Finding Turning Points of Morse Potential V(x)

[Logarythmic]

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

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

??

 

[HallsofIvy]

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

 

[Logarythmic]

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

 

[dextercioby]

Can you graph the Morse potential ?

Daniel.

 

[Logarythmic]

Yes I have graphed the potential.

 

[Logarythmic]

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

 

[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.

 

[Logarythmic]

Yes, but how do I determine the turning points?

 

[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.

 

[Logarythmic]

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

 

[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.

 

[Logarythmic]

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

 

[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.

 

[Logarythmic]

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

 

[dextercioby]

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

Daniel

 

[StatMechGuy]

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

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 $$\dot{x} = 0[\tex], and then a moment later the velocity changes signs (i.e. the particle goes from going to the right to going to the left), so then you do what Dexter suggests, and hopefully understand why you're doing it.$$

 

[Logarythmic]

I understand this now. I get the equation

$$D(e^{-2ax}-2e^{-ax})-E=0$$

Any tricks on how to solve this?

 

[dextercioby]

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

Daniel.