Particle in a triangular potential well

[masterjoda]

Homework Statement

Particle is in a potential well in the shape of a isosceles right-angled triangle. Need to find the wave function and allowed energies.

Homework 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.

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.

 

[ehild]

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

 

[masterjoda]

This plot of potential function

 

[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

 

[masterjoda]

Storry, like this

 

[ehild]

That is better. What is the potential function?

ehild

 

[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.

 

[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

 

[vela]

there is no strict point between potential equal to zero and infinity potential.

What is this supposed to mean?

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.

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.

 

[masterjoda]

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

Yes, exactly.

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

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

 

[vela]

Looks good. Now what?

 

[masterjoda]

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

 

[vela]

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

 

[masterjoda]

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

 

[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.

 

[vela]

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

 

[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.