Exploring Quantum Physics - Week 2 Question 9

Question:

Suppose we have the Hamiltonian $H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2$. What is the corresponding action?

Solution:

This problem is nothing but just substituting in the definitions. We have $E = \frac{p^2}{2m}$ and therefore $V = \frac{1}{2}m\omega^2 x^2$. Therefore the action would be $\int{E - V dt} = \int{ \frac{p^2}{2m} - \frac{1}{2}m\omega^2 x^2 dt}$. Now note that $\frac{p^2}{2m} = \frac{(mv)^2}{2m} = \frac{mv^2}{2}$, so the answer is $\int{ \frac{mv^2}{2} - \frac{1}{2}m\omega^2 x^2 dt}$.