### Sixth Assignment solutions

The median on the hour exam (returned Monday) was about 70, corresponding to a middle B. But I anticipated that you would have done better on question 1, and maybe worse on question 2. This is a bit disconcerting because it indicates that I don't have a good picture of what you know.
1. Conduct a detailed post-mortem of the exam: write an essay about your performance on this exam, focusing on the reasons for your errors, what you should have done differently, and how you will address any deficiencies. (It's OK to be facetious here, but try not to deceive yourself or me :) In particular, a surprisingly large number of you succumbed to two puzzling errors: (1) thinking that the condition for equilibrium is U = 0 (rather than grad U = 0) (2) differentiating l - x˛ to obtain l - x·˛ (instead of 2x x·). If you did this, please trace this back to its source -- where did you first get this idea, what reinforced it, etc.

2. Text, problem 13.12
3. Text, problem 13.23
4. Text, problem 13.25
5. This problem is about something you can do in phase space that you can't do in configuration space.
Also, most of it is talk, the actual steps you have to do are not hard.
Recall we defined the Poisson bracket of two functions in phase space as a "skew scalar product" of the gradient of two phase space functions. Take just 4D phase space with coordinates x, y, px, py. Then for two functions u, v of these coordinates our definition is
 [u, v] = ¶u ¶x ¶v ¶px - ¶u ¶px ¶v ¶x + ¶u ¶y ¶v ¶py - ¶u ¶py ¶v ¶y
(1)
We showed that the time development of any function on phase space, f(qi, pj) has the simple equation in this notation
 df/dt = [f, H]
We say that the Hamiltonian is the generator of the time development. Any other phase space function G will generate something as well. From the definition (df/dt = [f,G]) it does the following to x, y, px and py:
 dx/dt = ¶G/¶px     dy/dt = ¶G/¶py     dpx/dt = - ¶G/¶x     dpy/dt = - ¶G/¶y,
(2)
so that's how the generateor G acts on phase space. (Eq (2) gives the time derivative of its action, or its infintesimal action. That is, for small t, x ® x + t G/px etc.)
A conserved quantity K(qi, pi) has zero time derivative, 0 = [K, H]. To find such quantities, solve this equation! That's not so easy. Take the 2D SHO Hamiltonian with unit mass and unit spring constant:
 H = 1 2 (x2 + y2 + px2 + py2)
and check out this idea with angular momentum, L = xpy - ypx.
1. Verify that [L, H] = 0 (so L is indeed a conserved quantity).
We expect that because H is invariant under rotations. But what do rotations have to do with L? What does L generate?
2. Find equations like dx/dt = -y etc. for dy/dt, dpx/dt, dpy/dt from Eq (2), and verify that they describe a rotation - for examply by letting x = r cosf etc and noting dx/dt = - r cosfdf/dt = - y df/dt, putting df/dt = 1.
So the conserved L generates rotations, both in x, y and in px, py space - the rotated coordinates remain canonically conjugate. Note that indeed H is invariant under this simultaneous rotation in the x, y and px, py planes because it depends only on the "radius" in these planes.
But we could equally well pick two other planes in which to rotate, for example x, py and y, px, because all p's and x's appear equally in the Hamiltonian. This strange kind of rotation mixes x's and p's, it is not just a symmetry in the x's and p's separately, so it's something you cannot do with Lagrangians*. This kind of "hidden" symmetry will give further constants of the motion.
1. Take dx/dt = py, dpy/dt = -x, dy/dt = px,  dpx/dt = -y as your "strange rotation", insert in Eqs (2) and integrate for G. Verify that G is conserved (either by evaluating [G, H], or by substituting an explicit solution like x = A cos t ... ). The fact that you could integrate for a single function G means that the "strangely rotated" coordinates are still canonically conjugate.
2. If you did not know about the conserved L, you could use this conserved G to help solve the equations of motion: H + G and H - G are constants of the motion. Show that H + G involves only S = x + y, and H - G involves only D = x - y. You get a first order differential equation for S from H + G = const. that you can solve by separation of variables. (The constant value of H is of course E, give some name (like A) to the constant value of G.) Verify that your solution is equivalent to the usual SHO solution, x, y µ cos(t-t0)

*finally - most of the rest of this assignment would work as well or better with Lagrangians :)