Transmission over a finite square well

T-7
Messages
62
Reaction score
0
Hi,

My Quantum textbook loves to skip the algebra in its derivations. It claims that the solution for 'T', the transmission coefficient of the wave function (E>0, ie. unbound) is

T = \frac{2qke^{-2ika}}{2qkcos(2qa)-i(k^{2}+q^{2})sin(2qa)}

Its prior step is to offer two equations (which match my own derivation) for T in terms of R (the reflection coefficient):

<br /> e^{-ika}+R^{ika} = \frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}+(1-\frac{k}{q})e^{ika+2iqa})<br />

<br /> e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})<br />

My own attempt to fill in the gap between the two proceeds as follows, but I didn't quite arrive at that final expression...

<br /> 2e^{-ika} = T/2.e^{ika}[(1+k/q).e^{-2iqa} + (1-k/q).e^{2iqa} + (q/k-1).e^{-2iqa} - (q/k-1).e^{2iqa}]<br />

<br /> 4e^{-2ika} = T.[(1+k/q+q/k-1).e^{-2iqa} + (1-k/q-q/k+1).e^{2iqa}]<br />

<br /> 4qke^{-2ika} = T.[(k^{2}+q^{2}).e^{-2iqa} + (2kq-k^{2}-q^{2}).e^{2iqa}]<br />

<br /> 4qke^{-2ika} = T.[(k^{2}+q^{2}).(e^{-2iqa}- + e^{2iqa}) + 2kq.e^{2iqa}]<br />

<br /> 4qke^{-2ika} = T.[(k^{2}+q^{2}).(-2isin(2qa)) + 2kq.(cos 2qa + isin sqa)]<br />

<br /> T = \frac{4qke^{-2ika}}{2kq.(cos 2qa + isin 2qa)-2i(k^{2}+q^{2})sin(2qa)}<br /><br /> T = \frac{2qke^{-2ika}}{kq.(cos 2qa + isin 2qa)-i(k^{2}+q^{2})sin(2qa)}<br />

(Hopefully I've copied all that down correctly!).

Can anyone spot a false move, or what it is I need to do to make my derivation match the textbook formula.

Cheers!
 
Last edited:
Physics news on Phys.org
It seems like your answer would match theirs if:

e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})

was changed to:

e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})

which also matches the other line better. Are you sure you derived/copied this down right?
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top