Ladder operators and the momentum and position commutator

[kmchugh]

When using Fourier's trick for determining the allowable energies for stationary states, Griffiths introduces the a+- operators. When factoring the Hamiltonian, the imaginary part is assigned to the momentum operator versus the position operator. Is there a reason for this? If :

a-+ = k(ip + mwx)(-ip + mwx), and the commutator is (xp-px), Is

a+- = k(-ip = mwx)(ip + mwx) ?

If so, is the commutator (px-xp)?

Thanks in advance for your input.

 

[PhilDSP]

Hi kmchugh,

The Fourier transform of the nabla (or del) operator and partial time derivative operator is [itex] \ \ \ \ \hat F(\nabla) = ik \ \ \ \ \ [/itex] [itex]\hat F(\frac{\partial}{\partial t}) = -i\omega[/itex]

Where spatial variables [itex]x, y, z[/itex] are transformed into the wavenumber vector [itex]k[/itex] and the time variable is transformed into the angular frequency scalar [itex]\omega[/itex]

The first order nabla operator is associated with momentum and its transform contains [itex]i[/itex], meaning that it is an imaginary value when compared to the phase of position variables [itex]x, y, z[/itex].