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

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1. Jan 3, 2018

### SeM

Hi, I was wondering what is the physical meaning of these integrals:

\int_{-\infty}^{\infty} \psi \psi^* dx = 0

\int_{-\infty}^{\infty} \psi x \psi^* dx = 0

\int_{-\infty}^{\infty} \psi p \psi^* dx = NA

?

Last edited by a moderator: Jan 3, 2018
2. Jan 3, 2018

### Demystifier

Where did you find this expression?

3. Jan 3, 2018

### SeM

By a function I have been calculating on

4. Jan 3, 2018

### Demystifier

What is $t$? Time? 3-volume?

5. Jan 3, 2018

### SeM

Sorry, I should have used x. Corrected.

6. Jan 3, 2018

### Demystifier

(1) implies $\psi=0$.
(2) means that average position is zero.
(3) means that the average momentum is not well defined.

7. Jan 3, 2018

### SeM

This part is not clear to me. Evidently, Psi and Psi* are hermitian counterparts, so their multiplication with one another ends up with zero.

This is OK. But does that mean the wavefunction has not physical application?

This is fine. Does normalization solve this?

8. Jan 3, 2018

### Demystifier

No. For properly normalized wave functions you should have $\int \psi^*\psi dx=1$.

No, such wave functions have a lot of applications. For instance, the ground state of harmonic oscillator has zero average position.

Probably not. Consider, for example, the delta-function $\delta(x)$. It has undefined average momentum and change of normalization does not help.

9. Jan 3, 2018

### SeM

Is there any chance of solving this constant by setting the integral:

N^2 \int_0^L \psi \psi^* dx = 1

and solve for N?

I tried and got complex conjugates, in fact, not only one, but two.

Thanks for this interesting point! Do you know of a paper that reports such a wavefunction?

What does one do with such wavefunctions without a well-defined momentum and kinetic energy?

10. Jan 3, 2018

### Demystifier

Yes, provided that N is assumed to be real. (Otherwise, you should write $N^*N$ instead of $N^2$.)

Take into account that N is assumed to be real.

Any textbook on quantum mechanics (QM) should do. Which book on QM do you use?

Such a $\delta$-function is an idealization of a more realistic case of a very narrow Gaussian. Even though the idealization is not realistic, it is OK as long as you only consider positions and not momenta or energy.

11. Jan 3, 2018

### SeM

Tried and got two weird values.

\psi(x)= \frac{e^{-x(k_2-k_1)}\big(\sqrt{E-2\gamma^2}+\gamma i\big)}{2\sqrt{E-2\gamma^2}}-\frac{e^{-x(k_1+k_2)}\big(-\sqrt{E-2\gamma^2}+\gamma i\big)}{2\sqrt{E-2\gamma^2}}

Molecular Quantum Mechanics, there they normalize it, and get nice eigenfunctions. This function however, gave two horrible Normalization constants, of the length of half a page, and cannot really give a worse starting point for generating higher energy levels.

12. Jan 3, 2018

### Demystifier

In your wave function I see only one (not two) implicit values of N, and I see nothing strange with it.

13. Jan 3, 2018

### SeM

This function is without N. I had two initial conditions for the original ODE, first y(0)=1 and y'(0)=0 , and got this function. This function follows the integrals given in the first post.

Therefore I was looking for adding a normalization constant to the integrals.

14. Jan 3, 2018

### Demystifier

This function does not satisfy your Eq. (1).

15. Jan 3, 2018

### Demystifier

16. Jan 3, 2018

### SeM

Atkins and Friedman

17. Jan 3, 2018

### SeM

you need k_1 and k_2, these are:

k_1 = {\frac{\,\sqrt{E-2\,\gamma^2}}{\hbar}}

k_2 = \frac{\gamma i}{\hbar}

$\gamma = 5^{-28}$ and E is the zero point energy. If you still get that integral 1 is not satisfied, then MATLAB is crap

18. Jan 3, 2018

### Demystifier

Are you a chemist? I ask because it looks like you miss the basic foundations of QM. In any case, study first Chapters 1 and 2 of the book. For instance, in Fig. 2.27 (fourth edition) all wave functions have zero average position.

19. Jan 3, 2018

### SeM

Yes, but this model has not potential energy. and Yes I am a chemist.

What can I do to develop this model further as it does not follow the Hamiltonian on p 55, chapter 2, where fig 2.27 is.

Thanks!

20. Jan 3, 2018

### Demystifier

We must be using different editions of the book. Mine is the 4th edition. In any case, try to compute $\int \psi^*\psi dx$ again and show that it is a positive real number. You cannot make any progress before you do that.

21. 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 ?

22. Jan 3, 2018

### Demystifier

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

23. 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.

24. 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

25. 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...?