Solving a system of equations with 5 unknowns

[hmparticle9]

I am solving Schrödingers equation with a potential with two "delta" spikes: One at $a$ and another at $-a$, with magnitude $-\alpha$. In the process of looking for scattering solutions I have come across the following four equations:

$$A + B \beta = C + D \beta$$
$$C \beta + D = F \beta$$
$$ik(C - D\beta - A + B\beta) = \delta (A+B\beta)$$
$$ik(F\beta - C\beta + D) = \delta F \beta$$

where $\beta = e^{2ika}$ and $\delta = -\frac{2m\alpha}{h^2}$.


$A,B,C,D,F$ are (possibly) complex numbers. $k$ is a constant. $i$ is the square root of $-1$.

I need to find the transmission coefficient of the problem, which basically means I need the quantity:

$$\frac{|F|^2}{|A|^2}$$

which means I need to find $A$ in terms of $F$, or more so the other way around. $F$ in terms of $A$.

I have $F$ in terms of $C$ and $D$:

$$C \beta + D = F \beta$$

I have found $$C = (\frac{\delta}{2ik} +1) A + \frac{\delta \beta}{2ik}B$$
and $$D = -\frac{\delta}{2ik\beta} A + (1-\frac{\delta}{2ik}) B$$

I have put these values in the equation

$$A + B \beta = C + D \beta$$

to verify correctness. All I need to do now is find a relationship between $A$ and $B$.

This I am struggling with. Could someone give me a hand?

 

[fresh_42]

I would set up the linear equation system for $A,B,C,D$ and hopefully get a $4\times 4$ matrix. Then invert this matrix, either manually or with the help of https://www.symbolab.com/solver/matrix-calculator, which allows variable names. If you have the solutions for $A,B,C,D$, then you're done.

 

[hmparticle9]

My bad. I made a mistake in the title. I have 5 unknowns.

 

[fresh_42]

My bad. I made a mistake in the title. I have 5 unknowns.

Where is the fifth? $F$ is a linear combination of $C$ and $D$.

 

[hmparticle9]

How do I delete questions? I solved it by hand in the end. Quite simple but tedious work

 

[fresh_42]

How do I delete questions? I solved it by hand in the end. Quite simple but tedious work

You have an equation for $F$ so $A,B,C,D$ would have been enough. I think the website I linked to can also solve $5\times 5$ matrices.

 

[jedishrfu]

How do I delete questions? I solved it by hand in the end. Quite simple but tedious work

Why not answer your own question and show us the solution? It will help others who will read your thread.

 

[hmparticle9]

I have provided an outline. Two reasons. The algebra is intense, and my workings are now in the local skip.

 

[fresh_42]

I used for convenience the parameters $ik=n\, , \,b=\beta\, , \,d=\delta$ and set up the equation system
\begin{align*}
\begin{pmatrix}1&1&-1&-1\\-n-d&bn-db&n&-bn\\0&0&-db&2n-d\\0&0&0&1
\end{pmatrix}&\cdot
\begin{pmatrix}A\\B\\C\\D\end{pmatrix} =
\begin{pmatrix}0\\0 \\0\\ D\end{pmatrix}\\
\end{align*}
where I hopefully didn't make typos. Then I fed
https://www.symbolab.com/solver/matrix-calculator/\left(d-db+nb+n\right)\begin{pmatrix}1&1&-1&-1\\ -n-d&bn-db&n&-bn\\ 0&0&-db&2n-d\\ 0&0&0&1\end{pmatrix}^{\left(-1\right)}?or=input
to do me the inversion. This resulted in
\begin{align*}
\dfrac{1}{d-db+nb+n}
\begin{pmatrix}-db+nb&-1&\frac{db-nb-n}{db}&-\frac{d^2b^2-d^2b+nd+3ndb-2n^2-2n^2b}{db}\\ d+n&1&-\frac{1}{b}&-\frac{-nb^2-nb-db+d-2n}{b}\\ 0&0&-\frac{d-db+nb+n}{db}&\frac{\left(-d+2n\right)\left(d-db+nb+n\right)}{db}\\ 0&0&0&d-db+nb+n\end{pmatrix}
\end{align*}
so my result is
\begin{align*}
(d-db+nb+n)\begin{pmatrix}A\\B\\C\\D\end{pmatrix}=
\begin{pmatrix}-db+nb&-1&\frac{db-nb-n}{db}&-\frac{d^2b^2-d^2b+nd+3ndb-2n^2-2n^2b}{db}\\ d+n&1&-\frac{1}{b}&-\frac{-nb^2-nb-db+d-2n}{b}\\ 0&0&-\frac{d-db+nb+n}{db}&\frac{\left(-d+2n\right)\left(d-db+nb+n\right)}{db}\\ 0&0&0&d-db+nb+n\end{pmatrix}\begin{pmatrix}0\\0\\0\\D\end{pmatrix}
\end{align*}
I used $D$ as a parameter for the solution space because the algorithm handled it best. Since the first line says $A\sim B$ with an appropriate factor, we have also all solutions depending only on $A.$

 

[hutchphd]

My bad. I made a mistake in the title. I have 5 unknowns.

But one of the unknowns is normalization (for scattering this is usually normalized to incident flux.... hence the ratio requested) so your original statement is OK. I recommend looking at the limits of ......α= 0, and perhaps a=0 (single delta fn at origin)
Will return to this later.... don't have the time right now. Shouldn't be too hard methinks.

 

[fresh_42]

I have found
$$
\dfrac{F}{A}=\dfrac{2ik\delta(1-\beta)-2k^2(1+\beta)}
{\delta^2\beta(\beta-1)+ik\delta(1+3\beta)+2k^2(1+\beta)}
$$

If I were to find a mistake, I would look closer at the third equation. My suspicion is that the factor $\beta$ should be at $C$ and not at $D.$