Canonical Transformation

Poisson Bracket

The Poisson bracket is defined as $$\{F,G\}\equiv\sum_\beta\left(\frac{\partial F}{\partial q_\beta}\frac{\partial G}{\partial p_\beta}-\frac{\partial F}{\partial p_\beta}\frac{\partial G}{\partial q_\beta}\right)$$ Thus, the total time derivative of a function \(A(p_1,...,p_\alpha,q_1,...,q_\alpha,t)\) is \begin{align} \frac{dA}{dt} &= \sum_\alpha\left(\frac{\partial A}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial A}{\partial p_\alpha}\dot{p}_\alpha\right) + \frac{\partial A}{\partial t}\\ &=\sum_\alpha\left(\frac{\partial A}{\partial q_\alpha}\frac{\partial H}{\partial p_\alpha} - \frac{\partial A}{\partial p_\alpha}\frac{\partial H}{\partial q_\alpha}\right) + \frac{\partial A}{\partial t}\\ &= \{A,H\} + \frac{\partial A}{\partial t} \end{align}

Canonical Transformation

Consider a transformation $$(q_1,...,q_n,p_1,...,p_n)\to(Q_1,...,Q_n,P_1,...,P_n).$$ Suppose there exist a function \(K(Q,P,t)\) such that the Hamilton's equations $$\dot{Q}_i=\frac{\partial K}{\partial P_i}$$ and $$\dot{P}_i=-\frac{\partial K}{\partial Q_i}.$$ Since the Lagrangian can be obtained from Hamiltonian with Legendre Transformation, $$L = \sum_i \dot{q}_ip_i - H$$ and by the least action principle $$\delta \int^{t_2}_{t_1} L dt = 0,$$ we have $$\delta \int ^{t_2}_{t_1} \left(\sum_ip_i\dot{q}_i - H(q,p,t)\right)dt = 0.$$ Similarly for \(K(Q,P,t)\), $$\delta \int ^{t_2}_{t_1} \left(\sum_iP_i\dot{Q}_i - K(Q,P,t)\right)dt = 0.$$ As it can be shown that a Lagrangian differs by a total time derivative also satisfies the Lagrange's equations. The relation $$\lambda(\sum_i p_i\dot{q}_i-H) = \sum_i P_i\dot{Q}_i - K + \frac{dF}{dt}$$ will satisfy the above two statements. When \(\lambda = 1\), the transformation is called a canonical transformation. The new set of transformed coordinates also satisfy the Hamiltonian equations (with the new Hamiltonian). When \(\lambda\) is a constant not equal to 1, one can always find an intermediate transformation such that \(\lambda\) is equal to 1. Then, $$\sum_i p_i\dot{q}_i-H = \sum_i P_i\dot{Q}_i - K + \frac{dF}{dt}$$

Generating Function

Let \(F = F_1(q,Q,t)\). Then, \begin{align} \sum_i p_i\dot{q}_i - H &= \sum_i P_i\dot{Q}_i - K + \frac{dF_1}{dt} \\ &= \sum_i P_i\dot{Q}_i-K + \frac{\partial F_1}{\partial t} + \sum_i \frac{\partial F_1}{\partial q_i}\dot{q}_i + \sum_i \frac{\partial F_1}{\partial Q_i}\dot{Q}_i. \end{align} As \(q_i\) and \(Q_i\) are independent variables, we have \begin{align} p_i &= \frac{\partial F_1}{\partial q_i}\\ P_i &= -\frac{\partial F_1}{\partial Q_i}\\ K &= H+\frac{\partial F_1}{\partial t}. \end{align} Using \(F_1\) and Legendre transformation, one can define $$F_2(q,P,t)\equiv F_1(q,Q,t) + \sum_i Q_iP_i.$$ The total time derivative is \begin{align} \frac{dF_2}{dt} &=\frac{dF_1}{dt}+\sum_i(\dot{Q}_iP_i+Q_i\dot{P}_i \\ &=\sum_i(p_i\dot{q}_i-P_i\dot{Q}_i)+K-H+\sum_i(\dot{Q}_iP_i+Q_i\dot{P}_i) \\ &= K-H+\sum_ip_i\dot{q}_i + \sum_iQ_i\dot{P}_i \end{align} As $$\frac{dF_2}{d} = \sum_i\left(\frac{\partial F_2}{\partial q_i}\dot{q}_i+\frac{\partial F_2}{\partial P_i}\dot{P}_i\right)+\frac{\partial F_2}{\partial t}$$ Thus, we have \begin{align} p_i &= \frac{\partial F_2}{\partial q_i} \\ Q_i &= \frac{\partial F_2}{\partial P_i} \\ K &= H + \frac{\partial F_2}{\partial t}. \end{align} Similarly for \(F_3(p,Q,t)\) and \(F_4(p,P,t)\) generating functions.

Example

Use the generating function $$F_1(q,Q)=\frac{m}{2}\omega q^2\cot Q$$ to find the Hamilton's equation of a 1D harmonic oscillator.

Solution: The kinetic energy is $$T = \frac{1}{2}m\dot{q}^2.$$ The potential energy is $$V = \frac{1}{2}kq^2 = \frac{1}{2}m\omega^2q^2,$$ where $$\omega^2 = \frac{k}{m}.$$ Hence, the Lagrangian is $$L = \frac{1}{2}m\dot{q}^2-\frac{1}{2}m\omega^2q^2.$$ The generalized momentum is $$p = \frac{\partial L}{\partial \dot{q}}=m\dot{q}.$$ The Lagrangian is then $$L = \frac{1}{2m}p^2 - \frac{1}{2}m\omega^2q^2,$$ and the Hamiltonian is $$H = \frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$$ As the generating function is F_1(q,Q), it is of type 1 generating function. Thus, we have \begin{align} p &= \frac{\partial F_1}{\partial q} = m\omega q \cot Q \\ P &= -\frac{\partial F_1}{\partial Q} = \frac{m}{2}\frac{\omega q^2}{\sin^2Q}. \end{align} Then, \begin{align} q &= \sqrt{\frac{2}{m\omega}P}\sin Q\\ p&=\sqrt{2m\omega P}\cos Q. \end{align} The new Hamiltonian using the new set of coordinate \((P,Q)\) is $$K = P\omega(\cos^2Q+\sin^2Q) = \omega P.$$ Then, we have \begin{align} \dot{Q} &= \frac{\partial K}{\partial P} = \omega \\ \dot{P} &= -\frac{\partial K}{\partial Q} = 0. \end{align} So, \begin{align} Q &= \omega t + \phi \\ P &= P_0, \end{align} where \(P_0\) is a constant. Substituting these results back into the original coordinate set, we have \begin{align} q &= \sqrt{\frac{2E}{m\omega^2}}\sin(\omega t +\phi) \\ p &= \sqrt{2m\omega P_0}\cos(\omega t+\phi) \end{align}

Checkpoint

The Hamiltonian of a damped oscillator is $$H(q,p,t) = \frac{p^2}{2m}e^{-2\gamma t}+\frac{1}{2}m\omega^2e^{2\gamma t}q^2.$$ Use the generating function $$F_2(q,P,t)=e^{\gamma t}qP - \frac{1}{2}m\gamma e^{2\gamma t}q^2$$ to find the Hamilton's equations of a 1D damped harmonic oscillator.

$$F_2(q,P,t)=e^{\gamma t}qP - \frac{1}{2}m\gamma e^{2\gamma t}q^2$$ As this is type 1 generating function, we have \begin{align} p &= \frac{\partial F_2}{\partial q} = -m\gamma e^{2\gamma t}q + e^{\gamma t}P\\ Q &= \frac{\partial F_2}{\partial P} = e^{\gamma t}q. \end{align} Inverting \(p\), \(P\) is given by $$P = pe^{-\gamma t}+m\gamma e^{\gamma t}q.$$ So, the in terms of new set of \((P,Q)\), the old set of coordinates is \begin{align} q &= Qe^{-\gamma t} \\ p &= e^{\gamma t}P - m\gamma qe^{2\gamma t} \\ &= e^{\gamma t}P-m\gamma Qe^{\gamma t}. \end{align} The new Hamiltonian \(K\) is \begin{align} K &= H + \frac{\partial F_2}{\partial t}\\ &= \left(\frac{p^2}{2m}e^{-2\gamma t} + \frac{1}{2}m\omega^2e^{2\gamma t}q^2\right) \\ &+(\gamma e^{\gamma t}qP - m\gamma^2e^{2\gamma t}q^2)\\ &= \frac{p^2}{2m}+\frac{1}{2}m(\omega^2-\gamma^2)Q^2. \end{align} As \(\omega\) and \(\gamma\) are both constant, define $$\omega' = \pm \sqrt{\omega^2-\gamma^2}.$$ Then, the new Hamiltonian can be expressed as $$K = \frac{P^2}{2m}+\frac{1}{2}m\omega'^2Q^2.$$ The Hamilton's equations are \begin{align} \dot{P} &= -\frac{\partial K}{\partial Q}=-m\omega'^2Q\\ \dot{Q} &= \frac{\partial K}{\partial P} = \frac{P}{m}. \end{align} Substitute into \(Q\), we have $$\ddot{Q} = -\omega'^2 Q.$$ Hence, the solutions are \begin{align} Q(t) &= A\cos(\omega' t+\phi) \\ P(t) &= -mA\omega'\sin(\omega' t + \phi). \end{align} In terms of the old set of coordinates, the solutions are \begin{align} q &= Qe^{-\gamma t} = Ae^{-\gamma t}\cos(\omega't+\phi) \\ p&=-mAe^{\gamma t}(\omega'\sin(\omega't+\phi) + \gamma \cos(\omega't + \phi)) \end{align}