# Duplicate a plot - Figure 2.6 Griffiths QM page 55

## Homework Statement

Hello,
This is not a homework help problem but I believe this question belongs here more so than in the other forums.
I would like to (for personal benefit) duplicate / generate the plots in Figure 2.6 of Griffiths Quantum Mechanics book - page 55, 2nd edition.

## The Attempt at a Solution

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I apologize - I have no attempted solution (yet).

I do not want to buy any s/w if that is required.

I believe the big picture for the plots is to show solutions that "blow-up" for certain energy levels and not for the reader to generate the plots.

I see the x-axis is the dependent variable (not sure which symbol), is the y-axis psi? - plotting equation [2.72]?

What do I need to generate this plot on my own? Can it be done in Excel?

Thanks
Sparky_

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berkeman
Mentor
Can you post a (good quality) picture of a couple of the plots in that section?

DrClaude
Mentor
Can you post a (good quality) picture of a couple of the plots in that section?
Unfortunately, that won't be very helpful. You really need the book to understand what is being plotted in that figure.

What needs to be done is to solve the differential equation [2.72] by setting the value of $K$ to correspond to the desired $E$. The integration can be done by starting from $\xi=0$ with the initial conditions $\psi(\xi=0) = 1$ and $\psi'(\xi=0) = 0$ and integrating the solution both towards positive $\xi$ and negative $\xi$.

This can be done in Excel by setting up a column of $\xi$ from 0 to 4 with a step $\Delta \xi$ and two columns for $\psi$ and $\psi'$. The solution at $\xi + \Delta \xi$ can be obtained from the solution at $\xi$ using, e.g., the Euler method.

DrClaude
Mentor
This can be done in Excel by setting up a column of $\xi$ from 0 to 4
I forgot to say that the solution can also be calculated from 0 to -4, but since it is symmetric, one can simply use the same results to plot the negative $\xi$ points.

DrClaude and others,

is the solution to equation [2.72] not equation [2.77] or at least approximately eq [2.77]:

Ψ(ξ) = h(ξ)e2/2

I have been chasing a way to use this to generate the plot: Figure 2.6

The plot states the Energy E = .49 ħω or E = .51ħω

(and for E = 0.5ħω the plot behaves or rather doesn't go off to infinity)

I am thinking I could plug those values into equation [2.81] (coefficients of the power series) and somehow affix j and n - that I'm not sure about

and just generate the plot as if it were an simple algebraic equation

your thoughts? If I'm on a correct path - how do I pick a value for j and n for the upper limit?

Thanks
Sparky_