Isospin, Hypercharge and the Gell-Mann-Nishijima Relation

Derivation

Isospin, Hypercharge and the Gell-Mann-Nishijima Relation

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Statement

For the light hadrons built from up and down quarks, the electric charge in units of \(e\) is fixed by the third component of strong isospin \(I_3\) and the hypercharge \(Y\) through the Gell-Mann–Nishijima relation \(\;Q = I_3 + \tfrac{1}{2}Y\;\), where \(Y = B + S\) collects the additively conserved baryon number and strangeness (extended by charm, bottomness and topness for heavier flavours). This organises every isospin multiplet into equally spaced charge ladders whose centre is displaced by \(Y/2\).

Why it matters

Isospin is the first internal symmetry of the strong interaction: the near-degeneracy of the proton and neutron, and of the three pions, is explained by an approximate \(SU(2)\) acting on flavour that commutes with the strong Hamiltonian. The Gell-Mann–Nishijima relation is the bridge between that internal symmetry and the one truly exact quantum number, electric charge — it tells you that charge is not an independent label but a fixed linear combination of \(I_3\) and \(Y\).

Historically the relation was inferred phenomenologically from the observed hadron spectrum before quarks existed as physical objects. When Gell-Mann and Zweig introduced quarks, the same relation followed from the charge assignments of \(u\), \(d\), \(s\), and it becomes exact once \(Y\) is completed to include all flavour numbers. In the Standard Model the very same algebraic structure — a diagonal charge that is \(I_3\) plus half a \(U(1)\) charge — reappears as the electroweak \(Q = T_3 + \tfrac{1}{2}Y_{\text{w}}\), so the pattern learned here is the template for electroweak unification.

Assumptions

The strong interaction possesses an internal \(SU(2)_I\) isospin symmetry.If dropped, \(I\) and \(I_3\) are not good quantum numbers, multiplets need not be degenerate, and there is no ladder to organise; charge would have to be assigned state by state.

Electric charge \(Q\), baryon number \(B\) and strangeness \(S\) are additively conserved and simultaneously diagonalisable.If dropped, \(Q\), \(Y\) and \(I_3\) could not share eigenstates and no single linear relation among their eigenvalues could hold.

The charge operator commutes with the isospin raising and lowering operators only up to the \(U(1)\) shift, i.e. \([\hat Q,\hat I_\pm]=\pm\hat I_\pm\).If dropped, \(Q\) would not increase in unit steps along a multiplet and the linear-in-\(I_3\) form would fail.

Hypercharge \(Y=B+S+C+B'+T\) is an isoscalar: \([\hat Y,\hat I_a]=0\) for all \(a\).If dropped, \(Y\) would vary within a multiplet, the multiplet centre \(\langle Q\rangle\) would not be a single number \(Y/2\), and the relation could not be stated per multiplet.

Within a charge multiplet the members differ only by \(I_3\) (exact isospin limit; electromagnetic and up–down mass splittings neglected).If dropped, states of equal \((I,I_3)\) but different mass mix, and \(Q\) is only defined after diagonalising the small symmetry-breaking terms.

Derivation

\[ [\hat I_3,\hat I_\pm]=\pm\,\hat I_\pm,\qquad [\hat I_+,\hat I_-]=2\hat I_3 \]

Isospin generators obey the same \(SU(2)\) algebra as angular momentum (prior result: the \(SU(2)\) algebra), so \(\hat I_\pm=\hat I_1\pm i\hat I_2\) act as ladder operators on \(I_3\). A

\[ \hat I_\pm\,|I,I_3\rangle=\sqrt{(I\mp I_3)(I\pm I_3+1)}\;|I,I_3\pm 1\rangle \]

Standard \(SU(2)\) matrix elements: a multiplet of isospin \(I\) contains the \(2I+1\) states \(I_3=-I,\dots,+I\) in unit steps. A

\[ \hat Q\,|I,I_3\rangle = q(I_3)\,|I,I_3\rangle \]

By assumption \(\hat Q\) is diagonal in the same basis; within one multiplet its eigenvalue is some function \(q\) of \(I_3\) (and of the fixed labels \(I,Y\)). A

\[ [\hat Q,\hat I_\pm]=\pm\,\hat I_\pm \]

Charge raising accompanies isospin raising: converting a \(d\) to a \(u\) (the microscopic content of \(\hat I_+\)) raises \(Q\) by exactly one unit. This is the structural input that makes \(Q\) linear in \(I_3\). B

\[ \hat Q\,\hat I_+|I,I_3\rangle=\hat I_+\hat Q|I,I_3\rangle+[\hat Q,\hat I_+]|I,I_3\rangle=\big(q(I_3)+1\big)\hat I_+|I,I_3\rangle \]

Acting with \(\hat Q\) on the raised state and using step 4. But \(\hat I_+|I,I_3\rangle\propto|I,I_3+1\rangle\) has eigenvalue \(q(I_3+1)\). B

\[ q(I_3+1)-q(I_3)=1 \quad\Longrightarrow\quad q(I_3)=I_3+c \]

The finite difference is constant, so \(q\) is affine in \(I_3\) with unit slope; \(c\) is independent of \(I_3\) and therefore constant across the multiplet. B

\[ \sum_{I_3=-I}^{I} q(I_3)=\sum_{I_3=-I}^{I}\big(I_3+c\big)=(2I+1)\,c \]

Sum \(q\) over the multiplet; the \(I_3\) terms cancel in the symmetric range, isolating the offset \(c\) as the mean charge. This step fixes \(c\) without needing the microscopic quark content. C

\[ c=\bar Q\equiv\frac{1}{2I+1}\sum_{I_3=-I}^{I} q(I_3)=\frac{1}{2}Y \]

The mean charge of a multiplet is the isoscalar centre. Identifying the additively conserved, isospin-blind offset with \(\tfrac12 Y\), \(Y=B+S+C+B'+T\) — a definition consistent with all observed multiplets (the factor \(\tfrac12\) is a convention fixing the normalisation of \(Y\)). C

\[ \hat Q=\hat I_3+\tfrac{1}{2}\hat Y \]

Reassemble: the eigenvalue relation \(q=I_3+\tfrac12 Y\) holds on every basis state of every multiplet, so it is an operator identity on the hadron Hilbert space. B

Result

\[ \boxed{\,Q = I_3 + \frac{Y}{2}\,,\qquad Y = B+S+C+B'+T\,} \]

Reading. Electric charge is not an independent label of a hadron: it is the third isospin component shifted by half the hypercharge. Within one isospin multiplet, \(Y\) is fixed and \(Q\) climbs in unit steps as \(I_3\) runs from \(-I\) to \(+I\); the whole ladder is centred on \(\langle Q\rangle=Y/2\). Different multiplets are distinguished by their centre \(Y/2\), and only \(I_3\) moves you within one.

Units check. \(Q\), \(I_3\), \(B\), \(S\), \(Y\) are all dimensionless quantum numbers (charge measured in units of \(e\)). \(I_3\) is a half-integer or integer; \(Y\) is chosen so that \(Y/2\) supplies the correct half-integer or integer offset; their sum \(Q\) is quantised in units of \(e\). Both sides are pure numbers, and the equation is homogeneous.

Limiting cases

Nucleon doublet (\(I=\tfrac12\), \(B=1\), \(S=0\Rightarrow Y=1\)): \(p\) has \(I_3=+\tfrac12\Rightarrow Q=+\tfrac12+\tfrac12=1\); \(n\) has \(I_3=-\tfrac12\Rightarrow Q=0\).
Pion triplet (\(I=1\), \(B=S=0\Rightarrow Y=0\)): centre at \(Q=0\); \(\pi^+,\pi^0,\pi^-\) at \(I_3=+1,0,-1\Rightarrow Q=+1,0,-1\).
Lambda singlet (\(I=0\), \(B=1\), \(S=-1\Rightarrow Y=0\)): \(I_3=0\Rightarrow Q=0\), consistent with the neutral \(\Lambda^0\).
Kaon doublet (\(I=\tfrac12\), \(B=0\), \(S=+1\Rightarrow Y=1\)): \(K^+\) at \(I_3=+\tfrac12\Rightarrow Q=1\), \(K^0\) at \(I_3=-\tfrac12\Rightarrow Q=0\).
Single quarks: \(u\) has \(I_3=+\tfrac12\), \(B=\tfrac13\Rightarrow Y=\tfrac13\), \(Q=\tfrac12+\tfrac16=\tfrac23\); \(d\) has \(I_3=-\tfrac12\), \(Q=-\tfrac12+\tfrac16=-\tfrac13\).

Breaks when

Isospin is badly broken. The relation assumes \(I,I_3\) label states. Electromagnetism and the \(u\)–\(d\) mass difference break \(SU(2)_I\); where these dominate (small mass splittings, or states that mix across isospin like \(\pi^0\)–\(\eta\)), \(I_3\) is no longer sharp and the assignment must follow the physical mass eigenstates.
Hypercharge is not isoscalar for the object considered. The naive \(Y=B+S\) fails once heavier flavours enter unless \(Y\) is extended to \(B+S+C+B'+T\); for a charmed or bottom hadron the un-extended relation gives the wrong charge.
Beyond the flavour \(SU(2)\)/\(SU(3)\) hadron sector. For leptons, or in the full electroweak theory, the operative relation is \(Q=T_3+\tfrac12 Y_{\text{w}}\) with weak isospin and weak hypercharge — a different \(U(1)\) — and strong \(I_3\), \(Y\) no longer apply.
Fractionally charged free states or exotic charges. The unit-step argument (step 6) assumes \([\hat Q,\hat I_\pm]=\pm\hat I_\pm\). Any dynamics giving non-integer charge steps within a multiplet (not realised in QCD hadrons) would violate the linear form.

Failure modes

Using \(Y=B+S\) for charmed/bottom hadrons. Forgetting the extra flavour numbers gives, e.g., a wrong charge for the \(\Lambda_c^+\); \(Y\) must be \(B+S+C\).
Confusing strong isospin with weak isospin. Writing the electroweak \(Q=T_3+\tfrac12 Y_{\text{w}}\) with hadronic \(Y=B+S\) — two different \(U(1)\)s that happen to share an algebraic shape.
Sign of strangeness. Taking \(S(K^+)=-1\); the historical convention makes the \(s\) quark carry \(S=-1\), so \(K^+=u\bar s\) has \(S=+1\). A flipped sign misplaces the whole strange sector.
Treating \(Y\) as variable within a multiplet. Trying to change \(Q\) by changing \(Y\) inside one ladder; \(Y\) is the fixed multiplet label, only \(I_3\) moves.
Dropping the factor \(\tfrac12\). Writing \(Q=I_3+Y\) with the old "strong hypercharge" numbers; the \(\tfrac12\) is a normalisation convention that must match how \(Y\) is defined.
Assigning \(I_3\) from charge circularly. Reading \(I_3\) off \(Q\) and then "deriving" \(Q\); \(I_3\) must be fixed independently by the multiplet structure.

Discussion

The content of the derivation is that the diagonal charge operator of the strong-interaction flavour symmetry splits into a part living in the non-abelian \(SU(2)_I\) (the generator \(I_3\)) and a part living in an abelian \(U(1)\) that commutes with all of isospin (the hypercharge). The commutator \([\hat Q,\hat I_\pm]=\pm\hat I_\pm\) is doing all the work: it says \(\hat Q-\hat I_3\) commutes with the raising and lowering operators, hence is a Casimir-like isoscalar, hence constant on each multiplet. Everything after that is bookkeeping to name that constant \(Y/2\).

Embedding this in \(SU(3)\) flavour, isospin and hypercharge are two commuting diagonal generators — the Cartan subalgebra \(\{\lambda_3,\lambda_8\}\) — and hadron multiplets become weight diagrams (the octet, the decuplet). The Gell-Mann–Nishijima line is then literally a diagonal direction in the \((I_3,Y)\) weight plane: \(Q\) is constant along lines of slope \(-2\), and the physical particles sit at lattice points whose \(Q\) is read straight off the plane. The famous "eightfold way" hexagons are just contour plots of \(Q=I_3+Y/2\).

At the deepest level the same skeleton reappears in electroweak theory as \(Q=T_3+\tfrac12 Y_{\text{w}}\), where \(T_3\) is a weak-isospin generator and \(Y_{\text{w}}\) is weak hypercharge; the photon is the unbroken combination \(A_\mu=\cos\theta_W B_\mu+\sin\theta_W W_\mu^3\) precisely so that its coupling is \(\hat Q\). That electric charge is quantised — every hadron and lepton carries an integer multiple of \(\tfrac13 e\) or \(e\) — is, in grand-unified pictures, forced by embedding this \(U(1)\) inside a larger non-abelian group whose charges are automatically quantised. So the humble hadronic relation is a low-energy shadow of why charge comes in discrete units at all.

Common misconceptions. The relation is not a dynamical law and predicts no masses; it is a labelling identity fixed by symmetry, exact even though isospin itself is only approximate, because \(Q\), \(B\), \(S\) are separately (nearly) exactly conserved. It also does not "explain" charge quantisation on its own — it relates charge to other quantum numbers that are themselves quantised. And \(I_3\) is not "the charge"; the two coincide only for \(Y=0\) multiplets like the pions.

Worked examples

\[ \text{Charge of }\Sigma^{\pm,0}\text{ (isotriplet, }B=1,\;S=-1) \]

The \(\Sigma\) hyperons form an \(I=1\) triplet with baryon number \(B=1\) and strangeness \(S=-1\). Identify the fixed labels first. A

\[ Y=B+S=1+(-1)=0 \]

Hypercharge is isoscalar and fixed for the multiplet; here it vanishes, so the centre sits at \(Q=0\). A

\[ Q=I_3+\tfrac12 Y=I_3+0=I_3,\qquad I_3\in\{+1,0,-1\} \]

Apply the relation with numbers. The triplet spans \(I_3=+1,0,-1\). B

\[ Q(\Sigma^+)=+1,\quad Q(\Sigma^0)=0,\quad Q(\Sigma^-)=-1 \]

Reading. A \(Y=0\) triplet is charge-symmetric about zero, exactly like the pions but made of baryons; charge equals \(I_3\) here.

Units check. All entries integer multiples of \(e\); \(I_3\) integer because \(I=1\); consistent.

\[ \text{Assign }Y\text{ and charges to the }\Xi\text{ doublet }(\Xi^0,\Xi^-),\;B=1,\;S=-2 \]

The cascade hyperons form an \(I=\tfrac12\) doublet with \(S=-2\). Fix the labels. A

\[ Y=B+S=1+(-2)=-1 \]

Hypercharge for the multiplet; the centre is displaced to \(Q=Y/2=-\tfrac12\). A

\[ Q=I_3+\tfrac12(-1)=I_3-\tfrac12,\qquad I_3\in\{+\tfrac12,-\tfrac12\} \]

Insert numbers; the doublet ladder is shifted down by half a unit. B

\[ Q(\Xi^0)=+\tfrac12-\tfrac12=0,\qquad Q(\Xi^-)=-\tfrac12-\tfrac12=-1 \]

Reading. A doubly strange doublet is charge \(0\) and \(-1\); the half-integer \(I_3\) plus the half-integer offset \(Y/2=-\tfrac12\) combine to integer charges, as they must for a physical hadron.

Units check. \(-\tfrac12-\tfrac12=-1\): two half-units sum to an integer in units of \(e\); consistent.

Problems

Using \(Q=I_3+Y/2\), find the charges of the kaon doublet \((K^+,K^0)\), given \(I=\tfrac12\), \(B=0\), \(S=+1\).
Solution
\(Y=B+S=0+1=1\), so \(Q=I_3+\tfrac12\). With \(I_3=+\tfrac12\): \(Q(K^+)=\tfrac12+\tfrac12=1\). With \(I_3=-\tfrac12\): \(Q(K^0)=-\tfrac12+\tfrac12=0\). Charges \(+1,0\), matching \(K^+\) and \(K^0\). The antidoublet \((\bar K^0,K^-)\) has \(S=-1\Rightarrow Y=-1\), giving \(0\) and \(-1\).
The \(\Delta\) baryons form an \(I=\tfrac32\) quartet with \(B=1\), \(S=0\). List their \(I_3\) values and charges.
Solution
\(Y=B+S=1\), so \(Q=I_3+\tfrac12\). The quartet has \(I_3=+\tfrac32,+\tfrac12,-\tfrac12,-\tfrac32\), giving \(Q=2,1,0,-1\). These are \(\Delta^{++},\Delta^{+},\Delta^{0},\Delta^{-}\). Note the doubly charged \(\Delta^{++}\) is only possible because the half-integer \(I_3=+\tfrac32\) plus \(Y/2=\tfrac12\) reaches \(+2\).
A baryon has \(Q=+1\), \(I_3=0\), \(B=1\). Determine \(Y\) and its strangeness \(S\) (assume no charm or heavier flavour).
Solution
Rearrange: \(Y=2(Q-I_3)=2(1-0)=2\). Then \(S=Y-B=2-1=+1\). An \(I_3=0\), \(Q=+1\) baryon with \(B=1\) would need \(S=+1\); no such light baryon is observed, which is itself informative — it flags that the naive assignment is unphysical (real \(S=+1\) baryons, the pentaquark candidates, are exotic).
Verify the relation for the constituent quarks: \(u\) has \(I_3=+\tfrac12\), \(d\) has \(I_3=-\tfrac12\), \(s\) has \(I_3=0\); each has \(B=\tfrac13\), with \(S(s)=-1\) and \(S(u)=S(d)=0\). Compute all three charges.
Solution
For \(u\): \(Y=B+S=\tfrac13\), \(Q=\tfrac12+\tfrac16=\tfrac23\). For \(d\): \(Y=\tfrac13\), \(Q=-\tfrac12+\tfrac16=-\tfrac13\). For \(s\): \(Y=\tfrac13-1=-\tfrac23\), \(Q=0+\tfrac12(-\tfrac23)=-\tfrac13\). Charges \(+\tfrac23,-\tfrac13,-\tfrac13\), the standard quark charges. Building \(p=uud\) gives \(\tfrac23+\tfrac23-\tfrac13=1\), consistent.
The \(\Omega^-\) is an \(I=0\) baryon with \(B=1\), \(S=-3\). Predict its charge from the relation, and state why its \(I_3\) is unambiguous.
Solution
\(Y=B+S=1-3=-2\). Being an isosinglet, \(I=0\Rightarrow I_3=0\) with no ambiguity (a singlet has a single \(I_3=0\) member). Then \(Q=I_3+\tfrac12 Y=0+\tfrac12(-2)=-1\), giving \(\Omega^-\). Its prediction as an \(sss\) state — three \(s\) quarks each \(Q=-\tfrac13\) summing to \(-1\), \(S=-3\) — was a triumph of the eightfold-way classification before its discovery.