Consider a one dimensional oscillator in the ground state. Write the Hamiltonian for this oscillator. Assuming that there is a finite perturbation $V = \alpha*x$ acting on this oscillator, find the energy shift for the ground state. The parameter alpha is given and it is small. Discuss the applicability of the obtained result.

For the one dimensional oscillator the Hamiltonian is simply

$\hat H =\frac{\hat p}{2m} =-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}$

Satisfied by the wave functions $\psi(x)=A*sin(k x)$ so that

$-\frac{hbar^2}{2m}*\frac{d^2 \psi}{d x^2}= E\psi$

If we add a perturbation $V(x)= \alpha*x$ to the initial Hamiltonian $H$ in the first approximation the wave functions will be the same, but satisfying now the new equation

$-\frac{hbar^2}{2m}*\frac{d^2 \psi}{d x^2}+V(x)*\psi= E’\psi$

So that the shift in energy will be:

$E’-E=<\psi|V|\psi>=\alpha\int\psi x \psi^* d x= \alpha A^2*\int x (sin(k x))^2*d x$

If the linear dimension of the oscillator is L, for the ground state we have:

$A=\sqrt{2/L}$ and $k=\pi/L$ so that

$E’-E = \frac{2\alpha}{L}\frac{1}{8k^2}[2k^2x^2-cos(2kx)-2kx*six(2kx)]|_0^L =\frac{\alpha L}{4\pi^2}[2\pi^2-1+1] =\frac{\alpha L}{2}$

Which means that in the first approximation for a triangular well obtained from a square well the energy levels shift with half the energy of the perturbation at the well’s end.

Discussion:

In reality the wave function is also modified with a small correction from its ground state form. Therefore the energy shift computed above need to be corrected again with higher approximation terms.

The perturbation potential $V(x) =\alpha x$ is useful to be applied when obtaining the energy levels (and wave functions) of a triangular well from the energy levels of an infinite well.