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Forces in a Quantum Reflection
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[QUOTE="PhysicsandSuch, post: 4501769, member: 487816"] Hi folks, 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. [h2]Homework Statement [/h2] 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##. [h2]Homework Equations[/h2] $$V(x) = -V_0 (θ(x)-θ(x-a)) $$ $$F = -∇V$$ $$ψ_{region} = e^{i k x} + R e^{-i k x}$$ [h2]The Attempt at a Solution[/h2] [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! [/QUOTE]
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