Weyl's (and others') Unitary Basis Galitski's Exploring Quantum Mechanics says (on p.29) that

the number of (linearly) independent unitary ($N$-dimensional) matrices is also $N^2$. 

Since the set of unitary matrices does not form a vector space, I was curious about how to construct such a unitary basis of $F^{n\times n}$. So I ended up reading Nicholas Wheeler's note on unitary basis, where he refers to the Weyl's unitary basis;
$$E(\alpha,\beta) = e^{{i\over \hbar}({\alpha p + \beta x})}$$
with the property
$${1\over h}\text{Tr}(E(\alpha,\beta))=\delta(\alpha)\delta(\beta)$$
But I don't get


*

*How is the property derived from the BCH formula?
My attempt :
$$\text{Tr}(E(\alpha,\beta)) = \int dx\left<x\right|e^{{i\over \hbar}({\alpha p + \beta x})}\left|x\right>$$
$$=e^{-{i\over 2\hbar}\alpha\beta}\int dx \left<x\right|e^{{i\over \hbar}{\beta x}}e^{{i\over \hbar}{\alpha p}}\left|x\right>$$
by the BCH formula. But the 2 in the exponent's denominator seems  to do not match with the result.

*How can I construct $N$-dimensional basis from them, and

*Is there any other way to construct $N$-dimensional unitary basis?


Any kind of help will be appreciated!
 A: (1). The degenerate CBH formula is meant to clear the underbrush for you: 
$$\text{Tr}(E(\alpha,\beta)) = \int dx<x|e^{{i\over \hbar}({\alpha p + \beta x})}|x>\\
=e^{-{i\over 2\hbar}\alpha\beta}\int dx <x|e^{{i\over \hbar}{\beta x}}e^{{i\over \hbar}{\alpha p}}|x>= e^{-{i\over 2\hbar}\alpha\beta}\int dx ~e^{{i\over \hbar}{\beta x}}<x|e^{{i\over \hbar}{\alpha p}}|x> \\
=e^{-{i\over 2\hbar}\alpha\beta}\int dx ~e^{{i\over \hbar}{\beta x}}<x| x+ \alpha >=   e^{-{i\over 2\hbar}\alpha\beta} \delta(\alpha)\int dx ~e^{{i\over \hbar}{\beta x}}\\ =   e^{-{i\over 2\hbar}\alpha\beta} ~\delta(\alpha) ~\delta \left(\frac{\beta}{2\pi \hbar}\right)=h \delta(\alpha) \delta (\beta),  $$
where you use the action of the translation operator, the  Fourier representation of the delta function, the scaling property of it, and collapse the exponent of the term you were baffled about at its argument.
This is the large $N$ generalization of the celebrated trace-orthogonal  unitary clock-and-shift -matrix basis of U(N) utilized by Weyl in the context of QM in finite Hilbert spaces--it was actually invented in 1867 by J. J. Sylvester. $2\pi/N\leftrightarrow  \hbar$. 
You may explain the construction to yourself for N =3, Sylvester's "nonions" , constructed almost a century before Gell-Mann's su(3) Lie algebra matrices. (Clock, $Q\sim V\sim \Sigma_3$, shift, $P\sim U\sim \Sigma_1$, in evocation of generalizing Pauli matrices. Needless to say, I'd supplant "others" with "Sylvester" in your title. References to Schwinger, Werner, etc... are parochial and laughable.) 
(2). As the commentators point out, the Heisenberg Lie algebra has no finite dimensional unitary representations, but you might go the other way, $N\to \infty$, e.g. in Q10 of this exam; actually, the original Santhanam & Tekumalla paper is not that hard. 
In the finite N case, α and β  take discrete values on a 2-torus, and only become continuous for large N. Now,
1/h ~ N, that is, the Kronecker δs of the N×N identity matrix devolve to the δ-functions for large N.  But you can't get by with finite N.
(3) I don't think so. You might be tempted by the mirage of the formal Heisenberg group $H_3(R)$,
$$
\begin{bmatrix}
 1 & \beta & c\\
 0 & 1 & \alpha \\
 0 & 0 & 1\\
\end{bmatrix}= \exp   \begin{bmatrix}
 0 & \beta & c-\alpha\beta/2\\
 0 & 0 & \alpha \\
 0 & 0 & 0\\
\end{bmatrix}                           ,
$$
but note the central element in the Heisenberg algebra here, gotten from the logarithm at α=β=0, is not the identity! The trace of these upper triangular elements is always the identity, and this is evidently not a unitary basis.
