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4.3: Spherical Symmetry

Separation of Variables

Working in spherical coordinates is significantly more difficult than working in cartesian coordinates. So why do it? Because point-like particles are sources for spherically-symmetric potentials that affect other particles. Understanding how to work in spherical coordinates is essential for solving the hydrogen atom in particular.  First a reminder of the coordinates themselves:

Figure 4.3.1 – Spherical Coordinates

We begin with the stationary-state Schrödinger equation in three dimensions. We need to write the laplacian operator found in the hamiltonian (Equation 4.1.9) in spherical coordinates:

$\nabla^2 = \dfrac{1}{r^2}\left[\dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\right) + \left(\dfrac{1}{\sin\theta}\right)\dfrac{\partial}{\partial \theta}\left(\sin\theta\dfrac{\partial}{\partial \theta}\right) + \left(\dfrac{1}{\sin^2\theta}\right)\dfrac{\partial^2}{\partial \phi^2}\right]$

The stationary-state Schrodinger equation in spherical coordinates is therefore:

$-\dfrac{\hbar^2}{2mr^2}\left[\dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\right) + \left(\dfrac{1}{\sin\theta}\right)\dfrac{\partial}{\partial \theta}\left(\sin\theta\dfrac{\partial}{\partial \theta}\right) + \left(\dfrac{1}{\sin^2\theta}\right)\dfrac{\partial^2}{\partial \phi^2}\right]\psi_E\left(r,\theta,\phi\right) + V\left(r,\theta,\phi\right)\psi_E\left(r,\theta,\phi\right) = E\psi_E\left(r,\theta,\phi\right)$

Our goal is to collect similar variables, and before we use our usual trick, we can get a head start on that by removing the $$r^{-2}$$ factor from the second and third terms. Collecting all the terms with factors of $$r$$ on one side of the equation then gives:

$\left(\dfrac{1}{\sin\theta}\right)\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial}{\partial\theta}\right)\psi_E\left(r,\theta,\phi\right) + \left(\dfrac{1}{\sin^2\theta}\right)\dfrac{\partial^2}{\partial\phi^2}\psi_E\left(r,\theta,\phi\right) = \dfrac{2mr^2}{\hbar^2}\left[V\left(r,\theta,\phi\right)-E\right]\psi_E\left(r,\theta,\phi\right) - \dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\right)\psi_E\left(r,\theta,\phi\right)$

Now follow our two-step separation trick: Write the wave function as a product of single-variable functions...

$\psi_E\left(r,\theta,\phi\right) = R\left(r\right)\Theta\left(\theta\right)\Phi\left(\phi\right)$

...and divide the whole equation by the wave function:

$\dfrac{1}{\Theta}\left(\dfrac{1}{\sin\theta}\right)\dfrac{d}{d\theta}\left(\sin\theta\dfrac{d}{d\theta}\right)\Theta + \dfrac{1}{\Phi}\left(\dfrac{1}{\sin^2\theta}\right)\dfrac{d^2}{d\phi^2}\Phi = \dfrac{2mr^2}{\hbar^2}\left[V\left(r,\theta,\phi\right)-E\right] -\dfrac{1}{R} \dfrac{d}{dr}\left(r^2\dfrac{d}{dr}\right)R$

We haven't quite separated the angular variables yet, but let's come back to that in a bit. First, we note that the only thing keeping the right-hand side of the equation from containing only functions of $$r$$ is the potential. This method therefore only helps us when the potential is a function of $$r$$, which is the case for point-source potentials, as discussed above. With the left-hand side only a function of the angular coordinates, and the right hand side only functions $$r$$, they must both equal a common constant (which we will for the time being call "$$C_1$$"), giving us (after a bit of rearranging) what is known as the radial equation:

$-\dfrac{\hbar^2}{2mr^2}\dfrac{d}{dr}\left(r^2\dfrac{d}{dr}\right)R\left(r\right) + \dfrac{\hbar^2C_1}{2mr^2}R\left(r\right)+V\left(r\right)R\left(r\right)=E\;R\left(r\right)$

Plugging $$C_1$$ into the right-hand side of Equation 4.3.5, we now set out to separate the angular functions:

$\dfrac{1}{\Theta}\left(\dfrac{1}{\sin\theta}\right)\dfrac{d}{d\theta}\left(\sin\theta\dfrac{d}{d\theta}\right)\Theta + \dfrac{1}{\Phi}\left(\dfrac{1}{\sin^2\theta}\right)\dfrac{d^2}{d\phi^2}\Phi = C_1$

Multiply the equation by $$\sin^2\theta$$ and collect the functions of each variable to get:

$\dfrac{1}{\Theta}\left[\sin\theta\dfrac{d}{d\theta}\left(\sin\theta\dfrac{d}{d\theta}\right)\Theta-C_1\sin^2\theta\right] = -\dfrac{1}{\Phi}\dfrac{d^2}{d\phi^2}\Phi$

Separated at last, we can state that both sides of this equation equal a constant (which will will call "$$C_2$$"), giving us two more differential equations in single variables, the polar equation (in $$\theta$$) and the asimuthal equation (in $$\phi$$):

$\sin\theta\dfrac{d}{d\theta}\left(\sin\theta\dfrac{d}{d\theta}\right)\Theta\left(\theta\right) - \left(C_1\sin^2\theta+C_2\right)\Theta\left(\theta\right) = 0$

$\dfrac{d^2}{d\phi^2}\Phi\left(\phi\right)+C_2\Phi\left(\phi\right) = 0$

While the solution to the radial equation depends upon the details of the potential, the solutions to the polar and azimuthal equations do not (other than the fact that the potential is strictly a function of $$r$$). We have seen the azimuthal differential equation before, and we can write the solution this way:

$\Phi\left(\phi\right) = Ae^{\pm i\sqrt{C_2}\phi}$

We cannot go any further to determine constants in this solution without boundary conditions. In our previous encounters with this differential equation, we found our boundary conditions from the potential (e.g. the wave function must vanish at the walls of the infinite square well). That is not the case here, but we do have a condition we can impose: If $$r$$ and $$\theta$$ are held fixed and we change $$\phi$$ by sum multiple of $$2\pi$$, then we come back to the same point in space. We therefore insist that the function $$\Phi$$ returns to the same value every time $$\phi$$ changes by some multiple of $$2\pi$$. This is ensured by insisting that the constant $$\sqrt{C_2}$$ is an integer. For reasons that will become clear later (and with the ancillary benefit of not confusing it with the mass of the particle), we name this integer "$$m_l$$," giving us:

$\Phi_{m_l}\left(\phi\right) = Ae^{im_l\phi},\;\;\;\;\;\;\;\; m_l = 0,\;\pm1,\;\pm2,\;\dots$

The solution to the polar equation, while it doesn't depend upon the potential, is no picnic, and we will not delve into that bit of mathematics. Its solution is called a legendre polynomial, and like the azimuthal case, there is a periodic boundary condition involved, which once again introduces an integer-valued variable, this time related to the constant $$C_1$$, which we call "$$l$$." Specifically, it turns out:

$C_1 = -l\left(l+1\right), \;\;\;\;\;\;\;\; l=0,\;1,\;2,\;\dots$

Notice that the polar equation also includes the constant $$C_2$$, which means its solution also depends upon $$m_l$$. Indeed, this links the integers $$l$$ and $$m_l$$ intimately, which explains the subscript on $$m_l$$. For this reason, the functions $$\Theta_{lm_l}\left(\theta\right)$$ and $$\Phi_{m_l}\left(\phi\right)$$ are usually thrust together to make a single function (called a spherical harmonic function, which have been solved, and can simply be be looked-up) of both variables, and it includes both quantum numbers (typically the "$$l$$" subscript is suppressed in the variable representing this function, as the relationship between the two quantum numbers is understood, and in this context $$m$$ is not going to be confused with the mass):

$Y_{lm}\left(\theta,\phi\right) = \Theta_{lm_l}\left(\theta\right)\Phi_{m_l}\left(\phi\right)$

[Note: In the literature, the indices defining spherical harmonics are frequently written with the $$m_l$$ index raised: $$Y_l^m$$.]

For a given value of $$l$$, it so happens that $$m_l$$ can take on only a limited number of values:

$m_l = 0,\;\pm1,\;\pm2,\;\dots,\; \pm l$

That is, the absolute value of $$m_l$$ can never exceed $$l$$.

Plugging in for the value of $$C_1$$ removes the unknown constant from the radial equation:

$-\dfrac{\hbar^2}{2mr^2}\dfrac{d}{dr}\left(r^2\dfrac{d}{dr}\right)R\left(r\right) - \dfrac{\hbar^2l\left(l+1\right)}{2mr^2}R\left(r\right)+V\left(r\right)R\left(r\right)=E\;R\left(r\right)$

If the particle is bound, then this equation will give quantized partial wave functions $$R_{nl}\left(r\right)$$ (from the presence of the $$l\left(l+1\right)$$ in the radial equation, it seems clear that the radial wave function will have dependence on the quantum number $$l$$ as well as the quantum number $$n$$ that comes from its own differential equation), and we will once again have three quantum numbers: $$n$$, $$l$$, and $$m_l$$.

Just for completeness, we can put the wave function of the stationary state back together:

$\psi_{nlm_l}\left(r,\theta,\phi\right) = AR_{nl}\left(r\right)Y_{lm}\left(\theta,\phi\right)$

All that remains is to plug in the radial potential and use the boundary conditions to solve the differential equation for $$R_{nl\left(r\right)}$$.

The Free Particle

From our work with the free particle in cartesian coordinates, it is clear that it is the easiest coordinate system to use when one wishes to discuss plane waves. But as we have mentioned many times before, plane waves are not the only energy eigenstate wave functions. There is in fact a common energy eigenstate for which spherical coordinates are ideal. It is called a spherical wave, as it emanates radially outward from (or inward toward) a single point, which of course is the origin of our spherical coordinates. This is more often encountered when discussing light rather than massive particles (as light is frequently emitted from a point source), but as we have seen before, quantum mechanics treats both as "quanta."

For reasons we will see when we discuss the hydrogen atom next, a radially spreading (or contracting) wave function has an $$l$$-value equal to zero (and therefore a zero $$m_l$$-value as well), and it is left as an exercise to show that this wave function satisfies the free-particle ($$V\left(r\right)=0$$), radially-outward ($$l=0$$) radial equation:

$\psi_E\left(r\right)=A\dfrac{e^{\pm ikr}}{r},\;\;\;\;\;\;\;\; E=\dfrac{\hbar^2 k^2}{2m}$

The positive-valued exponent corresponds to the wave that moves radially-outward. Notice that the probability density is simply $$\frac{1}{r^2}$$, which reflects an inverse-square reduction in intensity when one converts this single particle wave function to a steady-state stream of particles. Also notice that like the plane wave, this wave function cannot be normalized. And finally, as is always the case with energy eigenstates, we can attach the time-dependence by multiplying the stationary-state solution by the usual $$e^{-i\omega t}$$.

Other Examples

There are other examples that we can solve with spherical symmetry, such as the spherical infinite potential well, and (interestingly) the isotropic harmonic oscillator discussed in Section 4.2, but we will be moving on now to the real prize – the hydrogen atom.