Daily Archives: Tue, 13 December 2016

Hamilton’s equations of motion under a regular canonical transformation

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.8; Exercise 2.8.5.

If the Hamiltonian is invariant under a regular canonical transformation and we can find a generator {g} such that an infinitesimal version of this transformation is given by

\displaystyle \bar{q}_{i} \displaystyle = \displaystyle q_{i}+\varepsilon\frac{\partial g}{\partial p_{i}}\equiv q_{i}+\delta q_{i}\ \ \ \ \ (1)
\displaystyle \bar{p}_{i} \displaystyle = \displaystyle p_{i}-\varepsilon\frac{\partial g}{\partial q_{i}}\equiv p_{i}+\delta p_{i} \ \ \ \ \ (2)

then {g} is conserved.

If we are dealing with a finite regular canonical transformation where we go from {\left(q,p\right)\rightarrow\left(\bar{q},\bar{p}\right)}, and the Hamiltonian is invariant under this transformation, then it turns out that if a trajectory {\left(q\left(t\right),p\left(t\right)\right)} satisfies Hamilton’s equations of motion:

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \dot{q}_{i}\ \ \ \ \ (3)
\displaystyle -\frac{\partial H}{\partial q_{i}} \displaystyle = \displaystyle \dot{p}_{i} \ \ \ \ \ (4)

then the trajectory obtained by transforming every point in the original trajectory {\left(q\left(t\right),p\left(t\right)\right)} to the barred system {\left(\bar{q}\left(t\right),\bar{p}\left(t\right)\right)} is also a solution of Hamilton’s equations in the sense that

\displaystyle \frac{\partial H}{\partial\bar{p}_{i}} \displaystyle = \displaystyle \dot{\bar{q}}_{i}\ \ \ \ \ (5)
\displaystyle -\frac{\partial H}{\partial\bar{q}_{i}} \displaystyle = \displaystyle \dot{\bar{p}}_{i} \ \ \ \ \ (6)

The proof of this is a bit subtle, but goes as follows. To begin, review the derivation of the conditions for a transformation to be canonical. This derivation applied to a passive transformation, in which the two sets of parameters {\left(q,p\right)\rightarrow\left(\bar{q},\bar{p}\right)} refer to the same point in phase space. The transformation we’re considering here is an active transformation, in which {\left(q,p\right)\rightarrow\left(\bar{q},\bar{p}\right)} actually moves the point in phase space. The original derivation (for passive transformations) relied on the fact that the numerical value of the Hamiltonian is the same in both coordinate systems, since both {\left(q,p\right)} and {\left(\bar{q},\bar{p}\right)} refer to the same point in phase space. However, for our active transformation, we’re assuming that the Hamiltonian is invariant under the transformation, that is {H\left(\bar{q},\bar{p}\right)=H\left(q,p\right)}, where {\left(q,p\right)} and {\left(\bar{q},\bar{p}\right)} now refer to different points in phase space. Since the assumption that the Hamiltonian satisfies {H\left(\bar{q},\bar{p}\right)=H\left(q,p\right)} was all that we used in the original derivation, the same derivation works both for passive transformations (always) and for active transformations (if the Hamiltonian is invariant under the active transformation). We therefore end up with the equations

\displaystyle \dot{\overline{q}}_{j} \displaystyle = \displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{q}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{q}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (7)
\displaystyle \dot{\overline{p}}_{j} \displaystyle = \displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{p}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{p}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (8)

Since the transformation is specified to be canonical, the conditions on the Poisson brackets apply here:

\displaystyle \left\{ \overline{q}_{j},\overline{q}_{k}\right\} \displaystyle = \displaystyle \left\{ \overline{p}_{j},\overline{p}_{k}\right\} =0\ \ \ \ \ (9)
\displaystyle \left\{ \overline{q}_{j},\overline{p}_{k}\right\} \displaystyle = \displaystyle \delta_{jk} \ \ \ \ \ (10)

The result is that the transformed trajectory also satisfies Hamilton’s equations 5 and 6.

We can now revisit the 2-d harmonic oscillator to show that a noncanonical transformation violates these results. The Hamiltonian is

\displaystyle H=\frac{1}{2m}\left(p_{x}^{2}+p_{y}^{2}\right)+\frac{1}{2}m\omega^{2}\left(x^{2}+y^{2}\right) \ \ \ \ \ (11)

and we consider the transformation where we rotate the coordinates but not the momenta. The transformation is

\displaystyle \bar{x} \displaystyle = \displaystyle x\cos\theta-y\sin\theta\ \ \ \ \ (12)
\displaystyle \bar{y} \displaystyle = \displaystyle x\sin\theta+y\cos\theta\ \ \ \ \ (13)
\displaystyle \bar{p}_{x} \displaystyle = \displaystyle p_{x}\ \ \ \ \ (14)
\displaystyle \bar{p}_{y} \displaystyle = \displaystyle p_{y} \ \ \ \ \ (15)

As we’ve seen, this is a noncanonical transformation. To see what happens, we’ll consider the initial conditions

\displaystyle x\left(0\right) \displaystyle = \displaystyle a\ \ \ \ \ (16)
\displaystyle p_{x}\left(0\right) \displaystyle = \displaystyle b\ \ \ \ \ (17)
\displaystyle y\left(0\right) \displaystyle = \displaystyle p_{y}\left(0\right)=0 \ \ \ \ \ (18)

The mass is started off at a point on the {x} axis with a momentum only in the {x} direction. In this case, the mass behaves like a one-dimensional harmonic oscillator, moving along the {x} axis only. To be precise, we can work out Hamilton’s equations of motion:

\displaystyle \dot{p}_{x} \displaystyle = \displaystyle -\frac{\partial H}{\partial x}=-m\omega^{2}x\ \ \ \ \ (19)
\displaystyle \dot{x} \displaystyle = \displaystyle \frac{\partial H}{\partial p_{x}}=\frac{p_{x}}{m} \ \ \ \ \ (20)

The equations for {y} and {p_{y}} are the same, with {x} replaced by {y} everywhere. We can solve these ODEs in the usual way, by differentiating the first one and substituting the second one into the first to get

\displaystyle \ddot{p}_{x}=-m\omega^{2}\dot{x}=-\omega^{2}p_{x} \ \ \ \ \ (21)

This has the general solution

\displaystyle p_{x}\left(t\right)=A\cos\omega t+B\sin\omega t \ \ \ \ \ (22)

We can do the same for {x} and get

\displaystyle x\left(t\right)=C\cos\omega t+D\sin\omega t \ \ \ \ \ (23)

Applying the initial conditions, we get

\displaystyle p_{x}\left(0\right) \displaystyle = \displaystyle A=b\ \ \ \ \ (24)
\displaystyle x\left(0\right) \displaystyle = \displaystyle C=a \ \ \ \ \ (25)

Plugging these into the equations of motion 19 and 20 and solving for {B} and {D} we get the final solution

\displaystyle p_{x}\left(t\right) \displaystyle = \displaystyle b\cos\omega t-m\omega a\sin\omega t\ \ \ \ \ (26)
\displaystyle x\left(t\right) \displaystyle = \displaystyle a\cos\omega t+\frac{b}{m\omega}\sin\omega t\ \ \ \ \ (27)
\displaystyle y\left(t\right) \displaystyle = \displaystyle p_{y}\left(t\right)=0 \ \ \ \ \ (28)

Now suppose we start off with {x\left(0\right)=0}, {y\left(0\right)=a}, {p_{x}\left(0\right)=b} and {p_{y}\left(0\right)=0}. That is, we have rotated the coordinates through {\frac{\pi}{2}}, but not the momenta. We now begin with the mass on the {y} axis, but moving in the {x} direction, so as time progresses, it will have components of momentum in both the {x} and {y} directions. Although it’s fairly obvious that this motion will not be simply the motion in the first case rotated through {\frac{\pi}{2}}, let’s go through the equations. By the same technique as above, we can solve the equations to get

\displaystyle p_{x}\left(t\right) \displaystyle = \displaystyle b\cos\omega t\ \ \ \ \ (29)
\displaystyle p_{y}\left(t\right) \displaystyle = \displaystyle -m\omega a\sin\omega t\ \ \ \ \ (30)
\displaystyle x\left(t\right) \displaystyle = \displaystyle \frac{b}{m\omega}\sin\omega t\ \ \ \ \ (31)
\displaystyle y\left(t\right) \displaystyle = \displaystyle a\cos\omega t \ \ \ \ \ (32)

If we look at the system at, say, {t=\frac{\pi}{2\omega}}, then {\cos\omega t=0} and {\sin\omega t=1}. The mass that started off on the {x} axis will be at position {\left(x,y\right)=\left(\frac{b}{m\omega},0\right)} and so will the mass that started off on the {y} axis. Since the two masses are in the same place, obviously one is not the rotated version of the other.

Another, probably easier, way to see this is that since the first mass moves only along the {x} axis, if the rotated version of the trajectory was also to be a solution, the rotated trajectory would have to lie entirely along the {y} axis, which is certainly not true for the mass that starts off on the {y} axis, but with a momentum {p_{x}\ne0}.

In the general case, if the transformation is noncanonical, then the Poisson brackets in 7 and 8 don’t satisfy the conditions 9 and 10, with the result that Hamilton’s equations aren’t satisfied in the {\left(\bar{q},\bar{p}\right)} coordinates. (There may be a deeper, physical interpretation that I’ve missed, but from a mathematical point of view, that’s what goes wrong.)

Infinitesimal rotations in canonical and noncanonical transformations

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.8; Exercises 2.8.3 – 2.8.4.

Here are a couple of examples of transformations of variables and their consequences with regard to conservation laws.

First, we look at the 2-d harmonic oscillator where the Hamiltonian is

\displaystyle H=\frac{1}{2m}\left(p_{x}^{2}+p_{y}^{2}\right)+\frac{1}{2}m\omega^{2}\left(x^{2}+y^{2}\right) \ \ \ \ \ (1)

If we rotate the system so that both the coordinates and momenta get rotated, then

\displaystyle \bar{x} \displaystyle = \displaystyle x\cos\theta-y\sin\theta\ \ \ \ \ (2)
\displaystyle \bar{y} \displaystyle = \displaystyle x\sin\theta+y\cos\theta\ \ \ \ \ (3)
\displaystyle \bar{p}_{x} \displaystyle = \displaystyle p_{x}\cos\theta-p_{y}\sin\theta\ \ \ \ \ (4)
\displaystyle \bar{p}_{y} \displaystyle = \displaystyle p_{x}\sin\theta+p_{y}\cos\theta \ \ \ \ \ (5)

We can show by direct calculation that {H} is invariant under this transformation, and we can verify that this is a canonical transformation. Shankar shows in his equation 2.8.8 that the generator of this transformation is the angular momentum {\ell_{z}=xp_{y}-yp_{x}}.

However, if we rotate only the coordinates and not the momenta, we get the transformation:

\displaystyle \bar{x} \displaystyle = \displaystyle x\cos\theta-y\sin\theta\ \ \ \ \ (6)
\displaystyle \bar{y} \displaystyle = \displaystyle x\sin\theta+y\cos\theta\ \ \ \ \ (7)
\displaystyle \bar{p}_{x} \displaystyle = \displaystyle p_{x}\ \ \ \ \ (8)
\displaystyle \bar{p}_{y} \displaystyle = \displaystyle p_{y} \ \ \ \ \ (9)

Again, we can show by direct calculation that

\displaystyle \bar{x}^{2}+\bar{y}^{2}=x^{2}+y^{2} \ \ \ \ \ (10)

so {H} is also invariant under this transformation. However, this transformation is noncanonical, as we can see by calculating one of the Poisson brackets:

\displaystyle \left\{ \bar{x},\bar{p}_{x}\right\} \displaystyle = \displaystyle \sum_{i}\left(\frac{\partial\overline{x}}{\partial q_{i}}\frac{\partial\bar{p}_{x}}{\partial p_{i}}-\frac{\partial\overline{x}}{\partial p_{i}}\frac{\partial\bar{p}_{x}}{\partial q_{i}}\right)\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \cos\theta\ne1 \ \ \ \ \ (12)

The other mixed brackets (with a coordinate and a momentum) are also not either 0 or 1 as would be required if the transformation were to be canonical.

In order for this transformation to give rise to a conservation law, we would need to find a generator {g} that satisfied, for an infinitesimal rotation {\varepsilon}:

\displaystyle \bar{q}_{i} \displaystyle = \displaystyle q_{i}+\varepsilon\frac{\partial g}{\partial p_{i}}\equiv q_{i}+\delta q_{i}\ \ \ \ \ (13)
\displaystyle \bar{p}_{i} \displaystyle = \displaystyle p_{i}-\varepsilon\frac{\partial g}{\partial q_{i}}\equiv p_{i}+\delta p_{i} \ \ \ \ \ (14)

For an infinitesimal rotation, the transformation 6 becomes

\displaystyle \bar{x} \displaystyle = \displaystyle x-\varepsilon y\ \ \ \ \ (15)
\displaystyle \bar{y} \displaystyle = \displaystyle y+\varepsilon x\ \ \ \ \ (16)
\displaystyle \bar{p}_{x} \displaystyle = \displaystyle p_{x}\ \ \ \ \ (17)
\displaystyle \bar{p}_{y} \displaystyle = \displaystyle p_{y} \ \ \ \ \ (18)

Therefore, the generator would have to satisfy

\displaystyle \frac{\partial g}{\partial p_{x}} \displaystyle = \displaystyle -y\ \ \ \ \ (19)
\displaystyle \frac{\partial g}{\partial p_{y}} \displaystyle = \displaystyle x\ \ \ \ \ (20)
\displaystyle \frac{\partial g}{\partial x} \displaystyle = \displaystyle 0\ \ \ \ \ (21)
\displaystyle \frac{\partial g}{\partial y} \displaystyle = \displaystyle 0 \ \ \ \ \ (22)

The last two conditions state that {g} cannot depend on {x} or {y}, but integrating the first two conditions, we get

\displaystyle g=-yp_{x}+xp_{y}+f\left(x,y\right) \ \ \ \ \ (23)

where {f} is a function that depends only on {x} and/or {y}. Thus there is no {g} that satisfies all four conditions, so there is no conservation law associated with a rotation of the coordinates only, even though the Hamiltonian is invariant under this transformation. Only canonical transformations that leave {H} invariant give rise to conservation laws.

As another example, suppose he have the one-dimensional system with

\displaystyle H=\frac{1}{2}\left(p^{2}+x^{2}\right) \ \ \ \ \ (24)

and perform a rotation in phase space, that is, in the {x-p} plane:

\displaystyle \bar{x} \displaystyle = \displaystyle x\cos\theta-p\sin\theta\ \ \ \ \ (25)
\displaystyle \bar{p} \displaystyle = \displaystyle x\sin\theta+p\cos\theta \ \ \ \ \ (26)

The Hamiltonian is invariant:

\displaystyle \bar{p}^{2}+\bar{x}^{2} \displaystyle = \displaystyle x^{2}\sin^{2}\theta+2xp\sin\theta\cos\theta+p^{2}\cos^{2}\theta+\ \ \ \ \ (27)
\displaystyle \displaystyle \displaystyle x^{2}\cos^{2}\theta-2xp\sin\theta\cos\theta+p^{2}\sin^{2}\theta\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle x^{2}+p^{2} \ \ \ \ \ (29)

The transformation is canonical as we can verify by calculating the Poisson bracket

\displaystyle \left\{ \bar{x},\bar{p}\right\} \displaystyle = \displaystyle \frac{\partial\overline{x}}{\partial x}\frac{\partial\bar{p}}{\partial p}-\frac{\partial\overline{x}}{\partial p}\frac{\partial\bar{p}}{\partial x}\ \ \ \ \ (30)
\displaystyle \displaystyle = \displaystyle \cos^{2}\theta-\left(-\sin^{2}\theta\right)\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle 1 \ \ \ \ \ (32)

An infinitesimal rotation gives the transformation

\displaystyle \bar{x} \displaystyle = \displaystyle x-\varepsilon p\ \ \ \ \ (33)
\displaystyle \bar{p} \displaystyle = \displaystyle p+\varepsilon x \ \ \ \ \ (34)

To find the generator, we need to solve 13 and 14:

\displaystyle \frac{\partial g}{\partial p} \displaystyle = \displaystyle -p\ \ \ \ \ (35)
\displaystyle \frac{\partial g}{\partial x} \displaystyle = \displaystyle -x \ \ \ \ \ (36)

These can be integrated to give

\displaystyle g\left(x,p\right)=-\frac{1}{2}\left(p^{2}+x^{2}\right)+C \ \ \ \ \ (37)

where {C} is a constant of integration. Thus the quantity that is conserved is (apart from the minus sign, which we could eliminate by rotating through {-\theta} instead of {\theta}) is just the original Hamiltonian, or total energy.