Quadrature operators and Hermite polynomials

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#1 Sun, 15/11/2009 - 20:25

Quadrature operators and Hermite polynomials

Hello everyone. I have the following problem.

I have The Hamiltonian of the 1D Harmonic Oscillator (hbar=m=w=1)

H= x^2+p^2

with the known solutions

psi(x)=exp(-x^2/2)*H_n(x)

where H_n are the Hermite polynomials of order n.

If I change the variables for the following ones (quadrature operators)

s=cos(y) x + sin(y) p
t=sin(y) x + cos(y) p

the new hamiltonian is H=s^2+t^2, so it is invariant. Now my question is: How can I change my old wavefunction psi(x) to the new space (s,t). I suposse that the new wavefunction must be something similar to the old ona, because the Hamiltonian is invariant, but I'm really don't sure.

Do you have any idea?

Thanks a lot.

Tue, 15/12/2009 - 09:33

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.

Tue, 29/12/2009 - 11:20
Anonymous (not verified)

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).