Identifying the Antitriplet via the Levi-Civita Tensor
An antisymmetric pair of two quarks transforms as the antiquark of the missing flavor. \[ ud - du \longleftrightarrow \bar s \] \[ us - su \longleftrightarrow \bar d \] \[ ds - sd \longleftrightarrow \bar u \] Here, $ u $, $ d $, and $ s $ denote the up, down, and strange quarks, respectively, while $ \bar{u} $, $ \bar{d} $, and $ \bar{s} $ denote the corresponding antiquarks.
This correspondence follows directly from a structural property of SU(3) flavor symmetry.
Contracting two quark fields with the completely antisymmetric Levi-Civita tensor \( \varepsilon_{ijk} \) produces an object that transforms as an antiquark:
\[ \boxed{ \varepsilon_{ijk}\, q^j q^k = \bar q_i } \]
Here \( q^j \) and \( q^k \) are quark fields transforming in the fundamental representation of SU(3).
The indices \( i,j,k \) take the values \(1,2,3 \), or equivalently the flavor labels \(u,d,s \) (up, down, strange).
The symbol \( \varepsilon_{ijk} \) denotes the completely antisymmetric Levi-Civita tensor, which satisfies three defining properties:
- \( \varepsilon_{ijk} = 0 \) whenever two indices coincide
- \( \varepsilon_{ijk} = +1 \) for even permutations of \( (u,d,s) \)
- \( \varepsilon_{ijk} = -1 \) for odd permutations of \( (u,d,s) \)
By standard convention, one fixes:
\[ \varepsilon_{uds} = +1 \]
For example, starting from \( \varepsilon_{uds} = +1 \), exchanging \( u \) and \( d \) yields \( \varepsilon_{dus} = -1 \), since the transformation $ (uds) \to (dus) $ corresponds to a single transposition, that is, an odd permutation. If instead two indices coincide, as in \( \varepsilon_{uus} \), the tensor vanishes: \[ \varepsilon_{uus} = 0 \]
A concrete example
Within SU(3) flavor symmetry, the quark flavors span the fundamental representation:
\[ 3 = \{ u, d, s \} \]
When two quarks are combined, the resulting state space is the tensor product:
\[ 3 \otimes 3 \]
There are $ 3 \times 3 = 9 $ possible two-quark basis states:
$$ uu, ud, us, du, dd, ds, su, sd, ss $$
Some of these states are symmetric, meaning they are invariant under exchange of the two quarks, while others are antisymmetric, meaning they change sign under such an exchange.
- The states $ uu, dd, ss $ consist of identical quarks. Exchanging the two quarks leaves the state unchanged, so these states are symmetric.
- The remaining states $ ud, us, du, ds, su, sd $ involve distinct quarks. Exchanging the two quarks maps the state into a different one, for example $ ud \to du $. Taken individually, these states are therefore neither symmetric nor antisymmetric.
Note. Exchanging the quarks in the state $ uu $ always gives $ uu $, so the state is symmetric: $$ uu \xrightarrow{\text{exchange}} uu \quad \text{(symmetric)} $$ By contrast, exchanging the quarks in the state $ ud $ yields $ du $, which is a different basis state. Hence, $ ud $ is neither symmetric nor antisymmetric: $$ ud \xrightarrow{\text{exchange}} du $$
At this stage, three symmetric states have been identified, but there are nine two-quark states in total. The remaining six must therefore be classified.
Symmetry is not a property of individual basis states taken in isolation, such as $ ud, du, us, \dots $, but can emerge through appropriate linear combinations of these states. In other words, symmetry is a property of the subspace spanned by the states, not of the basis vectors themselves.
However, not all linear combinations exhibit definite symmetry.
For instance, combinations such as $ ud + us $, $ ud - us $, or $ du + su $ are not invariant under exchange of the two quarks and are therefore neither symmetric nor antisymmetric. As an example, exchanging the quarks in $ ud + us $ gives: $$ ud + us \xrightarrow{\text{exchange}} du + su $$ which differs from the original combination.
Among all possible linear combinations, there are exactly three symmetric ones, which remain unchanged under exchange of the two quarks:
\[ ud+du \xrightarrow{\text{exchange}} du+ud = ud+du \]
\[ us+su \xrightarrow{\text{exchange}} su+us = us+su \]
\[ ds+sd \xrightarrow{\text{exchange}} sd+ds = ds+sd \]
There are also three antisymmetric combinations, whose overall form is preserved but whose sign reverses under exchange:
\[ ud-du \xrightarrow{\text{exchange}} du-ud = - (ud - du) \]
\[ us-su \xrightarrow{\text{exchange}} su-us = - (us - su) \]
\[ ds-sd \xrightarrow{\text{exchange}} sd-ds = - (ds - sd) \]
In summary, the tensor product decomposes into $ 6 $ symmetric states and $ 3 $ antisymmetric states, conventionally denoted by $ \bar{3} $:
\[ 3 \otimes 3 = 6 \oplus \bar{3} \]
We now focus on the three antisymmetric quark pairs.
According to the structural rule introduced above, each antisymmetric pair of two quarks transforms as the antiquark associated with the missing flavor:
\[ ud - du \longleftrightarrow \bar s \]
\[ us - su \longleftrightarrow \bar d \]
\[ ds - sd \longleftrightarrow \bar u \]
Note. In the antisymmetric combination \(ud - du\), the quarks \(u \) and \(d\) are present, while the quark \(s\) is absent. For this reason, the pair transforms as the strange antiquark \( \bar s \). Similarly, in \( us - su \) the quark \( d \) is missing, so the pair transforms as \( \bar{ d } \). Finally, in \( ds - sd \) the quark \( u \) is missing, and the pair transforms as \( \bar{ u } \).
This correspondence can be demonstrated explicitly. Consider the antisymmetric combination:
\[ ud - du \]
According to the structural rule, an antisymmetric quark pair corresponds to the missing antiquark through the contraction:
\[ \bar q_i = \varepsilon_{ijk} q^j q^k \]
In the combination $ ud-du $, the quark flavors involved are $ u $ and $ d $.
Accordingly, we choose the indices \( j=u \) and \( k=d \):
\[ \bar q_i=\varepsilon_{iud} \ q^u q^d+\varepsilon_{idu} \ q^d q^u \]
Both terms appear because the contraction includes contributions from both $ ud $ and $ du $.
Since the Levi-Civita tensor is antisymmetric, the relation
\[ \varepsilon_{idu}=-\varepsilon_{iud} \]
holds. Substituting this relation, we obtain:
\[ \bar q_i=\varepsilon_{iud} \ q^u q^d - \varepsilon_{iud} \ q^d q^u \]
Applying associativity then gives:
\[ \bar q_i=\varepsilon_{iud} \ (q^u q^d-q^d q^u) \]
or equivalently:
\[ \bar q_i=\varepsilon_{iud}\,(ud-du) \]
Thus, contraction with \( \varepsilon \) automatically projects out the antisymmetric combination \(ud-du \).
We now examine \(\varepsilon_{iud}\) and determine which flavor corresponds to the index $ i $. In SU(3), the possible values are \( u,d,s \):
- \( i = u \ \to \ \varepsilon_{uud}=0 \)
- \( i = d \ \to \ \varepsilon_{dud}=0 \)
- \( i = s \ \to \ \varepsilon_{sud}= \pm 1 \neq 0 \)
The only choice that yields a nonvanishing tensor component is $ i=s $:
\[ i=s \]
Substituting $ i=s $ into the expression gives:
\[ \bar q_s=\varepsilon_{sud} \ (ud-du) \]
By convention, $ \varepsilon_{uds} = +1 $. The permutation $ (uds) \to (sud) $ is a permutation of length 3, that is, a 3-cycle.
A 3-cycle is an even permutation, since it can be written as the product of two transpositions, for example:
$$ (uds) \to (sud) = (us)(ud) $$
Therefore, being an even permutation, the completely antisymmetric Levi-Civita tensor takes the value $ \varepsilon_{sud} = +1 $.
\[ \bar q_s= \underbrace{ \varepsilon_{sud}}_{+1} \ (ud-du) \]
\[ \bar q_s= (ud-du) \]
Since $ \bar q_s $ is the $ s $ component of the antitriplet, it is naturally identified with the strange antiquark $ \bar s $:
\[ \bar s= (ud-du) \]
For notational simplicity, we have written $ \bar s= (ud-du) $, but this expression is not mathematically literal, as it suggests an equality between two distinct objects. A more precise and conceptually correct notation is:
\[ \bar s \longleftrightarrow (ud-du) \]
This demonstrates that the antisymmetric combination $ ud-du $ transforms as a strange antiquark $ \bar{s} $.
Note. The overall sign depends on the convention adopted for \( \varepsilon_{uds} \), but it has no physical significance. Even if one were to obtain $ \varepsilon_{sud} = -1 $, the resulting state would represent the same physical state, differing only by a global minus sign: \[ \bar q_s= - (ud-du) \] In general, two states that differ by a nonzero global factor, in particular \( \pm 1 \), correspond to the same physical state.
The same reasoning applies to the remaining antisymmetric pairs, $ us - su $ and $ ds - sd $:
\[ ud - du \longleftrightarrow \bar s \]
\[ us - su \longleftrightarrow \bar d \]
\[ ds - sd \longleftrightarrow \bar u \]
In conclusion, the antisymmetric part of the tensor product of two quarks furnishes the \( \bar 3\ = \{ \bar u, \bar d, \bar s \} \) representation of SU(3).
Each antisymmetric pair of two quarks therefore transforms as the antiquark of the missing flavor.
And so on.
