Question

The Hamiltonian for a one-dimensional system with mass $m$, position $q$, and momentum $p$ is: $$H(p, q) = \frac{p^2}{2m} + q^2 A(q)$$ where $A(q)$ is a real function of $q$. If $$m \frac{d^2 q}{dt^2} = -5q A(q),$$ then $$\frac{d A(q)}{d q} = n \frac{A(q)}{q}.$$ The value of $n$ (in integer) is:

Hint:

For a system described by a Hamiltonian with a position-dependent potential, the equation of motion and the Hamiltonian's partial derivatives can be used to find relationships between the function $A(q)$ and its derivative.

Answer 1

1. Using the Hamiltonian:
The Hamiltonian of the system is given by:

$$H(p, q) = \frac{p^2}{2m} + q^2 A(q)$$

The Hamiltonian represents the total energy of the system, which is a sum of kinetic and potential energies.

2. Equations of motion:
The equations of motion are given by Hamilton's equations. For position $q$ and momentum $p$, we have:

$$\frac{dq}{dt} = \frac{\partial H}{\partial p} = \frac{p}{m}$$

and

$$\frac{dp}{dt} = -\frac{\partial H}{\partial q} = -2q A(q) - q^2 \frac{dA(q)}{dq}$$

3. Substitute the equation of motion:
According to the problem, we are given that:

$$m \frac{d^2 q}{dt^2} = -5q A(q)$$

Using $\frac{dq}{dt} = \frac{p}{m}$, we get:

$$m \frac{d^2 q}{dt^2} = \frac{d}{dt} \left( \frac{p}{m} \right) = \frac{dp}{dt}$$

Substituting the expression for $\frac{dp}{dt}$:

$$m \frac{d^2 q}{dt^2} = -2q A(q) - q^2 \frac{dA(q)}{dq}$$

Comparing this with the given equation:

$$-2q A(q) - q^2 \frac{dA(q)}{dq} = -5q A(q)$$

4. Solve for $\frac{dA(q)}{dq}$:
Simplifying:

$$-q^2 \frac{dA(q)}{dq} = -3q A(q) \quad \Rightarrow \quad \frac{dA(q)}{dq} = \frac{3A(q)}{q}$$

Comparing with $\frac{dA(q)}{dq} = n \frac{A(q)}{q}$, we find:

$$n = \boxed{3}$$