# Ladder operators and the momentum and position commutator

1. Jul 3, 2014

### 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)?

2. Jul 4, 2014

### PhilDSP

Hi kmchugh,

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

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

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

Last edited: Jul 4, 2014