Quantum Harmonic Oscillator with Additional Potential

Mr_Allod
Messages
39
Reaction score
16
Homework Statement
Apply and additional potential ##V'(x) = \alpha x## to the standard Hamiltonian of a harmonic oscillator. Find the solutions to the Schrodinger equation.
Relevant Equations
Harmonic Oscillator Hamiltonian: ##H = - \frac \hbar {2m} \frac {d^2}{dx^2}\psi + \frac {m\omega^2 x^2}{2}##
Hello there, I am trying to solve the above and I'm thinking that the solutions will be Hermite polynomials multiplied by a decaying exponential, much like the standard harmonic oscillator problem. The new Hamiltonian would be like so:

$$H = - \frac \hbar {2m} \frac {d^2}{dx^2}\psi + \frac {m\omega^2 x^2}{2} + \alpha x$$

Normally one would make the substitutions:

$$y = \sqrt{\frac {m\omega}{\hbar}}$$
$$\epsilon = \frac {2E}{\hbar\omega}$$

This would produce a solvable dimensionless differential equation:
$$\frac {d^2}{dy^2}\psi + (\epsilon-y^2)\psi = 0$$

Now I'm having trouble finding the correct substitution to make to reduce the new problem to a dimensionless one like above. I would appreciate it if someone could give me some suggestions, thank you!
 
Physics news on Phys.org
Rather think about the Hamiltonian a bit first.

Hint: Think about the classical case. What's the change of the potential due to the additional linear piece? Equivalently you can think about, what it changes for the force acting on the particle and how does it affect the solutions of the classical equations of motion?
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top