"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:[Only registered users see links. ]> wrote in
message news:itvbd.898$SW3.490@fed1read01...

Yes, the ends have to join up, no problems there-

I have equations of the type

phi1(x) = Ae^ik1x + Be^-ik1x (x<0)
C D (0<=x<=a)
E F (a<=x)

A,B etc are billed as arbitrary constants, which is very nice but
nowhere can I find any suggestions as how to choose them, which
implies I am missing the glaringly obvious. The math I have here
indicates that I only need A and F to get my transmission function out.

"Steve Ralph" <[Only registered users see links. ].uk> wrote in message news:<41740912$0$48002$[Only registered users see links. ].net>...

<snip>

The 'open ended' ranges are one key here: To be integrable, the
functions cannot grow without limit as one moves out into the open
range. This constrains both B and E to identically zero.

Another 'trick' is node counting: a node is a point where the wave
function vanishes. If C and D are of the same sign, there can be no
node in the interval 0<=x<=a, and because you have only one
coefficient in the unbounded open ranges, there can be no nodes there.
If they have opposite signs, then there will be a point at which they
will cancel each other out, producing a node.

The symmetry of the problem also tells us that |C| = |D|.

Result: you will have a node at x=a/2 when C = -D and no node when C =
+D. (The wave function *with* the node will have a higher energy.)

You now have only to integrate the two piece-wise wave functions (one
wothout the node and one with the node) and normalize the results to
evaluate the two unknowns: A = +/-F and C = +/-D, where the - sign
applies to the wave function with the node.

"tadchem" <[Only registered users see links. ]> wrote in message
news:130fe1c3.0410190531.48317602@posting.google.c om...
<snip>
Speaking of nodes, the document I'm working from is [Only registered users see links. ]
which is about the only description of the basic method I've managed to find
so far.
I've realised WKB looses resonances, which is no good - I'm very
much in instant cookbook mode at the moment :>)) I think it's finally
clicked that
I have expressions for d12 p2 etc in (17) so the top row of that 2x2 matrix
multiplication gives t11.
I'm starting to get my head round whats happening here, so thanks for some
most useful comments!
Doutless I'll be back in a day or so when I end up with infinite tunnelling
or
something equally silly

Thanx again. I graduated in physics in '76, and then got my doc in
turbulent combustion flow measurement. I had to struggle to get an equation
in
there at all :>)) It is something of a shock to find myself with my
trusty PC doing QM! It is like banging the rust off pipes.
All of a sudden the math sort of came into focus, (comments
from Tadchem and N:dlx wotsit helped ) so
feeling a lot more competent (until the next obstacle) its down to ordinary
programming and countering my tendancy to make dumb mistakes now. I
just hope I can put E = Vmax and get 1, and E=Vmax=0 and get 0. (fat
chance :>))

This is real fun. As an experimentalist I never expected to get involved in
theoretical work, let alone get to see new territory. (Provided I can get
my arithmetic right of course LOL)