# New Ground State Energy due to the change of Hamiltonian

Generally We know the potential of a Harmonic oscillator can be written as $V= \frac{1}{2} m \omega^2 x^2$. Whenever we get any changed to the potential, we can find the shifted energy. What if we write the potential as $V= \frac{1}{2} m \omega^2 (x-x_0)^2$ where the oscillator got a new equilibrium position but a shifted ground state energy. Now I want to find that energy using Perturbation or Variational method. I'm confused about that how the $x_o$ will act with the trial wave function. Can you give me insight how to find the shifted energy in this case?

• If you put $y = x - x_0$ then it's clear that the new wavefunction is found by replacing $x$ in the old wavefunction by $y$. To demonstrate the variational method, you need to replace $x$ by $\widetilde{y} = x - u$ and use $u$ as your variational parameter, you'll then find that minimization of the expected energy yields $u = x_0$. – Count Iblis Jan 21 '17 at 1:07
• so you are saying that x will be unchanged and the Hamiltonian will be applied on the variable u? If I'm wrong cud you please write the equation down.? :) – Numerical Person Jan 21 '17 at 1:25