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I've been thinking about the DOF of the $SU(n)$ group. I first consider the DOF of a unitary matrix. See if I get this right:

Any unitary matrix can be written in the form of $e^{iH}$, where $H$ is a hermitian matrix. So the DOF of a unitary matrix is equal to that of a hermitian matrix. For any $n\times n$ matrix, there are $n^2$ matrix elements and each one has two DOF, so the total DOF is $2n^2$. The diagonal elements should all be real so that the DOF is reduced by $n$. Once the upper triangle is determined, the lower triangle is also determined. This reduce the DOF by $\frac{n^2-n}{2}\times2=n^2-n$. So the total DOF of a unitary matrix is: $$2n^2-n-(n^2-n)=n^2$$

Now, consider a unitary matrix with unit determinant. I have problem understanding why this condition reduces one DOF.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Aug 28 '16 at 16:08
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$\det {e^A} = e^{tr A}$, thus your matrix $H $ must have vanishing trace: one of the $n $ real elements of its diagonal is fixed by the remaining elements. You therefore have $n^2-1$ real degrees of freedom.

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Lets start with the case of $SU(2)$ and at the end the results will be extended to $SU(N)$ in a straight forward way.

As with orthogonal matrices, the unitary groups can be defined in terms of quantities which are left invariant. Consider a general complex transformation in two dimensions, $x' = Ax$ which, in matrix form, reads: \begin{equation} \begin{pmatrix} x' \\y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\y \end{pmatrix}\tag{1} \end{equation}

where $a, b, c$ and $d$ are complex, so there are eight free parameters. Suppose we require the quantity $|x|^2+|y|^2$ to be an invariant under the transformations (1).

Demanding $|x'|^2+|y'|^2=|x|^2+|y|^2$, impose the following constraints on $a, b, c$, and $d$: $$|a|^2+|c|^2=1,\qquad|b|^2+|d|^2=1,\qquad ab^{*}+cd^{*}=0 \tag{2}$$.

These four conditions (the last equation provides two conditions because it involves complex quantities) means that the original eight free parameters are reduced to four.

If, in addition to the conditions above, we require that the determinant of the transformation is unity, the transformation matrix must have the \begin{equation} \begin{pmatrix} x' \\y' \end{pmatrix} = \begin{pmatrix} a & b \\ -b^{*} & a^{*} \end{pmatrix}\begin{pmatrix} x \\y \end{pmatrix}\tag{3} \end{equation} But with the cost of $$|a|^{2}+|b|^{2}=1\tag{4}$$ With (2) and (4) we ended up with five constraints on 8 parameters. Which left only $2^{2}-1=3$ as free parameter or DOF. For $SU(n)$ this simply extends to $n^2-1$ independent parameters.

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    $\begingroup$ To conclude that your reasoning fixes the dof you should prove that the constraints you are imposing are functionally independent as they, in fact, are. The approach which uses the exponential form is more easy in this regard as it transforms these contraints into linear constraints whose linear independence is evident. $\endgroup$ – Valter Moretti Aug 28 '16 at 17:12
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    $\begingroup$ @Valter Moretti, I agree with you. $\endgroup$ – sam Aug 28 '16 at 17:15

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