -axis yields the outcomes spin-up and spin-down with probabilities $\vert a\vert^2$ and $\vert b\vert^2$ respectively.

If we restrict $\hat{D}_n(\mu)$ to real entries, $n_x=n_z=0$ (in eq129) to eliminate their imaginary coefficients. This forces the unit rotation axis to point along the $y$ -axis, resulting in $n_y=1$ . Hence,

$\hat{D}_n(\gamma)=\begin{pmatrix}cos\gamma&-sin\gamma\\sin\gamma&cos\gamma\end{pmatrix}$

where $\gamma=\frac{\mu}{2}$ .

Here, $\hat{D}_n(\gamma)$ represents matrices that form a representation of SO(2), describing 2D rotations in the $xz$ -plane.

As mentioned earlier, each group element in the fundamental representation of SU(2) is represented by an $n\times n$ unitary matrix $U$ where $n=2$ . More broadly, a representation assigns each element of a group to an invertible square matrix in a way that preserves the group’s structure, such as the generator commutation relations $[J_a,J_b]=i\varepsilon_{abc}J_c$ . Therefore, besides the fundamental representation, SU(2) also has a one-dimensional $n=1$ trivial representation and higher-dimensional $n>2$ representations.

To derive these representations, we substitute the spin angular momentum operator $\hat{\bold{S}}$ in $\hat{D}_n(\mu)$ with the orbital angular momentum operator $\hat{\bold{L}}$ to give:

$\hat{D}_n(\mu)=e^{-i\frac{\mu}{\hbar}(\bold{n}\cdot\bold{\hat{L}})}=e^{-i\frac{\mu}{\hbar}(n_x\hat{L}_x+n_y\hat{L}_y+n_z\hat{L}_z)}$

It can be easily shown, by using the commutation relations of orbital angular momentum operators, that the generators $\hat{L}_x,\hat{L}_y,\hat{L}_z$ satisfy $[J_a,J_b]=i\varepsilon_{abc}J_c$ , where $J_k=\frac{\hat{L}_k}{\hbar}$ .

Consider a rotation by an angle $\mu$ around the $z$ -axis, with $\hat{D}_z(\mu)=e^{-i\frac{\mu}{\hbar}\hat{L}_z}$ , and where the eigenfunctions of $\hat{L}_z$ are the spherical harmonics $Y_l^{m_l}$ , satisfying $\hat{L}_zY_l^{m_l}=m_l\hbar Y_l^{m_l}$ . Using the relation $A\psi=a\psi\implies e^A\psi=e^a\psi$ (see Property 15 of this article for proof), where $A=-i\frac{\mu}{\hbar}\hat{L}_z$ , the action of the rotation operator on $Y_l^{m_l}$ is $\hat{D}_z(\mu)Y_l^{m_l}=e^{-im_l\mu}Y_l^{m_l}$ . Therefore, the transformation can be expressed as

$\hat{D}_z(\mu)Y_l^{m_l}=\begin{bmatrix}e^{-il\mu}&0&\cdots&0\\0&e^{-i(l-1)\mu}&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&e^{il\mu}\end{bmatrix}\begin{pmatrix}Y_l^l\\Y_l^{l-1}\\\vdots\\Y_L^{-l}\end{pmatrix}$

For $l=0$ , this gives a $1\times 1$ identity rotation matrix, corresponding to the trivial representation of SU(2). For $l=1$ and $l=2$ , the representations have dimensions 3 and 5 respectively. These illustrate that irreducible representations of SU(2) can have dimensions other than the two-dimensional fundamental representation

In conclusion, SU(2) is deeply embedded in any system involving quantum spins, rotations, or two-state dynamics. It describes quantum angular momentum and spin operators, and plays an essential role in the Standard Model, which explains three of the four known fundamental interactions. For example, isospin uses SU(2) symmetry to treat protons and neutrons as two states of the same particle, the nucleon, under the strong nuclear interaction, while neglecting their small electromagnetic differences. These examples illustrate how symmetry can reveal deep connections between physical states that might otherwise appear distinct.

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