0
$\begingroup$

This is from chapter 4 of David Tong's notes on Classical Dynamics (Hamiltonian Formalism).

Let's say you make an infinitesimal canonical transformation (with $\alpha$ as the infinitesimal parameter) as follows $$q_i\rightarrow Q_i=q_i+\alpha F_i(q,p) \qquad p_i\rightarrow P_i=p_i+\alpha E_i(q,p) \qquad \qquad(1)$$

By some mumbo jumbo of using the Jacobian, he finds out that the functions $F_i(q,p)$ and $E_i(q,p)$ are generated by a "generating function" $G(q,p)$ such that $$F_i=\frac{\partial {G}}{\partial p_i} \qquad E_i=-\frac{\partial {G}}{\partial q_i} \qquad\qquad(2)$$

He then goes on to conclude that from the above two equations, I quote him "the tangent vectors to these lines are given by" $$\frac{ dq_i}{d\alpha}=\frac{\partial {G}}{\partial p_i} \qquad \frac{dp_i}{d\alpha}=-\frac{\partial {G}}{\partial q_i}$$

And he claims that these look just like Hamilton’s equations, with the Hamiltonian replaced by the function $G$ and time replaced by the parameter $\alpha$. So every one-parameter family of canonical transformations can be thought of as “Hamiltonian flow” on phase space for an appropriately chosen “Hamiltonian” $G$.

Now I can't understand how he deduced the above results, especially the last one from the equations (1) and (2). Also, I have no clue how we can conclude the result of (2) without involving the Jacobian. I am completely clueless about this. Any sort of help is appreciated.

$\endgroup$
  • $\begingroup$ Why do you not want to use the Jacobian? $\endgroup$ – probably_someone Jun 12 at 5:52
  • $\begingroup$ Ah. Actually, I am studying Quantum Mechanics from R. Shankar, and according to him, knowing a bit of Classical Mechanics is necessary. In his book, he derived the same results but without any use of the Jacobian. So, honestly, it's because I am not really sure how Jacobian comes into the picture for this. $\endgroup$ – Tachyon209 Jun 12 at 6:05
2
$\begingroup$

I am not entirely sure of what objections you have, but I will try to explain what Tong is getting at from a different angle, and perhaps then the route he takes through his notes will make more sense. The idea he's trying to get across is roughly that one can view time evolution as being a canonical transformation which is generated by the Hamiltonian function - however, different functions (substituted in place of $H$) will generate different canonical transformations, which can have very deep implications.


1. Hamiltonian Vector Fields

To any smooth function $F(q_i,p_i)$ on the phase space, we can associate the Hamiltonian vector field (not to be confused with the Hamiltonian function) $\mathbf X_F$ given by

$$\mathbf X_F = \pmatrix{\frac{\partial F}{\partial p_1}\\ \vdots \\ \frac{\partial F}{\partial p_N}\\ -\frac{\partial F}{\partial q_1}\\ \vdots \\-\frac{\partial F}{\partial q_N}} = J\nabla F$$ where $J= \pmatrix{0& 1_{N}\\-1_N&0}$ is called the symplectic form, $1_N$ is the $N\times N$ identity matrix, and $\nabla$ is the phase space gradient operator. Don't worry too much about the motivation for this - just note that we can follow this procedure to convert a smooth function into a vector field, which looks like the gradient except (i) the $q$'s and $p$'s are switched, and (ii) the $q$'s get a minus sign.

For a 2D phase space, which I'll use for the remainder of this answer, this takes the form $$X_F = \pmatrix{\frac{\partial F}{\partial p}\\-\frac{\partial F}{\partial q}} = J\nabla F$$ where $J=\pmatrix{0&1\\-1&0}$.


2. Integral Curves

An integral curve of a vector field $\mathbf X$ is what you get when you basically follow the arrows of $\mathbf X$ at every point to define a path. More precisely, $\mathbf r(\lambda) = \big(q(\lambda),p(\lambda)\big)$ is called an integral curve of a vector field $\mathbf X$ if the tangent vector to the curve at a particular point is equal to the vector of $\mathbf X$ attached to that point:

$$\mathbf r'(\lambda) = \pmatrix{q'(\lambda)\\q'(\lambda)} =\pmatrix{X^1\big(q,p\big)\\X^2\big(q,p\big)}$$

An example is in order. Consider the vector field $\mathbf X(q,p) = \pmatrix{-q\\p}$. We will construct an integral curve from the starting point $(q_0,p_0)$. We seek some $\mathbf r(\lambda)=\pmatrix{q(\lambda)\\p(\lambda)}$ such that $$\mathbf r'(\lambda) = \pmatrix{q'(\lambda)\\p'(\lambda)}=\pmatrix{-q\\p}$$ So we have two equations: $$q'(\lambda) = -q \implies q(\lambda)=c_1 e^{-\lambda}$$ $$p'(\lambda) = p \implies p(\lambda) = c_2e^{\lambda}$$ Applying the initial condition $\mathbf r(0) =\pmatrix{q_0\\p_0}$ yields the integral curve $$\mathbf r(\lambda) = \pmatrix{q_0e^{-\lambda}\\p_0e^{\lambda}}$$ I've plotted the integral curves of this vector field for several choices of initial point.

enter image description here


3. Flows

With this idea of integral curves in mind, we can define a map which takes points in the phase space and "pushes" them along the flow to new points in phase space. The flow $\Phi_{\mathbf X}^\lambda$ is a function which eats a starting point $(q_0,p_0)$ and sends it to the point $\big(q(\lambda),p(\lambda)\big)$ which you get by following the integral curve of $\mathbf X$ for a distance $\lambda$. Explicitly, using the vector field from the last example,

$$\Phi_\mathbf{X}^\lambda (q_0,p_0) = (q_0 e^{-\lambda},p_0e^{\lambda})$$


4. Hamiltonian Flows

Now we can chain those two ideas together. Given a smooth function $F(q,p)$, we can obtain the Hamiltonian vector field $$\mathbf X_F = \pmatrix{\frac{\partial F}{\partial p}\\-\frac{\partial F}{\partial q}}$$

We can then use the integral curves of this vector field to define the associated Hamiltonian flow $\Phi_F^\lambda$. We say that the function $F$ generates the flow. The flow we obtained in part 4 is an example of a Hamiltonian flow generated by the smooth function $F(q,p)=-qp$, as you can quickly check.

The rate of change of some other function $K$ as you follow the flow is seen to be

$$\frac{d}{d\lambda} K\big(q(\lambda),p(\lambda)\big) = \frac{\partial K}{\partial q} q'(\lambda) + \frac{\partial K}{\partial p}p'(\lambda)$$ but, since the flow is generated by the function $F$, we have $$\pmatrix{q'(\lambda)\\p'(\lambda)} = \pmatrix{\frac{\partial F}{\partial p}\\-\frac{\partial F}{\partial q}}$$ so

$$\frac{d}{d\lambda} K\big(q(\lambda),p(\lambda)\big) = \frac{\partial K}{\partial q}\frac{\partial F}{\partial p} - \frac{\partial K}{\partial p} \frac{\partial F}{\partial q} \equiv \{K,F\}$$

where $\{\bullet,\bullet\}$ is the Poisson bracket.


5. The Punchline

The point that Tong is making is that every smooth function $F$ on phase space generates a Hamiltonian flow $\Phi_F^\lambda$, and that these Hamiltonian flows are, in fact, canonical transformations for every choice of $\lambda$. Therefore, smooth functions generate entire families of canonical transformations which you get by following the flow for some chosen distance.

Though it may not seem like it, this is a deep and beautiful idea which provides the entire structure of Hamiltonian mechanics. First and foremost, note that the flow generated by the Hamiltonian function itself pushes points in phase space forward in time (that is, the parameter $\lambda$ should be interpreted as the time). So right off the bat, we see that once we have defined this structure, we can say simply that $H(q,p)$ generates the flow corresponding to time evolution.

There is more to the story, however. Note that if we choose the function $P(q,p)=p$ rather than the Hamiltonian, the resulting flow gives (exercise for the reader) $\big(q(\lambda),p(\lambda)\big) = (q_0+\lambda,p_0)$. We therefore say that the momentum function $P$ generates spatial translations. It is a good exercise to show that $L_z$, the $z$-component of the angular momentum, generates spatial rotations about the $z$-axis.

If you've taken a quantum mechanics course, this should sound familiar. The standard formulation of quantum mechanics uses precisely the same structure (the Hamiltonian generating time evolution, momentum operators generating spatial translation, angular momentum operators generating rotations, etc).

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

It might be clearer if we reduce to the one-dimensional case.

Suppose we have just one coordinate $q$ and one momentum $p$. Hamilton's equations are:

$$\dot{q}=\frac{\partial H}{\partial p}$$

$$\dot{p}=-\frac{\partial H}{\partial q}$$

Suppose we apply the following transformation:

$$Q=q+\alpha F(q,p)$$

$$P=p+\alpha E(q,p)$$

Now let's suppose we want to write Hamilton's equations in terms of $P$ and $Q$. We can do this using the chain rule:

$$\dot{Q}=\frac{\partial Q}{\partial q}\dot{q}=\frac{\partial Q}{\partial q}\frac{\partial H}{\partial p}=\frac{\partial Q}{\partial q}\frac{\partial H}{\partial P}\frac{\partial P}{\partial p}$$

$$\dot{P}=\frac{\partial P}{\partial p}\dot{p}=-\frac{\partial P}{\partial p}\frac{\partial H}{\partial q}=-\frac{\partial P}{\partial p}\frac{\partial H}{\partial Q}\frac{\partial Q}{\partial q}$$

Now, we want our transformation to be a canonical transformation, which by definition, leaves Hamilton's equations unchanged. Comparing our coordinate-changed Hamilton's equations to the ones above, we see that the following must be true:

$$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}=1$$

or in other words:

$$\left(1+\alpha\frac{\partial F}{\partial q}\right)\left(1+\alpha\frac{\partial E}{\partial p}\right)=1$$

Simplifying, and dropping terms of order $\alpha^2$ (remember, $\alpha$ is small), we arrive at:

$$\frac{\partial F}{\partial q}=-\frac{\partial E}{\partial p}$$

The condition above is met if there exists a function $G$ such that $\partial G/\partial q=-E$ and $\partial G/\partial p=F$. After all, then the above equation reduces to

$$\frac{\partial^2 G}{\partial p \partial q}=\frac{\partial^2 G}{\partial q\partial p}$$

which, by Clairaut's theorem, is true for any continuously differentiable function. The idea of this generating function is essentially an insight from inspecting the equation, seeing one partial derivative on one side, a different one on the other, and remembering that you can swap around the order of partial derivatives without consequence as long as your function is continuously differentiable.


Now, if you want to do this with multiple variables, you'll find that the calculations get tedious extremely quickly. Fortunately, we have tools that deal with systems of linear equations extremely efficiently, namely, matrices. If you inspect our work above, you'll find that all of it can be rewritten in matrix form. The original Hamilton's equations can be written as:

$$\begin{pmatrix}\dot{q}\\\dot{p}\end{pmatrix}=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}\frac{\partial H}{\partial q}\\\frac{\partial H}{\partial p}\end{pmatrix}$$

while the coordinate-changed ones can be written as:

$$\begin{pmatrix}\dot{Q}\\\dot{P}\end{pmatrix}=\begin{pmatrix}\frac{\partial Q}{\partial q} & \frac{\partial Q}{\partial p}\\\frac{\partial P}{\partial q} & \frac{\partial P}{\partial p}\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}\frac{\partial Q}{\partial q} & \frac{\partial P}{\partial q}\\\frac{\partial Q}{\partial p} & \frac{\partial P}{\partial p}\end{pmatrix}\begin{pmatrix}\frac{\partial H}{\partial Q}\\\frac{\partial H}{\partial P}\end{pmatrix}\equiv\mathcal{J}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\mathcal{J}^T\begin{pmatrix}\frac{\partial H}{\partial Q}\\\frac{\partial H}{\partial P}\end{pmatrix}$$

where $\mathcal{J}$ is precisely the Jacobian of the coordinate transformation.

This means that a canonical transformation must satisfy:

$$\mathcal{J}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\mathcal{J}^T=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$

from which you will extract the same result. This way, the definition of a canonical transformation is a single condition on the Jacobian of that transformation, instead of $2n$ separate conditions on $2n$ separate equations, which very rapidly grows unworkable as $n$ becomes large.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But I still don't understand how did this happen- $\frac{ dq_i}{d\alpha}=\frac{\partial {G}}{\partial p_i}$ and $\frac{dp_i}{d\alpha}=-\frac{\partial {G}}{\partial q_i}$ $\endgroup$ – Tachyon209 Jun 13 at 2:33
  • $\begingroup$ @Tachyon209 Can you be more specific about what part of Tong's reasoning you don't understand? There's a paragraph that talks about how that comes from looking at the situation completely differently - as an "active" transformation that moves points around on a fixed coordinate grid, rather than a "passive" transformation that moves the grid around while the points are fixed. $\endgroup$ – probably_someone Jun 13 at 2:48
  • $\begingroup$ The line- "In this “active” interpretation, as we vary the parameter $\alpha $ we trace out lines in phase space. Using the results (4.106) and (4.109) (the eqns (1) and (2) ) , the tangent vectors to these lines are given by the equations" I just wrote in the comment. I don't understand what tangent vectors and what lines he is referring to. $\endgroup$ – Tachyon209 Jun 13 at 2:57
  • $\begingroup$ @Tachyon209 If you have a curve $\vec{r}(s)$ parametrized by some arclength parameter $s$, then the tangent vector to that curve is given by $\partial \vec{r}/\partial s$. $\endgroup$ – probably_someone Jun 13 at 3:04
  • $\begingroup$ @Tachyon209 Write $p=P-\alpha E$ and $q=Q-\alpha F$ and take the derivative with respect to $\alpha$, and you should see immediately how those relate to $G$. $\endgroup$ – probably_someone Jun 13 at 3:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.