# Homework Help: Particle in a triangular potential well

1. Feb 9, 2012

### masterjoda

1. The problem statement, all variables and given/known data
Particle is in a potential well in the shape of a isosceles right-angled triangle. Need to find the wave function and allowed energies.

2. Relevant equations
How to determinate a boundaries for the potential when that line is in a form of some linear function, there is no strict point between potential equal to zero and infinity potential.

3. The attempt at a solution
I tried potential as $kx$ for $0<x<\frac{a}{2}$, and $-kx$ for $\frac{a}{2}<x<a$, but seems that is not right.

2. Feb 12, 2012

### ehild

Plot your potential function. Is it a well? Write up a potential function which has minimum at x=0.

ehild

Last edited: Feb 12, 2012
3. Feb 13, 2012

### masterjoda

This plot of potential function

#### Attached Files:

• ###### V(x).jpg
File size:
4.7 KB
Views:
254
4. Feb 13, 2012

### ehild

Your function is not the same as the plot. -kx is negative in the range a/2<x<a

By the way, the drawing shows a hill instead of a well.

ehild

5. Feb 13, 2012

### masterjoda

Storry, like this

#### Attached Files:

• ###### V(x).jpg
File size:
4.1 KB
Views:
249
6. Feb 14, 2012

### ehild

That is better. What is the potential function?

ehild

7. Feb 14, 2012

### masterjoda

That is my main problem, I don't know how to write it, I know that potential is zero inside and ∞ outside, that green area is unreachable for the particle.

8. Feb 14, 2012

### ehild

You know the equation of a straight line? You have a straight line for 0<x<a/2 which intersects the x axis at x=a/2, and makes -45° angle with it. And you have an other equation for a/2<x<a, that encloses a positive angle of 45° with the x axis.

ehild

9. Feb 14, 2012

### vela

Staff Emeritus
What is this supposed to mean?

The potential isn't zero inside. It varies with x.

If you want to make things a bit simpler, take ehild's suggestion to center the well at x=0.

10. Feb 15, 2012

### masterjoda

Yes, exactly.

If I center the well at x=0 then I have a function

$V(x)= \begin{cases} -x & \text{for } -\frac{a}{2}<x < 0 \\ x & \text{for } 0 < x < \frac{a}{2} \\ 0 & \text{for } x=0\\ \infty & \text{elsewhere} \end{cases}$

Last edited: Feb 15, 2012
11. Feb 15, 2012

### vela

Staff Emeritus
Looks good. Now what?

12. Feb 16, 2012

### masterjoda

I need to find ψ and allowed energies inside the well.

13. Feb 16, 2012

### vela

Staff Emeritus
Well, we're not here to do your homework for you. Show us your work. What do you have so far?

14. Feb 16, 2012

### masterjoda

I've just realized that the right potential is $V(x)=\left | x \right | \text{for } x\in \left \{ -\infty, \infty \right \}$

Last edited: Feb 16, 2012
15. Feb 16, 2012

### masterjoda

So far I have this:

$-\frac{\hbar^2}{2m}\frac{\mathrm{d^2}\psi }{\mathrm{d} x^2} + |x|\psi=E\psi$

$c=\frac{2m}{\hbar^2}$ and $k=\frac{\sqrt{2mE}}{k}$

solution is $\psi(x)=C_1\text{Ai}(\frac{cx-k^2}{c^{\frac{2}{3}}})+C_2\text{Bi}(\frac{cx-k^2}{c^{\frac{2}{3}}})$ because x is always > 0, Bi goes to ∞ so C_2 have to be 0. And final solution is $\psi(x)=C_1\text{Ai}((\frac{2m}{\hbar^2})^{\frac{1}{3}}(x-E))$. I've determined the energy from conditions $\psi^{(1)}_n=\psi^{(2)}_n$ and $\psi'^{(1)}_n=\psi'^{(2)}_n$. And energies are: for even $E_n=-a'_{n+1}$ and for the odd $E_n=-a_{n+1}$. Normalization is a bit tricky $C_1^2\int_{-\infty}^{\infty} \text{Ai}^2(x-E_n)dx=1$, I don't know how to solve this integral.

Last edited: Feb 16, 2012
16. Feb 16, 2012

### vela

Staff Emeritus
Your solutions don't work out unit-wise. For example, you can't subtract E from x. They don't have the same units.

17. Feb 16, 2012

### masterjoda

Yes they do, potential at x have the same value as the x coordinate, V(x)=|x|, unit of that x is the same as the unit of E.