9.4 Tensor products of representations

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie algebra representation

Given two representations of a group a new representation can be defined, the tensor product representation, by:

Definition (Tensor product representation of a group)

For $(π_{V}, V)$ and $(π_{W}, W)$ representations of a group $G$ , there is a tensor product representation $(π_{V \otimes W}, V \otimes$ W) defined by
$(π_{V \otimes W} (g)) (v \otimes w) = π_{V} (g) v \otimes π_{W} (g) w$

One can easily check that $π_{V \otimes W}$ is a homomorphism.

To see what happens for the corresponding Lie algebra representation, compute (for $X$ in the Lie algebra)

π_{V \otimes W}^{'} (X) (v \otimes w) = \frac{d}{d t} π_{V \otimes W} (e^{tX}) (v \otimes w)_{t = 0} = \frac{d}{d t} (π_{V} (e^{tX}) v \otimes π_{W} (e^{tX}) w)_{t = 0} = ((\frac{d}{d t} π_{V} (e^{tX}) v) \otimes π_{W} (e^{tX}) w)_{t = 0} + (π_{V} (e^{tX}) v \otimes (\frac{d}{d t} π_{W} (e^{tX}) w))_{t = 0} = (π_{V}^{'} (X) v) \otimes w + v \otimes (π_{W}^{'} (X) w)

which could also be written

π_{V \otimes W}^{'} (X) = (π_{V}^{'} (X) \otimes 1_{W}) + (1_{V} \otimes π_{W}^{'} (X))

9.4.1 Tensor products of SU(2) representations

Concept links · terms present in this machine draft; source roles are unverified: irreducible representation

Given two representations $(π_{V}, V)$ and $(π_{W}, W)$ of a group $G$ , we can decompose each into irreducibles. To do the same for the tensor product of the two representations, we need to know how to decompose the tensor product of two irreducibles. This is a fundamental and non-trivial question, with the answer for $G = S U (2)$ as follows:

Theorem 9.1 (Clebsch-Gordan decomposition)

Theorem 9.1 (Clebsch-Gordan decomposition).

The tensor product $(π_{V^{n_{1}} \otimes V^{n_{2}}}, V^{n_{1}} \otimes V^{n_{2}})$ decomposes into irreducibles as

(π_{n_{1} + n_{2}}, V^{n_{1} + n_{2}}) \oplus (π_{n_{1} + n_{2} - 2}, V^{n_{1} + n_{2} - 2}) \oplus \dots \oplus (π_{∣ n_{1} - n_{2} ∣}, V^{∣ n_{1} - n_{2} ∣})

Proof

One way to prove this result is to use highest weight theory, raising and lowering operators, and the formula for the Casimir operator. We will not try and show the details of how this works out, but in the next section give a simpler argument using characters. However, in outline (for more details, see for instance section 5.2 of [71]), here’s how one could proceed:

One starts by noting that if $v_{n_{1}} \in V_{n_{1}}, v_{n_{2}} \in V_{n_{2}}$ are highest weight vectors for the two representations, $v_{n_{1}} \otimes v_{n_{2}}$ will be a highest weight vector in the tensor product representation $(i.e.$ , annihilated by $π_{n_{1} + n_{2}}^{'} (S_{+}))$ , of weight $n_{1} + n_{2}$ So $(π_{n_{1} + n_{2}}, V^{n_{1} + n_{2}})$ ) will occur in the decomposition. Applying $π_{n_{1} + n_{2}}^{'} (S_{-})$ to $v_{n_{1}} \otimes v_{n_{2}}$ one gets a basis of the rest of the vectors in $(π_{n_{1} + n_{2}}, V^{n_{1} + n_{2}})$ . However, at weight $n_{1} + n_{2} - 2$ one can find another kind of vector, a highest weight vector orthogonal to the vectors in $(π_{n_{1} + n_{2}}, V^{n_{1} + n_{2}})$ . Applying the lowering operator to this gives $(π_{n_{1} + n_{2} - 2}, V^{n_{1} + n_{2} - 2})$ . As before, at weight $n_{1} + n_{2} - 4$ one finds another, orthogonal highest weight vector, and gets another representation, with this process only terminating at weight $∣ n_{1} - n_{2} ∣$ □

9.4.2 Characters of representations

Concept links · terms present in this machine draft; source roles are unverified: irreducible representation

A standard tool for dealing with representations is that of associating to a representation an invariant called its character. This will be a conjugation invariant function on the group that only depends on the equivalence class of the representation. Given two representations constructed in very diferent ways, it is often possible to check whether they are isomorphic by seeing if their character functions match. The problem of identifying the possible irreducible representations of a group can be attacked by analyzing the possible character functions of irreducible representations. We will not try and enter into the general theory of characters here, but will just see what the characters of irreducible representations are for the case of $G = S U (2)$ These can be used to give a simple argument for the Clebsch-Gordan decomposition of the tensor product of $S U (2)$ representations. For this we don’t need general theorems about the relations of characters and representations, but can directly check that the irreducible representations of $S U (2)$ correspond to distinct character functions which are easily evaluated.

Definition (Character)

The character of a representation $(π, V)$ of a group $G$ is the function on $G$ given by
$χ_{V} (g) = T r (π (g))$

Since the trace of a matrix is invariant under conjugation, $χ_{V}$ will be a complex-valued, conjugation invariant function on $G$ . One can easily check that it will satisfy the relations

χ_{V \oplus W} = χ_{V} + χ_{W}, χ_{V \otimes W} = χ_{V} χ_{W}

For the case of $G = S U (2)$ , any element can be conjugated to be in the $U (1)$ subgroup of diagonal matrices. Knowing the weights of the irreducible representations $(π_{n}, V^{n})$ of $S U (2)$ , we know the characters to be the functions

χ_{V^{n}} ((e^{i θ} 0 0 e^{- i θ})) = e^{in θ} + e^{i (n - 2) θ} + \dots + e^{- i (n - 2) θ} + e^{- in θ} (9.2)

As n gets large, this becomes an unwieldy expression, but one has Theorem (Weyl character formula).

χ_{V^{n}} ((e^{i θ} 0 0 e^{- i θ})) = \frac{e ^{i (n + 1) θ} - e ^{- i (n + 1) θ}}{e ^{i θ} - e ^{- i θ}} = \frac{sin (( n + 1 ) θ )}{sin ( θ )}

Proof. One just needs to use the identity

(e^{in θ} + e^{i (n - 2) θ} + \dots + e^{- i (n - 2) θ} + e^{- in θ}) (e^{i θ} - e^{- i θ}) = e^{i (n + 1) θ} - e^{- i (n + 1) θ}

and equation 9.2 for the character.

To get a proof of 9.1, compute the character of the tensor product on the diagonal matrices using the Weyl character formula for the second factor (ordering things so that $n_{2} > n_{1})$ 1

χ_{V^{n_{1}} \otimes V^{n_{2}}} = χ_{V^{n_{1}}} χ_{V^{n_{2}}} = (e^{i n_{1} θ} + e^{i (n_{1} - 2) θ} + \dots + e^{- i (n_{1} - 2) θ} + e^{- i n_{1} θ}) \frac{e ^{i (n_{2} + 1) θ} - e ^{- i (n_{2} + 1) θ}}{e ^{i θ} - e ^{- i θ}} = \frac{( e ^{i (n_{1} + n_{2} + 1) θ} - e ^{- i (n_{1} + n_{2} + 1) θ} ) + \dots + ( e ^{i (n_{2} - n_{1} + 1) θ} - e ^{- i (n_{2} - n_{1} + 1) θ} )}{e ^{i θ} - e ^{- i θ}} = χ_{V^{n_{1} + n_{2}}} + χ_{V^{n_{1} + n_{2} - 2}} + \dots + χ_{V^{n_{2} - n_{1}}}

$So .$ when we decompose the tensor product of irreducibles into a direct sum of irreducibles, the ones that must occur are exactly those of theorem 9.1.

9.4.3 Some examples

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · irreducible representation · symmetry group

Some simple examples of how this works are:

• Tensor product of two spinors:

V^{1} \otimes V^{1} = V^{2} \oplus V^{0}

This says that the four complex dimensional tensor product of two spinor representations (which are each two complex dimensional) decomposes into irreducibles as the sum of a three dimensional vector representation and a one dimensional trivial (scalar) representation.

Using the basis $(0 1), (1 0)$ for $V^{1}$ , the tensor product $V^{1} \otimes V^{1}$ has a basis

(0 1) \otimes (0 1), (0 1) \otimes (1 0), (1 0) \otimes (0 1), (1 0) \otimes (1 0)

The vector

\frac{1}{2} ((0 1) \otimes (1 0) - (1 0) \otimes (0 1)) \in V^{1} \otimes V^{1}

is clearly antisymmetric under permutation of the two factors of $V^{1} \otimes V^{1}$ One can show that this vector is invariant under $S U (2)$ , by computing either the action of $S U (2)$ or of its Lie algebra $su (2)$ . So, this vector is a basis for the component $V^{0}$ in the decomposition of $V^{1} \otimes V^{1}$ into irreducibles.

The other component, $V^{2}$ , is three dimensional, and has a basis

(0 1) \otimes (0 1), \frac{1}{2} ((0 1) \otimes (1 0) + (1 0) \otimes (0 1)), (1 0) \otimes (1 0)

These three vectors span one dimensional complex subspaces of weights $q = 2, 0, - 2$ under the $U (1) \subset S U (2)$ subgroup

(e^{i θ} 0 0 e^{- i θ})

They are symmetric under permutation of the two factors of $V^{1} \otimes V^{1}$

We see that if we take two identical quantum systems with $H = V^{1} =$ $C^{2}$ and make a composite system out of them, if they were bosons we would get a three dimensional state space $V^{2} = S^{2} (V^{1})$ , transforming as a vector (spin one) under $S U (2)$ . If they were fermions, we would get a one dimensional state space $\dot{V}^{0} = Λ^{2} (V^{1})$ of spin zero (invariant under $S U (2))$ . Note that in this second case we automatically get an entangled state, one that cannot be written as a decomposable product.

• Tensor product of three or more spinors:

V^{1} \otimes V^{1} \otimes V^{1} = (V^{2} \oplus V^{0}) \otimes V^{1} = (V^{2} \otimes V^{1}) \oplus (V^{0} \otimes V^{1}) = V^{3} \oplus V^{1} \oplus V^{1}

This says that the tensor product of three spinor representations decomposes as a four dimensional (“spin $3/2")$ representation plus two copies of the spinor representation.

This can be generalized by considering $N$ -fold tensor products $(V^{1})^{\otimes N}$ of the spinor representation. This will be a sum of irreducible representations, including one copy of the irreducible $V^{N}$ , giving an alternative to the construction using homogeneous polynomials. Doing this however gives the irreducible as just one component of something larger, and a method is needed to project out the desired component. This can be done using the action of the symmetric group $S_{N}$ on $(V^{1})^{\otimes N}$ and an understanding of the irreducible representations of $S_{N}$ . This relationship between irreducible representations of $S U (2)$ and those of $S_{N}$ coming from looking at how both groups act on $(V^{1})^{\dot{\otimes} \dot{N}}$ is known as “Schur-Weyl duality”. This generalizes to the case of $S U (n)$ for arbitrary $n,$ where one can consider $N$ -fold tensor products of the defining representation of $S U (n)$ matrices on $C^{n}$ . For $S U (n)$ this provides perhaps the most straightforward construction of all irreducible representations of the group.

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9.4 Tensor products of representations