Good day,
From my reading, I understand that there are 15 parameters needed to produce an arbitrary element of the SU(4) Lie group. This is because this group contains 15 generators.
My question is: if I select one or two parameters to produce one such element in SU(4), and then multiply it with its transpose, why don't I get the identity matrix?.
Here the example.
Given SU(4) paramterization:
$$U=e^{i\lambda_3\alpha_1}e^{i\lambda_2\alpha_2}e^{i\lambda_3\alpha_3}e^{i\lambda_5\alpha_4}e^{i\lambda_3\alpha_5}e^{i\lambda_{10}\alpha_6}e^{i\lambda_3\alpha_7}e^{i\lambda_2\alpha_8}e^{i\lambda_3\alpha_9}e^{i\lambda_5\alpha_{10}}e^{i\lambda_3\alpha_{11}}e^{i\lambda_2\alpha_{12}}e^{i\lambda_3\alpha_{13}}e^{i\lambda_8\alpha_{14}}e^{i\lambda_{15}\alpha_{15}}$$
My problem is: I want to manipulate the value of parameter $\alpha_2$ with the rest parameter constant by substitute the parameter with 00(make them Identity matrix). Hence, I expand the Exponential with generator $\lambda_2$ given as below,
$$\lambda_2=\begin{pmatrix} 0& -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
and I get the expansion :
$$cos\alpha_2\begin{pmatrix} 1& 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + i sin\alpha_2\begin{pmatrix} 0& -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
and get the matrix of $e^{i\lambda_2\alpha_2}$
$$\begin{pmatrix} cos\alpha_2& sin\alpha_2 & 0 & 0 \\ -sin\alpha_2 & cos\alpha_2 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
When I make $UU^{\dagger}$, it suppose to get identity matrix. But I cant get it. Why? please help me where the mistake I expansion.
I use this expansion: $$e^{i\alpha_2\lambda_2}=1+i\alpha_2\lambda_2+\frac{(i\alpha_2\lambda_2)^2}{2!}+\frac{(i\alpha_2\lambda_2)^3}{3!}. . . $$ $$=(1-\frac{(i\alpha_2)^2}{2!}+...)\lambda^2_2 + i (\alpha_2-\frac{(i\alpha_2)^3}{3!}+..)\lambda_2$$ $$= cos\alpha_2\begin{pmatrix} 1& 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + i sin\alpha_2\begin{pmatrix} 0& -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
why the two end row not equal to 1. Please correct me the expansion. I'm really stuck here.
Thank you very much
Thank you
$\hskip 1in$
You replaced $1$ (blue) with $1\cdot\lambda_2^2$ (green). That's why you're missing a $2\times 2$ identity matrix in the lower right block. You also have incorrect signs in the second line (first is in red). Since you left the powers of $i$ uncomputed you shouldn't have any minus signs yet.