Step Potential Question (Quantum Mechanics)

[Qbit42]

Homework Statement

Write out the spatial portion of the wavefunction for a particle coming from the right with energy (E) greater than the step potential (U) and passing over the step.

Homework Equations

The Attempt at a Solution

I just want to know if this is the right set of equations for this situation.

$$\psi_{1}$$ = A$$e^{ik_{1}x}$$ + B$$e^{-ik_{1}x}$$
$$\psi_{2}$$ = C$$e^{-ik_{2}x}$$ (Free particle)

 

[vela]

Can you explain why you think those equations apply?

 

[Qbit42]

Can you explain why you think those equations apply?

Umm well before the step down the particle is under the influence of a nonzero potential. I thought the general form for this type of situation (in 1D) was the one I had listed for the first wavefunction. After the step the potential is zero so I just used the equation of a free particle propagating to the left.

 

[vela]

Not quite. The fact that the potential is zero or non-zero doesn't affect the wave function. The zero of potential energy is arbitrary, so you could always subtract U from the potential so that the potential is 0 for x>0 and -U for x<0, and the wave function should remain the same.

So why does your wave function for x>0 have the form it does? It is correct, by the way. That's the general form for a free particle in a constant potential. But why does the x>0 function have two terms and the x<0 only one?

 

[Qbit42]

Well I guess I could add a $$De^{ik_{2}x}$$ term to $$\psi_{2}$$ but wouldn't that be representative of a particle going over the barrier and then heading back towards the right (I'm thinking in terms of incident, transmitted, and reflected)?
My prof wants us to eliminate all but one constant term using everything but the normalization condition for the wavefunction. So I'd have to get rid of either C or D using boundary conditions. I could evaluate $$\psi_{2}$$ at negative infinity and force the result equal to zero by getting rid of C I suppose. But again I don't like that leaves me with a particle passing over the barrier and then returning to the right.

 

[vela]

I'm just asking you to explain why you came up with the wave functions you did, not necessarily change them. If you understand why they should have a specific form, rather than just guessing at the form, you don't have to wonder if they're right or not.

 

[Qbit42]

Well here's what I know about the topic. The spatial portion of the time independent schrodinger equation in 1D satisfies the differential equation

$$\frac{-h(bar)}{2m}\frac{d^{2}\psi}{dx^{2}} + U(x)\psi = E\psi$$.

When U(x) is constant then the general solution for this equation is

$$\psi = Ae^{ikx} + Be^{-ikx}$$ where k = $$\sqrt{\frac{2m(E-U)}{h(bar)^{2}}}$$.

Using this I wrote down the original equations. I tried using the general solution for region 2 but the boundary condition that $$\psi_{2}$$ had to be finite at - infinity caused me to discard my C term and I wasn't confident that I should.

 

[vela]

Using this I wrote down the original equations. I tried using the general solution for region 2 but the boundary condition that $$\psi_{2}$$ had to be finite at - infinity caused me to discard my C term and I wasn't confident that I should.

You don't want that term, but not for the reason you stated. After all, don't you have the same problem with all the other terms not vanishing at infinity? The free-particle wave functions aren't normalizable.

 

[Qbit42]

You don't want that term, but not for the reason you stated. After all, don't you have the same problem with all the other terms not vanishing at infinity? The free-particle wave functions aren't normalizable.

I've actually always had trouble understanding why they didn't all just vanish, thanks for the clarification there. But without the finite condition that leaves me with only 2 boundary conditions (continuity of $$\psi$$ and its derivative at the boundary). I won't be able to eliminate 3 unknowns with 2 boundary conditions.

 

[vela]

It gets back to what you said about the incident, transmitted, and reflected waves. There's just no physical reason why you should have a wave traveling to the right in the x<0 region. You just have the transmitted wave in this region. In the x>0 region, on the other hand, you have the incident wave going to the left which partially reflects at the discontinuity, causing a wave moving to the right.

 

[Qbit42]

So if my initial equations were correct then I can't figure out why I'm having problems actually computing my transmission and reflection coefficients. Using my 2 boundary conditions I get

A = $$\frac{C}{2}(1-\frac{k_{2}}{k_{1}})$$
B =$$\frac{C}{2}(1+\frac{k_{2}}{k_{1}})$$.

Solving for Probability Currents

$$J_{T} = \frac{-h(bar)k_{2}}{m}C^{*}C$$
$$J_{I} = \frac{-h(bar)k_{1}}{4m}C^{*}C (1+\frac{k_{2}}{k_{1}})^{2}$$
$$J_{R} = \frac{h(bar)k_{1}}{4m}C^{*}C(1-\frac{k_{2}}{k_{1}})^{2}$$

Therefore

T = $$\frac{J_{T}}{J_{I}} = \frac{4k_{2}}{k_{1}(1+\frac{k_{2}}{k_{1}})^{2}}$$

R = $$\frac{J_{R}}{J_{I}} = \frac{-(1-\frac{k_{2}}{k_{1}})^{2}}{(1+\frac{k_{2}}{k_{1}})^{2}}$$

These do not add up to 1, however if the negative on R were gone they would

Edit: Never mind, I think that in calculating the Transmission and Reflection coefficients you use the absolute values of the probability currents

 

[vela]

That's right. You need to use the absolute values.