A particle of mass m moves on the x-axis under the influence of the potential $$V(x)=\frac{1}{2}m\omega^2x^2+Fx$$ Can anyone help me, using Schrödinger's equation in one dimension that the energy levels are: $$E_n=(n+\frac{1}{2})h\omega-\frac{1}{2}\frac{F^2}{m\omega^2}$$ Where n is a non negative integer?
2 Answers
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Try a change of coordinates $x\rightarrow x-x_0$, where $x_0$ is an appropriate constant.
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$V(x)=\frac{1}{2}m\omega^2x^2+Fx=\frac{1}{2}m\omega^2(x^2+\frac{2F}{m\omega^2}x)=\frac{1}{2}m\omega^2((x+\frac{2F}{m\omega^2})^2-\frac{F^2}{m^2\omega^4})=\frac{1}{2}m\omega^2x'^2-\frac{1}{2}\frac{F^2}{m\omega^2}$ and you have, that it is potential of the oscillator minus constant. So energy levels offset by this constant.