Isospin and the SU(2) Symmetry of Up and Down Quarks

We speak of a symmetry between up (u) and down (d) quarks because, within the strong interaction, they have nearly identical masses, they couple to the strong force in the same way, and exchanging one for the other leaves the underlying theory unchanged. In other words, replacing an up quark with a down quark does not modify the strong interaction.

To formalize this invariance under transformations that mix the up and down quarks, physicists introduce the concept of isospin.

Isospin is not a physical spin, although it follows the same $ SU(2) $ algebra.

• Spin is defined in physical space and describes how a particle transforms under actual spatial rotations. It is a physical rotation in real space.
• Isospin belongs to an internal space with no direct spatial interpretation. In this internal space, the up and down quarks form a two component vector, and a rotation of this vector mixes the two states. This rotation has no relation to spatial motion.

Mathematically, both spin and isospin are governed by the group $ SU(2) $, which gives them a shared algebraic structure even though they describe distinct physical concepts.

Because isospin does not act in real space, it is classified as an internal symmetry.

Note. Isospin offers a concise and powerful framework for describing all transformations that mix up and down quarks without altering the dynamics of the strong interaction. This unified viewpoint clarifies several physical observations: particles with different charges can have almost identical masses, certain strong processes occur with the same probability, and specific decays are forbidden when they break isospin symmetry. These features arise directly from the symmetry structure of the theory.

Isospin is therefore not a vector in ordinary space with $ x, y, z $ components. Instead, it is characterized by abstract components $ I_1 $, $ I_2 $, $ I_3 $.

Two quantum numbers specify the isospin of a state:

• $ I $ is the total isospin
  It determines the size of the multiplet to which the particle belongs.
• $ I_3 $ is the third component
  It identifies the specific member of the multiplet.

If the strong interaction is invariant under $ SU(2) $ rotations in isospin space, Noether's theorem requires that isospin be conserved in every strong interaction process.

Note. This is the exact analogue of angular momentum conservation in systems that respect rotational symmetry in physical space.

The $ SU(2) $ symmetry organizes states into multiplets of definite isospin:

• doublets with $ I = \frac{1}{2} $
• triplets with $ I = 1 $
• quartets with $ I = \frac{3}{2} $

Each multiplet corresponds to a fixed value of $ I $, and each component of the multiplet is distinguished by its value of $ I_3 $.

The total isospin \( I \) can be deduced simply by counting the distinct values assumed by \( I_3 \).

Since \( I_3 \) ranges from \( -I \) to \( +I \) in integer steps, a multiplet with isospin \( I \) contains exactly \( 2I + 1 \) states.

This multiplet structure follows directly from the underlying $ SU(2) $ symmetry.

The Origin of Isospin

Isospin was introduced by Werner Heisenberg in 1932.

Heisenberg observed that the proton and neutron have nearly identical masses and exhibit virtually the same behavior under the strong interaction.

To explain this near degeneracy, he proposed a new quantum number modeled on the mathematical structure of ordinary spin but acting on internal degrees of freedom rather than spatial ones. This became known as isotopic spin, later shortened to isospin.

In Heisenberg's formulation, the proton and neutron represent two eigenstates of a single underlying particle: the nucleon \( N \).

$$ N = \frac{\alpha}{\beta} $$

The proton (p) and neutron (n) correspond to orthonormal basis states in this internal doublet:

$$ p = \frac{1}{0} $$

$$ n = \frac{0}{1} $$

The nucleon forms a two state system with total isospin \( I = \tfrac12 \):

$$ I = \frac{1}{2} $$

The third component \( I_3 \) distinguishes the two states, taking the eigenvalue \( +\tfrac12 \) for the proton and \( -\tfrac12 \) for the neutron:

$$ p = \left| \tfrac12, +\tfrac12 \right\rangle $$

$$ n = \left| \tfrac12, -\tfrac12 \right\rangle $$

Thus the proton and neutron appear as two projections of the same nucleon, analogous to the spin up and spin down states of a spin one half particle. The quantum number that differentiates them is the isospin projection \( I_3 \).

A rotation of 180° in isospin space exchanges the proton and neutron.

The isospin concept extends naturally to quark flavors.

The up (u) and down (d) quarks form an isospin doublet closely paralleling the nucleon doublet:

$$ u = \left| \tfrac12, +\tfrac12 \right\rangle $$

$$ d = \left| \tfrac12, -\tfrac12 \right\rangle $$

Both quarks have total isospin \( I = \tfrac12 \), differing only in their value of \( I_3 \):

• the up quark has \( I_3 = +\tfrac12 \)
• the down quark has \( I_3 = -\tfrac12 \)

The remaining quark flavors (s, c, b, t) all carry isospin zero, since they lie outside the approximate \( SU(2) \) flavor symmetry of the light quarks.

The Two Nucleon System

Isospin is not merely a classification scheme. It also determines which combined states are allowed by the symmetry.

Because the proton and neutron each carry isospin \( I = \tfrac12 \), the isospin of a two nucleon system is obtained using the angular momentum addition rule:

$$ \tfrac12 \otimes \tfrac12 = 1 \oplus 0 $$

Note. To add two isospins \( \tfrac12 \): $$ I_1 = \tfrac12 \quad I_2 = \tfrac12 $$ The allowed total isospins follow the familiar angular momentum rule: $$ I = |I_1 - I_2|,\ |I_1 - I_2| + 1,\ \dots,\ I_1 + I_2 $$ Thus: $$ I = \left|\tfrac12 - \tfrac12\right| = 0 $$ $$ I = \tfrac12 + \tfrac12 = 1 $$ The possible total isospins are therefore: $$ I = 1 \quad (\text{triplet}) $$ $$ I = 0 \quad (\text{singlet}) $$ This is precisely what is expressed by $$ \tfrac12 \otimes \tfrac12 = 1 \oplus 0 $$

This decomposition yields the following two nucleon states:

• the symmetric triplet with \( I = 1 \): $$ |1,1\rangle = |pp\rangle \\[4pt] |1,0\rangle = \tfrac{1}{\sqrt{2}} \big(|pn\rangle + |np\rangle\big) \\[4pt] |1,-1\rangle = |nn\rangle $$
• the antisymmetric singlet with \( I = 0 \): $$ |0,0\rangle = \tfrac{1}{\sqrt{2}} \big(|pn\rangle - |np\rangle\big) $$

These exhaust all possible isospin configurations of the two nucleon system.

Which of these states occurs in nature?

The only bound two nucleon system observed experimentally is the deuteron (pn), consisting of a proton and a neutron.

Neither a bound pp system nor a bound nn system exists in nature.

For this reason, the deuteron cannot belong to the triplet \( I = 1 \). If it did, the symmetry would require the triplet to appear in its entirety (pp, pn, nn), since the three are connected by \( SU(2) \) rotations in isospin space.

The only consistent assignment is that the deuteron belongs to the singlet \( I = 0 \).

How the (u, d) Quark Doublet Generates All Isospin Multiplets

Light quarks carry isospin

$$ I = \frac{1}{2} $$

with the up and down quarks forming the two components:

$$ u = \left| \tfrac12, +\tfrac12 \right\rangle $$

$$ d = \left| \tfrac12, -\tfrac12 \right\rangle $$

This (u, d) doublet is the fundamental representation from which all isospin multiplets of light hadrons are constructed.

1] Mesons (q\bar q)

Combining two isospin \( \tfrac12 \) representations yields

$$ \tfrac12 \otimes \tfrac12 = 1 \oplus 0 $$

meaning that mesons built from \(u\bar u\) and \(d\bar d\) appear either as an isospin singlet with \( I = 0 \) or as an isospin triplet with \( I = 1 \). These are the only possibilities allowed by the tensor product of two doublets.

Example 1. A standard example of an isospin triplet is the pion system: $$ \pi^+ = \left| 1, +1 \right\rangle $$ $$ \pi^0 = \left| 1, 0 \right\rangle $$ $$ \pi^- = \left| 1, -1 \right\rangle $$ Example 2. The only isospin singlet arising from \(u\bar u\) and \(d\bar d\) is the antisymmetric combination, usually associated with the \( \eta_0 \) state in the \((u, d)\) flavor sector.

2] Baryons (qqq)

The isospin structure of three light quarks begins with the tensor product

$$ \tfrac12 \otimes \tfrac12 \otimes \tfrac12. $$

Since

$$ \tfrac12 \otimes \tfrac12 = 1 \oplus 0, $$

we rewrite the product as

$$ (1 \oplus 0) \otimes \tfrac12. $$

Using distributivity,

$$ (1 \otimes \tfrac12) \oplus (0 \otimes \tfrac12). $$

The standard $SU(2)$ decomposition rules give

$$ 1 \otimes \tfrac12 = \tfrac32 \oplus \tfrac12, \qquad 0 \otimes \tfrac12 = \tfrac12, $$

so the full decomposition becomes

$$ \left( \tfrac32 \oplus \tfrac12 \right) \oplus \left( \tfrac12 \right). $$

Associativity leads to the final result:

$$ \tfrac32 \oplus \tfrac12 \oplus \tfrac12. $$

Thus, three light quarks can form exactly one quartet with \( I = \tfrac32 \) and two doublets with \( I = \tfrac12 \). No additional isospin representations appear in the three quark sector.

Example 1. The nucleons form an isospin doublet with \( I = \tfrac12 \): $$ p = uud = \left| \tfrac12, +\tfrac12 \right\rangle $$$$ n = udd = \left| \tfrac12, -\tfrac12 \right\rangle $$ These are the only light baryons with isospin \( \tfrac12 \).

Example 2. The \( \Delta \) baryons form an isospin quartet with \( I = \tfrac32 \): $$ \Delta^{++} = uuu = \left| \tfrac32, +\tfrac32 \right\rangle $$ $$ \Delta^{+} = uud = \left| \tfrac32, +\tfrac12 \right\rangle $$ $$ \Delta^{0} = udd = \left| \tfrac32,  -\tfrac12 \right\rangle $$ $$ \Delta^{-} = ddd = \left| \tfrac32, -\tfrac32 \right\rangle $$ These states are fully symmetric in flavor, as required by the \( I = \tfrac32 \) representation.

The central point is that isospin multiplets follow directly from the representation theory of \( SU(2) \). If QCD treats u and d quarks as nearly identical, then all hadrons composed solely of u and d quarks must organize themselves into the allowed $SU(2)$ irreducible representations.

A Practical Example

The \( \Delta \) baryons appear in four charge states: \( \Delta^{++}, \Delta^{+}, \Delta^{0}, \Delta^{-} \).

$$ 2I + 1 = 4 $$

so the corresponding isospin is

$$ I = \frac{4 - 1}{2} = \tfrac32. $$

Thus, the \( \Delta \) family forms an isospin quartet.

The values of the third component are

$$ I_3 = +\tfrac32,\ +\tfrac12,\ -\tfrac12,\ -\tfrac32. $$

The associated physical states are:

• \( \Delta^{++} \) with \( I_3 = +\tfrac32 \)
• \( \Delta^{+} \) with \( I_3 = +\tfrac12 \)
• \( \Delta^{0} \) with \( I_3 = -\tfrac12 \)
• \( \Delta^{-} \) with \( I_3 = -\tfrac32 \)

In ket notation:

• \( \Delta^{++} = \left| \tfrac32, +\tfrac32 \right\rangle \)
• \( \Delta^{+} = \left| \tfrac32, +\tfrac12 \right\rangle \)
• \( \Delta^{0} = \left| \tfrac32, -\tfrac12 \right\rangle \)
• \( \Delta^{-} = \left| \tfrac32, -\tfrac32 \right\rangle \)

In terms of quark flavor:

• \( \Delta^{++} = uuu \)
• \( \Delta^{+} = uud \)
• \( \Delta^{0} = udd \)
• \( \Delta^{-} = ddd \)

Example. The \( \Delta^{++} \) sits at the top of the \( I = \tfrac32 \) multiplet with \( I_3 = +\tfrac32 \). Because each up quark contributes \( I_3 = +\tfrac12 \), the only combination that yields $$ I_3 = \tfrac12 + \tfrac12 + \tfrac12 = \tfrac32 $$ is $$ \Delta^{++} = uuu. $$ This quark composition does not determine the total isospin by itself. The total isospin is fixed by the requirement that the state belong to the fully symmetric flavor representation of \( SU(2) \). The quark content simply reflects why the \( \Delta^{++} \) occupies the highest component of the quartet.

In summary, the \( \Delta \) baryons form an isospin quartet because they arise from the tensor product of three isospin doublets:

$$ \tfrac12 \otimes \tfrac12 \otimes \tfrac12 $$

which decomposes as

$$ \tfrac32 \oplus \tfrac12 \oplus \tfrac12. $$

The \( I = \tfrac32 \) representation is the fully symmetric one in flavor space.

This representation contains four states, each with a distinct value of

$$ I_3 = +\tfrac32,\ +\tfrac12,\ -\tfrac12,\ -\tfrac32. $$

These correspond precisely to

$$ \Delta^{++},\ \Delta^{+},\ \Delta^{0},\ \Delta^{-}. $$

This structure follows directly from \( SU(2) \) symmetry: if the u and d quarks form an isospin doublet, then any three quark system built from them must fall into the allowed irreducible representations, including the fully symmetric \( I = \tfrac32 \) representation that produces the \( \Delta \) quartet.

Why Exchanging Up and Down Quarks Leaves the Strong Interaction Unchanged

Consider the isospin doublet of the light quarks:

\[ \begin{pmatrix} u \\ d \end{pmatrix} \]

Now apply an $ SU(2) $ transformation, which acts as a rotation in an internal two dimensional space:

\[ \begin{pmatrix} u' \\ d' \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} u \\ d \end{pmatrix} \]

This rotation mixes the up and down quark fields:

• $ u' = \cos\theta\, u + \sin\theta\, d $
• $ d' = -\sin\theta\, u + \cos\theta\, d $

The key point is that QCD is invariant under this transformation. The rotated fields $ u' $ and $ d' $ interact with the strong force exactly as $ u $ and $ d $ do. Nothing in the interaction Lagrangian changes.

This is what defines an internal symmetry: the up and down quarks can be mixed by any $ SU(2) $ rotation without altering the physics of the strong interaction.

Therefore, hadrons formed from $ u' $ and $ d' $ have the same properties and strong interaction behavior as those formed from the original $ u $ and $ d $. This captures the essence of isospin symmetry: the two quarks represent different states of a single internal degree of freedom.

For example, replacing an up quark with a down quark turns a proton ($ p = uud $) into a neutron ($ n = udd $) and vice versa. Treated as an isospin doublet, they transform as \[ \begin{pmatrix} p' \\ n' \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} p \\ n \end{pmatrix} \] QCD is blind to this transformation, which is why the strong force does not distinguish between the two nucleons.

A Practical Illustration

We can make this structure explicit by writing proton and neutron fields as components of a single isospin doublet.

The nucleon doublet is

$$ N = \begin{pmatrix} p \\ n \end{pmatrix} $$

Here \( p \) is the proton and \( n \) the neutron. The field \( N \) represents the nucleon with its two isospin states.

These states form the two dimensional representation corresponding to total isospin \( I = \tfrac12 \).

Now apply an $ SU(2) $ rotation within this internal space.

Consider the real representation of an $ SU(2) $ rotation matrix:

\[ R(\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \]

This matrix satisfies the defining $ SU(2) $ relations

\[ R^\dagger R = \mathbb{1}, \quad \det R = 1 \]

where \( R^\dagger \) is the Hermitian conjugate of \( R \). Any $ 2\times2 $ unitary matrix with unit determinant belongs to \( SU(2) \).

Applying the rotation yields

\[ \begin{pmatrix} p' \\ n' \end{pmatrix} = R(\theta)\, \begin{pmatrix} p \\ n \end{pmatrix} \]

\[ \begin{pmatrix} p' \\ n' \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} p \\ n \end{pmatrix} \]

so that

\[ \begin{cases} p' = p\,\cos\theta + n\,\sin\theta \\[4pt] n' = -p\,\sin\theta + n\,\cos\theta \end{cases} \]

Choose a specific angle, say \( \theta = \frac{\pi}{2} \), corresponding to a 90 degree rotation:

\[ \begin{cases} p' = p\,\cos\frac{\pi}{2} + n\,\sin\frac{\pi}{2} \\[4pt] n' = -p\,\sin\frac{\pi}{2} + n\,\cos\frac{\pi}{2} \end{cases} \]

Using \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), we obtain

\[ \begin{cases} p' = n \\[4pt] n' = -p \end{cases} \]

This shows that a 90 degree isospin rotation effectively exchanges the two nucleon states, up to an overall minus sign.

The minus sign corresponds to a global phase factor \( e^{i\pi} = -1 \), which has no observable consequences. Thus, the state \( n' = -p \) is physically equivalent to \( p \).

The transformation can be summarized as

\[ R\left(\frac{\pi}{2}\right)\begin{pmatrix} p \\ n\end{pmatrix} = \begin{pmatrix} n \\ -p \end{pmatrix} \]

In this formulation, the proton and neutron appear as components of a two dimensional internal vector, which can be rotated just like a vector in ordinary space, except that the rotation occurs in isospin space rather than physical space.

Nota. This transformation is not a literal spatial rotation. It acts only in the internal isospin space. The significance of the symmetry is that the strong interaction remains invariant under such rotations and therefore does not distinguish between protons and neutrons in strong interaction processes.

Notes

Additional remarks on isospin and internal symmetries

• Electric charge and isospin
  The electric charge \( Q \) of a hadron is linked to its isospin through the Gell Mann - Nishijima relation $$ Q = I_3 + \tfrac12(A + S) $$ where \( A \) is the baryon number and \( S \) the strangeness quantum number. This relation applies to hadrons built from the light quarks \( u, d, s \). Within a given multiplet, the state with the largest electric charge is assigned \( I_3 = I \), and the remaining states follow in descending order of charge.
• Limits of proton neutron symmetry
  The symmetry between proton and neutron applies to the strong interaction Hamiltonian, not to the particles considered individually. Exchanging a single proton with a neutron does not preserve the many body quantum state because of the Pauli exclusion principle. Only a collective exchange of all protons with all neutrons in a system preserves the symmetry.

 And so on.