# Harmonic oscillator

1. Nov 22, 2013

### aaaa202

Let a+,a- be the ladder operators of the harmonic oscillators. In my book I encountered the hamiltonian:

H = hbarω(a+a-+½) + hbarω0(a++a-)
Now the first term is just the regular harmonic oscillator and the second term can be rewritten with the transformation equations for x and p to the ladder operators as:
hbarω0(a++a-) = x/(√(2hbar/mω))
My question is: Does this last term just represent a translation in the origin of the harmonic oscillator i.e. the potential is mω2(x-x0)^2 where x0 is determined by ω0? If so how do I see that algebraically?

2. Nov 22, 2013

### hilbert2

If the potential energy function in the usual harmonic oscillator Hamiltonian is $V(x)=kx^{2}$, and in the case of this problem it is $V(x)=kx^{2}+k'x$, you can complete the square to write the potential in form $V(x)=a(x-x_{0})^{2}+b$. To find $a$,$x_{0}$ and $b$, you just require that like powers of $x$ in both sides of equation $a(x-x_{0})^{2}+b=kx^{2}+k'x$ have the same coefficients.