Turning points

1. Nov 27, 2006

Logarythmic

How can I find the turning points for the one dimensional Morse potential

$$V(x) = D(e^{-2ax}-2e^{-ax})$$

??

2. Nov 27, 2006

HallsofIvy

Staff Emeritus
It would be a good start to state the DEFINITION of "turning point"!

3. Nov 27, 2006

Logarythmic

That's probably my problem then, what is the definition of "turning point"?

4. Nov 27, 2006

dextercioby

Can you graph the Morse potential ?

Daniel.

5. Nov 27, 2006

Logarythmic

Yes I have graphed the potential.

6. Nov 27, 2006

Logarythmic

Should I invert the function and find the min and max for x(V) ?

7. Nov 27, 2006

dextercioby

Turning points are related to the "classically forbidden regions", they are boundaries for these regions. It's easier to see if the problem is unidimensional and you can graph the potential.

Daniel.

8. Nov 27, 2006

Logarythmic

Yes, but how do I determine the turning points?

9. Nov 27, 2006

dextercioby

By solving an impossible equation. Joking, just solve

$$V(x)=E$$

.The eqn is not impossible. It can be brought to an algebraic one, a quadratic one, even.

Daniel.

10. Nov 27, 2006

Logarythmic

But isn't
$$E = \frac{1}{2}m \dot{x}^2 + V(x)$$?

11. Nov 27, 2006

dextercioby

Nope and yes. I assumed you wish to find the classical turning points of the Morse potential and for that you need to solve the eqn i wrote. At these turning points the classical KE is zero.

Daniel.

12. Nov 27, 2006

Logarythmic

Yes of course. But then I've only got V(x) = V(x) ??

13. Nov 27, 2006

dextercioby

Nope, V(x) is equal to E, the total, given, energy. The E is the a generic notation for the the variable used to index the spectrum of the quantum Hamiltonian.

Daniel.

14. Nov 27, 2006

Logarythmic

Ok, but I'm studying classical mechanics so I think I have to use another approach...

15. Nov 27, 2006

dextercioby

Alright, then, find the solution from other source and compare to what i said and suggested.

Daniel

16. Nov 27, 2006

I'm not aware of another approach. You start off with a certain amount of energy, which is a constant in a conserved system. But the classical turning points are when $$\dot{x} = 0[\tex], and then a moment later the velocity changes signs (i.e. the particle goes from going to the right to going to the left), so then you do what Dexter suggests, and hopefully understand why you're doing it. 17. Nov 27, 2006 Logarythmic I understand this now. I get the equation [tex]D(e^{-2ax}-2e^{-ax})-E=0$$

Any tricks on how to solve this?

18. Nov 28, 2006

dextercioby

Yes, i already did tell you b4. Just substitute $e^{-ax} =t$ and then c what you get.

Daniel.