Physics 441 Homework IV solutions

35. By direct computation, with $Q = a^{†}$

F_{2} (q, P) = \frac{i}{2} (ω q^{2} + a^{2} + 2 \sqrt{2 ω} q a) = \frac{i}{2} (ω q^{2} + P^{2} + 2 \sqrt{2 ω} q P)

You could ﬁnd a type-1 if you wish, by integrating the differential relations for a type-1;

F_{1} (q, Q) = F_{1} (q, a^{†}) = A {(a^{†})}^{2} + B q a^{†} + C q^{2}, p = \frac{\partial F_{1}}{\partial q} = B a^{†} + 2 C q, P = a = - \frac{\partial F_{1}}{\partial a^{†}} = - 2 A a^{†} - B q

a^{†} = \frac{p - 2 C q}{B}, q = \frac{a + 2 A a^{†}}{- B}

compare with Eq. 5.34 and you get $2 C = - i ω$ , $B = \sqrt{2 ω}$ , and $A = \frac{i}{2}$

F_{1} (q, Q) = F_{1} (q, a^{†}) = \frac{i}{2} {(a^{†})}^{2} + \sqrt{2 ω} q a^{†} - \frac{i}{2} ω q^{2}

Use the direct or symplectic tests, but to do so you must pick a pair of independent variables, and re-write the other two variables (which become functions of the independent pair). For example, lets use the direct test

(\frac{\partial p}{\partial Q})_{q} = - (\frac{\partial P}{\partial q})_{Q}, for which q, Q, are independent

You must re-write $p$ and $P$ in terms of $q$ and $Q$ only;

p = \sqrt{2 ω} Q - i ω q, P = \frac{(\sqrt{2 ω} Q - i ω q) - i ω q}{\sqrt{2 ω} i} = - i Q - \sqrt{2 ω} q

which passes the direct test

(\frac{\partial p}{\partial Q})_{q} = \sqrt{2 ω}, - (\frac{\partial P}{\partial q})_{Q} = \sqrt{2 ω}

You can show that all direct tests are passed (but passing one is sufficient), but you must write the dependent variables in terms of the independent variables speciﬁed by the chosen test. Why? You know that a transformation is canonical if it has a generating function. You will show in problem 46 that each direct test comes from one of the four types of generating functions. If you can ﬁnd a generating function of any type, you can Legendre transform it into a generating function of any other type.
Start with type-1, convert to type-3;

\begin{array}{rcl} F_{1} (q, Q) & = & \frac{i}{2} Q^{2} + \sqrt{2 ω} q Q - \frac{i}{2} ω q^{2} = F_{3} (p, Q) + p q, q = \frac{\sqrt{2 ω} Q - p}{i ω} \\ F_{3} (p, Q) & = & \frac{i}{2} Q^{2} + \sqrt{2 ω} (\frac{\sqrt{2 ω} Q - p}{i ω}) Q - \frac{i ω}{2} (\frac{\sqrt{2 ω} Q - p}{i ω})^{2} - p (\frac{\sqrt{2 ω} Q - p}{i ω}) \\ = & - \frac{i}{2} Q^{2} + i \sqrt{\frac{2}{ω}} p Q - \frac{i}{2 ω} p^{2} & (1) \end{array}

Lets check it;

q = - (\frac{\partial F_{3}}{\partial p})_{Q} = \frac{i}{ω} p - i \sqrt{\frac{2}{ω}} Q, \Rightarrow Q = \frac{p + i ω q}{\sqrt{2 ω}}

It works as expected. Since the transformation is generated by a type-3 generating function, we can be assured that it passes the direct test $(\frac{\partial P}{\partial p})_{Q} = (\frac{\partial q}{\partial Q})_{p}$ .

36. Solve the new equations of motion;

Q (t) = Q_{0}^{'} e^{i ω t}, P (t) = P_{0}^{'} e^{- i ω t}

put them back into the variable transformations

q = \frac{a^{†} - i a}{\sqrt{2 ω} i} = \frac{Q - i P}{\sqrt{2 ω} i} = \frac{Q_{0}^{'} e^{i ω t} - i P_{0}^{'} e^{- i ω t}}{\sqrt{2 ω} i}

and choose the constants $Q_{0}^{'}, P_{0}^{'}$ any way you wish to make the solution exactly match the one given in the problem.

37. We did this in class, ﬁrst we will need the “old differential relations” for $F_{1}$ ;

p = (\frac{\partial F_{1}}{\partial q})_{Q}, P = - (\frac{\partial F_{1}}{\partial Q})_{q}

The take differentials of

F_{1} (q, Q, t) = F_{2} (q, P, t) - Q P, (\frac{\partial F_{1}}{\partial q})_{Q} d q + (\frac{\partial F_{1}}{\partial Q})_{q} d Q = (\frac{\partial F_{2}}{\partial q})_{P} d q + (\frac{\partial F_{2}}{\partial P})_{q} d P - Q d P - P d Q

now eliminate “old variables” between this relation and the “old” differential relations

p d q - P d Q = (\frac{\partial F_{2}}{\partial q})_{P} d q + (\frac{\partial F_{2}}{\partial P})_{q} d P - Q d P - P d Q

p d q = (\frac{\partial F_{2}}{\partial q})_{P} d q + (\frac{\partial F_{2}}{\partial P})_{q} d P - Q d P

and now invoke independence of the new variables $q$ and $P$ (remember that over the phase space manifold the co-vectors $d q$ and $d P$ are linearly independent, so their coefficients in this equation must be zero)

p = (\frac{\partial F_{2}}{\partial q})_{P}, Q = (\frac{\partial F_{2}}{\partial P})_{p}

Play the same game with $F_{3}$ , we again need the $F_{1}$ differential relations, and we take differentials of

F_{1} (q, Q, t) = F_{3} (p, Q, t) + q p

in order to get a vector relation over the phase-manifold, so that we can deploy linear independence of the new basis for one-forms (co-vectors)

(\frac{\partial F_{1}}{\partial q})_{Q} d q + (\frac{\partial F_{1}}{\partial Q})_{q} d Q = (\frac{\partial F_{3}}{\partial p})_{Q} d p + (\frac{\partial F_{3}}{\partial Q})_{p} d Q + q d p + p d q

insert the “old differential relations”

p d q - P d Q = (\frac{\partial F_{3}}{\partial p})_{Q} d p + (\frac{\partial F_{3}}{\partial Q})_{p} d Q + q d p + p d q

eliminate “old” variables

- P d Q = (\frac{\partial F_{3}}{\partial p})_{Q} d p + (\frac{\partial F_{3}}{\partial Q})_{p} d Q + q d p

invoke linear independence of co-vectors $d p$ and $d Q$

q = - (\frac{\partial F_{3}}{\partial p})_{Q}, P = - (\frac{\partial F_{3}}{\partial Q})_{p}

You can see that canonical transformation concepts employ the manifold version of tangent vectors and their co-vector duals.

From

F (Q, p) = (r sin θ cos φ) p_{x} + (r sin θ sin φ) p_{y} + (r cos θ) p_{z}

which is clearly of type-3, we get

\begin{array}{rcl} P_{r} & = & - (\frac{\partial F_{3}}{\partial r})_{p} = - (sin θ cos φ p_{x} + sin θ sin φ p_{y} + cos θ p_{z}) \\ P_{φ} & = & - (\frac{\partial F_{3}}{\partial φ})_{p} = r sin θ sin φ p_{x} - r sin θ cos φ p_{y} \\ P_{θ} & = & - (\frac{\partial F_{3}}{\partial θ})_{p} = - r cos θ cos φ p_{x} - r cos θ sin φ p_{y} + r sin θ p_{z} \end{array}

Multiply the ﬁrst by $r sin θ$ and the third by $cos θ$ and add

\frac{r P_{r} sin θ + P_{θ} cos θ}{r} = - (cos φ p_{x} + sin φ p_{y}), \frac{P_{φ}}{r sin θ} = sin φ p_{x} - cos φ p_{y}

Multiply the second of these by $i$ and add or subtract it from the ﬁrst, and use Euler $e^{\pm i φ} = cos φ \pm i sin φ$

\frac{r P_{r} sin θ + P_{θ} cos θ}{r} + i \frac{P_{φ}}{r sin θ} = - (cos φ p_{x} + sin φ p_{y}) + i (sin φ p_{x} - cos φ p_{y}) = - e^{- i φ} (p_{x} + i p_{y})

p_{x} + i p_{y} = - e^{i φ} (P_{r} sin θ + P_{θ} \frac{cos θ}{r} + i \frac{P_{φ}}{r sin θ})

and so forth.

38.

F = α_{11} q_{1} P_{1} + α_{1, 2} q_{1} P_{2} + α_{2, 1} q_{2} P_{1} + α_{2, 2} q_{2} P_{2} = cos ε (q_{1} P_{1} + q_{2} P_{2}) + sin ε (q_{1} P_{2} - q_{2} P_{1})

which is of type-2;

Q_{1} = \frac{\partial F_{2}}{\partial P_{1}} = cos ε q_{1} - sin ε q_{2}, Q_{2} = \frac{\partial F_{2}}{\partial P_{2}} = cos ε q_{2} + sin ε q_{1}

which looks exactly like a counter-clockwise rotation by angle $ε$ .

39. This is a type-2, so

Q_{1} = \frac{\partial F_{2}}{\partial P_{1}} = q_{1}, Q_{2} = \frac{\partial F_{2}}{\partial P_{2}} = q_{2}, p_{1} = \frac{\partial F_{2}}{\partial q_{1}} = P_{1}, p_{2} = \frac{\partial F_{2}}{\partial q_{2}} = P_{2}

which is a “do nothing” transformation, the new conjugate variables are the same as the old.

40. This is a type-2, so with $F = q_{1} P_{1} + q_{1} P_{2} + ε (q_{1} P_{2} - q_{2} P_{1})$ we obtain new coordinates

Q_{1} = \frac{\partial F_{2}}{\partial P_{1}} = q_{1} - ε q_{2}, Q_{2} = \frac{\partial F_{2}}{\partial P_{2}} = q_{2} + ε q_{1}

If we let $ε \to 0$ in problem $38$ and use $sin ε \to ε$ , $cos ε \to 1$ in this limit, we arrive at the same new variable set, and so this transformation (problem 40) is the “inﬁnitismal generator” of the transformation of problem 38.

41. This is a type-2, so with $F = q_{1} P_{1} + q_{1} P_{2} + (a P_{1} + b P_{2})$ we obtain new coordinates

Q_{1} = \frac{\partial F_{2}}{\partial P_{1}} = q_{1} + a, Q_{2} = \frac{\partial F_{2}}{\partial P_{2}} = q_{2} + b

representing a shift of the coordinate origin, a “translation” generated by what is clearly the momenta.

42. Start with the deﬁnition

d q d p = J d Q d P = |\begin{matrix} \frac{\partial p}{\partial P} & \frac{\partial p}{\partial Q} \\ \frac{\partial q}{\partial P} & \frac{\partial q}{\partial Q} \end{matrix}| d Q d P

and repeat the example given for type-1, or demonstrate that you have learned your lessons well by employing differential forms;

d Q \land d P = ((\frac{\partial Q}{\partial q})_{p} d q + (\frac{\partial Q}{\partial p})_{q} d p) \land ((\frac{\partial P}{\partial q})_{p} d q + (\frac{\partial P}{\partial p})_{q} d p)

= ((\frac{\partial Q}{\partial q})_{p} (\frac{\partial P}{\partial p})_{q} - (\frac{\partial Q}{\partial p})_{q} (\frac{\partial P}{\partial q})_{p}) d q \land d p

In a few days you will learn that the object in the parenthesis is a Poisson bracket

{Q, P}_{p b} \equiv ((\frac{\partial Q}{\partial q})_{p} (\frac{\partial P}{\partial p})_{q} - (\frac{\partial Q}{\partial p})_{q} (\frac{\partial P}{\partial q})_{p})

and therefore we will be able to make a connection between preservation of the Lebesque measure $d q \land d p = d Q \land d P$ under canonical transformation, and the values of Poisson brackets such as ${Q, P}_{p b} = 1$ .
If you want to procede with the problem using this equation as a starting point, ﬁrst solve number $46$ . Problem $46$ says this about type two generating functions;

Type- 2; (\frac{\partial p}{\partial P})_{q} = (\frac{\partial Q}{\partial q})_{P} = (\frac{\partial Q}{\partial q})_{p, P}, F_{2} = F_{2} (q, P) \Rightarrow q, P, independent \Rightarrow (\frac{\partial P}{\partial q})_{x} = 0

(because when you take partial derivatives, you hold all variables ﬁxed except for the one that you differentiate with respect to). $P$ and $q$ independent means that it does not matter what is held ﬁxed, $(\frac{\partial P}{\partial q})_{x} = 0$ for any $x$ . Then

{Q, P}_{p b} = ((\frac{\partial Q}{\partial q})_{p} \frac{1}{(\frac{\partial Q}{\partial q})_{p}} - (\frac{\partial Q}{\partial p})_{q} \cdot 0) = 1

and so we have $d q \land d p = d Q \land d P$ . Play the game with type-3;

Type- 3; (\frac{\partial P}{\partial p})_{Q} = (\frac{\partial q}{\partial Q})_{p} = (\frac{\partial P}{\partial p})_{Q, q}, F_{3} = F_{3} (p, Q) \Rightarrow p, Q, independent \Rightarrow (\frac{\partial Q}{\partial p})_{x} = 0

Then

{Q, P}_{p b} = (\frac{1}{(\frac{\partial P}{\partial p})_{q}} (\frac{\partial P}{\partial p})_{q} - 0 \cdot (\frac{\partial P}{\partial q})_{p}) = 1

Let’s do type-4 since we are on a roll;

Type- 4; - (\frac{\partial q}{\partial P})_{p} = (\frac{\partial Q}{\partial p})_{P} = (\frac{\partial Q}{\partial p})_{q, P} F_{4} = F_{4} (p, P) \Rightarrow p, P, independent \Rightarrow (\frac{\partial P}{\partial p})_{x} = 0

{Q, P}_{p b} = ((\frac{\partial Q}{\partial q})_{p} \cdot 0 - \frac{- 1}{(\frac{\partial P}{\partial q})_{p}} (\frac{\partial P}{\partial q})_{p}) = 1

and there we have it, for all canonical transformations ${Q, P}_{p b} = J = 1$ . The Poisson bracket of the new variables with respect to the old is the Jacobian of the variable transformation.

44.

p = \frac{\partial F_{2}}{\partial q} = P + t \frac{\partial F}{\partial q}, P = p - t \frac{\partial F_{2}}{\partial q}, Q = \frac{\partial F_{2}}{\partial P} = q + t \frac{\partial F}{\partial P}, F = H (q, P)

This is quite deep, it means the the Hamiltonian acts like the generator of canonical transformations from variables representing initial conditions $(q_{0}, p_{0})$ to the variables $(q (t), p (t))$ . The Hamiltonian generates time evolution.

45. Take differentials of $F_{1} (q, Q, t) = q p - Q P + F_{4} (p, P, t)$ ;

(\frac{\partial F_{1}}{\partial q})_{Q} d q + (\frac{\partial F_{1}}{\partial Q})_{q} d Q = (\frac{\partial F_{4}}{\partial p})_{P} d p + (\frac{\partial F_{4}}{\partial P})_{p} d P - Q d P - P d Q + q d p + p d q

insert the differential relationd for type-1 and eliminate

p d q - P d Q = (\frac{\partial F_{4}}{\partial p})_{P} d p + (\frac{\partial F_{4}}{\partial P})_{p} d P - Q d P - P d Q + q d p + p d q

0 = (\frac{\partial F_{4}}{\partial p})_{P} d p + (\frac{\partial F_{4}}{\partial P})_{p} d P - Q d P + q d p

and employ linear independence of the co-vectors $d p$ and $d P$ ;

q = - (\frac{\partial F_{4}}{\partial p})_{P}, Q = (\frac{\partial F_{4}}{\partial P})_{p}, K = H + \frac{\partial F_{4}}{\partial t}

46.

Type 2; (\frac{\partial p}{\partial P})_{q} = (\frac{\partial Q}{\partial q})_{P}, Type 1; - (\frac{\partial P}{\partial q})_{Q} = (\frac{\partial p}{\partial Q})_{q}

Type 3; (\frac{\partial P}{\partial p})_{Q} = (\frac{\partial q}{\partial Q})_{p}, Type 4; - (\frac{\partial q}{\partial P})_{p} = (\frac{\partial Q}{\partial p})_{P}