Schrodinger's Equation - Step Potential

[irony of truth]

In the left side of the barrier, the potential energy V(x)= 0, while on the right side of the barrier, V(x) = V. Given that the total energy of the particle in such a system has a total energy E < V..

a. What are my acceptable solutions?
On the left side:

Should I include the cos kx and sin kx alone?

b. How do I show that /A/ + /B/ = 1? Are these absolute values of the expressions A and B or /A/ = A*A and /B/ = B*B?

I got A = C/2 (1 + iq/p) and B = C/2 ( 1 - iq/p)

 

[dextercioby]

On the lest side,it's just a combination of complex exponentials...What do you mean "include ...alone"...?

As for the second,i assume you refer to the coefficients of reflection & transmission.Great.But please,use this CORRECT expression:
[tex] A^{*}A=|A|^{2} [/tex]

I don't know who "C" is.Could you be more specific...?

Daniel.

 

[irony of truth]

Hello sir...

what I meant about cos px and sin px (not k) alone is that there are 3 solutions... first is that expression A cos px + B sin px, second is cosp x and third is sin px on the left side of the barrier... the right side involves only e^(-qx).

Hmm... I am referring to my 2 general solutions A cos px + B sin px ( or Ae^(ipx) + Be^(-ipx) ) and Ce^(-px) using then the continuity requirements... I imposed the \psi(x) and the first derivative of \psi at x = 0.

Thus resulting to A + B = C and ipA - ipb = -qC

then I tried to find the expressions of A and B in terms of C.. which are A = C/2 (1 + iq/p) and B = C/2 ( 1 - iq/p) respectively.

But I don't know why [tex] A^{*}A=|A|^{2} [/tex] + [tex] B^{*}B=|B|^{2} [/tex] don't equal to one?

 

[dextercioby]

Hello sir...

what I meant about cos px and sin px (not k) alone is that there are 3 solutions... first is that expression A cos px + B sin px, second is cosp x and third is sin px on the left side of the barrier... the right side involves only e^(-qx).

Hmm... I am referring to my 2 general solutions A cos px + B sin px ( or Ae^(ipx) + Be^(-ipx) ) and Ce^(-px) using then the continuity requirements... I imposed the \psi(x) and the first derivative of \psi at x = 0.

Thus resulting to A + B = C and ipA - ipb = -qC

then I tried to find the expressions of A and B in terms of C.. which are A = C/2 (1 + iq/p) and B = C/2 ( 1 - iq/p) respectively.

But I don't know why [tex] A^{*}A=|A|^{2} [/tex] + [tex] B^{*}B=|B|^{2} [/tex] don't equal to one?

Why would they...?What' the physical justification behind your requirement...?
Now,that u've managed to express 2 coefficients in terms of a third,u can resort to the definition of the probablilty currents and transmission and reflection coefficients...

Isn't this what you're looking for,or what else...?

Daniel.

 

[irony of truth]

So I will not add both A = C/2 (1 + iq/p) and B = C/2 ( 1 - iq/p) then.

I got this T = 4(q/p)/(1 + iq/p)^2 and R = /(1 - iq/p) / (1 + iq/p)/^2 ...

My problem is here is that will they add up to 1?

 

[dextercioby]

In square modulus,they MUST,viz.

[tex] |T|^{2}+|R|^{2}=1 [/tex]

Else,it would mean your calculations are incorrect.

Daniel.

 

[reilly]

Any, I repeat any QM text does this problem. It will be spelled out right in front of your eyes if you simply take the trouble to read. We are not talking rocket science here, just very basic stuff.
Regards,
Reilly Atkinson

 

[Eye_in_the_Sky]

to irony of truth

Hi, I'm just passing through, and I don't have much time (just enough to quickly browse).

If you are still having some questions, irony of truth, you may want to look at the thread below, in particular posts #2 and #10:

https://www.physicsforums.com/showthread.php?t=36076

If look, I hope it helps.
________

EDIT:
... oops!
I see you're looking at the E < V case. My posts were for E > V.