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Under $\mathrm{SU}(2)$ group, a doublet transforms like: $$\phi \rightarrow \exp\left(i\frac{\sigma_i}{2}\theta_i\right)\phi.$$ The doublet looks like $$\binom{a}{b} ,$$ which is easy to understand because the matrix representation of $\mathrm{SU}(2)$ is a second order square matrix. What is a $\mathrm{SU}(2)$ triplet? What does it look like? And what's the matrix representation? Is it the $\mathrm{O}(3)$ group? If their is a $\mathrm{SU}(2)$ doublet, is there a $\mathrm{SU}(2)$ singlet?

I'm not very familiar with group theories, so I would much appreciate a detailed explanation. Also, what do these doublets and triplets mean physically?

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The $SU(2)$ triplet results from the Adjoint Representation $\mathrm{Ad}: SU(2)\to SO(3)$ of $SU(2)$, whereby $SU(2)$ acts on its own Lie algebra. As a $2\times2$ matrix, an element of the Lie algebra $\mathfrak{su}(2)$ can be written:

$$X=\begin{pmatrix}i\,z&i\,x + y\\i\,x - y&-i\,z\end{pmatrix}=i\,(x\,\sigma_x+y\,\sigma_y + z\,\sigma_z)\tag{1}$$

where $\sigma_j$ are the Pauli matrices and $\gamma\in SU(2)$ acts on this entity through the so-called spinor map $X\mapsto\gamma\,X\gamma^{-1}$. So the triplet is a 3 element, real vector of the co-efficients of the Pauli matrices in (1) and, to find the matrix of the triplet transformation, you need to find the matrix of the linear, homogeneous transformation $X\mapsto\gamma\,X\gamma^{-1}$.

If you have the $SU(2)$ member in the form you write, i.e. $\gamma =\exp\left(i\frac{\sigma_i}{2}\theta_i\right)$ then there is an easy way to find the matrix of the triplet transformation since we can show that (look up the relationship between the adjoint representation of the Lie algebra and that of the group) $X\mapsto \exp\left(i\frac{\sigma_i}{2}\theta_i\right)\, X\, \exp\left(-i\frac{\sigma_i}{2}\theta_i\right)$ is the transformation:

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)\mapsto\exp\left(\frac{1}{2}\theta_i \,\mathrm{ad}(i\,\sigma_i)\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{2}$$

where $\mathrm{ad}(Y)$ is the $3\times3$ matrix of the linear mapping $X\mapsto[Y,\,X]$. So you need to work out the $\mathrm{ad}(i\,\sigma_i)$ from the commutation relationships for the Pauli matrices. The result is the triplet transformation law:

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)\mapsto\exp\left(\left(\begin{array}{ccc}0&-\theta_z&\theta_y\\\theta_z&0&-\theta_x\\-\theta_y&\theta_x&0\end{array}\right)\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{3}$$

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In quantum mechanics, operators $\{J_x,J_y,J_z\}$ measuring the angular momentum of a state are required to obey the commutation relations \begin{equation} [J_i,J_j]=i \sum_k \epsilon_{ijk} J_k. \end{equation}

If we only care about the spin of a particle, which does not know about the wavefunction, the state of a particle becomes a length $n$ vector (we do not know what n is yet, it depends on the spin of the particle), and the operators measuring spin angular momentum become $n\times n$ matrices $\{S_x,S_y,S_z\}$ satisfying the same commutation relations \begin{equation} [S_i,S_j]=i \sum_k \epsilon_{ijk} S_k. \end{equation}

A set of three matrices obeying these commutation relations is called a representation of the Lie algebra $\frak{su}_2$. To know what the different possible spins are, we need to know what the possible values of $n$ are.

It turns out that there is such a set of matrices for each integer $n=1,2,3,...$ and they are unique in some sense. Physically we have decided that the spin of a particle is $(n-1)/2$. There is a general procedure for constructing these matrices, and you can find it in the quantum mechanics book by Cohen-Tannoudji for example. I will give the answer for $n=1,2,3$.

The singlet state has $n=1$, and we choose $S_i=[0]$. The three angular momentum operators are the $1\times 1$ zero matrix, and the only state in the system has $(n-1)/2=0$ angular momentum.

The doublet state has $n=2$, and we choose $S_i=\sigma_i/2$. You can check that this has the correct commutation relations. The matrices $S_i$ all have eigenvalues $\pm 1/2$, so the possible angular momenta components are $\pm 1/2$ (spin up or down) and we say the particle has spin $(n-1)/2=1/2$.

The triplet state has n=3, and we choose \begin{equation} S_x=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix},S_y=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{bmatrix},S_z=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}.\end{equation} These matrices all three have eigenvalues $1,0,-1$, so the components of the angular momentum can have these three values and we say the spin is $(n-1)/2=1$.

The matrices get uncomfortably large for larger n, and I think you can see the pattern, so I won't give more examples.

Now, what is a representation of the Lie group $SU_2?$ If we have an $n\times n$ representation of $\frak{su}_2$, we can use the matrix exponential to make a set of matrices $\exp(i S_k \theta_k)$ for $\theta_i$ three real numbers. By using the Campbell-Baker-Hausdorff formula, we can work out how to multiply two of these matrices from the commutation relations. Because of the way this multiplication behaves, we say that the set of matrices $\exp(i S_k \theta_k)$ together with the multiplication formula give a representation of $SU_2$.

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The group elements are in principle abstract objects defined by the way they act on some structure. For example, the rotation group in three dimensions is formed by elements that rotate coordinate systems in some appropriate way. In order to make things easier to understand and visualize we assign linear representations, i.e. matrices to the elements of these groups. All elements of a Lie group in a given $d$-dimensional representation are then mapped to matrices of dimension $d$.

Among all elements of a Lie group there are special ones that can be used to generate any other. These are called generators of the group and they satisfy a particular structure, called Lie algebra. For example, the group $\mathrm{SU}(2)$ has a Lie algebra $\mathfrak{su}(2)$ whose generators are $T_a$, $a=1,2,3$, satisfying $$[T_a,T_b]=i\epsilon_{abc}T_c.$$ A representation $R$ of these abstract elements has to preserve this structure, i.e., $$[R(T_a),R(T_b)]=i\epsilon_{abc}R(T_c),$$ where $R(T)$ shall be understood as a $d$-dimensional matrix.

From the Lie algebra one can obtain all possible representations. For instance, the $\mathfrak{su}(2)$ algebra has $d$-dimensional representations for any integer $d$. The singlet is the one dimensional representation, i.e. they are only numbers. Notice that the only possibility numbers can satisfy a non-trivial algebra is that they are all zero. The doublet is a two dimensional representation and the matrices representing the generators $T_a$ are just the Pauli matrices. We normally call the $n$-dimensional representation of $\mathfrak{su}(n)$ the defining representation. The group obtained, as the exponential of the algebra, is then $\mathrm{SU}(2)$. The triplet has dimension three which equals the number of generators of the algebra. This representation is called the adjoint representation and the explicit matrix form of the generators is given by the $l=1$ angular momentum matrices. There is although some subtleties. Starting from a given Lie algebra and assigning a given representation one can ends up with different Lie groups. So for adjoint representation of $\mathfrak{su}(2)$ the group generated turns out to be $\mathrm{SO}(3)$ instead of $\mathrm{SU}(2)$.

Regarding the physical meaning of a doublet or a triplet, it depends actually on the physics you want to describe. For instance, the spin in quantum mechanics is associated to a Lie algebra $\mathfrak{su}(2)$. If you are going to describe the spin of an electron you need the two dimensional representation of this algebra, since the electron has two states of spin, $\pm 1/2$. On the other hand if you interested in the spin of a $\rho$ meson, then you shall describe it through a triplet, since this vector meson has three states of spin, $0,\, \pm1$.

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  • $\begingroup$ I'm studying neutrino physics, my book says "$\nu_{L}$ is part of the $SU(2)_{L}$ doublet field and has lepton number $+1$, the neutrino mass term transforms as an $SU(2)_L$ triplet". In this case, how can a neutrino be a doublet and a triplet at the same time? $\endgroup$ – LY3000 Jul 22 '16 at 14:10
  • $\begingroup$ A given "piece" of the standard model can have different symmetries. Take a quark for instance. The spin states transforms as a doublet of $\mathfrak{su}(2)$. But at same time, to take into account the strong interaction, this quark transform as triplet of the $\mathfrak{su}(3)$. These groups are acting on different spaces and related to different symmetries. $\endgroup$ – Diracology Jul 22 '16 at 14:14
  • $\begingroup$ See if I get this right. $\nu_{L}$ transforms like a $SU(2)$ doublet under lepton number symmetry but transforms like a $SU(2)$ in the mass basis. Is this correct? $\endgroup$ – LY3000 Jul 22 '16 at 14:21
  • $\begingroup$ @ Daniel, Lepton number conservation is based on global $U(1)$ symmetry. Well it is wrong then if we say $\nu_{L}$ belongs to a doublet. What we can say is: both left-handed and neutrino belongs to weak isospin gauge group. Which is in fact local symmetry of the SM not global. $\endgroup$ – AMS Jul 22 '16 at 16:37
  • $\begingroup$ i guess neutrino mass term invovles two spin-1/2 fields whose combination forms a spin-1 object which transform as $\mathfrak{su}(2)$ triplet(due to its spin being 1). hence your question can be answered after knowing the form of the mass term. $\endgroup$ – Liberty May 8 '18 at 1:38

protected by Qmechanic Jul 22 '16 at 18:35

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