TISE questions regarding solutions and energy

[weguihapzi]

Homework Statement

a) For a region, the potential energy experienced by a particle is:

\frac{-e^2}{4\pi \varepsilon _{0}x}

where e is electric charge.For this region, by substituting into the Schrodinger equation, show the wave function:

u(x)=Cxe^{-\alpha x}

can be a satisfactory solution. And determine the unique expression for α in terms of m, e, and other fundamental constants.b) Show the energy of the particle represented by u(x) is given by:

E=\frac{-me^4}{2(4\pi \varepsilon _{0}hbar)^2}

Homework Equations

TISE for this potential:

\frac{-hbar^2}{2m}\frac{d^2\Psi }{dx^2}-\frac{e^2}{4\pi \varepsilon _{0}x}\Psi = E\Psi

The Attempt at a Solution

It's been a long time since I've done any quantum exercises, and just trying to get myself back into it, so please bear with me if this seems like total nonsense.

For part a) so far, I've tried differentiating the given wave function twice, and plugging that in the TISE, along with the undifferentiated wavefunctions.

I've then tried rearranging this equation to get in terms of α, but end up E (not sure how to get rid of this) with other non-fundamental constants in the answer, so I was wondering if I'm on the right track, or whether I'm way off.As for part b), I'm really not sure. Is that simply a case of rearranging the equation to get E? Or would I need to use an operator to get this?

Again, I can only apologise for my (very) rusty knowledge, so any help is very much appreciated! Thank you.

 

[TSny]

Hello, weguihapzi. Welcome to PF!

You have the right idea. Note that the right hand side of the SE is just a multiple ψ. So, the left hand side is going to have to reduce to a multiple of ψ. In general, you will find that this is impossible unless α is chosen to have a certain value that makes some terms cancel out on the left side.

 

[weguihapzi]

Ok thank you, that helps to get my head around it a little more.

I'm still a little unsure of how I can get rid of the E term from the RHS? Can I replace that with something?

 

[TSny]

You don't want to get rid of the E term. You want to get rid of some terms on the left side so that when you combine the remaining terms on the left, you get a constant times ψ.

 

[weguihapzi]

Ah right ok, so would this be right?

Differentiate wave function twice and plug into Schrodinger to get:

\frac{-hbar^2}{2m}\alpha ^2Cxe^{-\alpha x}-\frac{e^2Cxe^{-\alpha x}}{4\pi \varepsilon _{0}x}=ECxe^{-\alpha x}

Reduce to:

\frac{-hbar^2}{2m}\alpha^2 =E\frac{e^2}{4\pi \varepsilon _{0}x}

The question says this potential is for x greater than or equal to 0, so if I ignore x (which I'm almost certain I can't do):

\alpha=\sqrt{}\frac{2me^2}{4\pi \varepsilon _{0}hbar^2}

Can you please point out where I've gone wrong? Thank you.

 

[weguihapzi]

Sorry, that second equation should be E + e^2/4pie0x

 

[Dick]

Ah right ok, so would this be right?

Differentiate wave function twice and plug into Schrodinger to get:

\frac{-hbar^2}{2m}\alpha ^2Cxe^{-\alpha x}-\frac{e^2Cxe^{-\alpha x}}{4\pi \varepsilon _{0}x}=ECxe^{-\alpha x}

Reduce to:

\frac{-hbar^2}{2m}\alpha^2 =E\frac{e^2}{4\pi \varepsilon _{0}x}

The question says this potential is for x greater than or equal to 0, so if I ignore x (which I'm almost certain I can't do):

\alpha=\sqrt{}\frac{2me^2}{4\pi \varepsilon _{0}hbar^2}

Can you please point out where I've gone wrong? Thank you.

You don't ignore the x! And you aren't differentiating $x e^{-\alpha x}$ correctly. You need to use the product rule. If you do that you will get two terms that aren't proportional to $\Psi$. Forcing them to cancel will tell you what $\alpha$ has to be.

 

[weguihapzi]

You don't ignore the x! And you aren't differentiating $x e^{-\alpha x}$ correctly. You need to use the product rule. If you do that you will get two terms that aren't proportional to $\Psi$. Forcing them to cancel will tell you what $\alpha$ has to be.

Ah of course, thank you very much! It seems I need to brush up on the basics...