Contradicting solutions in the Fourier space.

Question

Consider the following differential equations $$ \begin{cases} \partial_t X(t) = -i \alpha X(t) + \beta Y(t), \\ \partial_t Y(t) = -i \alpha Y(t) - \beta X(t). \end{cases}\label{1}\tag{1} $$ with $\alpha$, $\beta$ some complex constants (independent of $t$).

Case-1 [Time domain]: The solution (with initial condition, say $X(0) = 1, Y(0) = 0$) reads $$ \begin{cases} X(t) = \frac{1}{2}[e^{-ist} + e^{-idt}], \\ Y(t) = -\frac{1}{2} [e^{-ist} - e^{-id t}], \end{cases}\label{2}\tag{2} $$ where $s = \alpha + \beta$ and $d = \alpha - \beta$. Now, the Fourier transform $$ \mathscr{F}[A(t)] = A(\omega) = \int_{-\infty}^{\infty} dt e^{i\omega t} A(t) $$ of \eqref{2} reads: $$ \begin{cases} X(\omega) = \frac{1}{2}[ \delta (\omega - s) + \delta (\omega + s)], \\ Y(\omega) = -\frac{i}{2}[ \delta (\omega - s) - \delta (\omega + s)] \end{cases}\label{3}\tag{3} $$ Case-2 [Fourier domain]: Taking Fourier transform of \eqref{1}, we have $$ \begin{cases} -i \omega X(\omega) = -i \alpha X(\omega) + \beta Y(\omega),\\ -i\omega Y(\omega) = -i\alpha Y(\omega) - \beta X(\omega). \end{cases}\label{4}\tag{4} $$ The solution of \eqref{4} (using Mathematica) turns out to be $$ \begin{cases} X(\omega) = 0, \\ Y(\omega) = 0. \end{cases}\label{5}\tag{5} $$ Clearly solution \eqref{5} is not the same as solution \eqref{3}. Am I missing too many things here?

Edit: My question is simple. How to start with equation \eqref{4} and obtain equation \eqref{3}?

Answer 1

Equation \eqref{4} is actually $$ \begin{bmatrix}-i\alpha & \beta \\-\beta & -i\alpha\end{bmatrix} \begin{bmatrix}X \\Y \end{bmatrix}=-i\omega\begin{bmatrix}X \\Y \end{bmatrix} $$ so besides the zero vector, there are 2 eigenvalues. This means that the solution to equation \eqref{4} is zero except when $\omega$ is an eigenvalue of the matrix equation. This will yield your equation \eqref{3}.

Answer 2

It is possible to disentangle/diagonalize the system of equations $(4)$ in the Fourier domain by defining $Z_\pm = X \pm iY$, hence $i(\omega - s_\pm) Z_\pm(\omega) = 0$, where $s_\pm = \alpha \pm \beta$. However, you cannot conclude $Z_\pm(\omega) = 0$ straightforwardly, because there is an implicit convention when taking the Fourier transform that we are working in a distributional space. Actually, the linear equation $\omega f(\omega) = 0$ is solved by $f(\omega) \propto \delta(\omega)$, because $\omega\delta(\omega) \equiv 0$. In other words, the multiplicative operator $\omega$ is not invertible anymore and its one-dimensional kernel is generated by the Dirac delta function.

In consequence, one has $Z_\pm(\omega) = A_\pm \delta(\omega - s_\pm)$, with $A_\pm$ some constants. How to find these constants ? They have to be determined from the initial conditions ! One has $Z_\pm(t=0) = 1$. From there, two strategies can be adopted. Either you apply the initial conditions after inverse Fourier transforming $Z_\pm$, i.e. $Z_\pm(t) = A_\pm e^{-is_\pm t}$, hence $A_\pm = 1$. Alternatively, you may take advantage of the relation $f(t=0) = \int_\Bbb{R} F(\omega) \,\mathrm{d}\omega$, which will lead to the same conclusion.

Finally, you switch back to the initial variables by writing $$ \begin{cases} X = \,\frac{1}{2}(Z_+ + Z_-) \\ Y = \frac{1}{2i}(Z_+ + Z_-) \end{cases}, $$ either after or before having inverted the Fourier transform, depending on the strategy you have followed, and you will obtain the same solutions as derived in the time domain only. In fact, the whole procedure could have been carried with $X$ and $Y$ instead of $Z_\pm$, but it permitted to avoid to consider manipulations over matrices, determinant, eigenvalues, etc.