- #1

PhysicsandSuch

- 39

- 2

So everyone at some point studying QM deals with reflections from 1D finite potential wells using the TISE, but something that I never covered was how to talk about or calculate the forces involved in turning around that particle. I tried to work something out below, but I'm not sure if it's a valid approach and was hoping for some advice.

## Homework Statement

Consider a 1D finite potential well starting at position ##x=0##, length ##a##, and depth ##-V_0##. A particle is incident on the well from the left with energy ##E>0##. Calculate the expectation of the force on the particle in that region given a reflection coefficient ##R##.

## Homework Equations

$$V(x) = -V_0 (θ(x)-θ(x-a)) $$

$$F = -∇V$$

$$ψ_{region} = e^{i k x} + R e^{-i k x}$$

## The Attempt at a Solution

[tex]

\begin{align}

F &= V_0 ( δ(x)-δ(x-a)) \\

<F> &= ∫_{-∞}^0 ψ^*Fψ dx \\

&= ∫_{-∞}^0 (e^{- i k x} + R e^{i k x}) V_0 ( δ(x)-δ(x-a)) (e^{i k x} + R e^{-i k x}) dx \\

&\text{Expanding and noting only the unshifted delta function will survive the integral} \\

& =∫_{-∞}^0 V_0 δ(x) (1 + R(e^{2i k x} + e^{-2i k x}) + R^2) dx \\

& =V_0 (R+1)^2

\end{align}

[/tex]

If this is right, then at full reflection, can we expect a force of 4 times the potential strength, while at no reflection, the particle still experiences a force equal to the potential? Also, my dimensions seem weird here: after integrating out the delta function, I'm just left with energy, not force...right? It seems proportionally correct, but I can't figure out where am I going wrong...

I'm not sure if it's a calculation error or a conceptual misunderstanding on how to deal with finding this force.

If this is a sound line of thinking or could be massaged into one, could it also be applied to a wavefunction undergoing exponential decay as it tunnels through a finite barrier?

Thanks for the help!