# Confused on a Certain Approximation/Expansion

1. Aug 25, 2008

### ChrisJM

Hi, I need a quick bit of help for an approximation/expansion used in a paper by Rolf Landauer (1950). The paper deals with the WKB method (Quantum Mechanics) but that is irrelevant to my question. There is an approximation presented as follows:

A = 1 - (K/2a)

K = a - b

If a>>K, then A will be given by:

A = (b/a)^(1/2), to the first order in K

How is this approximation made? This is the entirety of the information provided: it is at the very start of the paper.

I've change a couple variable names (a and b are infact k1 and k2, the propogation constants in a potential well, if anyone is interested). A is the transmission coefficent of a wavefunction. I appreciate any help!

Cheers

Chris

2. Aug 26, 2008

### CompuChip

Hmm smells like one of those fishy physicists' "if f(x) = g(x) are equal to first order, they're equal" tricks :)

Where I got, is:
if a >> K then $\epsilon = K/a$ is very small.
And then indeed, in first order,
$$\sqrt{1 - \epsilon} = 1 - \frac12 \epsilon + \mathcal O(\epsilon^2) = A + \mathcal O(\epsilon^2)$$.
Now if there were a plus instead of a minus,
[tex]1 + \frac{K}{a} = \frac{a + b - a}{a} = \frac{b}{a}[/itex]
and it would have been explained (sort of).
Unfortunately, I think it should be a minus... but maybe this puts you on a track.

3. Aug 26, 2008

### ChrisJM

Hmm that could be the right direction.

Another bit I neglected to mention (but may help with the puzzle!), deals with another value B.

B = K/2a

And following the approximation that yields A = (a/b)^(1/2), it implies that this means that B>>A.

This obviously suggests that K/2a is very small, thus A is approximately 1 and B is approximately 0 (which fits with the following discussion). However, I'm still not sure how the approximation of A is done!