(Apologies if HW questions are not allowed -- I couldn't really find a definite answer on this)


Let $Q^1 = (q^1)^2, Q^2 = q^1+q^2, P_{\alpha} = P_{\alpha}\left(q,p \right), \alpha = 1,2$ be a CT in two freedoms.

(a) Complete the transformation by finding the most general expression for the $P_{\alpha}$.

(b) Find a particular choice for the $P_{\alpha}$ that will reduce the Hamiltonian

$$H = \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2$$


$$K = P_1^2 + P_2.$$


I have shown that

$$P_1 = \frac{1}{2q^1} \left( p_1 + \frac{\partial F}{\partial q^1} - p_2 - \frac{\partial F}{\partial q^2} \right), $$

$$P_2 = p_2 + \frac{\partial F}{\partial q^2}$$

is the most general canonical transformation for the momenta, where $F=F(q^1, q^2)$. This is consistent with the solution manual. For part b, however, the answer I get for an intermediate step is inconsistent with the solutions manual, and I don't understand why. Given that the transformation is canonical, all I need to do to find the transformed Hamiltonian K is find the inverse transformation and plug it in to the Hamiltonian H. The inverse transformation is

$$p_2 = P_2 - \frac{\partial F}{\partial q^2},$$ $$p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.$$

Plugging this into H, and renaming H to K since it's in terms of the transformed coordinates we have

$$K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.$$

Since we want K to be

$$K = P_1^2 + P_2,$$

this means

$$\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.$$ $$F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.$$

Plugging this into the general transformation I derived I find that

$$P_1 = \frac{1}{2q^1} \left(p_1-p_2-(q^1)^2 \right),$$ $$P_2 = (q^1+q^2)^2+p_2.$$

My equation for $P_2$ is consistent with the solutions manual, but my equation for $P_1$ is not. According to the solutions manual


My question is, is my methodology essentially correct, and if so did I go wrong in the algebra or did I make some sort of mistake in how I solved the problem.

  • 1
    $\begingroup$ Homework questions are frowned upon because they are generally of interest only to one person (the one doing the homework) and we prefer questions that will be interesting to a large audience. $\endgroup$ – John Rennie Dec 10 '13 at 10:12
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    $\begingroup$ I think this is conceptual enough, actually. $\endgroup$ – Abhimanyu Pallavi Sudhir Dec 10 '13 at 15:18

I think your solution is basically correct.

Part (a)

To find the missing transformations of the momenta, we first try to find a generating function $\cal F_2(q, P)$ that generates the known transformations of the coordinates. Then, we use this generating function $\cal F_2(q, P)$ to compute the relations regarding the momenta.

The transformation of coordinates $Q^i = Q^i(q)$ can be conveniently generated by the generating function of type 2 as \begin{align} \cal F_2(q, P) &=\sum_i P_i \, Q^i(q) + F(q), \end{align} where $F(q)$ is arbitrary function of $q$.

In this way, the requirement \begin{align} \frac{ \partial \cal F_2(q, P) }{ \partial P_i} = Q^i(q). \end{align} is automatically satisfied.

In our case \begin{align} \cal F_2(q, P) &= P_1 \, Q^1(q^1, q^2) + P_2 \, Q^2(q^1, q^2) - F \\ &= P_1 \, (q^1)^2 + P_2 \, (q^1 + q^2) - F, \end{align} where $F \equiv F(q^1, q^2)$ is an arbitrary function of $q^1$ and $q^2$.

So \begin{align} p_1 &= \frac{ \partial \cal F_2(q, P) }{ \partial q^1 } = 2 P_1 \, q^1 + P_2 - \frac{\partial F }{\partial q^1}, \\ p_2 &= \frac{ \partial \cal F_2(q, P) }{ \partial q^2 } = P_2 - \frac{\partial F }{\partial q^2}. \end{align}

Or \begin{align} P_1 &= \frac{1}{2q^1} \left( p_1 + \frac{ \partial F } { \partial q^1 } -p_2 - \frac{ \partial F } { \partial q^2 } \right) \tag{1} \\ P_2 &= p_2 + \frac{ \partial F } { \partial q^2 }. \tag{2} \end{align}

Part (b)

Basically we need to find an $F$ such that $K$ matches $H$, because $$ d{\cal F}_2 = p \, dq + Q dP + (K - H) \, dt, $$ and our $F_2$ does not depend on time explicitly (so $K-H$ must vanish).

Now by the solution of part (a), we have

\begin{align} H &= \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2, \\ K &= P_1^2 + P_2 \\ &= \left( \frac{p_1 - p_2 + \partial F/\partial q^1 - \partial F/\partial q^2}{2q^1} \right)^2 + p_2 + \partial F / \partial q^2. \end{align}

It would be nice if \begin{align} \partial F/\partial q^1 &= \partial F/\partial q^2, \\ \partial F/\partial q^2 &= (q^1 + q^2)^2. \end{align}

A simple solution would be \begin{align} F = \frac{1}{3} (q^1 + q^2)^3. \end{align}

Then Eq. (1) and (2) means \begin{align} P_1 &= \frac{1}{2q^1} \left( p_1 - p_2 \right) \tag{1} \\ P_2 &= p_2 + (q^1 + q^2)^2. \tag{2} \end{align}

The result $P_1 = (p_1 + p_2)/(2q^1)$ doesn't make sense, because it implies $\dot P_1 \ne 0 = -\partial K/\partial Q^1$. So the plus sign might be a typo.


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