How is adiabatic invariant proved in a simple dynamic system?

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maverick6664
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I'm reading a book on quontum mechanics in japanese (Quontum Mechanics by Shinichiro Tomonaga) and am stuck in proving the action variable "J" is constant in a one dimensional cyclic movement. i.e.

The action variable "J" created by the trajectory of

H(p(t),q(t),a(t/T)) = E(t)

doesn't change. This trajectory won't make a closed region when a(t/T) changes, but when a(t/T) is fixed or changes very slowly the trajectory is assumed to be closed.

Will anyone give me good online references on it, or recommend nice English books on Quontum Mechanics (not so thick or thin) ? Japanese books don't look nice to me. I'm good at math and have knowledge on electromagnetic mechanics and special/general theories of relativity, but not quontum mechanics.

Thanks in advance!
 
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To prove that the action variable "J" is constant in a one dimensional cyclic movement, it is necessary to first understand the definition of the action variable. The action variable is a measure of the energy expended by a particle in a given period of time. It is associated with the total energy of a system over a single period of time, and is related to the momentum and angular momentum of the particle. The action variable "J" can be defined as:J = Integral(0 to T) (p(t)dq(t)-H(p(t),q(t),a(t/T))dt)Where p(t) is the momentum of the particle, q(t) is the position of the particle, a(t/T) is the angular velocity of the particle, and H(p(t),q(t),a(t/T)) is the Hamiltonian of the system. To prove that the action variable J is constant in a one dimensional cyclic movement, it is necessary to demonstrate that the integral of the Hamiltonian over a single period of time is always equal to the total energy of the system. This can be shown by using the Euler-Lagrange equation, which states that the integral of the Lagrangian over a single period of time is equal to the total energy of the system. The Lagrangian for this system is given by:L(p,q,a) = p(t)dq(t)-H(p(t),q(t),a(t/T))Using the Euler-Lagrange equation, we can show that the integral of the Lagrangian over a single period of time is equal to the total energy of the system:Integral(0 to T) (p(t)dq(t)-H(p(t),q(t),a(t/T))dt) = E(t)where E(t) is the total energy of the system. Therefore, the action variable J is constant in a one dimensional cyclic movement. For further information on the action variable and quontum mechanics, you may wish to consult an English language textbook on the subject