If I have a group $G=SU(2)$, the $s=\frac{1}{2}$ irreducible representation is given by matrices

$$ U(G)=\begin{pmatrix} \alpha & - \beta^* \\ \beta & \alpha^* \end{pmatrix}\;\;\;\; : \alpha \alpha^* + \beta \beta^* =1, $$ with the two states written as $$\begin{pmatrix} 1\\0 \end{pmatrix}=|0\rangle,\;\;\;\;\begin{pmatrix} 0\\1 \end{pmatrix}=|1\rangle.$$

My confusion is over the terminology used.

If a Hilbert space $\mathcal{H}$ is said to carry the irreducible representation $U(G)$, does that mean that the vector space $\mathcal{H}$ only contains states that can be written as $$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} |0\rangle + \begin{pmatrix}-\beta^* \\ \alpha^* \end{pmatrix} |1\rangle\;\;\;\;\;:\alpha \alpha^* + \beta \beta^* =1 \tag{I.} $$

and that no other linear combinations are allowed? That is, given solely the conditions in equation (I.), is it correct to infer that the space of states carries the irreducible representation $U(G)$? My issue is that vector spaces are usually defined over a field $K$ which in this case, the field is that of the complex numbers, however, it is incorrect to say that any linear combination is allowed in the above example. I was hoping someone could clear up my confusion in terminology.


From the answers here is my updated question.

  1. 'If the Hilbert space $\mathcal{H}$ carries a representation $\Pi(G)$ of a group $G$', does this only require the existence of a map $$ \Pi(G): \mathcal{H} \rightarrow \mathcal{H} $$ such that $\Pi(G)$ is a representation of the group $G$ with no other requirements? In other words, is that all we can say from the statement that is parenthesized?

  2. Next, my confusion is over spin and other similar observables. Is it correct to say that for non-relativistic spin, $\mathcal{H}$ carries all the unitary irreducible representations of $SU(2)$? Following this, when describing the state space of the Pauli Hamiltonian, would I say that the Hilbert space carries an irreducible unitary representation of the $s=\frac{1}{2}$ representation of $SU(2)$?


You have a matrix representation $\lambda$ when the matrices satisfy $T^\lambda(g_1)T^{\lambda}(g_2)=T^\lambda(g)$ when $g_1\circ g_2=g$ in the group, for all $g_1,g_2,g$.

If a set of basis vectors (usually an orthonormal set) $\{\vert\psi_k\rangle\}$ is such that \begin{align} T^\lambda(g_1)T^{\lambda}(g_2)\vert\psi_k\rangle=T^\lambda(g)\vert\psi_k\rangle\tag{1} \end{align} then the vector space spanned by these basis vector "carries" the representation $\lambda$. Basically this follows from the way the matrix elements of $T^\lambda(g)$ are constructed, i.e. \begin{align} \langle \psi_i\vert T^\lambda(g)\vert\psi_j\rangle := T^\lambda_{ij}(g)\, .\tag{2} \end{align}

It is not always trivial to find the $T^{\lambda}_{ij}$. For instance, the spherical harmonics with $\ell=1$ span a 3-dimensional representation of $so(3)$. Using \begin{align} Y_{11}(\theta,\phi)\mapsto (1,0,0)^\top\, ,\qquad Y_{10}(\theta,\phi)\mapsto (0,1,0)^\top\, ,\qquad Y_{1,-1}(\theta,\phi)\mapsto (0,0,1)^\top \end{align} the usual $L_z\mapsto -i\frac{\partial}{\partial \phi}$ etc, you can first obtain a representation of the algebra $su(2)$ using the standard inner product \begin{align} (L_k)_{ij}=\int \sin\theta d\theta d\phi Y_{1i}^*(\theta,\phi) \hat L_k Y_{1j}(\theta,\phi) \end{align} and then using $R_k(\beta)=e^{-i\beta L_k}$ plus the Euler factorization $R_z(\alpha)R_y(\beta)R_z(\gamma)$ obtain the 3-dimensional representation of SO(3) with $Y_{1,m}(\theta,\phi)$ as basis vectors.

Arbitrary normalizable linear combinations of these vectors are permissible in the vector space, and (in the physics parlance) the vectors "carry" the irreducible representation.

As - for instance - you can have a 3-dimensional representation of $su(2)$, any linear combination of the basis vectors is allowed provided that $T^1(g_1)T^1(g_2)=T^1(g)$ as $3\times 3$ matrices, and where I've used $\lambda=j=1$ since the SU(2) irreps with $j=1$ has dimension $3$.

In particular, the combination $\vert\phi\rangle = (1,1,1)^\top /\sqrt{3}$ is allowed, which is clearly not of the form you suggest. To find how such a vector transform under $SU(2)$, one would write \begin{align} \vert\phi\rangle&=\sum_k a_k\vert\psi_k\rangle\, ,\\ T^1(g)\vert\phi\rangle&= \sum_k a_k T^1(g)\vert\psi_k\rangle\, ,\\ &= \sum_{km} a_k \vert\psi_m\rangle T^1_{mk}(g) \, . \end{align}

  • $\begingroup$ But the equation $T^\lambda(g_1) T^\lambda(g_2) | \psi_k \rangle = T^\lambda(g) | \psi_k \rangle$ is true for any vector $| \psi_k \rangle$ with the same dimension as $T^\lambda(g_i)$. $\endgroup$
    – DIRAC1930
    Aug 24 at 1:33
  • $\begingroup$ Yes but (2) must hold, and it’s nontrivial to find $T_{ij}(g)$ that will then make (1) and (2) hold. $\endgroup$ Aug 24 at 3:09
  • $\begingroup$ Why isn't spin represented on $L^2$ functions like $SO(3)$ is? $\endgroup$
    – DIRAC1930
    Aug 24 at 14:22
  • $\begingroup$ see physics.stackexchange.com/a/594553/36194 and associated reference: Gatland, I.R., 2006. Integer versus half-integer angular momentum. American journal of physics, 74(3), pp.191-192. $\endgroup$ Aug 24 at 14:54
  • $\begingroup$ Thanks for this. I've updated my question (see Edit). Could you clear up my confusions in the edit if possible? $\endgroup$
    – DIRAC1930
    Aug 24 at 15:21

A representation is, strictly speaking, a homomorphism $\rho \colon G \to \textrm{GL}(V)$ from the group $G$ into the linear group $\textrm{GL}(V)$ of some vector space $V$. This is a technical way of saying "A representation is a way to write the group in terms of matrices acting on a vector space".

Let me exemplify with your case. When we think about $\textrm{SU}(2)$, it is in principle an abstract group. A representation of it will be a collecting of matrices (well, linear operators if you want to be rigorous) that "copy" the group structure. So when we say $\mathcal{H}$ carries a representation of $G$, we mean that there is a collection of linear operators acting on $\mathcal{H}$ that has the same group structure. In your example, this just means that there is a collection of $2 \times 2$ matrices which are unitary and of unit determinant.

So why isn't this obvious? We could pick a different Hilbert space. For example, $\mathcal{H} = \mathbb{C}^3$. In this case, it isn't immediate whether $\mathcal{H}$ does carry a representation of $\textrm{SU}(2)$. Nevertheless, it does: it is known as the adjoint representation.

In summary, you can still add vectors without extra restrictions, $\mathcal{H}$ is still a vector space over $\mathbb{C}$. The meaning of "carries a representation" is that there are some matrices acting on the space that copy the group structure.

I should add that this copy doesn't need to be injective nor surjective. I could just choose $\rho \colon G \to \textrm{GL}(V)$ to be $\rho(g) = \mathbb{1}$ for all $g \in G$. This is still a representation.

  • $\begingroup$ So for the case of any 2 state system e.g. spin, it is correct to say that $\mathcal{H}$ carries a representation of any group that acts on a 2-dimensional complex vector space e.g. the 2-dimensional representations of SO(2), Spin(2), etc.? $\endgroup$
    – DIRAC1930
    Aug 23 at 23:16
  • $\begingroup$ @DIRAC1930 for groups build up of $2 \times 2$ matrices, that is correct, though not necessarily useful. For spin, $\textrm{SU}(2)$ is particularly interesting because it can be related to the rotation properties of the system. While the same vector space also has a representation of $\textrm{SO}(2)$, it not necessarily has any interesting physical meaning $\endgroup$ Aug 23 at 23:26
  • 2
    $\begingroup$ In the observables. The system is invariant under rotations, for example, if the Hamiltonian commutes with the angular momenta (which are the generators of rotations). While the vectors are the same, the observables need not be, and as a consequence the interpretation of each vector as a state changes. While $(1,0)^\intercal$ might be an energy eigenstate for some Hamiltonian, it might not be for a different one $\endgroup$ Aug 24 at 1:30
  • 1
    $\begingroup$ @DIRAC1930 The previous comment is important because observables are often the motivation to choose one symmetry group over another, and as operators the observables often form a vector space. In particular, sets of operators can often be used a basis vectors for a representation. For instance, $L_+$, $L_+$ and $L_z$ transform amongst themselves under the SO(3) group action $T(g)L_k T^{-1}(g)$ in the same way as the spherical harmonics $Y_{1k}$ transform under $T(g)Y_{1k}$. As a result, $L_\pm$ and $L_z$ carry the 3-dimensional representation of the $SO(3)$. $\endgroup$ Aug 24 at 12:13
  • 1
    $\begingroup$ The symmetry in not in the vectors but in the way they transform. $\endgroup$ Aug 24 at 15:05

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