# Transmission over a finite square well

1. Aug 17, 2007

### T-7

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):

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

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

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

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

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

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

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

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

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

$$T = \frac{2qke^{-2ika}}{kq.(cos 2qa + isin 2qa)-i(k^{2}+q^{2})sin(2qa)}$$

(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: Aug 17, 2007
2. Aug 18, 2007

### StatusX

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