Starting from this post, we are going to apply the techniques we learned in QM to solve some real-world problems. In this post, we are going to describe the hydrogen atom with Quantum Mechanics!

QM in 3-D

We’ve been doing QM in 1-D for a long time now and that assumption has almost become explicit. However, the hydrogen atom is a three dimensional structure that obviously requires more sophisticated treatment. It necessitates an understanding of QM in 3-D. After that, we will write out the Hamiltonian for a hydrogen atom and solve the Schrodinger equation in 3-D. This will give us the energy eigenfunctions, which describe the shape of the electron orbitals we know well from chemistry.

Thankfully, adding two additional dimensions is not a hard adjustment to make. We just need to make sure that every 1-D quantity/expression/operation finds its 3-D correspondence. Here is a table that summarizes all the essential changes.

Hamiltonian for Two Particles in Center of Mass Frame

Now we try to write out the Hamiltonian of the hydrogen atom. The main components of a hydrogen atom are a proton and an electron. When analyzing two particle systems like this, it is convenient to transform to the center of mass frame and express things there.

Consider the most general case where we have particles 1 and 2. We have:

In the “normal” frame (lab frame), we can easily write out the Hamiltonian of the two particle system as:

Where we have expressed the kinetic energy simply as the sum of the kinetic energy of the two particles. In the center of mass frame, we seek a similar approach. The only difference is that we think of kinetic energy as coming from two components: the kinetic energy of the motion of the center of mass and the kinetic energy of relative motion. The kinetic energy from the motion of the center of mass is easily expressed as:

where:

The kinetic energy from relative motion is a bit tricky. We start from consider the momentum of particle 1 in the center of mass frame:

Comparing the final form of this expression with $\pmb p = m\pmb v$, we can define a new mass called the reduced mass:

and call:

Relative velocity and relative position.

Thus, we can write the Hamiltonian in a new way:

Essentially, we have converted the momentum variables from $(\pmb{p_1}, \pmb{p_2})$ to $(\pmb{p_{rel}}, \pmb P)$ and correspondingly converted the position variables from $(\pmb{r_1}, \pmb{r_2})$ to $(\pmb r, \pmb{R_{CM}})$.

To check that this transformation is valid even with all the operator symbols, we just need to check the commutator relationships between the various position and momentum variables/operators. I won’t go through that derivation here since it’s simple but tedious.

We can do a couple more simplifications to out center of mass frame Hamiltonian. Assuming we are in the center of mass frame, the first term describing the kinetic energy of the center of mass goes to 0. Furthermore, with the Coulomb potential present in hydrogen atoms, the potential energy term is only dependent on the separation of the two particles (such potentials are called central potentials). With these information in mind, we can write our Hamiltonian as:

Separation of Variables

We now want to solve the Schrodinger equation:

The first step in solving the Schrodinger equation is to expand out the del operator in the Hamiltonian. We know that the hydrogen atom is a spherically symmetric structure, so it is best to expand out del in spherical coordinates:

Substitute del into the Schrodinger equation, we will get a single equation of three unknowns ($r, \theta, \phi$). A common approach to clean up this mess is to use a technique called separation of variables. The purpose of this technique is to manipulate the equation algebraically such that each term in the equation is only a function of one variable. Then, we can gather terms that dependent on the same variable on either sides of the equation. To have the two sides that depend on different variables equal to each other, the only way is to set both sides equal to a constant. This gives us to much simpler differential equations to solve. Let’s try to apply this technique to practice and see what happens.

First, we separate the wave function into a radial part and an angular part. Let:

Substitute this form into the Schrodinger equation, we get:

Now we need to take out the Y’s in the r dependent terms and take out the R’s in the angle dependent terms. To do so, we multiply everything by $-\frac {2\mu r^2}{\hbar^2}$, divide by RY (which is just the wave function $\psi$ itself), and gather like terms. We get:

Now we see that the LHS is only dependent on r, and the RHS is only dependent on the angles. This means the only these two sides are equal to each other is that they are both equal to a constant. We define that constant to be $l(l+1)$ and separate our result into two equations:

The first equation is known as the Radial Equation, and the second equation is known as the Angular Equation. We proceed to solve these two equations one at a time.

Solving the Angular Equation

We focus on the Angular Equation first. Note that this differential equation is still challenging to solve because there are two variables in it. Now you should know what to do. Use separation of variables once more!

Let:

Substitute in, divide by $\Theta\Phi$, and rearrange, we get:

Again, both sides need to be equal to a constant. let that constant be $m^2$, and look at the $\phi$ dependent part. we have:

Now we finally have something that’s easy to solve. The solution to this differential equation is obviously:

If you still remember how we defined the wave function, you will see that this is only one-third of the solution. We carry on with the $\theta$ dependent part:

The solution to this differential equation is:

Where $A$ is the normalization factor and $P^m_l$ is the Legendre function. The formulae for the Legendre functions and the Legendre polynomials can be found in Townsend’s textbook.

Note that $l, m$ in this case serves as a numbering system for the wave functions. The Legendre function imposes that:

From the angular wavefunction, we often two quantum numbers: $m$ the magnetic quantum number as it is related to the magnetic properties of the hydrogen atom, $l$ the azimuthal quantum number because it is obtained from the azimuthal equation.

Now we can put the two parts together and obtain the complete solution to the angular part of the Schrodinger equation. Wave functions in 3-D need to be normalized in the angular and the radial part separately, so we already have enough information to compute the normalization factor for the angular wave function. After much algebra, we can obtain the result, which we call spherical harmonics ($Y_{lm}$):

The relationship between $l$ and $m$ should let you recall post 2 where we discussed angular momentum. Indeed, $l(l+1)$ is also the eigenvalue of the total orbital momentum operator $\hat{L}^2$ (remember that we mysteriously set the constant to be $l(l+1)$ in the derivation), and $m$ is the eigenvalue of orbital momentum operator $\hat{L_z}$.

You can check the above claims by applying the operators to the spherical harmonics in the position space. In the position space,

Solving the Radial Equation

Now we turn our attention to solving the radial equation. The first step is to substitute the form of the Coulomb potential for the proton-electron interaction into the radial equation we obtained earlier. At the same time, we make the substitution $u_{(r)} = rR_{(r)}$ to simplify. We have:

where e is the charge of an electron.

The part

is known as the effective potential.

First, let’s try to solve this differential equation when $l=0$. We know from solving the angular equation that l is a quantum number that “labels” wave functions at different energy levels, so finding the solution when $l=0$ should give us information about the ground state wave function. This will help us understand the most general solutions that we will get later.

Subbing $l=0$ into the Schrodinger equation and using $R$ instead of $u$, we get:

Even without looking at this equation, we know that R has to approach 0 when r approaches infinity, otherwise the wave function cannot be properly normalized. We also know that R needs to be well-behaved at the origin. Keeping these two criteria in mind, it’s sensible to try:

we get:

This equation is true at all $r$, so we need to set the two $r$ dependent terms to cancel out each other, which means:

and

Solving the first equation, we get:

Note that the unit of $\kappa$ is $\frac{1}{[distance]}$. We define $a_0$ to be:

and consequently:

$a_0$ is called the Bohr’s radius. It describes the size of a ground state hydrogen atom. In fact, Bohr’s radius is the average (expectation value of) distance between the proton and the electron at the ground state. The electron exist as a probability cloud, so it only makes sense to talks about the average distance.

Using values obtained from delicate experiments, we now know that:

This energy is called the Rydberg constant, it is the energy of a ground state hydrogen atom. This energy is negative because the hydrogen atom is a bound system. The electron needs to absorb at least 1 Rydberg of energy to become unbound.

Now let’s deal with the most general case ($l\ne 0$). Before manipulating the radial equation, we need to first determine the form of $u_{(r)}$ by looking at its asymptotic behaviors.

Remember our radial equation:

First consider the case when $r\to 0$. The $\frac{1}{r^2}$ term dominate over the $\frac{1}{r}$ tern, and we have:

Try $u=r^p$ as a solution:

The RHS goes to 0 as $r\to 0$. So we need the expression inside the square bracket equal to 0. Which means:

We eliminate the $p=-l$ solution because it blows up at the origin. So from this $r\to 0$ case, we know that:

Now let’s consider the case when $r\to \infty$. We are only left with:

We know that E is less than 0 because the hydrogen atom is a bound system. This criterion allows us to quickly solve this differential equation and get:

Where $\frac{\hbar^2\kappa^2}{2\mu} = |E|$. We can eliminate the $+\frac{\kappa r}{2}$ solution because it goes to infinity at large $r$.

Now we have two “parts” to the solution of u. In order to make sure that u has the characteristic of both parts, we assume that the “complete” u is a product of the two “parts” and some other arbitrary function that is yet to be determined.

To determine the unknown function, we need to return to our radial equation and find its solution. Let’s remind ourselves what it looks like:

To simplify things, we make the following definitions:

Then the differential equation becomes:

Now sub in the form of $u_{(r)}$ we obtained earlier and solve using a power series expansion on F, the unknown function. After MUCH algebra, we get:

and the power series terminates when

n is called the primary quantum number and is used together with the azimuthal and magnetic quantum number to label a state/wave function. In fact, you can check that the Hamiltonian $\hat H$ commutes with $\hat{L}^2$ and $\hat{L_z}$, so that we can label the state of a hydrogen atom by

For a given $n$, $l=0,1,2, \cdots, n-1$, and for a given $l$, $m=-l,-l+1, \cdots, l-1,l$, which means for a given $n$, there are $n^2$ possible energy states.

Labeling using these three quantum numbers is equivalent to the naming conventions for the orbitals in chemistry. $l=0$ corresponds to the s-orbital, $l=1$ corresponds to the p-orbital, and so on. The three dimensional shapes of the orbitals are determined by the wave functions we solves (three parts multiplied together).

Lastly, from the equality between n and $\lambda$, we can get:

Which means for all energy states, we have:

References

[1] Zajc, William. PHYS GU4021-4022 Quantum Mechanics I-II, Department of Physics, Columbia University. 2019-2020.

[2] Townsend, John. A Modern Approach to Quantum Mechanics. 2nd Edition.