In between the Sakurai problems, I decided to tackle a small problem I set for myself.

The Sakurai quantum mechanics book is directed at about graduate student level, meaning that it explicitly overlooks problems that it deems too ‘undergraduate.’ When I started into the next problem in the chapter, which deals with the Wigner-Eckert relation, I decided to direct myself at a ‘lower level’ problem that demands practice from time to time. I worked in early January solving the angular component of the hydrogen atom by deriving the spherical harmonics and much of my play time since has been devoted to angular and angular momentum type problems. So, I decided it would be worth switching up a little and solving the radial portion of the hydrogen atom electron central force problem.

One of my teachers once suggested that deriving the hydrogen atom was a task that any devoted physicist should play with every other year or so. Why not, I figured; the radial solution is actually a bit more mind boggling to me than the angular parts because it requires some substitutions that are not very intuitive.

The hydrogen atom problem is a classic problem mainly because it’s one of the last exactly solvable quantum mechanics problems you ever encounter. After the hydrogen atom, the water gets deeper and the field starts to focus on tools that give insight without actually giving exact answers. The only atomic system that is exactly solvable is the hydrogen atom… even helium, with just one more electron, demands perturbation in some way. It isn’t exactly crippling to the field because the solutions to all the other atoms are basically variations of the hydrogen atom and all, with some adjustment, have hydrogenic geometry or are superpositions of hydrogen-like functions that are only modified to the extent necessary to make the energy levels match. Solving the hydrogen atom ends up giving profound insight to the structure of the periodic table of the elements, even if it doesn’t actually solve for all the atoms.

As implied above, I decided to do a simplified version of this problem, focusing only on the radial component. The work I did on the angular momentum eigenstates was not in context of the hydrogen electron wave function, but can be inserted in a neat cassette to avoid much of the brute labor of the hydrogen atom problem. The only additional work needed is solving the radial equation.

A starting point here is understanding spherical geometry as mediated by spherical polar coordinates.

A hydrogen atom, as we all know from the hard work of a legion of physicists coming into the turn of the century, is a combination of a single proton with a single electron. The proton has one indivisible positive charge while the electron has one indivisible negative charge. These two charges attract each other and the proton, being a couple thousand times more massive, pulls the electron to it. The electron falls in until the kinetic energy it gains forces it to have enough momentum to be unlocalized to a certain extent, as required by quantum mechanical uncertainty. The system might then radiate photons as the electron sorts itself into a stable orbiting state. The resting combination of proton and electron has neutral charge with the electron ‘distributed’ around the proton in a sort of cloud as determined by its wave-like properties.

The first approximation of the hydrogen atom is a structure called the Bohr model, proposed by Niels Bohr in 1913. The Bohr model features classical orbits for the electron around the nucleus, much like the moon circles the Earth.

This image, from duckster.com, is a crude example of a Bohr atom. The Bohr atom is perhaps the most common image of atoms in popular culture, even if it isn’t correct. Note that the creators of this cartoon didn’t have the wherewithall to make a ‘right’ atom, giving the nucleus four plus charges and the shell three minus… this would be a positively charged ion of Beryllium. Further, the electrons are not stacked into a decent representation for the actual structure: cyclic orbitals would be P-orbitals or above, where Beryllium has only S-orbitals for its ground state, which possess either no orbital angular momentum, or angular momentum without any defined direction. But, it’s a popular cartoon. Hard to sweat the small stuff.

The Bohr model grew from the notion of the photon as a discrete particle, where Bohr postulated that the only allowed stable orbits for the electron circling the nucleus is at integer quantities of angular momentum delivered by single photons… as quantized by Planck’s constant. ‘Quantized’ is a word invoked to mean ‘discrete quantities’ and comes back to that pesky little feature Deepak Chopra always ignores: the first thing we ever knew about quantum mechanics was Planck’s constant –and freaking hell is Planck’s constant small! ‘Quantization’ is the act of parsing into discrete ‘quantized’ states and is the word root which loaned the physics field its name: Quantum Mechanics. ‘Quantum Mechanics’ means ‘the mechanics of quantization.’

Quantum mechanics, as it has evolved, approaches problems like the hydrogen atom using descriptions of energy. In the classical sense, an electron orbiting a proton has some energy describing its kinetic motion, its kinetic energy, and some additional energy describing the interaction between the two masses, usually as a potential source of more kinetic energy, called a potential energy. If nothing interacts from the outside, the closed system has a non-varying total energy which is the sum of the kinetic and potential energies. Quantum mechanics evolved these ideas away from their original roots using a version of Hamiltonian formalism. Hamiltonian formalism, as it appears in quantum, is a way to merely sum up kinetic and potential energies as a function of position and momentum –this becomes complicated in Quantum because of the restriction that position and momentum cannot be simultaneously known to arbitrary precision. But, Schrodinger’s equation actually just boils down to a statement of kinetic energy plus potential energy.

Here is a quick demonstration of how to get from a statement of total energy to the Schrodinger equation:

After ‘therefore,’ I’ve simply multiplied in from the right with a wave function to make this an operator equation. The first term on the left is kinetic energy in terms of momentum while the second term is the Gaussian CGS form of potential energy for the electrical central force problem (for Gaussian CGS, the constants of permittivity and permeability are swept under the rug by collecting them into the speed of light and usually a constant of light speed appears with magnetic fields… here, the charge is in statcoulombs, which take coulombs and wrap in a scaling constant of 4*Pi.) When you convert momentum into its position space representation, you get Schrodinger’s time independent equation for an electron under a central force potential. The potential, which depends on the positional expression of ‘radius,’ has a negative sign to make it an attractive force, much like gravity.

Now, the interaction between a proton and an electron is a central force interaction, which means that the radius term could actually be pointed in any direction. Radius would be some complicated combination of x, y and z. But, because the central force problem is spherically symmetric, if we could move out of Cartesian coordinates and into spherical polar, we get a huge simplification of the math. The inverted triangle that I wrote for the representation of momentum is a three dimensional operator called the Laplace operator, or ‘double del.’ Picking the form of del ends up casting the dimensional symmetry of the differential equation… as written above, it could be Cartesian or spherical polar or cylindrical, or anything else.

A small exercise I sometimes put myself through is defining the structure of del. The easiest way that I know to do this is to pull apart the divergence theory of vector calculus in Spherical polar geometry, which means defining a differential volume and differential surfaces.

Well, that turned out a little neater than my usual meandering crud.

This little bit of math is defining the geometry of the coordinate variables in spherical polar coordinates. You can see the spherical polar coordinates in the Cartesian coordinate frame and they consist of a radial distance from the origin and two angles, Phi and Theta, that act at 90 degrees from each other. If you pick a constant radius in spherical polar space, you get a spherical surface where lines of constant Phi and Theta create longitude and latitude lines, respectively, making a globe! You can establish a right handed coordinate system in spherical polar space by picking a point and considering it to be locally Cartesian… the three dimensions at this point are labeled as shown, along the outward radius and in the directions in which each of the angles increases.

If you were to consider an infinitesimal volume of these perpendicular dimensions, at this locally cartesian point, it would be a volume that ‘approaches’ cubic. But then, that’s the key to calculus: recognizing that 99.999999 effectively approaches 100. So then, this framework allows you to define the calculus occurring in spherical polar space. The integral performed along Theta, Phi and Rho would be adding up tiny cubical elements of volume welded together spherically, while the derivative would be with respect to each dimension of length as locally defined. The scaling values appear because I needed to convert differentials of angle into linear length in order to calculate volume, which can be accomplished by using the definition of the radian angle, which is arc length per radius –a curve is effectively linear when an arc becomes so tiny as to be negligible when considering the edges of an infinitesimal cube, like thinking about the curvature of the Earth effecting the flatness of the sidewalk outside your house.

The divergence operation uses Green’s formulas to say that a volume integral of divergence relates to a surface integral of flux wrapping across the surface of that same volume… and then you simply chase the constants. All that I do to find the divergence differential expression is to take the full integral and remove the infinite sum so that I’m basically doing algebra on the infinitesmal pieces, then literally divide across by the volume element and cancel the appropriate differentials. There are three possible area integrals because the normal vector is in three possible directions, one each for Rho, Theta and Phi.

The structure becomes a derivative if the volume is in the denominator because volume has one greater dimension than any possible area, where the derivative is with respect to the dimension of volume that doesn’t cancel out when you divide against the areas. If a scaling variable used to convert theta or phi into a length is dependent on the dimension of the differential left in the denominator, it can’t pass out of the derivative and remains inside at completion. The form of the divergence operation on a random vector field appears in the last line above. The value produced by divergence is a scalar quantity with no direction which could be said to reflect the ‘poofiness’ of a vector field at any given point in the space where you’re working.

I then continued by defining a gradient.

Gradient is basically an opposite operation from divergence. Divergence creates a scalar from a vector which represents the intensity of ‘divergence’ at some point in a smooth function defined across all of space. Gradient, on the other hand, creates a vector field out of a scalar function, where the vectors point in the dimensional direction where the function tends to be increasing.

This is kind of opaque. One way to think about this is to think of a hill poking out of a two dimensional plane. A scalar function defines the topography of the hill… it says simply that at some pair of coordinates in a plane, the geography has an altitude. The gradient operation would take that topography map and give you a vector field which has a vector at every location that points in the direction toward which the altitude is increasing at that location. Divergence then goes backward from this, after a fashion: it takes a vector map and coverts it into a map which says ‘strength of change’ at every location. This last is not ‘altitude’ per se, but more like ‘rate at which altitude is changing’ at a given point.

The Laplace operator combines gradient with divergence as literally the divergence of a gradient, denoted as ‘double del,’ the upside-down triangle squared.

In the last line, I’ve simply taken the Laplace operator in spherical polar coordinates and dropped it into its rightful spot in Schrodinger’s equation as shown far above. Here, the wave equation, called Psi, is a density function defined in spherical polar space, varying along the radius (Rho) and the angles Theta and Phi (the so-called ‘solid angle’). Welcome to greek word salad…

What I’ve produced is an explicit form for Schrodinger’s equation with a coordinate set that is conducive to the problem. This differential equation is a multivariate second order partial differential equation. You have to solve this by separation of variables.

Having defined the hydrogen atom Schrodinger equation, I now switch to the more simple ‘radial only’ problem that I originally hinted at. Here’s how you cut out the angular parts:

You just recognize that the second and third differential terms are collectively the square of the total angular momentum and then use the relevant eigenvalue equation to remove it.

The L^2 operator comes out of the kinetic energy contained in the electron going ‘around.’ For the sake of consistency, it’s worth noting that the Hamiltonian for the full hydrogen atom contains a term for the kinetic energy of the proton and that the variable Rho refers to the distance between the electron and proton… in its right form, the ‘m’ given above is actually the reduced mass of that system and not directly the mass of the electron, which gives us a system where the electron is actually orbiting the center of mass, not the proton.

Starting on this problem, it’s convenient to recognize that the Psi wave function is a product of a Ylm (angular wave function) with a Radial function. I started by dividing out the Ylm and losing it. Psi basically just becomes R.

The first thing to do is take out the units. There is a lot of extra crap floating around in this differential equation that will obscure the structure of the problem. First, take the energy ‘E’ down into the denominator to consolidate the units, then make a substitution that hides the length unit by setting it to ‘one’. This makes Rho a multiple of ‘r’ involving energy. The ‘8’ wedged in here is crazily counter intuitive at this point, but makes the quantization work in the method I’ve chosen! I’ll point out the use when I reach it. At the last line, I substitute for Rho and make a bunch of cancellations. Also, in that last line, there’s an “= R” which fell off the side of the picture –I assure you it’s there, it just didn’t get photographed.

After you clean everything up and bringing the R over from the behind the equals sign, the differential equation is a little simpler…

The ‘P’ and ‘Q’ are quick substitutions made so that I don’t have to work as hard doing all this math; they are important later, but they just need to be simple to use at the moment. I also make a substitution for R, by saying that R = U/r. This converts the problem from radial probability into probability per unit radius. The advantage is that it lets me break up the complicated differential expression at the beginning of the equation.

The next part is to analyze the ‘asymptotic behavior’ of the differential equation. This is simply to look at what terms become important as the radius variable grows very big or very small. In this case, if radius gets very big, certain terms become small before others. If I can consider the solution U to be a separable composition of parts that solve different elements of this equation, I can create a further simplification.

If you consider the situation where r is very very big, the two terms in this equation which are 1/r or 1/r^2 tend to shrink essentially to zero, meaning that they have no impact on the solution at big radii. This gives you a very simple differential equation at big radii, as written at right, which is solved by a simple exponential with either positive or negative roots. I discard the positive root solution because I know that the wave equation must suppress to zero as r goes far away and because the positive exponential will tend to explode, becoming bigger the further you get from the proton –this situation would make no physical sense because we know the proton and electron to be attractive to one another and solutions that have them favor being separated don’t match the boundaries of the problem. Differential equations are frequently like this: they have multiple solutions which fit, but only certain solutions that can be correct for a given situation –doing derivatives loses information, meaning that multiple equations can give the same derivative and in going backward, you have to cope with this loss of information. The modification I made allows me to write U as a portion that’s an unknown function of radius and a second portion that fits as a negative exponent. Hidden here is a second route to the same solution of this problem… if I considered the asymptotic behavior at small radii. I did not utilize the second asymptotic condition.

I just need now to find a way to work out the identity of the rest of this function. I substitute the U back in with its new exponentially augmented form…

With the new version of U, the differential equation rearranges to give a refined set of differentials. I then divide out the exponential so that I don’t have it cluttering things up. All this jiggering about has basically reduced the original differential equation to a skin and bones that still hasn’t quite come apart. The next technique that I apply is the Frobenius method. This technique is to guess that the differential equation can be solved by some infinite power series where the coefficients of each power of radius control how much a particular power shows up in the solution. It’s basically just saying “What if my solution is some polynomial expression Ar^2 -Br +C,” where I can include as many ‘r’s as I want. This can be very convenient because the calculus of polynomials is so easy. In the ‘sum,’ the variable n just identifies where you are in the series, whether at n=0, which just sets r to 1, or n=1000, which has a power of r^1000. In this particular case, I’ve learned that the n=0 term can actually be excluded because of boundary conditions since the probability per unit radius will need to go to zero at the origin (at the proton), and since the radius invariant term can’t do that, you need to leave it out… I didn’t think of that as I was originally working the problem, but it gets excluded anyway for a second reason that I will outline later.

The advantage of Frobenius may not be apparent right away, but it lets you reconstruct the differential equation in terms of the power series. I plug in the sum wherever the ‘A’ appears and work the derivatives. This relates different powers of r to different A coefficients. I also pull the 1/r and 1/r^2 into their respective sums to the same affect. Then, you rewrite two of the sums by advancing the coefficient indices and rewriting the labels, which allows all the powers of r to be the same power, which can be consolidated all under the same sum by omitting coefficients that are known to be zero. This has the effect of saying that the differential equation is now identically repeated in every term of the sum, letting you work with only one.

The result is a recurrence relation. For the power series to be a solution to the given differential equation, each coefficient is related to the one previous by a consistent expression. The existence of the recurrence relation allows you to construct a power series where you need only define one coefficient to immediately set all the rest. After all those turns and twists, this is a solution to the radial differential equation, but not in closed form.

Screwing around with all this math involved a ton of substitutions and a great deal of recasting the problem. That’s part of why solving the radial equation is challenging. Here is a collection of all the important substitutions made…

As you can see, there is layer on layer on layer of substitution here. Further, you may not realize it yet, but something rather amazing happened with that number Q.

If you set Q/4 = -n, the recurrence relation which generates the power series solution for the radial wave function cuts off the sequence of coefficients with a zero. This gives a choice for cutting off the power series after only a few terms instead of including the infinite number of possible powers, where you can choose how many terms are included! Suddenly, the sum drops into a closed form and reveals an infinite family of solutions that depend on the ‘n’ chosen as to cut off. Further, Q was originally defined as a function of energy… if you substitute in that definition and solve for ‘E,’ you get an energy dependent on ‘n’. These are the allowed orbital energies for the hydrogen atom.

This is an example of Quantization!

Having just quantized the radial wave function of the hydrogen atom, you may want to sit back and smoke a cigarette (if you’re into that sort of thing).

It’s opaque and particular to this strategy, but the ‘8’ I chose to add way back in that first substitution that converts Rho into r came into play right here. As it turns out, the 4 which resulted from pulling a 2 out of the square root twice canceled another 2 showing up during a derivative done a few dozen lines later and had the effect of keeping a 2 from showing up with the ‘n’ on top of the recurrence relation… allowing the solutions to be successive integers in the power series instead of every other integer. This is something you cannot see ahead, but has a profound, Rube Goldbergian effect way down the line. I had to crash into the extra two while doing the problem to realize it might be needed.

At this point, I’ve looked at a few books to try to validate my method and I’ve found three different ways to approach this problem, all producing equivalent results. This is only one way.

The recurrence relation also gives a second very important outcome:

The energy quantum number must be bigger than the angular momentum quantum number. ‘n’ must always be bigger than ‘l’ by at least 1. And secondarily, and this is really important, the unprimed n must also always be bigger than ‘l.’ This gives:

n’ = n > l

This constrains which powers of n can be added in the series solution. You can’t just start blindly at the zero order power; ‘n’ must be bigger than ‘l’ so that it never equals ‘l’ in the denominator and the primed number is always bigger too. If ‘l’ and ‘n’ are ever equal, you get an undefined term. One might argue that maybe you can include negative powers of n, but these will produce terms that are 1/r, which are asymptotic at the origin and blow up when the radius is small, even though we know from the boundary conditions that the probability must go to zero at the origin. There is therefore a small window of powers that can be included in the sum, going between n = l+1 and n = n’.

I spent some significant effort thinking about this point as I worked the radial problem this time; for whatever reason, it has always been hazy in my head which powers of the sum are allowed and how the energy and angular momentum quantum numbers constrained them. The radial problem can sometimes be an afterthought next to the intricacy of the angular momentum problem, but it is no less important.

For all of this, I’ve more or less just told you the ingredients needed to construct the radial wave functions. There is a big amount of back substitution and then you must work the recurrence relation while obeying the quantization conditions I’ve just detailed.

A general form for the radial wave equations appears at the lower right, fabricated from the back-substitutions. The powers of ‘r’ in the series solution must be replaced with the original form of ‘rho’ which now includes a constant involving mass, charge and Plank’s constant which I’ve dubbed the Bohr radius. The Bohr radius a_{o} is a relic of the old Bohr atom model that I started off talking about and it’s used as the scale length for the modern version of the atom. The wave function, as you can see, ends up being a polynomial in radius multiplied by an exponential, where the polynomial is further multiplied by a single 1/radius term and includes terms that are powers of radial distance between l+1, where l is the angular momentum quantum number, and n’, the energy quantum number.

Here is how you construct a specific hydrogen atom orbital from all the gobbledigook written above. This is the simplest orbital, the S-orbital, where the energy quantum number is 1 and the angular momentum is 0. This uses the Y_{00} spherical harmonic, the simplest spherical harmonic, which more or less just says that the wave function does not vary across any angle, making it completely spherically symmetric.

The ‘100’ attached in subscript to the Psi wave function is a physicist shorthand for representing the hydrogen atom wave functions: these subscripts are ‘nlm,’ the three quantum numbers that define the orbital, which are n=1, l=0 and m=0 in this case. All I’ve done to produce the final wave function is take my prescription from before and use it to construct one of an infinite series of possible solutions. I then perform the typical Quantum Mechanics trick of making it a probability distribution by normalizing it. The process of normalization is just to make certain that the value ‘under the curve’ contained by the square of the wave function, counted up across all of space in the integral, is 1. This way, you have a 100% chance of finding the particle somewhere in space as defined by the probability distribution of the wave function.

You can use the wave function to ask questions about the distribution of the electron in space around the proton –for instance, what’s the average orbital radius of the electron? You just look for the expectation value of the radius using the wave function probability distribution:

For the hydrogen atom ground state, which is the lowest energy state for a 1 electron, 1 proton atom, the electron is distributed, on average, about 1 and a half Bohr radii from the nucleus. Bohr radius is about 0.52 angstrom (1×10^-10 meters), which means that the electron is on average distributed 0.78 angstroms from the nucleus.

Right now, this is all very abstract and mathematical, so I’ll jump into the more concrete by including some pictures. Here is a 3D density plot of the wave function performed using Mathematica.

Definitely anticlimactic and a little bit blah, but this is the ground state wave function. We know it doesn’t vary in any angle, so it has to be spherically symmetric. The axes are distance in units of Bohr’s radius. One thing I can do to make it a little more interesting is to take a knife to it and chop it in half.

This is just the same thing bisected. The legend at left just shows the intensity of the wave function as represented in color.

As you can see, this is a far cry from the atomic model depicted in cartoon far above.

For the moment, I’m going to hang up this particular blog post. This took quite a long time to construct. Some of the higher energy, larger angular momentum hydrogenic wave functions start looking somewhat crazy and more beautiful, but I really just had it in mind to show the math which produces them. I may produce another post containing a few of them as I have time to work them out and render images of them. If the savvy reader so desires, the prescriptions given here can generate any hydrogenic wave function you like… just refer back to my Ylm post where I talk some about the spherical harmonics, or by referring directly to the Ylm tables in wikipedia, which is a good, complete online source of them anyway.

Edit:

Because I couldn’t leave it well enough alone, I decided to do images of one more hydrogen atom wave function. This orbital is 210, the P-orbital. I won’t show the equation form of this, but I did calculate it by hand before turning it over to Mathematica. In Mathematica, I’m not showing directly the wave function this time because the density plot doesn’t make clear intuitive sense, but I’m putting up the probability densities (which is the wave function squared).

Mr. Peanut is the P-orbital. Here, angular momentum lies somewhere in the x-y plane since the z axis angular momentum eigenstate is zero. You can kind of think of it as a propeller where you don’t quite know which direction the axle is pointed.

Here’s a bisection of the same density map, along the long axis.

Edit 5-18-16

I keep finding interesting structures here. Since I was just sitting on all the necessary mathematical structures for hydrogen wave function 21-1 (no work needed, it was all in my notebook already), I simply plugged it into mathematica to see what the density plot would produce. The first image, where the box size was a little small, was perhaps the most striking of what I’ve seen thus far…

I knew basically that I was going to find a donut, but it’s oddly beautiful seen with the outsides peeled off. Here’s more of 21-1…

The donut turned out to be way more interesting than I thought. In this case, the angular momentum is pointing down the Z-axis since the Z-axis eigenstate is -1. This orbital shape is most similar qualitatively to the orbits depicted in the original Bohr atom model with an electron density that is known to be ‘circulating’ clockwise primarily within the donut. This particular state is almost the definition of a magnetic dipole.

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