In order to expose this, consider the definition of the quantum mechanical energy of a system consisting of 10 particles in three dimensions. The energy is the expectation value of the Hamiltonian \( H \) and reads $$ \begin{equation*} E=\frac{\int d\mathbf{R}_1d\mathbf{R}_2\dots d\mathbf{R}_N \Psi^{\ast}(\mathbf{R_1},\mathbf{R}_2,\dots,\mathbf{R}_N) H(\mathbf{R_1},\mathbf{R}_2,\dots,\mathbf{R}_N) \Psi(\mathbf{R_1},\mathbf{R}_2,\dots,\mathbf{R}_N)} {\int d\mathbf{R}_1d\mathbf{R}_2\dots d\mathbf{R}_N \Psi^{\ast}(\mathbf{R_1},\mathbf{R}_2,\dots,\mathbf{R}_N) \Psi(\mathbf{R_1},\mathbf{R}_2,\dots,\mathbf{R}_N)}, \end{equation*} $$ where \( \Psi \) is the wave function of the system and \( \mathbf{R}_i \) are the coordinates of each particle. If we want to compute the above integral using for example Gaussian quadrature and use for example ten mesh points for the ten particles, we need to compute a ten-dimensional integral with a total of \( 10^{30} \) mesh points. As an amusing exercise, assume that you have access to today's fastest computer with a theoretical peak capacity of more than one Petaflops, that is \( 10^{15} \) floating point operations per second. Assume also that every mesh point corresponds to one floating operation per second. Estimate then the time needed to compute this integral with a traditional method like Gaussian quadrature and compare this number with the estimated lifetime of the universe, $T\approx 4.7 \times 10^{17}$s. Do you have the patience to wait?