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Quadrature operators and Hermite polynomials
#1
Sun, 15/11/2009 - 20:25
Hi Here is my opinion ,I
Hi
Here is my opinion ,I don‘t think the new hamiltonian is H=s^2+t^2, because both x and p are q-numbers. After a transformation (and your matrix is not a unitary matrix), the formula of the new hamiltonian will not identical with the old one.
Oscillator
Surely you want a minus sign for one of the sin(y) terms.
Then it is just the time evolution of the classical oscillator system. and at the same time the solution of the Heisenberg picture time evolution in the quantum case. This works for all Hamiltonians quandratic in p,q, because then the time evolution of the quadrature operators is linear with scalar coefficients. (If you want some fancy language, this is called the metaplectic representation of the affine symplectic group on phase space).
Since you already diagonalized your oscillator Hamiltonian
it is clear that you get your wave functions just pick up a phase \exp(iy \omega_n).