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

Correct Answer: 3

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 the 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 \( \frac{dp}{dt} \) from Hamilton’s equations: \[ m \frac{d^2 q}{dt^2} = -2q A(q) - q^2 \frac{dA(q)}{dq}. \] Comparing this with the given equation \( m \frac{d^2 q}{dt^2} = -5q A(q) \), we have: \[ -2q A(q) - q^2 \frac{dA(q)}{dq} = -5q A(q). \] 4. Solve for \( \frac{dA(q){dq} \):}
Simplifying this equation: \[ -q^2 \frac{dA(q)}{dq} = -3q A(q), \] which gives: \[ \frac{dA(q)}{dq} = \frac{3A(q)}{q}. \] Comparing this with the given relation \( \frac{dA(q)}{dq} = n \frac{A(q)}{q} \), we find that: \[ n = 3. \] Thus, the value of \( n \) is \( \boxed{3} \).