Daily Archives: Sat, 3 December 2016

Cyclic coordinates and Poisson brackets

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 2.7.1 – 2.7.2.

Hamilton’s canonical equations are:

\displaystyle   \frac{\partial H}{\partial p_{i}} \displaystyle  = \displaystyle  \dot{q}_{i}\ \ \ \ \ (1)
\displaystyle  -\frac{\partial H}{\partial q_{i}} \displaystyle  = \displaystyle  \dot{p}_{i} \ \ \ \ \ (2)

If a coordinate {q_{i}} is missing in the Hamiltonian (that is, {H} is indepedent of {q_{i}}), then

\displaystyle  \dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}=0 \ \ \ \ \ (3)

Thus the conjugate momentum {p_{i}} is conserved. Such a missing coordinate {q_{i}} is known as a cyclic coordinate. [I’m not sure of the origin of this term. Again Google doesn’t provide a definitive answer.]

There is a general method for calculating the rate of change of some function {\omega\left(p,q\right)} that depends on the momenta and coordinates, but not explicitly on the time ({\omega} is allowed to depend implicitly on time since {p} and/or {q} can depend on time). The time derivative can then be written using the chain rule:

\displaystyle   \frac{d\omega}{dt} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\dot{q}_{i}+\frac{\partial\omega}{\partial p_{i}}\dot{p}_{i}\right)\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (5)
\displaystyle  \displaystyle  \equiv \displaystyle  \left\{ \omega,H\right\}  \ \ \ \ \ (6)

where in the second line we used Hamilton’s equations 1 and 2. The last line defines the Poisson bracket of the function {\omega} with the Hamiltonian {H}. We can see that if {\left\{ \omega,H\right\} =0}, the function {\omega} is conserved.

Since {\left\{ H,H\right\} =0} automatically, the total energy (represented by the Hamiltonian) is conserved, provided there is no explicit time dependence. Such a time dependence can arise if the system is subject to some external force, for example.

From the definition 5 we can derive a few fundamental properties of Poisson brackets. We’ll consider a general Poisson bracket between two arbitrary functions {\omega\left(p,q\right)} and {\lambda\left(p,q\right)}. Then

\displaystyle   \left\{ \omega,\lambda\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -\sum_{i}\left(\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}-\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}\right)\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -\sum_{i}\left(\frac{\partial\lambda}{\partial q_{i}}\frac{\partial\omega}{\partial p_{i}}-\frac{\partial\lambda}{\partial p_{i}}\frac{\partial\omega}{\partial q_{i}}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\left\{ \lambda,\omega\right\} \ \ \ \ \ (10)

A Poisson bracket is distributive, in the sense that

\displaystyle   \left\{ \omega,\lambda+\sigma\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\left[\frac{\partial\lambda}{\partial p_{i}}+\frac{\partial\sigma}{\partial p_{i}}\right]-\frac{\partial\omega}{\partial p_{i}}\left[\frac{\partial\lambda}{\partial q_{i}}+\frac{\partial\sigma}{\partial q_{i}}\right]\right)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (14)

One more identity is useful, which we can derive using the product rule:

\displaystyle   \left\{ \omega,\lambda\sigma\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\sigma\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\lambda\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (17)

The Poisson brackets involving the coordinates {q_{i}} and momenta {p_{i}} turn up frequently, so it’s worth deriving them in detail. We have

\displaystyle  \left\{ q_{i},q_{j}\right\} =\sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial q_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial q_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (18)

This follows because, in the Hamiltonian formalism, the {q_{i}}s and {p_{i}}s are independent variables, so {\frac{\partial q_{j}}{\partial p_{k}}=\frac{\partial p_{j}}{\partial q_{k}}=0} for all {j} and {k}. For the same reason, we have

\displaystyle  \left\{ p_{i},p_{j}\right\} =\sum_{k}\left(\frac{\partial p_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial p_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (19)

The mixed Poisson bracket is a different story, however:

\displaystyle   \left\{ q_{i},p_{j}\right\} \displaystyle  = \displaystyle  \sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \sum_{k}\delta_{ik}\delta_{jk}-0\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \delta_{ij} \ \ \ \ \ (22)

Hamilton’s equations 1 and 2 can be written using Poisson brackets by setting {\omega} equal to {q_{i}} and {p_{i}} respectively in 6:

\displaystyle   \dot{q}_{i} \displaystyle  = \displaystyle  \left\{ q_{i},H\right\} \ \ \ \ \ (23)
\displaystyle  \dot{p}_{i} \displaystyle  = \displaystyle  \left\{ p_{i},H\right\} \ \ \ \ \ (24)

Example In two dimensions, we have a Hamiltonian:

\displaystyle  H=p_{x}^{2}+p_{y}^{2}+ax^{2}+by^{2} \ \ \ \ \ (25)

If {a=b}, then in polar coordiantes, the only coordinate appearing in {H} is the radial distance from the origin {r=\sqrt{x^{2}+y^{2}}}, which means that the polar angle {\theta} is a cyclic coordinate. This means that the conjugate momentum {p_{\theta}} must be conserved. That is,

\displaystyle  \dot{p}_{\theta}=\left\{ p_{\theta},H\right\} =0 \ \ \ \ \ (26)

However, {p_{\theta}} is the angular momentum {\ell_{z}}, so this just says that angular momentum is conserved.

To see this explicitly, it’s easier to convert to polar coordinates. From Hamilton’s equations

\displaystyle   \dot{x} \displaystyle  = \displaystyle  \frac{\partial H}{\partial p_{x}}=2p_{x}\ \ \ \ \ (27)
\displaystyle  \dot{y} \displaystyle  = \displaystyle  2p_{y}\ \ \ \ \ (28)
\displaystyle  p_{x}^{2}+p_{y}^{2} \displaystyle  = \displaystyle  \frac{1}{4}\left(\dot{x}^{2}+\dot{y}^{2}\right)\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{v^{2}}{4}\ \ \ \ \ (30)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) \ \ \ \ \ (31)

where in the fourth line, {v} is the linear velocity and in the fifth line we converted this to polar coordinates. Thus the Hamiltonian becomes, in the case where {a=b}:

\displaystyle  H=\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)+ar^{2} \ \ \ \ \ (32)

To find the conjugate momenta in polar coordinates, we can write out the Lagrangian. We use {p_{x}\dot{x}=\frac{\dot{x}^{2}}{2}} and {p_{y}\dot{y}=\frac{\dot{y}^{2}}{2}} and get

\displaystyle   L \displaystyle  = \displaystyle  \sum_{i}p_{i}\dot{q}_{i}-H\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)-\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2}\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2} \ \ \ \ \ (35)

The conjugate momenta are thus

\displaystyle   p_{\theta} \displaystyle  = \displaystyle  \frac{\partial L}{\partial\dot{\theta}}=\frac{1}{2}r^{2}\dot{\theta}\ \ \ \ \ (36)
\displaystyle  p_{r} \displaystyle  = \displaystyle  \frac{\partial L}{\partial\dot{r}}=\frac{\dot{r}}{2} \ \ \ \ \ (37)

From this we can see that {p_{\theta}} is indeed angular momentum as it’s proportional to the product of {r} and the tangential velocity {v_{\theta}=r\dot{\theta}}. (‘Real’ momentum and angular momentum must, of course, also contain a factor of a mass, but from the definition of the Hamiltonian above, we see that the mass has been incorporated into the momentum parameters.)

Plugging these back into 32 we get

\displaystyle  H=p_{r}^{2}+p_{\theta}^{2}+ar^{2} \ \ \ \ \ (38)

We can now calculate the Poisson brackets easily:

\displaystyle   \left\{ p_{\theta},H\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial p_{\theta}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{\theta}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (39)
\displaystyle  \displaystyle  = \displaystyle  0-\frac{\partial p_{\theta}}{\partial p_{\theta}}\frac{\partial H}{\partial\theta}=0\ \ \ \ \ (40)
\displaystyle  \left\{ p_{r},H\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial p_{r}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{r}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (41)
\displaystyle  \displaystyle  = \displaystyle  0-\frac{\partial p_{r}}{\partial p_{r}}\frac{\partial H}{\partial r}\ \ \ \ \ (42)
\displaystyle  \displaystyle  = \displaystyle  -2ar \ \ \ \ \ (43)

Thus {p_{\theta}} (the angular momentum) is conserved, while {p_{r}<0}, so that the object is always being pulled in towards the origin.

Hamiltonian for the electromagnetic force

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.6.

Here we derive the equations of motion for the electromagnetic force using the Hamiltonian formalism.

The Hamiltonian is given by

\displaystyle  H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)

where the velocities {\dot{q}_{i}} are expressed in terms of the positions {q_{i}} and momenta {p_{i}}. The electromagnetic Lagrangian is

\displaystyle  L=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}-q\phi+\frac{q}{c}\mathbf{v}\cdot\mathbf{A} \ \ \ \ \ (2)

where {\phi} is the electric potential and {\mathbf{A}} is the magnetic potential, with {\mathbf{v}} the velocity of the charge {q} with mass {m}. To convert to the Hamiltonian, we need the momentum, defined as

\displaystyle  p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}

In this case, the generalized velocity is given by

\displaystyle  \dot{q}_{i}=v_{i} \ \ \ \ \ (3)

so we have

\displaystyle  p_{i}=mv_{i}+\frac{q}{c}A_{i} \ \ \ \ \ (4)

or, in vector notation

\displaystyle   \mathbf{p} \displaystyle  = \displaystyle  m\mathbf{v}+\frac{q}{c}\mathbf{A}\ \ \ \ \ (5)
\displaystyle  \mathbf{v} \displaystyle  = \displaystyle  \frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A} \ \ \ \ \ (6)

The Lagrangian is therefore

\displaystyle  L=\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}-q\phi+\frac{q}{c}\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\cdot\mathbf{A} \ \ \ \ \ (7)

The first sum in the Hamiltonian is

\displaystyle  \sum_{i}p_{i}\dot{q}_{i}=\mathbf{p}\cdot\mathbf{v}=\mathbf{p}\cdot\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right) \ \ \ \ \ (8)

The Hamiltonian is then

\displaystyle   H \displaystyle  = \displaystyle  \mathbf{p}\cdot\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)-\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi-\frac{q}{c}\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\cdot\mathbf{A}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\left(\mathbf{p}-\frac{q}{c}\mathbf{A}\right)-\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi \ \ \ \ \ (11)