In Landau & Lifshitz "Quantum Mechanics" there's a statement:
If the Hamiltonian of the system is the sum of two (or more) parts, $\hat H=\hat H_1+\hat H_2$, one of which contains only the coordinates $q_1$ and the other only the coordinates $q_2$, then the eigenfunctions of the operator $\hat H$ can be written down as products of the eigenfunctions of the operators $\hat H_1$ and $\hat H_2$, and the eigenvalues of the energy are equal to the sums of the eigenvalues of these operators.
This is indeed true when one uses position representation where the Hamiltonian is a differential operator. But this statement seems to be limited to only position representation, since e.g. in momentum (Fourier) space "product" of wavefunctions corresponds to a convolution, not a usual product:
$$\mathcal{F}\left\{\Psi(x_1)\Psi(x_2)\right\}=\Psi(k_1)*\Psi(k_2),$$
so that statement doesn't seem to hold.
Now, if we take $\hat H$ as a general Hermitian operator, not necessarily quantum Hamiltonian operator, this rule seems to be incorrect in general.
So, is it true that the statement in Landau & Lifshitz is only correct for differential operators? Or have I mistaken it somehow?
Best Answer
There's actually no problem, because you're looking at functions of the form $f(x_1,x_2) = \Psi_1(x_1)\Psi_2(x_2)$, so that $$ \mathcal{F}\{\Psi_1(x_1)\Psi_2(x_2)\}(k_1,k_2) = \int_{\mathbb{R}^{n_1+n_2}}\Psi_1(x_1)\Psi_2(x_2)e^{2\pi i (x_1,x_2) \cdot (k_1,k_2)} d^{n_1}x_1 d^{n_2}x_2\\ = \left(\int_{\mathbb{R}^{n_1}} \Psi_1(x_1) e^{2\pi i x_1 \cdot k_1} d^{n_1}x_1\right)\left(\int_{\mathbb{R}^{n_2}} \Psi_2(x_2) e^{2\pi i x_2 \cdot k_2} d^{n_2}x_2\right)\\ = \mathcal{F}\{\Psi_1(x_1)\}(k_1) \mathcal{F}\{\Psi_2(x_2)\}(k_2), $$ as you want.
Abstractly, to clarify L&L's abuse of notation, what's really going on is that the composite of non-interacting Hamiltonians
is given by $\hat{H}_1 \otimes I + I \otimes \hat{H}_2$ on $$ L^2(\mathbb{R}^{n_1},d^{n_1}x_1) \otimes L^2(\mathbb{R}^{n_2},d^{n_2}x_2) \cong L^2(\mathbb{R}^{n_1+n_2},d^{n_1}x_1 d^{n_2}x_2), $$ and that the Fourier transform $$ \mathcal{F} : L^2(\mathbb{R}^{n_1+n_2},d^{n_1}x_1 d^{n_2}x_2) \cong L^2(\mathbb{R}^{n_1},d^{n_1}x_1) \otimes L^2(\mathbb{R}^{n_2},d^{n_2}x_2)\\ \to L^2(\mathbb{R}^{n_1},(2\pi)^{-n_1/2}d^{n_1}k_1) \otimes L^2(\mathbb{R}^{n_2},(2\pi)^{-n_2/2}d^{n_2}k_2) \cong L^2(\mathbb{R}^{n_1+n_2},(2\pi)^{-(n_1+n_2)/2}d^{n_1}k_1 d^{n_2}k_2) $$ is given by $\mathcal{F}_1 \otimes \mathcal{F}_2$, where $\mathcal{F}_j$ is the Fourier transform $$ \mathcal{F}_j : L^2(\mathbb{R}^{n_j},d^{n_j}x_j) \to L^2(\mathbb{R}^{n_j},(2\pi)^{-n_j/2}d^{n_j}k_j) $$ so that $$ \mathcal{F} (\hat{H}_1 \otimes I + I \otimes \hat{H}_2) \mathcal{F}^{-1} = (\mathcal{F}_1 \hat{H}_1 \mathcal{F}_1^{-1}) \otimes I + I \otimes (\mathcal{F}_2 \hat{H}_2 \mathcal{F}_2^{-1}) $$ on $$ L^2(\mathbb{R}^{n_1},(2\pi)^{-n_1/2}d^{n_1}k_1) \otimes L^2(\mathbb{R}^{n_2},(2\pi)^{-n_2/2}d^{n_2}k_2) \cong L^2(\mathbb{R}^{n_1+n_2},(2\pi)^{-(n_1+n_2)/2}d^{n_1}k_1 d^{n_2}k_2). $$
Given all this, what L&L are saying (modulo abuse of notation) is that if $$ \hat{H}_1 f_1(x_1) = E_1 f_1(x_1), \quad \hat{H}_2 f_2(x_2) = E_2 f_2(x_2), $$ then $(f_1 \otimes f_2)(x_1,x_2) := f_1(x_1)f_2(x_2)$ satisfies $$ (\hat{H}_1 \otimes I + I \otimes \hat{H}_2)(f_1 \otimes f_2)(x_1,x_2) = (E_1+E_2)(f_1 \otimes f_2)(x_1,x_2); $$ using $\mathcal{F} = \mathcal{F}_1 \otimes \mathcal{F}_2$, then, you get that $$ \mathcal{F}(f_1 \otimes f_2)(k_1,k_2) = \mathcal{F}_1\{f_1\} \otimes \mathcal{F}_2\{f_2\}(k_1,k_2) = \mathcal{F}_1\{f_1\}(k_1)\mathcal{F}_2\{f_2\}(k_2), $$ as you wanted.