Physical Chemistry, P.W.Atkins, the Schrödinger equation

[Dj pie safety]

Homework Statement

Im doing an A-level project on the Schrödinger equation and am unsure on the mathematics used to obtain the following results:

The Schrödinger for a particle in no potential field (=0) has the solution:

psi(x)=e^ikx. i is defined below, I haven't really a clue as to how you get this result.

Looking at it I would guess that its a double integral to remove the second derivative of psi with respect to x, can anyone help me out here? I know how the rest of the equations goes after and using Eulers relationship to express it in tems of cos kx + i sin kx. You can also re-arrange k to show de Broglie's relationship.

any help is greatly appreciated

Homework Equations

The Attempt at a Solution

??

 

[dextercioby]

Do you know how to solve 2-nd order ODE-s ?

 

[Dj pie safety]

ordinary differential equations?

i can solve second order derivatives but only relatively simple ones. This is a bit above my head. Would you be able to point me in the right direction of what to start looking at to get the correct result?

Thanks for the reply so far.

Regards, Simon.

 

[Astronuc]

What dextercioby was getting at is reforming the differential equation into a workable or recognizable form.

e.g. take $$\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,=\,E\psi$$ and rewrite it as

$$\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,+\,E\psi = 0$$ and then


$$\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi = 0$$

and letting $$k^2\,=\,\frac{2mE}{\hbar^2}$$, see what that form looks like and what the possible solutions are.

 

[Dj pie safety]

Ok i think i have got this now:

$$\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi = 0$$

as you said: let:

$$k^2\,=\,\frac{2mE}{\hbar^2}$$

$$(1.) \frac{d^2\psi}{dx^2} = -k^2\psi$$

let:

$$\psi = e^mx$$

$$\frac{d\psi}{dx} = me^{mx}$$

$$\frac{d^2\psi}{dx^2} = m^2e^{mx}$$

Substitute into (1.):

$$m^2e^{mx} = -k^2e^{mx}$$

$$m^2 = -k^2$$

$$m = -k$$
if:
$$-k < 0$$
then:
$$m = ik$$
so:
$$\psi = e^{mx}$$
Therefore:
$$\psi = e^{ikx}$$

How does that look? it was a big effort but worth it :D

only thing I am not too sure about is the sign of k at the end of the process, + or -?

Thanks for the help so far!

Regards, Si

 

[Gokul43201]

There's an error here:

$$m^2 = -k^2$$

$$m = -k$$
if:
$$-k < 0$$
then:
$$m = ik$$

$$m^2 = -k^2 ~\implies m=ik$$
It doesn't matter if k>0 or k<0.