Kronig Penney Model Matrix Determinant I have been struggling a bit with the determinant of the Kronig-Penney model for an infinite series of finite wells. The main issue is finding a way to nicely simplify the output from Mathematica. If you are wondering how the problem is stated, a solution to the problem is linked here. Now, the four equations that result from applying the boundary conditions to the Schrodinger equation as well as Bloch's Theorem are as follows:
$$Ae^{iKb}+Be^{-iKb}-Ce^{{\gamma}b}-De^{-{\gamma}b}=0$$
$$iK(Ae^{iKb}-Be^{-iKb})-{\gamma}(Ce^{{\gamma}b}-De^{-{\gamma}b})=0$$
$$Ae^{iKa}+Be^{-iKa}-(C+D)e^{ika}=0$$
$$iK(Ae^{iKa}-Be^{-iKa})-(C-D){\gamma}e^{ika}=0$$
The solution that we are interested in is the one for which the determinant of the coefficients of A, B, C & D are equal to zero. In other words:
$$
        det\begin{vmatrix}
        e^{iKb} & e^{-iKb} & -e^{{\gamma}b} & -e^{-{\gamma}b} \\
        iKe^{iKb} & -iKe^{-iKb} & -{\gamma}e^{{\gamma}b} & {\gamma}e^{-{\gamma}b} \\
        e^{iKa} & e^{-iKa} & -e^{ika} & -e^{ika} \\
        iKe^{iKa} & -iKe^{-iKa} & -{\gamma}e^{ika} & {\gamma}e^{ika} \\
        \end{vmatrix}=0
$$
Now according to the solutions that you will see online we should get the following result if we're determined:
$$\left(\frac{\gamma^2-K^2}{2{\gamma}K}\right)sinh({\gamma}b)sin[K(a-b)]cosh({\gamma}b)cos[K(a-b)]=cos(ka)
 $$
The problem that I am having is arriving at this result from the determinant of our matrix above. I've tried doing it in Mathematica and no matter how many times I Simplify, FullSimplify, TrigToExp etc. I can't seem to get the answer to drop out. If you have any suggestions on how I should go about simplifying this determinant or if there are any clever identity tricks that I can use to make this problem simpler please tell me. I love seeing a good math trick, so hit me with your best one.
Following the suggestion given below I now have the following matrix:
$$
        \begin{pmatrix}
       Ccos(bK) & iSin(bK) & -Cosh({\gamma}b) & -Sinh({{\gamma}b}) \\
        -KSin({bK}) & iKCos({bK}) & -{\gamma}Sinh({{\gamma}b}) & -{\gamma}Cosh({{\gamma}b}) \\
        Cos(aK) & iSin(aK) & -2e^{ika} & 0 \\
        -KSin(aK) & -iKCos(aK) & 0 & 2e^{ika} \\
        \end{pmatrix}
$$
Which then gives the following result for the determinant (after some simplification):
$$2 e^{i a k} \left((\gamma -1) K \cosh (b \gamma ) \cos (K (a-b))-\left(\gamma +K^2\right) \sinh (b \gamma ) \sin (K (a-b))\right)+K \left(-\gamma +4 e^{2 i a k}\right)=0$$
I am not sure how to simplify past this point now to arrive at the desired result.
Thank you
 A: I found myself in the same problem when following the discussion about energy bands from Kittel. I had to use the cofactors method performing the following operations:


*

*$C_0 \rightarrow C_0 + C_1$

*$C_2 \rightarrow C_2 + C3$

*$C_0 \rightarrow 1/2C_0$

*$C_2 \rightarrow 1/2C_2$

*$C_3 \rightarrow -C_3$

*$C_3 \rightarrow -i\frac{K}{\gamma}C_3$

*$C_3 \rightarrow C_3 + C_1$

*Develop in minors by row 1

*$C_2 \rightarrow \frac{\gamma}{\gamma-iK}C_2$

*$C_2 \rightarrow C_2-C_0$

*$C_1 \rightarrow C_1-C_0$

*After 11 and developing by minor in row 1 you will end with a 2x2 determinant, which you need to work until the final expression.


Note that these steps worked for the matrix I obtained following the discussion by Kittel.  
A: A hint: Let $C_k$ denote the columns of the corresponding matrix.
By right-multiplying this matrix by this one:
$$\dfrac12\begin{pmatrix}
       1 & \ \ 1 & 0 & 0 \\
       1 & -1 & 0 & 0 \\
       0 & 0 & 1 & \ \ 1 \\
       0 & 0 & 1 & -1 \\
\end{pmatrix}$$
which has a non-zero determinant, we obtain: 
$$\det[C_1,C_2,C_3,C_4]=0  \ \iff \   \det[\dfrac{C_1+C_2}{2},\dfrac{C_1-C_2}{2},\dfrac{C_3+C_4}{2},\dfrac{C_3-C_4}{2}]=0 $$
which will give you the 2 first columns with $\sin$ and $\cos$ functions, the 2 last with $\cosh$ and $\sinh$ functions.
