Quantum Harmonic Oscillator With a Linear "Perturbation"

Question

It is well known that the energy solutions for the unidimensional quantum harmonic oscillator $V(x) = \frac{1}{2}m\omega^2x^2$ are $E_n = (n + \frac{1}{2})\hbar\omega, n \in \mathbb{N}$. In particular, the energy of the ground state is $E_0 = \frac{1}{2}\hbar \omega$.

I am attempting to solve a different version of the problem. The objective is to determine the ground state energy considering a modified potential $V(x) = \frac{1}{2}m\omega^2x^2 + \lambda x$, i.e. there's an additional linear term on the potential, where $\lambda \in \mathbb{R}$. I tried to complete the square and execute a change of variables.

$$ V(x) = \frac{1}{2}m\omega^2 x^2 + \lambda x + \frac{\lambda^2}{2m\omega^2} - \frac{\lambda^2}{2m\omega^2} = \left(\sqrt{\frac{m}{2}}\omega x + \sqrt{\frac{1}{2m}} \frac{\lambda}{\omega}\right)^2 - \frac{\lambda^2}{2 m \omega^2} = y^2 - \frac{\lambda^2}{2 m \omega^2} = \frac{2}{m\omega^2}\left(\frac{1}{2}m\omega^2 y^2 - \frac{\lambda^2}{4}\right)$$

$$y \equiv \sqrt{\frac{m}{2}}\omega x + \sqrt{\frac{1}{2m}} \frac{\lambda}{\omega} \implies \frac{dy}{dx} = \sqrt{\frac{m}{2}}\omega$$

Now, Schrödinger's Equation tells us that

$$ \hat{H}\psi = -\frac{h²}{2m}\frac{d^2\psi}{dx²} + V(x)\psi = E\psi$$

However, since we've done a change of variables, I recall that

$$\frac{d\psi}{dx} = \frac{dy}{dx}\frac{d\psi}{dy} \implies \frac{d^2\psi}{dx²} = \left(\frac{dy}{dx}\right)^2\frac{d^2\psi}{dy^2} = \frac{1}{2}m\omega²\frac{d^2\psi}{dy^2}$$

given that $\frac{dy}{dx}$ is a constant, as we've found. Therefore,

$$ -\frac{\hbar²}{2m}\left(\frac{1}{2}m\omega^2\right) \frac{d^2\psi}{dy^2} + \frac{2}{m\omega^2}\left(\frac{1}{2}m\omega^2 y^2 - \frac{\lambda^2}{4}\right)\psi = E\psi, \quad (1)$$

from here I thought of maybe using the fact that for the ordinary quantum harmonic oscillator it holds (does it?) that

$$ -\frac{\hbar²}{2m}\frac{d² \psi}{dy²} + \frac{1}{2}m\omega² y² \psi = \frac{1}{2}\hbar \omega \psi$$

since we are interested in the ground state only, and then the desired energy would be this one with a constant difference. However, due to some constants in (1) that I wasn't able to get rid, I am not able to use this directly. From here, I have two questions:

1. Suppose I was able to identify on my expression the term $-\frac{\hbar²}{2m}\frac{d² \psi}{dy²} + \frac{1}{2}m\omega² y² \psi$. If $\psi$ represents the ground state of the system with the modified potential, would I really be able to express this term as $\frac{1}{2}\hbar \omega \psi$? That is, utilize the result from the original quantum harmonic oscillator, considering that $\psi$ is the ground state of the modified system and not the original one? If so, why?

2. I am not sure whether (1) leads me anywhere, since I wasn't able to use the result I desired, as explained. Am I missing something conceptual here or is it just a matter of math?

Answer 1

Your main idea is right: it is a matter of making a change of variables. But I think you made a change of variables that could be slightly improved.

If I didn't mess up any calculations (please double check my algebra), we could define $$z = x + \frac{\lambda}{m \omega^2}$$ to get to \begin{align} V(x) &= \frac{m \omega^2}{2} \left(x + \frac{\lambda}{m \omega^2}\right)^2 - \frac{\lambda}{2 m \omega^2}, \\ &= \frac{m \omega^2}{2} z^2 - \frac{\lambda}{2 m \omega^2}. \end{align}

The advantage of this substitution in comparison to yours is that $\frac{\mathrm{d}z}{\mathrm{d}x} = 1$ and the formulae are translates more easily: we just redefined the origin, without doing anything else. The Schrödinger equation reads now $$- \frac{\hbar^2}{2m} \frac{\mathrm{d}^2\psi}{\mathrm{d}z^2} + V(z) \psi = E \psi$$ with $$V(z) = \frac{m \omega^2}{2} z^2 - \frac{\lambda}{2 m \omega^2}.$$

Define $$V_0(z) = \frac{m \omega^2}{2} z^2.$$ We know the eigenenergies for $$- \frac{\hbar^2}{2m} \frac{\mathrm{d}^2\psi}{\mathrm{d}z^2} + V_0(z) \psi = E^0 \psi$$ are just $E^0_n = \hbar \omega\left(n + \frac{1}{2}\right)$, because the problem is just a regular harmonic oscillator.

The problem we want to solve is has the potential $V(z) = V_0(z) - \frac{\lambda}{2 m \omega^2}$. Now here's a different exercise for you to do:

Exercise: Let $E_n$ be the eigenergies for a potential $V(z)$. Then the eigenergies for the potential $V(z) + V_0$ are $E_n + V_0$.

Using this quick theorem (which is easy to prove) you conclude that the eigenergies you're looking for are $E_n = \hbar \omega\left(n + \frac{1}{2}\right) - \frac{\lambda}{2 m \omega^2}$. Notice this makes sense: the potential is just a harmonic oscillator with an overall energy shift.

Here's the summary: you change the origin of the spatial axis so it is aligned with the minimum of the potential. When you do so, you notice the minimum of the potential is not zero, but a bit lower. The eigenergies are just the ones you'd have for the regular harmonic oscillator, but now a bit lower because the minimum of the potential is not zero.