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I am trying to rewrite a Schrödinger equation using dimensionless quantities but here, the potential is perturbed by $\lambda x^4$:

\begin{equation}V(x) = \frac{m\omega^2}{2}x^2 + \lambda x^4\end{equation}

$\lambda$ and $\omega$ are parameters of the system.

The Schrödinger equation reads then: \begin{equation}-\frac{\hbar}{2m}\psi''(x) + \frac{m\omega^2}{2}x^2 \psi(x) + \lambda x^4 \psi(x) = E \psi(x)\end{equation}

If you try to do the same with the unperturbed version, what one would do is, to find a length scale by combining $\hbar$, $\omega$ and $m$ such that their units cancel to length $\big($ this would be $a = \sqrt{\frac{\hbar}{m\omega}}\big)$.

How can I do the same here?

$[\hbar] = \frac{kg m^2}{s}, [m] = kg , [\omega] = \frac{1}{s}, [\lambda] = \frac{kg}{s^2m^6}$ $\big($Units for $\lambda$ are chosen like this, otherwise it wouldn't cancel with the $x^4$ to units of $\frac{kgm^2}{s^2}$$\big)$.

Now trying to get a length scale: \begin{equation}M^{\alpha}L^{2\alpha}T^{-\alpha} \,\,\, M^{\beta}\,\,\, T^{-\gamma} \,\,\, M^{\delta}T^{-2\delta}L^{-6\delta} = L\end{equation}

This leads to: \begin{equation}\alpha + \beta + \delta = 0\end{equation} \begin{equation}2\alpha - 6\delta = 1\end{equation} \begin{equation}-\alpha - \gamma - 2\delta = 0\end{equation}

These are 3 equations with 4 unknowns, meaning one of the $\alpha,\beta,\gamma,\delta$ is freely choosable. Does it make sense that their are different length scales and what difference does it make whatever I choose?

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    $\begingroup$ Why do you want to choose your length scale to depend on the size of th perturbation $\lambda$? $\endgroup$ – ZeroTheHero May 20 '18 at 16:57
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You already have perfectly good characteristic scales for the problem. If you non-dimensionalize the rest of the equation as you did for the unperturbed oscillator, you'll end up with a dimensionless quantity $\epsilon \propto \lambda$ multiplying the (now dimensionless) $x^4$ term. The size of $\epsilon$ determines whether the perturbation can be considered small.

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