Vibrating Nuclei_potential energy

[terp.asessed]

Homework Statement

Calculate <K>, the expectation value of the kinetic energy for the ground state of a pair of vibrating nuclei (assume internuclear distance--hence -infinite to +infinite)

Homework Equations

K = -h²/2(mu) d²/dx²

where (mu) = reduced mass; m₁m₂/(m₁+m₂)
and
wave(x) = (a/pi)^1/4 e^(-ax2/2)
where a = (mu) (omega)/h

(Assume h = Planck's constant with dash)

The Attempt at a Solution

<K> = integral (x = -infinite to infinite) wave²(x) K(x) dx
= integral (a/pi)^1/2 e^(-ax2) -h²/2(mu) d²/dx² dx
= -ah²/(2pi(mu)) integral e^(-ax2) -h²/2(mu) d²/dx² dx
= -a^5/4h²/(2pi^5/4(mu)) integral {-ae^-3ax2/2 + a²x²e^(-3ax2/2)}dx
...beyond, I don't know how to simplify the integral--have I done something wrong in the midway? I don't know how to integrate anymore. Any suggestions would be welcome!

 

[Simon Bridge]

You'll notice that the operator K does not make sense without some function in front of it. $$\langle K \rangle = \int_{-\infty}^\infty \psi^\star(x)\hat K \psi(x)\; dx$$... would be more usual.

 

[terp.asessed]

That's what I did--wave(x) is a psi (x)...
So, that's why I did psi²(x) which is wave²(x), and I still have trouble getting to the solution...

Plus, what's with asterisk on one psi? Could you pls explain? Thanks!

 

[BvU]

Asterisk = complex conjugate

And ${d\over dx}e^{-ax^2}$ is not something with $e^{-2ax^2}$ !

 

[terp.asessed]

Is there a way to solve integral {-ae^-3ax2/2 + a²x²e^(-3ax2/2)}dx?
(where integral is from x = -infinite to + infinite)​

 

[BvU]

It looks suspiciously identical to the Schroedinger equation for the harmonic oscillator, only (if you do it right) a factor 1/2 different...

 

[terp.asessed]

Could you please explain what you mean by:

a factor 1/2 different...

 

[BvU]

work out $\left ({a\over \pi}\right )^{1\over 2}\;\int_{-\infty}^{+\infty}\;\left (e^{-ax^2/2}\right )\; \left (-{h^2\over 2\mu}{d^2\over dx^2}\right ) e^{-ax^2/2}\ dx \$ first and see what you get.

 

[Simon Bridge]

That's what I did--wave(x) is a psi (x)...

I got that, it's the next bit I wanted to direct you attention to.
BvU is doing a decent job so I'll just reinforce and sit back for a bit:

So, that's why I did psi2(x) which is wave2(x),...

Which is not correct.
The $\hat K$ operator has a second derivative in it - that means the $\psi$ to the right of it gets altered.
You should have some course notes about how to apply operators to wavefunctions.
$$\hat K \psi(x) = -\frac{\hbar^2}{2\mu}\left(\frac{d^2}{dx^2}\psi\right)$$

Plus, what's with asterisk on one psi? Could you pls explain?

The star refers to the complex conjugate.
If $\psi = a+ib$ then $\psi^\star = a-ib$ (a,b both real and i²=-1)
In your case, $\psi$ is real - but what I wrote is the formal definition and it's a step you missed out and it is important to your thinking.

Aside: you are also reaching the stage where you will need to learn to use LaTeX ;)