Harmonic Oscillator: Let a+,a- be the Ladder Operators

[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?

 

[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.