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The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ where the electrons are denoted 1 and 2, and $r_i$ is the distance to the nucleus at the origin and $r_{12}$ the distance between the electrons. Since the ...


8

This is essentially the definition of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically. If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is ...


4

You shouldn't do a Fourier in the time coordinate just in the position coordinates (i.e. x y z). And you know by simple calculations that $\hat{\Delta \psi}$ is just $-k^2\psi(t, \vec{k})$ where $\vec{k}$ is (x, y, z) after the Fourier transform. And the by substituting this to the equation you get a simple ode for $\hat{\psi}(t, \vec{k})$ because $\hat{\...


3

The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities. In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E_K+V$ in ...


1

What you've done seems fine, it's just that you added an extra conjugate by accident. Using your notation, if $$F[\psi, \psi^*] = \int \psi^* \psi \, d \vec{r}$$ then we have $$F[\psi + t h, \psi^* + t h^*] = \int (\psi^* + t h^*) (\psi + t h) \, d \vec{r}.$$ Taking the derivative with respect to $t$ at $t = 0$ gives $$\frac{\partial F[\psi + t h, \psi^* + ...


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