# I Feynman Lectures: negative alpha for solving Schrödinger equation

1. Jul 16, 2016

### forcefield

2. Jul 16, 2016

### Simon Bridge

Dorsnt eq19.14 say that alpha is positive?

3. Jul 17, 2016

### forcefield

How ? I don't see it.

4. Jul 18, 2016

### Simon Bridge

Isn't that where alpha is introduced?
Physicists seldome define new variables explicitly the way mathematitians do, relying on context.
If alpha, in that equation, could take on negative values, then it does not nake sence to write $-\alpha$ in that equation does it?
But you can check to see if it makes sense: what role does alpha play in that equation? What properties does the equation need to have in order to work as intended? If alpha were negative, would the equation have the needed properties?

Mind you, I could have misread.
What I want to draw your attention to is this concept of an implicit definition.

5. Jul 18, 2016

### forcefield

Yes.
That's right but what if it's a mistake and $\alpha$ could take negative values ?
I don't see why $\alpha$ couldn't be negative in that context.

I'm just not getting the point that there can't be bound state solutions with negative $\alpha$ from that text.

6. Jul 18, 2016

### Simon Bridge

Why does Feynman bother to introduce eqn 19.14 in the first place? What is it supposed to do?

7. Jul 18, 2016

### forcefield

It's just a mathematical trick that happens to be useful.

8. Jul 18, 2016

### Simon Bridge

Yes but what is the nature of the trick?

9. Jul 18, 2016

### forcefield

That is not helpful. I get the feeling that you know my mistake but are not telling it, right ?

10. Jul 18, 2016

### Simon Bridge

The idea is to get you to think about what you have read.
Why b other with that particular equation in the first place? Why not some other useful trick: there are lots?

11. Jul 18, 2016

### forcefield

12. Jul 18, 2016

### Simon Bridge

13. Jul 19, 2016

### Demystifier

The wave function is proportional to $e^{-\alpha\rho}$. If $\alpha$ were negative, then wave function would be infinite for $\rho\rightarrow\infty$. Such a wave function could not be normalized to unity, so could not be interpreted as probability density amplitude. Therefore such a wave function would not be physical, which is why the case of negative $\alpha$ is discarded.

14. Jul 19, 2016

### forcefield

No because it gets coupled with $e^{+2\alpha\rho}$

15. Jul 20, 2016

### Demystifier

Read the next paragraph of the Feynman lectures! As explained there, the above is true for most values of $\alpha$, but not for all values. For some special values (more precisely, for $\alpha=1/n$ where $n$ is any positive integer), instead of $e^{+2\alpha\rho}$ one obtains a polinomial in $\rho$. The product of polinomial and $e^{-\alpha\rho}$ vanishes at infinity. This means that only those special values of $\alpha$ are physical. Indeed, this explains the quantization of the hydrogen spectrum, i.e. the fact that energy can only take some special values. See equations (19.24)-(19.30).

This shows that positive $\alpha$ is OK. But what is wrong with negative $\alpha$? The function $e^{+2\alpha\rho}$ is only an approximative sum of the infinite series. The approximation is good for large $\rho$, provided that $\alpha$ is positive. For negative $\alpha$, the series cannot be approximated by $e^{+2\alpha\rho}$. (In this case the series is alternating, which makes it very problematic to deal with.) Therefore, for negative $\alpha$, it is incorrect to argue that the wave function behaves well at infinity because $e^{+2\alpha\rho}$ behaves so.

Last edited: Jul 20, 2016
16. Jul 20, 2016

Thanks!