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# 5.2: Applying 3D Quantum Theory to Hydrogen

#### The Principal Quantum Number

We have inched our way up to this moment, solving everything we could with the stationary-state Schrödinger equation for a general radially-symmetric potential, and all that remains is to plug in the potential for the hydrogen atom. Plugging this potential (Equation 5.1.5) into our radial equation (Equation 4.3.16), we have:

$-\dfrac{\hbar^2}{2mr^2}\dfrac{d}{dr}\left(r^2\dfrac{d}{dr}\right)R_{nl}\left(r\right) - \dfrac{\hbar^2l\left(l+1\right)}{2mr^2}R_{nl}\left(r\right) - \dfrac{e^2}{4\pi\epsilon_o r}R_{nl}\left(r\right) = E_n\;R_{nl}\left(r\right)$

The radial part of the wave function has been labeled with the two quantum numbers that define it. As with the case of the legendre polynomials, we will not go through the process of solving this differential equation, but there is one very interesting result that arises. For a general potential, the energy spectrum does depend upon both the $$n$$ and $$l$$ quantum numbers, but surprisingly, the $$\frac{1}{r}$$-dependence of the coulomb potential leads to a result where only the quantum number $$n$$ (called the principal quantum number) plays a role in the energy level. Indeed, this quantum number factors into the energy spectrum of hydrogen in exactly the same way that $$n$$ did for the Bohr model, and this explains why there is only an $$n$$ subscript on the energy in the above equation.

#### Degeneracy

Recall that whenever multiple quantum states result in the same energy, we refer to them as "degenerate." We found that most degeneracies result from symmetries in the physical system, and in fact in this case the spherical symmetry does result in degeneracy – there are $$2l+1$$ different possible $$m_l$$ values for a given energy. But the independence of the energy on the $$l$$ quantum number is not a result of spherical symmetry – it would not occur for any other radial potential function. This is the most famous example of an "accidental degneracy." We will see this degeneracy broken when we get to atoms with additional electrons, as the net potential affecting a single electron ceases to behave like a simple coulomb potential.

We found that the magnetic quantum number was limited by the value of the orbital quantum number. It turns out that there is a similar limitation on the orbital quantum number imposed by the principal quantum number. Namely, the value of $$l$$ must be strictly less than the value of $$n$$. Putting all these quantum number limitations together, we get the following degeneracies:

$\begin{array}{c} \text{energy level} & & \text{principal} & & \text{orbital} & & \text{magnetic} & & \text{degeneracy} \\ \text{ground} & & n=1 & & l=0\;(only) & & m_o=0\;(only) & & \text{non-degenerate} \\ 1^{st}\text{-excited} & & n=2 & & l=0,\;1 & & m_o=0, \;\;m_1=0,\;\pm 1 & & \text{4-fold} \\ 2^{nd}\text{-excited} & & n=3 & & l=0,\;1,\;2 & & m_o=0, \;\;m_1=0,\;\pm 1, \;\;m_2=0,\;\pm 1,\;\pm 2 & & \text{9-fold} \\ \vdots & & \vdots & & \vdots & & \vdots & & \vdots \\ n^{th}\text{-excited} & & n & & l=0,\;1,\dots n-1 & & m_o=0, \;\;m_1=0,\;\pm 1, \;\;\dots \;\;m_l=0,\;\pm 1\dots,\; \pm l & & n^2\text{-fold} \end{array}$

There is a rather arcane, but still universally-used, notation that describes energy levels and total angular momentum orbitals, called spectroscopic notation. It consisted of two characters, one for each the principal and orbital quantum numbers. The first character is the integer that is the principal quantum number, and the second is a letter. The letter for each value of $$l$$ is given below:

$\begin{array}{l} \text{letter} & & s & & p & & d & & f & & g & & ...\text{(alphabetical from here)} \\ l\text{-value} & & 0 & & 1 & & 2 & & 3 & & 4 & & ... \end{array}$

So for example, the $$l=2$$ orbital of the 4th excited state is written "$$5d$$."

#### The Wave Function

We have written the differential equations for the various separated parts of the wave function of the hydrogen atom, but aside from the azimuthal portion, we have not written down any solutions. We know that the full solution is the radial wave function multiplied by the spherical harmonic function, and while the full wave function is what needs to satisfy a normalization condition, it is standard practice to normalize each of these pieces individually. For an extensive list of normalized spherical harmonics, go here.

The solution to the radial equation bears some resemblance to what we found for the harmonic oscillator in one dimension. In that case, we found that the "foundation" of the wave function (ground state) was a gaussian curve, and higher energy level states came from multiplying this gaussian by hermite polynomials, the orders of which assured the right number of antinodes. For the radial part of the hydrogen atom wave function, the foundation function is different, and the polynomials are as well, but the idea is the same. Let's look at the differential equation for the ground state, to see what this foundation function is. For the ground state, there can be no orbital angular momentum, so $$l=0$$, simplifying the differential equation to:

$-\dfrac{\hbar^2}{2mr^2}\dfrac{d}{dr}\left(r^2\dfrac{d}{dr}\right)R_{10}\left(r\right) - \dfrac{e^2}{4\pi\epsilon_o r}R_{10}\left(r\right) = E_1\;R_{10}\left(r\right)$

Like the one dimensional harmonic oscillator, the ground state wave function is a decaying exponential, but the dependence on position in the exponential is linear rather than quadratic. Remarkably, the constant that appears in the exponential is the principal quantum number  (in the case of the ground state, this is just 1) multiplied by the Bohr radius. The (unnormalized) ground state radial wave function is:

$R_{10}\left(r\right) = Ae^{-\dfrac{r}{a_o}}$

Plugging this in above, performing the derivatives, and doing the algebra confirms that this is the form of the ground state wave function. We'll discuss the normalization of this wave function shortly. As mentioned above, the higher energy levels (and the effect of different orbital angular momenta) is a matter of multiplying the exponential by a polynomial (which is written in powers of $$\rho \equiv \frac{r}{a_o}$$). We will not go into detail about this polynomial, except that it is related to what is called the associated laguerre polynomial, which depends on both the $$n$$ and $$l$$ quantum numbers. Below is a list of the first few normalized radial wave functions:

$R_{nl}\left(\rho\right) = 2\left(na_o\right)^{-\frac{3}{2}} e^{-\rho/n} \times \left\{ \begin{array}{l} 1 & & & n=1,\;l=0 \\ \\ 1-\dfrac{1}{2}\;\rho & & & n=2,\;l=0 \\ \dfrac{1}{2\sqrt{3}}\;\rho & & & n=2,\;l=1 \\ \\ 1-\dfrac{2}{3}\;\rho +\dfrac{2}{27}\;\rho^2 & & & n=3,\;l=0 \\ \dfrac{2\sqrt{2}}{9}\;\rho-\dfrac{\sqrt{2}}{27}\;\rho^2 & & & n=3,\;l=1 \\ \dfrac{\sqrt{2}}{27\sqrt{5}}\;\rho^2 & & & n=3,\;l=2 \end{array} \right\} \;\;,\;\;\;\;where:\;\; \rho \equiv \dfrac{r}{a_o}$

#### Normalization and Orthogonality

We already stated the normalization condition for the wave function in three dimensions in spherical coordinates, in Equation 4.1.5. With our separated wave function, we have normalized each of the parts, which means that each individual integral applied to the separated partial wave function yields zero. Namely:

$\begin{array}{l} 1 = \int \limits_0^{2\pi} \Phi^*_{m_l}\left(\phi\right)\;\Phi_{m_l}\left(\phi\right)\;d\phi \\ 1 = \int \limits_0^{\pi} \Theta^*_{lm_l}\left(\theta\right)\;\Theta_{lm_l}\left(\theta\right)\;\sin\theta\;d\theta \\ 1 = \int \limits_0^\infty R^*_{nl}\left(r\right)\;R_{nl}\left(r\right)\;r^2\;dr \end{array}$

The first of these is simple. The product of the azimuthal wave function with its complex conjugate is just 1, so the integral comes out to $$2\pi$$, making the normalization constant $$\frac{1}{\sqrt{2\pi}}$$:

$\Phi_{m_l}\left(\phi\right)=\dfrac{1}{\sqrt{2\pi}} e^{-im_l\phi}$

This function is multiplied by $$\Theta_{lm_l}$$ to construct the spherical harmonic function $$Y_{lm}$$, and indeed a glance at a list of the spherical harmonic functions shows this.

Equation 4.1.7 expresses that two quantum eigenstates with any equivalent quantum numbers that are different have a vanishing inner product. In position space, this inner product turns into an overlap integral (throw the identity operator between the bra and ket), which means that for the azimuthal wave function, we should see:

$\int \limits_0^{2\pi} \Phi^*_{m_l}\left(\phi\right)\;\Phi_{m'_l}\left(\phi\right)\;d\phi = \delta_{m_lm'_l}$

We already showed the normalization, so let's confirm orthogonality by plugging-in for the wave functions with $$m_l \ne m'_l$$.

$\int \limits_0^{2\pi} \left[\dfrac{1}{\sqrt{2\pi}} e^{im_l\phi}\right]\left[\dfrac{1}{\sqrt{2\pi}} e^{-im'_l\phi}\right]d\phi = \dfrac{1}{2\pi}\int \limits_0^{2\pi} e^{i\left(m_l-m'_l\right)\phi}\;d\phi = \left[\dfrac{1}{2\pi i\left(m_l-m'_l\right)}e^{i\left(m_l-m'_l\right)\phi}\right]_0^{2\pi} = \dfrac{1}{2\pi i\left(m_l-m'_l\right)}\left(1-1\right) = 0$

The same orthogonality and normalization conditions hold for the other two partial wave functions – if the quantum numbers are identical, then the normalization integral holds, and if any do not match, then the integral vanishes. Of course, the normalization integrals are far less trivial for the polar and radial parts, as they are real-valued, and don't result in a simple cancelation to one, as we saw for the azimuthal part.

The standard computation of probability of position is to start with the probability of the particle being in an infinitesimal volume, as described in Equation 4.1.2. Then integrating this over limits that define a specific finite volume gives the probability that the particle is found within that volume. Suppose the volume is a spherical shell centered at the origin. Then the integral over the volume will entail an integral over all possible azimuthal and polar angles, which, thanks to the fact that those parts of the wave function are normalized, comes out to just 1. The probability of the particle lying in a shell of thickness $$dr$$ (from $$r$$ to $$r+dr$$) can therefore be found using just the radial part of the wave function:

$Prob\left(r\rightarrow r+dr\right) = r^2\;R^2_{nl}\left(r\right)\;dr$

Notice the presence of the $$r^2$$ here. This comes from the radial part of the infinitesimal volume element, and can sometimes cause confusion if one is too accustomed to the separated probability densities we found for cartesian coordinates. There is one place where the presence of this factor of $$r^2$$ obviously has an important impact. The radial wave function for $$n=1,\;l=0$$ is actually non-zero at the origin, which we would normally think means that the probability of finding the electron at the position of the proton is non-zero. But the probability density requires a product with $$r^2$$, so while the radial part of the wave function doesn't vanish at the origin, the probability of finding the electron there does.

When we use the radial wave function to compute things like expectation values, we have to remember to keep this factor in the integral (though the units not working out properly will be a clue if it is forgotten).  That is, suppose we are computing an expectation value of a quantity that is purely a function of $$r$$, $$f\left(r\right)$$. The integrals over the angles will again result in 1, which means that the expectation value is $$f\left(r\right)$$ integrated with the probability density:

$\left<\;f\left(r\right)\;\right> = \int \limits_0^\infty f\left(r\right) r^2\;R^2_{nl}\left(r\right)\;dr$

It is useful at this point to compare look at graphs of the probability density functions and see if they make sense, using what we know about wave functions in one-dimensional wells. In this case, the "one dimensional" well is bound on one side ($$r=0$$) with what is effectively an infinite barrier – certainly there can be no wave function in the $$r<0$$ region (how can the electron be a negative distance from the proton!?). On the other side of the well is the coulomb potential.  We will look at the lowest three energy levels in this well.

Let's first consider the $$s$$ ($$l=0$$) states. In these cases, no part of the electron's kinetic energy comes from motion going around the origin, as there is no angular momentum. That is, all of the momentum of the particle is radial. This corresponds nicely to what we would expect from one-dimensional wells, as can be seen in the graphs of these probability densities below. An antinode is added with each increase in energy level, just as in the case of one-dimensional potentials.

Figure 5.2.1 – The 1s, 2s, and 3s Radial Probability Densities

Now consider what happens if we allow the electron to have some angular momentum about the origin. For the same energy level, less of the kinetic energy will be in the form of radial motion, so our "one-dimensional" radial well should reflect this. We know from our study of one-dimensional wells that we are very limited by our boundary conditions when it comes to representing a certain amount of energy. The energy is quantized, and the only way to change it is to change the number of antinodes. So we would expect that the more angular momentum the electron has, the less of its kinetic energy comes from radial motion, which means its radial probability density must have fewer antinodes.  The graphs of the three $$n=3$$ probability densities below bears this out:

Figure 5.2.2 – The 3s, 3p, and 3d Radial Probability Densities

Note that there are three antinodes for the state with zero angular momentum, two for the state with $$l=1$$, and one for the state with $$l=2$$. The probability density loses an antinode with each increment of the $$l$$ quantum number, and runs out of antinodes just as $$l$$ hits its limit (recall $$l<n$$). The "accidental degeneracy" here occurs because the kinetic energy loss that corresponds to a lost antinode is exactly balanced by the increase in the kinetic energy do to the angular motion of the electron.

There is one other feature we can extract if we compare the three states of maximal angular momentum: $$1s$$, $$2p$$, and $$3d$$. If we think about orbits classically, it's clear that for a fixed amount of kinetic energy, the angular momentum is a maximum for circular orbits (because none of the energy is in the radial mode). Of course, quantum mechanically the electron is not following a specific path, so we can't say that maximum angular states are circular orbits, but it's not a stretch – given the fact that we are requiring maximal angular momentum – that some probabilistic measure of the orbit will bear a resemblance to circular. Indeed, it turns out that the most probable positions of the electron for these maximal-angular-momentum orbits correspond to the orbits computed in the Bohr model (which, you recall, assumed circular orbits). See the graph below for confirmation of this.

Figure 5.2.3 – The 1s, 2p, and 3d Radial Probability Densities