Understanding the radial equation, why is the left hand side different from the right hand side?

Question

Im studying the hydrogen atom and Ive realized that one side of the radial differential equation isnt equal to the other. What am I getting wrong?

Knowing that the potential for the hydrogen atom is $$PE=-\frac{k_ee^2}{r}$$ and $$V_{eff}=\frac{ħ^2}{2m_e}\frac{l(l+1)}{r^2}.$$ And then knowing $$L^2=l(l+1)ħ^2=m_e^2v^2r^2$$ we get that $$V_{eff}=\frac{L^2}{2m_er^2}=\frac{m_e^2r^2v^2}{2m_er^2}=\frac{m_ev^2}{2}=KE$$ so the effective potential is equal to the kinetic energy plus the potential energy which should be equal to the total energy so the equation would look like this: $$-\frac{\hbar^2}{2m_e}\frac{d^2u}{dr^2}+[PE+KE]u=Eu.$$

If the second part of the sum is already equal to the total energy what is the first part doing? The unique solution would be that the second derivative of the radial wave function is 0 but that isnt the case. What did I got wrong?

Answer 1

The answer is that $d^2u/dr^2$ is a part of the kinetic energy that you've neglected. You can think of this as coming from the radial movement of the particle. The effective potential accounts for the kinetic energy arising from the angular movement of the particle. Essentially, the first term is analogous to the $m/2 (dr/dt)^2$ term in the kinetic energy you would write in classical mechanics.

The derivation of the kinetic energy in the Schrodinger equation comes from the term $\hat{\mathbf{P}}^2/2m$, where $\hat{\mathbf{P}}=-i\hbar\nabla$, so that $\hat{\mathbf{P}}^2=-i\hbar\nabla^2$. Here, $\nabla^2$ is the Laplacian which can be written in spherical coordinates. The radial term of the Laplacian, $d^2/dr^2$, leads to the radial kinetic energy term. See any standard quantum mechanics book for details (e.g. Griffiths.)

As for why this term is nonzero for an electron in orbit, you should understand that an electron whizzing around in a circle is just a classical picture. Quantum mechanically, the position or momentum of this electron is undefined. Rather, we only have the electron's wavefunction $\psi$, which allows us to obtain probabilities for where the electron might appear (or its momentum, or any other observable) when we measure it. Handwavingly, one might say (for the sake of intuition) that the electron is whizzing around in a circle, but there is a chance that it sometimes has a radial component to its momentum. This leads to a contribution to its radial kinetic energy. This is directly analogous to the zero-point motion of the particle one discusses in the quantum harmonic oscillator.

In that example of a harmonic oscillator, where the potential goes as $V(x) = kx^2/2$, one would expect classically that a particle would simply sit at the bottom of the well, at $x=0$, with no momentum. The energy then would be zero (no potential energy or kinetic energy.) However, quantum mechanically, this turns out not to be true. Both the kinetic and potential energy are nonzero in the ground state of the harmonic oscillator, and this is also due to the zero-point motion. (again, see textbooks like Griffiths for more details/derivations.)