I have the expressions :
H = $p^2/2 - 1/(2q^2)$
D = (pq)/2 - Ht
And I want to calculate the Poisson bracket [H,D] and show that it is -H = $-p^2/2 + 1/(2q^2)$
It just isn't giving me the result I want to show because, in the terms ∂D/∂p (and ∂D/∂q, but I'll just write the former) I am doing:
∂D/∂p = q/2 - t(∂H/∂p)
and I end up with this extra t term in the end.
I've tried justifying why I should be able to just remove this term but I am not being able to, so I'd really appreciate any help!
We have that $$\partial_q H=\frac{1}{q^3}$$ $$\partial_p H=p$$ $$\partial_q D=\frac{p}{2}-t\frac{1}{q^3}$$ $$\partial_p D=\frac{q}{2}-tp$$ So \begin{align}\{H,D\} &=\frac{1}{q^3}\left(\frac{q}{2}-tp\right)-p\left(\frac{p}{2}-t \frac{1}{q^3}\right)\\ &=\frac{1}{2q^2}-\frac{tp}{q^3}-\frac{p^2}{2}+\frac{tp}{q^3}\\ &=\frac{1}{2q^2}-\frac{p^2}{2}\\ &=-H \end{align} Sidenote: to be mathematically correct, this is Hamiltonian mechanics, so everything is a function of the time $t$, generalised coordinate $q$ and generalised momentum $p$, so we are doing the derivatives according to this.