# I What does a zero-value of the Born-condition mean?

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1. Jan 3, 2018

### blue_leaf77

Sorry to barge in, but your wavefunction model is proportional to $\cosh (k_1x)$, which is not square integrable for $x\in (-\infty,\infty)$. Moreover, the wavefunction is also proportional to an oscillatory function whose wave number is extremely large $~10^{-28}/10^{-34} \approx 10^6$. It can be the reason why MATLAB returns zero. Have you tried executing "format long" command to extend the decimal places in the output ?

2. Jan 3, 2018

### Demystifier

His wave function probably needs to be defined in a box of finite length $L$, as suggested in post #9.

3. Jan 3, 2018

### Demystifier

One of the first rules in any numerical computation (with MATLAB or anything else) is to first introduce new dimensionless variables (which are certain ratios of the dimensional ones) such that all relevant parameters are of the order of unity. Very big and very small numbers are always potentially dangerous in numerical computations.

4. Jan 3, 2018

### SeM

Thanks! Can you possibly scan the page or mention any of the subttiles of the page or preceeding pages, so I may look for it in the book?

I will recompute the integral tomorrow and see if it gives another value in the meantime.

Cheers

5. Jan 3, 2018

### SeM

Defining it in a box of length L, implies that I do this:

Psi(x) = 0 , L=0

Psi(x) = 0 , L=L

and I add a normalization constant N^2 to the Born intergral, and then use L=L and L=0 as the limits?

I then solve for this integral, but x will still be there...?

6. Jan 3, 2018

### Demystifier

Search for the section The harmonic oscillator - The solutions.

7. Jan 3, 2018

### Demystifier

You mean x=0 and x=L. When you integrate over x from 0 to L, then x is no longer there.

8. Jan 3, 2018

### SeM

Indeed! Sorry for that mistake. I will try this tomorrow,

All the best

9. Jan 4, 2018

### SeM

Dear Demystifier, I have gone through the Further information on chap. 2.2 Harmonic Oscillator, and it appears as that I have to solve the Schrödinger eqn using the annhalation and creation operators, instead of solving it numerically, as I did and got that strange wavefunction.

Thanks for your tip. This becomes an entirely new project, because it's currently huge.

Thanks!

10. Jan 5, 2018

### Demystifier

Well, if you want to learn quantum chemistry, you cannot avoid learning more about harmonic oscillator.

11. Jan 5, 2018

### SeM

I look forward to it.