I have $H(q,p,t)$. I also know that $q$ and $p$ depend on $\alpha$.

Why $\frac{\partial H}{\partial \alpha}$=$\frac {\partial H}{\partial q} \frac {\partial q}{\partial \alpha}+ \frac {\partial H}{\partial p} \frac {\partial p}{\partial \alpha}$ ?

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In problems like this there's a tendency to use the same name for different functions, which I think causes confusion. Let's state very carefully that $H:\mathbb R^3 \to \mathbb R$, and that $p$ and $q$ are functions from $\mathbb R$ to $\mathbb R$.

Let $\hat{H}(\alpha,t) = H(p(\alpha),q(\alpha),t)$. Note that $\hat{H}$ is a different function than $H$ ! We wish to compute $\frac{\partial \hat{H}(\alpha,t)}{\partial \alpha}$.

We can use the chain rule to write down the answer directly. But here is a computation that shows some intuition behind the chain rule. We will use the approximations $$H(P + \Delta P,Q + \Delta Q,t) \approx H(P,Q,t) + \frac{\partial H(P,Q,t)}{\partial P} \Delta P + \frac{\partial H(P,Q,t)}{\partial Q} \Delta Q$$ and $$p(\alpha + \Delta \alpha) \approx p(\alpha) + p'(\alpha) \Delta \alpha.$$

Here is how we can derive the chain rule in this situation:

\begin{align*} \hat{H}(\alpha + \Delta \alpha,t) &= H(p(\alpha + \Delta \alpha),q(\alpha + \Delta \alpha),t) \\ &\approx H(p(\alpha) + p'(\alpha) \Delta \alpha,q(\alpha) + q'(\alpha) \Delta \alpha,t)\\ &\approx H(p(\alpha),q(\alpha),t) + \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial P} p'(\alpha) \Delta \alpha + \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial Q} q'(\alpha) \Delta \alpha \\ &= \hat{H}(\alpha,t) + \left( \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial P} p'(\alpha) + \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial Q} q'(\alpha) \right) \Delta \alpha. \end{align*}

Comparing this with the equation $$\hat{H}(\alpha + \Delta \alpha,t) \approx \hat{H}(\alpha,t) + \frac{\partial \hat{H}(\alpha,t)}{\partial \alpha} \Delta \alpha$$ we see that $$\frac{\partial \hat{H}(\alpha,t)}{\partial \alpha} = \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial P} p'(\alpha) + \frac{\partial H(p(\alpha),q(\alpha),t)}{\partial Q} q'(\alpha) .$$

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Thanks a lot, littleO! – sunrise Nov 15 '12 at 17:04

If you think of $H(q,p)$ as a function on a two dimensional plane with coordinates $(q,p)$, then $$\vec \pi(\alpha)=(q(\alpha),p(\alpha))$$ is a curve, it's tangent vector a 2x2 vector $$\tfrac{\partial}{\partial\alpha}\vec \pi(\alpha)=(q'(\alpha),p'(\alpha)).$$ Your expression is the scalar product with the gradient of $$\tfrac{\partial}{\partial\alpha}H=\vec \nabla H\cdot\tfrac{\partial}{\partial\alpha}\vec \pi(\alpha)$$ imposed by the chain rule.

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