How to solve a matrix differential equation $\frac{d X}{dt} = i(X Y- Y^\dagger X)$?

Question

In eq (21) on page-5 of this article, we have matrix

HEP = (\[Omega]/2)*{{-2*I, 1}, {1, 0}}; 

and I am interested in finding matrix G (given in eq (23)) which satisfies eq (B1) i.e.

D[G, t] == I*(G . HEP - ConjugateTranspose[HEP] . G)

How can one achieve this in Mathematica?

Answer 1

A brute force solution can be obtained by, first, defining

x = {{x11[t], x12[t]}, {x21[t], x22[t]}};
y = {{-2*I, 1}, {1, 0}};
f = {{1, 2 t w}, {-I, -I (2 t w - 4)}};
g = {{g11, g12}, {g21, g22}};

Then, the components of x are obtained from

sx = DSolve[D[x, t] == I w/2 (x . y - ConjugateTranspose[y] . x), 
    Flatten[x], t] // Simplify // Flatten
(* {x11[t] -> 1/4 E^(t w) ((2 + t w)^2 C[1] + 
    I t w ((2 + t w) C[2] - (2 + t w) C[3] - I t w C[4])), 
    x12[t] -> 1/4 E^(t w) (4 C[2] + 2 I t w C[1] - C[4]) + 
    t^2 w^2 (I C[1] - C[2] + C[3] + I C[4])), 
    x21[t] -> 1/4 E^(t w) (4 C[3] - 2 I t w (C[1] - C[4]) + 
    t^2 w^2 (-I C[1] + C[2] - C[3] - I C[4])), 
    x22[t] -> 1/4 E^(t w) (t^2 w^2 C[1] + I t w (-2 + t w) C[2] - 
    I t w (-2 + t w) C[3] + (-2 + t w)^2 C[4])} *)

Expressing this result in terms of f and g is accomplished by

Solve[Thread[Simplify[
     Exp[t w] f . g . ConjugateTranspose[f], {t, w} \[Element] 
     Reals] == x /. sx], {C[1], C[2], C[3], C[4]}] // Flatten
(* {C[1] -> g11, C[2] -> I (g11 - 4 g12), C[3] -> -I (g11 - 4 g21), 
    C[4] -> g11 - 4 g12 - 4 g21 + 16 g22} *)

x /. Simplify[sx /. %]
(* {{E^(t w) (g11 + 2 t w (g12 + g21 + 2 g22 t w)), 
     I E^(t w) (g11 + 2 g12 (-2 + t w) + 2 t w (g21 + 2 g22 (-2 + t w)))}, 
    {-I E^(t w) (g11 + 2 g21 (-2 + t w) + 2 t w (g12 + 2 g22 (-2 + t w))), 
     E^(t w) (g11 + 2 (-2 + t w) (g12 + g21 - 4 g22 + 2 g22 t w))}} *)

Answer 2

The Lie bracket is an order respecting, non commutave 'derivative' $$\left[X, Y\right] =\nabla_X Y$$ by linearity and product formula $$\left[ X ,Y Z\right] = X Y Z - Y Z X = Y \left[ X , Z\right] + \left[ X , Y\right] Z = Y(XZ-ZX)+ (XY-YX)Z $$.

$$\nabla_X (Y \ Z) = Y \nabla_X Z + (\nabla_X Y) \ Z$$

So you can try to solve differential equations generally, parallel to commuative calculus

$$\nabla_X Y = \lambda Y : \quad Y=Y_0 e^{\lambda X} $$

Simply think of diagonal matrices $X_d$ and normal, diagonalizable matrices with $X = U^* X_d U$, such that in power series and analytic functions $f$ one has

$$ (U^*\ X \ U)^2 = U^*\ X^2 \ U , \quad U^*\ f(X) \ U = f(U^* \ X \ U)$$

by cancellation of all inner products $U^*U=1$.