SU(3) group

The SU(3) group, or special unitary group of degree 3, is defined by its fundamental representation of $3\times 3$ unitary matrices $U$ with determinant 1, with the generators $H_k$ of all its $n\times n$ representations satisfying the commutation relation $[H_a,H_b]=i\sum_{c=1}^8f_{abc}H_c$ .

An element of SU(3) can be expressed as

$U=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$

where $det(U)=1$ and each entry is a complex number.

Since each complex element contains two real variables, the matrix initially contains eighteen real variables. The unitarity condition $U^{\dagger}U=I$ imposes nine independent real constraints, while the condition $det(U)=1$ removes one further degree of freedom. Consequently, every element of SU(3) depends on eight independent real parameters. In other words, the matrices of SU(3) describe an eight-dimensional manifold.

Question

Show that $U^{\dagger}U=I$ imposes nine independent real constraints.

Answer

$U^{\dagger}U=I$ becomes

$U=\begin{pmatrix}a^*&d^*&g^*\\b^*&e^*&h^*\\c^*&f^*&i^*\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

Carrying out the multiplication gives three equations corresponding to diagonal entries

$\vert a\vert^2+\vert d\vert^2+\vert g\vert^2=1$

$\vert b\vert^2+\vert e\vert^2+\vert h\vert^2=1$

$\vert c\vert^2+\vert f\vert^2+\vert i\vert^2=1$

and six equations corresponding to the off-diagonal entries

$\begin{matrix}a^*b+d^*e+g^*h=0\\a^*c+d^*f+g^*i=0\\b^*a+e^*d+h^*g=0\\\vdots\end{matrix}$

Therefore, these nine simultaneous equations impose nine independent real constraints.

As with SU(2), every element of SU(3) can be generated by exponentiating a linear combination of the group’s generators. However, instead of the three 2×2 Pauli matrices, SU(3) requires the generators to be 3×3 Hermitian matrices, each with zero trace. Futhermore, the number of independent real parameters specifying a group element must equal the number of generator matrices (see this article for details), resulting in eight linearly independent 3×3 generator matrices with the following general form:

$\lambda_i=\begin{bmatrix}a&x-iy&u-iv\\x+iy&b&s-it\\u+iv&s+it&-(a+b)\end{bmatrix}$

Another important observation is that the SU(3) rotation operator $\hat{U}$ transforms the basis vectors $\phi_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\phi_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\phi_3=\begin{pmatrix}0\\0\\1\end{pmatrix}$ , which span the three-dimensional complex vector space, into linear combinations of the basis vectors. The simplest of such a transformation mixes a pair of basis vectors; that is, it transforms each vector in the pair into a linear combination of the two while leaving the third basis vector unchanged:

$\hat{U}\phi_1=a\phi_1+b\phi_2$

$\hat{U}\phi_2=c\phi_1+d\phi_2$

$\hat{U}\phi_3=\phi_3$

Consequently, it is natural to choose generators that mix the three pairs of basis vectors, $(\phi_1,\phi_2)$ , $(\phi_1,\phi_3)$ and $(\phi_2,\phi_3)$ . The most basic traceless Hermitian generator matrix that mixes $(\phi_1,\phi_2)$ is

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}$

To see how this symmetric generator matrix is related to an SU(3) transformation, we rewrite it as

$\lambda_1=\begin{pmatrix}\sigma_x&0\\0&0\end{pmatrix}$

where $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is one of three Pauli matrices.

The Taylor expansion of the rotation operator $\hat{U}=e^{-i\frac{\mu}{2}\lambda_1}$ is

$e^{-i\frac{\mu}{2}\lambda_1}=I+\left$-i\frac{\mu}{2}\lambda_1\right$+\frac{1}{2!}\left$-i\frac{\mu}{2}\lambda_1\right$^2+\cdots$

Since every power of $\lambda_1$ has the same block structure $\lambda_1^n=\begin{pmatrix}\sigma_x^n&0\\0&0\end{pmatrix}$ , every term in the Taylor series acts only on the first two components a basis vector, leaving the third component unchanged. Furthermore, the identity matrix is $I=\begin{pmatrix}I_{2\times 2}&0\\0&1\end{pmatrix}$ . Therefore, the SU(3) rotation operator becomes

$\hat{U}=\begin{pmatrix}e^{-i\frac{\mu}{2}\sigma_x}&0\\0&1\end{pmatrix}=e^{-i\frac{\mu}{2}\lambda_1}$

From eq129, where $n_y=n_z=0$ , we have the one-parameter subgroup of SU(3):

$\hat{U}\phi_1=\begin{pmatrix}cos\frac{\mu}{2}&-isin\frac{\mu}{2}&0\\-isin\frac{\mu}{2}&cos\frac{\mu}{2}&0\\0&0&1\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}cos\frac{\mu}{2}\\-isin\frac{\mu}{2}\\0\end{pmatrix}=cos\frac{\mu}{2}\phi_1-isin\frac{\mu}{2}\phi_2$

$\hat{U}\phi_2=-isin\frac{\mu}{2}\phi_1+cos\frac{\mu}{2}\phi_2$

$\hat{U}\phi_3=\phi_3$

demonstrating how $\lambda_1$ is related to an SU(3) transformation.

Question

What is the geometric interpretation of the transformation $\hat{U}(\mu)=\begin{pmatrix}cos\frac{\mu}{2}&-isin\frac{\mu}{2}&0\\-isin\frac{\mu}{2}&cos\frac{\mu}{2}&0\\0&0&1\end{pmatrix}$ ?

Answer

In general, an element of SU(3), which is an 8-dimensional manifold, can be described locally by eight independent real parameters:

$\hat{U}=e^{i\theta_1H_1+i\theta_2H_2+\cdots+i\theta_8H_8}$

where $H_k=\frac{\lambda_k}{2}$ .

If $\theta_2=\cdots=\theta_8=0$ and $\theta_1=\mu$ , then $\hat{U}=\hat{U}(\mu)$ . As $\mu$ varies, $\hat{U}(\mu)$ traces out a one-dimensional curve within the 8-dimensional manifold.

Another basic traceless Hermitian generator matrix that mixes $(\phi_1,\phi_2)$ is

$\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}$

which is the antisymmetric counterpart of $\lambda_1$ .

Notably, $\lambda_2$ contains the Pauli matrix $\sigma_y$ , in the same way that $\lambda_1$ contains $\sigma_x$ . In the early 1960s, Murray Gell-Mann introduced a particularly convenient set of SU(3) generators known as the Gell-Mann matrices. The first three generators contain the Pauli matrices in their upper left blocks, producing SU(3) transformations that mix $(\phi_1,\phi_2)$ :

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$

It follows that the corresponding symmetric and antisymmetric generators that mix $(\phi_1,\phi_3)$ and $(\phi_2,\phi_3)$ are

$\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\;\;\;\;\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}$

and

$\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\;\;\;\;\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}$

respectively.

All generators must be mutually orthogonal with respect to the Frobenius inner product $Tr(\lambda_i\lambda_j)=h\delta_{ij}$ , where $h$ is a scaling constant, to satisfy $U=e^{-\frac{i}{2}\bold{\mu}\cdot\bold{\lambda}}$ . The first seven generators are conveniently chosen to meet this criterion. The final matrix

$\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$

is then scaled to be orthogonal to the first seven such that $Tr(\lambda_i\lambda_j)=2\delta_{ij}$ . These eight linearly independent Hermitian generators are known as the Gell-Mann matrices:

$\lambda_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\;\;\;\;\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}$

$\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\;\;\;\;\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\;\;\;\;\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\;\;\;\;\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$

Notably, these matrices satisfy the commutation relation $[H_a,H_b]=i\sum_{c=1}^8f_{abc}H_c$ , where $H_k=\frac{\lambda_k}{2}$ , and $f_{abc}$ are the structure constants of SU(3), whose values are determined by the specific combination of indices $a,b,c$ (see this article for details).

The significance of the Gell-Mann matrices extends beyond providing a mathematical basis for SU(3). In particle physics, the same SU(3) symmetry appears as the flavour symmetry of quarks and the colour symmetry of the strong interaction.

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