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} $.
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} $.
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