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[?] Solve the Kronig Penney model determinant?

Posted 7 years ago

So I have been trying very hard for the last day or so to solve the Kronig-Penney model for finite barriers to no avail with Mathematica. The unfortunate bit is that it doesn't seem like my 4x4 matrix is wrong and that I just can't seem to get Mathematica to simplify out the result nicely. It is particularly infuriating because I have exactly the same steps written down at this point as a solution guide (Kronig-Penney Solution). But take note of what's said in the first paragraph of page 2. The only interesting solutions that exist are ones for which the coefficients of A, B, C, D (for equations C5 to C8) have a determinant that is equal to zero. They then quote the result below stating that you can arrive at the result if you are determined.

After setting all of our equations equal to zero the matrix that we are to take the determinant of should look something like this: (K = x, gamma = y)

m = {{Exp[I*x*b], 
   Exp[-I*x*b], -Exp[y*b], -Exp[-y*b]}, {I*x*Exp[I*x*b], -I*x*
    Exp[-I*x*b], -y*Exp[I*b], y*Exp[-I*b]}, {Exp[I*x*a], 
   Exp[-I*x*a], -Exp[I*k*a], -Exp[I*k*a]}, {I*x*Exp[I*x*a], -I*x*
    Exp[-I*x*a], -y*Exp[I*k*a], y*Exp[I*k*a]}}

But no matter how many times I take the determinant, FullSimplify, expand, reduce or divide out any terms that are outside the parentheses because the det(m) = 0 I just can't seem to get the result quoted in the solution. I really don't want to just quote a result from online so I would really appreciate a hand in understanding how I can solve this.

Thank you

POSTED BY: John Tiessen
8 Replies

Try this

m = ( {
   {1, 1, -1, -1},
   {I K, -I K, -Q, Q},
   {E^(I K a), 
    E^(-I K a), -E^(-Q b + I k (a + b)), -E^(Q b + I k (a + b))},
   {I K E^(I K a), -I K E^(-I K a), -Q E^(-Q b + I k (a + b)), 
    Q E^(Q b + I k (a + b))}
  } )

Be sure to use Mathematica's symbols for I and e. Then try

Simplify[TrigExpand[ExpToTrig[Det[m] == 0]]]

It should give you the result you are looking for (with some extra terms you can cancel out).

POSTED BY: Dan Lawrence

Hi, Sorry for bringing back to life this old thread. I am also struggling with getting the right simplification. Can someone help me? The question is exactly the same. Thanks, Arturo

POSTED BY: Arturo Rodriguez

The determinant is not going to be zero independent of the variables present, specifically {a,b,k,x,y}. Is the goal to set it to zero and solve for one of those in terms of the others?

POSTED BY: Daniel Lichtblau
Posted 7 years ago

The goal (according to the solution) is to set the determinant equal to zero to arrive at a non-trivial solution. The determinant should then simplify to an equation of the following form:

[((y^2)-(x^2))/(2xy)]Sinh[yb]Sin[x(a-b)]+Cosh[yb]Cos[x(a-b)] = Cos[k*a]

Or, in the prettier Mathematica format:

((y^2 - x^2)/(2*x*y))*Sinh[y*b]*Sin[x (a - b)] + 
  Cosh[y*b]*Cos[x*(a - b)] == Cos[k*a]
POSTED BY: John Tiessen
Posted 7 years ago

In Mathematica imaginary number is I not i.

POSTED BY: Okkes Dulgerci
Posted 7 years ago

I switched it to lowercase 'i' for the post so that it would show up more clearly. The matrix determinant still does not simplify out nicely.

POSTED BY: John Tiessen

That switch from I to i means the code will not cut-and-paste to an actual functional example of the problem. It is expected that posts requesting assistance will contain code one can use to replicate the issue at hand.

I will add that Ix is not the same as I x (note space between, which is an implicit Times infix operator). Whether that's the source of trouble I do not know but it's something to check.

POSTED BY: Daniel Lichtblau
Posted 7 years ago

Alright, I'll just copy-paste the code that I was using then. Thank you for you assistance by the way.

POSTED BY: John Tiessen
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