Schrodinger Equation/verify solution

[kwagz]

Homework Statement

Trying to use change of variables to simplify the schrodinger equation. I'm clearly going wrong somewhere, but can't see where.

Homework Equations

[/B]
Radial Schrodinger:

-((hbar)²)/2M * [(1/r)(rψ)'' - l(l+1)/(r^2) ψ] - α(hbar)c/r ψ = Eψ

The Attempt at a Solution

We're first told to replace rψ(r) with U(r/a). For this I got the following:

-((hbar)²)/2M * [(1/r)(d/dr)²(U(r/a)) - l(l+1)/(r³) *U(r/a)] - α(hbar)c/r² *U(r/a) = (E/r)*U(r/a)

The next step is to use x=r/a to change variables to x. a=hbar/α*M*c This leads me to:

-((hbar)²)/2M * [(1/xa)(d/dxa)²(U(x)) - l(l+1)/(xa)³) *U(x)] - α(hbar)c/(x*a²) *U(x) = (E/xa)*U(x)

Then we replace E by ε=-2E/(α² *M*c²). This gives the final form (after some simplifying):

(d/d(ax))²)U(x)=U(x)(ε/a² + l(l+1)/(xa)² -2/x*a²)Then we're to check that (x²)*e^(-(x2)) is a solution to the equation.

Plugging that in gives

(d/d(ax))²)(x²)*e^-(x2)=(x²)*e^-(x2)(ε/a² + l(l+1)/(xa)² -2/x*a²)

After taking the second derivative (which I got as (x⁴ -5x² +2)*e^-(x2))/a²), I ended up with:

e^-(x2)(x⁴ -5x²+2)=e^-(x2)(εx² + l(l+1) -2x)

I'm pretty sure this means I went wrong somewhere, as I think I should have an equivalent expression on the left and right. If anyone can see where I might have made a mistake, it'd be very helpful.

 

[Simon Bridge]

Hmm... best format equations using LaTeX - this is very difficult to read.

You started from:
$$-\frac{\hbar^2}{2m}\left[ \frac{1}{r} \frac{d^2}{dr^2}\left(r\psi \right) - \frac{l(l+1)}{r^2}\psi \right] -\frac{\alpha\hbar c}{r}\psi = E\psi$$
... and you followed instructions to substitute $U(r/a)=r\psi$ ... the r/a part tells you to measure r in units of a.
This means that: $\frac{d}{dr}U(r/a) = \frac{1}{a}\frac{d}{dx}U(x) : x=\frac{r}{a}$ ... a can be picked to be some convenient unit: which seems to have been chosen to cancel out a lot of messy constants.

I'm having a hard time following the working: but you seem to be following instructions without understanding them. Have you seen: https://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential

You need to check that $U(x)=x^2e^{-x^2}$ is a solution ... why not substitute back $x=kr$ (k=1/a=constant) and see if this is a solution to the original equation before you did all that substituting and rearranging? (recall: $\psi(r)=\frac{1}{r}U(r/a)$)

 

[kwagz]

Ah sorry, didn't know how to Latex. I fixed it up below. Hopefully it's a bit easier to follow now. I added numbers in front of the steps, those are ones that were stated for us to follow in the homework. I'll try verifying the solution before the substitutions, and if that's works out, then it would confirm that I did something mathematically wrong in the substitution/rearranging.

Radial Schrodinger:

$-\frac{\hbar^2}{2M} [\frac{1}{r}(rψ)'' - \frac{l(l+1)}{r^2} ψ] - \frac{α\hbar c}{r} ψ = Eψ$

The Attempt at a Solution

1) We're first told to replace rψ(r) with $U(\frac{r}{a})$. For this I got the following:

$-\frac{\hbar^2}{2M}[(\frac{1}{r}\frac{d^2}{dr^2}U(\frac{r}{a}) - \frac{l(l+1)}{r^3}U(\frac{r}{a})] - \frac{α\hbar c}{r^2}U(\frac{r}{a}) = \frac{E}{r}U(\frac{r}{a})$

2) The next step is to use $x=\frac{r}{a}$ to change variables to x. $a=\frac{\hbar}{αMc}$ This leads me to:

$-\frac{\hbar^2}{2M}[(\frac{1}{xa}\frac{d^2}{d(xa)^2}U(x) - \frac{l(l+1)}{(xa)^3}U(x)] - \frac{α\hbar c}{(xa)^2}U(x) = \frac{E}{xa}U(x)$

3) Then we replace E by $ε=-\frac{2E}{α^2 Mc^2}$. This got me the final form (after some simplifying):

$\frac{d^2}{d(ax)^2}U(x)=U(x)[\frac{ε}{a^2} + \frac{l(l+1)}{(xa)^2} -\frac{2}{xa^2}]$4) Then we're to check that $x^2 e^\frac{-x^2}{2}$ is a solution to the equation.

Plugging that in gives

$\frac{d^2}{d(ax)^2}x^2 e^\frac{-x^2}{2}=x^2 e^\frac{-x^2}{2}[\frac{ε}{a^2} + \frac{l(l+1)}{(xa)^2} -\frac{2}{xa^2}]$

After taking the second derivative (which I got as $\frac{(x^4 -5x^2 +2)e^\frac{-x^2}{2} }{a^2}$), I ended up with:

$e^\frac{-x^2}{2}(x^4 -5x^2 +2)=e^\frac{-x^2}{2}(εx^2 + l(l+1) -2x)$

I suppose the "-2x" term could be zero, which would mean it's a spherical shell in ground state (particle free to move inside)? But I think that would be contradicted by the fact that "l" can be equal to nonzero integers here, so it shouldn't be just ground state?