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)=<br /> \begin{cases}<br /> -x &amp; \text{for } -\frac{a}{2}&lt;x &lt; 0 \\<br /> x &amp; \text{for } 0 &lt; x &lt; \frac{a}{2} \\<br /> 0 &amp; \text{for } x=0\\<br /> \infty &amp; \text{elsewhere}<br /> \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&#039;^{(1)}_n=\psi&#039;^{(2)}_n. And energies are: for even E_n=-a&#039;_{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.