When can I substitute transformations into the Hamiltonian?

[Coffee_]

1. If I know that ##H(q_i,p_i,t)## is a valid Hamiltonian for which the hamilton equations hold. Now we are given that ##Q_j(q_i,p_i)## and ##P_j(q_i,p_i)## are canonical transformations. This means that there is a function ##K(Q_j,P_j)##, the new hamiltonian, for which the Hamilton equations hold in the new variables.

QUESTION : What is this new function ##K##? Can I just substitute the transformations into the old ##H## to find ##K##? If not, when is this allowed?

Reason for confusion: My notes from class tell me this is allowed. However any source on the net I find says that in general canonical transformations do not preserve the Hamiltonian and thus ##H## and ##K## are not always equal.

 

[MisterX]

If P and Q depend only p and q, and the there is no additional time dependence in the coordinate transformation, then
$$K(Q, P) = H(q(Q,P), p(Q, P))$$

Otherwise it can take forms such as
$$K(Q, P) = H(q(Q,P), p(Q, P)) + \frac{\partial F_1(q(Q, P), Q, t)}{\partial t} $$
Where for this type of transformation
$$p = \frac{\partial F_1}{\partial q} $$
$$P = -\frac{\partial F_1}{\partial Q} $$
So in this case if ##F_1## has no time dependence then ##K ##is just ##H##at the corresponding coordinates.

 

[Coffee_]

If P and Q depend only p and q, and the there is no additional time dependence in the coordinate transformation, then
$$K(Q, P) = H(q(Q,P), p(Q, P))$$

Otherwise it can take forms such as
$$K(Q, P) = H(q(Q,P), p(Q, P)) + \frac{\partial F_1(q(Q, P), Q, t)}{\partial t} $$
Where for this type of transformation
$$p = \frac{\partial F_1}{\partial q} $$
$$P = -\frac{\partial F_1}{\partial Q} $$
So in this case if ##F_1## has no time dependence then ##K ##is just ##H##at the corresponding coordinates.

Thanks for the answer. This is exactly what I have thought up myself by now. One more question since you seem to know the subject if you don't mind?

I'm a bit confused on the 4 ''types'' of generating functions and how their forms are found. For example for ##F(q,Q,p)## the form makes sense because it's just ##F=F(q,Q,p)##, but how does one knows to add the ##-QP## when dealing with the ##F(q,P,t)## type? It seems to work out alright but I seem to not grasp how one can make these 4 distinctions so clear and so strict? What if I take ##F=F(q,P,t)## as the generating function without the ##-QP## part?