For $\a=1/2$ and $H=\D$ the Laplacian on $\R^d$ we have (cf. exam): $$ P_t^{1/2}f(x)=C_t*f(x), \quad\mbox{where}\quad C_t(x) =\frac{\G(\frac{d+1}2)}{\pi^{\frac{d+1}2}} \frac{t}{(t^2+\Vert x\Vert^2)^{\frac{d+1}2}}~. $$ is the Cauchy kernel.
By definition $$ P_t^{1/2}f(x) =\int_0^\infty P_sf(x)\,\mu_t(ds) =\int_{\R^d}\int_0^\infty (4\pi s)^{-d/2}t(4\pi s^3)^{-1/2} e^{-(t^2+\norm{y-x}^2)/4s} f(y)\,ds\,dy $$ and the result follows from the formula $$ \int_0^\infty s^{-a-1}e^{-\l/4s}\,ds=(4/\l)^a\G(a)~. $$