Daily Archives: Thu, 1 December 2016

Hamiltonian for the two-body problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercise 2.5.4.

Here we derive the equations of motion of the two-body problem 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}}. In this case, we start with the Lagrangian in terms of the centre of mass position {\mathbf{r}_{CM}} and the relative position {\mathbf{r}} of mass 2 to mass 1.

\displaystyle L \displaystyle = \displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right)\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle \frac{M}{2}\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{\mu}{2}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right) \ \ \ \ \ (3)

where {M=m_{1}+m_{2}} is the total mass and {\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}} is the reduced mass.

There are potentially 6 velocity components and 6 coordinate components in the Lagrangian, but the 3 components of {\mathbf{r}_{CM}} do not appear, which simplifies things a bit. To convert to a Hamiltonian, we need the momenta

\displaystyle p_{i}=\frac{\partial L}{\partial\dot{q}_{i}} \ \ \ \ \ (4)

The {x} component of momentum of the centre of mass is

\displaystyle p_{CM,x}=\frac{\partial L}{\partial\dot{r}_{CM,x}}=M\dot{r}_{CM,x} \ \ \ \ \ (5)

The other two components of the centre of mass velocity, and of the relative velocity, have a similar form, and in general we can write

\displaystyle p_{CM,i} \displaystyle = \displaystyle M\dot{r}_{CM,i}\ \ \ \ \ (6)
\displaystyle p_{i} \displaystyle = \displaystyle \mu\dot{r}_{i} \ \ \ \ \ (7)

In vector notation, this becomes

\displaystyle \dot{\mathbf{r}}_{CM} \displaystyle = \displaystyle \frac{\mathbf{p}_{CM}}{M}\ \ \ \ \ (8)
\displaystyle \dot{\mathbf{r}} \displaystyle = \displaystyle \frac{\mathbf{p}}{\mu}\ \ \ \ \ (9)
\displaystyle \left|\dot{\mathbf{r}}_{CM}\right|^{2} \displaystyle = \displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{M^{2}}\ \ \ \ \ (10)
\displaystyle \left|\dot{\mathbf{r}}\right|^{2} \displaystyle = \displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu^{2}} \ \ \ \ \ (11)

The Lagrangian thus becomes

\displaystyle L=\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right) \ \ \ \ \ (12)

The Hamiltonian is

\displaystyle H \displaystyle = \displaystyle \mathbf{p}\cdot\dot{\mathbf{r}}+\mathbf{p}_{CM}\cdot\dot{\mathbf{r}}_{CM}-L\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu}+\frac{\left|\mathbf{p}_{CM}\right|^{2}}{M}-\left[\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right)\right]\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}+V\left(\mathbf{r}\right) \ \ \ \ \ (15)

Once we’ve got the Hamiltonian, we can apply Hamilton’s canonical equations to get the equations of motion.

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \dot{r}_{i}\ \ \ \ \ (16)
\displaystyle -\frac{\partial H}{\partial r_{i}} \displaystyle = \displaystyle \dot{p}_{i} \ \ \ \ \ (17)

Since {\mathbf{r}_{CM}} does not appear in the Hamiltonian, we have

\displaystyle \dot{\mathbf{p}}_{CM} \displaystyle = \displaystyle 0\ \ \ \ \ (18)
\displaystyle \mathbf{p}_{CM} \displaystyle = \displaystyle \mbox{constant} \ \ \ \ \ (19)

so the momentum of the centre of mass does not change, as expected.

For {\mathbf{r}}, we have

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \frac{p_{i}}{\mu}=\dot{r}_{i}\ \ \ \ \ (20)
\displaystyle \frac{\partial H}{\partial r_{i}} \displaystyle = \displaystyle \frac{\partial V}{\partial r_{i}}=-\dot{p}_{i} \ \ \ \ \ (21)

The first equation tells us nothing new, while the second is just Newton’s law for a central force: {\mathbf{\dot{p}}=-\nabla V}.

Hamiltonians for harmonic oscillators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercises 2.5.2 – 2.5.3.

Here are a couple of examples of equations of motion using the Hamiltonian formalism. First, we look at the simple harmonic oscillator, in which we have a mass {m} sliding on a frictionless horizontal surface. The mass is connected to a spring with constant {k}, with the other end of the spring connected to a fixed support.

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}}. In this case, we have, using the coordinate {x} as the displacement from equilibrium

\displaystyle L\left(x,\dot{x}\right) \displaystyle = \displaystyle \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2}\ \ \ \ \ (2)
\displaystyle p \displaystyle = \displaystyle \frac{\partial L}{\partial\dot{x}}=m\dot{x}\ \ \ \ \ (3)
\displaystyle \dot{x} \displaystyle = \displaystyle \frac{p}{m}\ \ \ \ \ (4)
\displaystyle L\left(x,\dot{x}\left(x,p\right)\right) \displaystyle = \displaystyle \frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\ \ \ \ \ (5)
\displaystyle H \displaystyle = \displaystyle \frac{p^{2}}{m}-\left(\frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2} \ \ \ \ \ (7)

We can now apply Hamilton’s canonical equations:

\displaystyle \frac{\partial H}{\partial p} \displaystyle = \displaystyle \dot{x}\ \ \ \ \ (8)
\displaystyle -\frac{\partial H}{\partial x} \displaystyle = \displaystyle \dot{p} \ \ \ \ \ (9)

We get

\displaystyle \frac{\partial H}{\partial p} \displaystyle = \displaystyle \frac{p}{m}=\dot{x}\ \ \ \ \ (10)
\displaystyle -\frac{\partial H}{\partial x} \displaystyle = \displaystyle -kx=\dot{p} \ \ \ \ \ (11)

We thus get a pair of first order ODEs which can be solved in the usual way, given {x\left(0\right)} and {p\left(0\right)}. The second order ODE that we got by using the Lagrangian method can be obtained by differentiating the first equation and plugging it into the second:

\displaystyle \ddot{x} \displaystyle = \displaystyle \frac{\dot{p}}{m}\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle -\frac{k}{m}x \ \ \ \ \ (13)

From 7 we see that, since in the absence of external force, the total energy {H=T+V=E} is a constant,

\displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2}=E=\mbox{constant} \ \ \ \ \ (14)

This can be written as the equation of an ellipse:

\displaystyle \frac{p^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}=1 \ \ \ \ \ (15)

where

\displaystyle a^{2} \displaystyle = \displaystyle \frac{2E}{k}\ \ \ \ \ (16)
\displaystyle b^{2} \displaystyle = \displaystyle 2mE \ \ \ \ \ (17)

We can use the Hamiltonian formalism to get the equations of motion of the coupled harmonic oscillator. From our Lagrangian treatment, we had

\displaystyle L=\frac{1}{2}m\left(\dot{x}_{1}^{2}+\dot{x}_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (18)

Converting to coordinates and momenta, we have

\displaystyle p_{i} \displaystyle = \displaystyle \frac{\partial L}{\partial\dot{x}_{i}}=m\dot{x}_{i}\ \ \ \ \ (19)
\displaystyle \dot{x}_{i} \displaystyle = \displaystyle \frac{p_{i}}{m}\ \ \ \ \ (20)
\displaystyle H \displaystyle = \displaystyle \sum_{i}p_{i}\dot{x}_{i}-L\left(x,\dot{x}\right)\ \ \ \ \ (21)
\displaystyle \displaystyle = \displaystyle \frac{1}{m}\left(p_{1}^{2}+p_{2}^{2}\right)-\left[\frac{1}{2m}m\left(p_{1}^{2}+p_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right)\right]\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{1}{2m}\left(p_{1}^{2}+p_{2}^{2}\right)+k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (23)

Applying the canonical equations gives

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \frac{p_{i}}{m}=\dot{x}_{i}\ \ \ \ \ (24)
\displaystyle -\frac{\partial H}{\partial x_{1}} \displaystyle = \displaystyle -2kx_{1}+kx_{2}=\dot{p}_{1}\ \ \ \ \ (25)
\displaystyle -\frac{\partial H}{\partial x_{2}} \displaystyle = \displaystyle -2kx_{2}+kx_{1}=\dot{p}_{2} \ \ \ \ \ (26)

Again, by taking the derivative of the first line and substituting into the last two lines, we get back the previous equations of motion:

\displaystyle \ddot{x}_{1} \displaystyle = \displaystyle -2\frac{k}{m}x_{1}+\frac{k}{m}x_{2}\ \ \ \ \ (27)
\displaystyle \ddot{x}_{2} \displaystyle = \displaystyle \frac{k}{m}x_{1}-2\frac{k}{m}x_{2} \ \ \ \ \ (28)