Mathematica Asked by Nikolaos Petropoulos on July 13, 2021
So I am trying to use DSolve to solve the following system of equations:
eqns = {c1'[
t] == -I/
h (H12 c2[t] E^(-I W12 t ) +
H13 c3[t] E^(-I W13 t ) +
H14 c4[t] E^(-I W14 t )), c2'[t] == -I/
h (H12 c1[t] E^(I W12 t ) +
H23 c3[t] E^(-I W23 t ) +
H24 c4[t] E^(-I W24 t )), c3'[t] == -I/
h (H23 c2[t] E^(I W23 t ) +
H13 c1[t] E^(I W13 t ) +
H34 c4[t] E^(-I W34 t )), c4'[t] == -I/
h (H24 c2[t] E^(I W24 t ) +
H34 c3[t] E^(I W34 t ) +
H14 c1[t] E^(I W14 t )), c1[0] == a, c2[0] == b, c3[0] == c, c4[0] == d}; DSolve[eqns, {c1, c2, c3, c4}, t]
But Mathematica returns the input and doesn’t do anything. Can someone help me out?
Thanks! (Sorry for the bad layout, I don’t know how to seperate the equations here without getting them out of the code section)
It looks like you are trying to propagate the time-dependent Schrödinger equation in the interaction picture. If this is indeed the case, could it be that your W
parameters can be written as $W_{ij}=w_i-w_j$? If yes, then the solution consists of going back to the Schrödinger picture, where the Hamiltonian becomes time-independent:
W12 = w1 - w2;
W13 = w1 - w3;
W14 = w1 - w4;
W23 = w2 - w3;
W24 = w2 - w4;
W34 = w3 - w4;
c1[t_] = a1[t] E^(-I w1 t);
c2[t_] = a2[t] E^(-I w2 t);
c3[t_] = a3[t] E^(-I w3 t);
c4[t_] = a4[t] E^(-I w4 t);
Your equations can now be written as a time-dependent Schrödinger equation with a time-independent Hamiltonian $H$ for the $a_i(t)$,
y[t_] = {a1[t], a2[t], a3[t], a4[t]};
H = {{-h w1, H12, H13, H14 },
{H12, -h w2, H23, H24 },
{H13, H23, -h w3, H34 },
{H14, H24, H34, -h w4}};
y'[t] == -(I/h) H . y[t] // Thread
(* four differential equations equivalent to yours *)
The solution is now found by matrix exponentiation,
Y[t_] = MatrixExp[-(I/h) H t, {a, b, c, d}]
which contains RootSum
terms that you can convert to radicals with ToRadicals
:
Y[t] // ToRadicals
(* huge result *)
From this solution for the $a_i(t)$ you can compute the $c_i(t)$ with the aforementioned formulas.
Alternatively, with the time-independent Hamiltonian $H$ you can go the more traditional route of using the time-independent Schrödinger equation and simply look for the (energy) eigenvalues of $H$,
Eigenvalues[H]
or
Eigenvalues[H, Quartics -> True]
Answered by Roman on July 13, 2021
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