I Error of the WKB approximation

[phyQu]

TL;DR Summary

How to calculate the error made using the WKB approximation?

hello everyone

I tell you a little about my situation.
I already found the approximate wavefunctions for the schrodinger equation with the potential $V(x) = x^2$, likewise, energy, etc.
I have the approximate WKB solution and also the exact numeric solution.

What I need to do is to calculate the error of the approximate solution with respect to the exact solution,
but I need help to start, because I am lost with this topic.

Will it be possible to make an error analysis?
Will it be possible apply that is defined as $e = |y_{exact} - y_{approx}|$?

Do you know if someone has already worked out the error for the WKB approximation?
Do you know of any recent article dealing with this?
I've searched the internet and I can't find anything that tells me specifically about the error of this method, please, can someone help me.

Thanks in advance

 

[DrClaude]

I am not aware of any previous results concerning the error of the WKB approximation. Since the quality of the approximation depends on the variation of the potential with respect to the oscillation (wave number) of the wave function, I would be surprised if there was a general answer to this.

I would calculate the relative error on the eigenenergy as well as the L2 norm of the difference in wave functions,
$$
\int_{-\infty}^\infty \left| \psi_\textrm{exact}(x) - \psi_\textrm{approx}(x) \right|^2 \, dx
$$
However, this might be too much influenced by the complex phase, which is not physically relevant, so I would also calculate the difference in probability
$$
\int_{-\infty}^\infty \left| \psi_\textrm{exact}(x)\right|^2 - \left|\psi_\textrm{approx}(x) \right|^2 \, dx
$$

 

[phyQu]

Thank you very much, by the way, do you know a book that has this topic you are talking about? I would help me to reference my work.

 

[vanhees71]

The latter integral is not a good measure for the "distance of states", because the overall factor is just determined by the normalization condition, i.e., that integral should be 0 when normalizing both wave functions properly.

 

[gentzen]

However, this might be too much influenced by the complex phase, which is not physically relevant, so I would also calculate the difference in probability
$$
\int_{-\infty}^\infty \left| \psi_\textrm{exact}(x)\right|^2 - \left|\psi_\textrm{approx}(x) \right|^2 \, dx
$$

The latter integral is not a good measure for the "distance of states", because the overall factor is just determined by the normalization condition, i.e., that integral should be 0 when normalizing both wave functions properly.

Well, you could integrate over the absolute value of the differences:
$$
\int_{-\infty}^\infty \left|\left| \psi_\textrm{exact}(x)\right|^2 - \left|\psi_\textrm{approx}(x) \right|^2\right| \, dx
$$

 

[DrClaude]

The latter integral is not a good measure for the "distance of states", because the overall factor is just determined by the normalization condition, i.e., that integral should be 0 when normalizing both wave functions properly.

Obviously . I should've thought a bit more about it. I was after a measure of the difference in local probability distribution. Maybe something like
$$
\max_x \left( \left| \psi_\textrm{exact}(x)\right|^2 - \left|\psi_\textrm{approx}(x) \right|^2 \right)
$$