 Welcome to quantum mechanics 6, hydrogen. Hydrogen is the first element in the periodic table and the most common element by far in the universe. It is the simplest atom consisting of a single proton with positive charge e and a single electron with negative charge minus e. The electron is much lighter than the proton, so we expect that, approximately, the proton stays put and it's the electron that moves around. We want to find out what predictions the Schrödinger equation makes for hydrogen. Specifically, we're interested in the stationary states, the states of definite energy, so we limit our consideration to the corresponding form of the Schrödinger equation. The potential energy v is minus e squared over r, r being the distance between proton and electron. This tells us that when the two are infinitely far apart, the potential energy is zero. That's our point of reference. As they get closer together, the potential energy becomes more negative. A negative energy means we have to put positive energy into the system to separate the two particles back to infinitely far apart, the state of zero energy. In the Schrödinger equation, the first term looks the most intimidating, but recall that the so-called Laplacian is simply the difference of the field at a point from the average field value on a very small surrounding sphere times some factor. It's just a measure of the non-uniformity of the field. In fact, using this last expression, you can rearrange things to show that Schrödinger's equation is simply the requirement that the field at each point is proportional to the average of neighboring points, where the proportionality factor depends on the energy e and the distance r from the proton. Interestingly, you can find a field meeting this requirement only for certain specific values of the energy e. It's convenient to use so-called atomic units in which Planck's constant divided by two pi, that's h bar, the mass of the electron and the fundamental charge are all equal to one. And Schrödinger's equation is a bit less messy. We're looking for three-dimensional solutions. Although we're generally most comfortable with rectangular coordinates, x, y, and z, they're not very convenient for actually turning the math crank in this case. It's much more efficient to use the distance between proton and electron, r, as one coordinate, since that explicitly appears in the Schrödinger equation. To complete the so-called spherical coordinate system, we employ the two angles, theta and phi, shown here. Now it's simply a matter of expressing the Schrödinger equation in these coordinates and solving it. What Schrödinger found was that only certain distinct solutions exist for a discrete set of values of the energy e. And these can be indexed by three integers, n, l, and m. So we have a set of solutions, psi n, l, m of r, theta, phi. The integer indices n, l, and m are called quantum numbers. n can take on any positive value, 1, 2, 3, and so on. And for a given value of n, l can take on values 0, 1, so on, up to n minus 1. Finally, for a given value of l, m can take on the values minus l, minus l, plus 1, so on, up to plus l. No additional assumptions are required to arrive at this result. It follows directly from the form of Schrödinger's equation in spherical coordinates. The energy turns out to be fixed by the quantum number n exactly as in the Bohr formula. E sub n equals some constant over n squared. In normal units, the constant is minus 13.6 electron volts, just as in the Bohr model. Now we're ready to visualize the hydrogen atom as described by the Schrödinger equation. We'll use two methods. On the left we move a plane through the wave function and show the electron probability density in false color. Black is the lowest density, up through blue, green, red, and finally white, which is the highest. This is like a CAT scan of the atom. On the right we show a surface of some constant probability density. The electron is equally likely to be anywhere on this surface. The first energy level, the so-called ground state of hydrogen corresponding to n equals 1, has only a single solution. We call this solution n orbital by analogy with the quasi-classical orbits of the Bohr model. The quantum number l can only have the value 0, which corresponds to a so-called s orbital. The quantum number m is likewise limited to 0. So this is the quantum state n equals 1, l equals 0, m equals 0. In the ground state, the one-ass orbital, the electron probability density has a compact, spherically symmetric cloud centered on the nucleus. Here the scale bar labeled 30 has a length of 30 atomic length units, where one unit corresponds to the Bohr radius of hydrogen about 5.3 times 10 to the minus 11 meters. The second energy level, n equals 2, has an s orbital, but also has solutions with quantum number l equals 1. These are called p orbitals. For l equals 1, the quantum number m can take on values minus 1, 0, and 1, so there are three p orbitals. The two-ass orbital is larger than the one-ass orbital and it consists of two concentric spherical lobes. The two p orbitals are rotated versions of the same two lobes structure. For the third energy level, quantum number n equals 3, we see a pattern in which each energy level has versions of the previous energy levels orbitals, plus a new family corresponding to a larger value of quantum number l. In addition to one-ass orbital and three p orbitals, there are five d orbitals corresponding to l equals 2 and m equals minus 2 through 2. We first look at the s and p orbitals. Note the scale bar is now 45 units long. These are qualitatively similar to the n equals 2 orbitals, but they are larger and have more lobes, more variation in space. Four of the five new d orbitals are rotated versions of the same four lobes structure, what looks a bit like a clutch of four eggs. The fifth orbital corresponding to n equals 0 has sort of a donut between two balls appearance. Continuing to the fourth energy level, we have versions of all the n equals 3 orbitals and we add a new family corresponding to l equals 3, the f orbitals. There are seven of these with quantum numbers m ranging from minus 3 to 3. The s and p orbitals are again similar to those of the n equals 2 and n equals 3 energy levels, but are larger and have more complicated structure. Note the scale bar is now 60 units long. The four d orbitals continue with the four lobe clutch of eggs structure, but now we have two nested groups of these. The m equals 0 four d orbital, one of my personal favorites, looks something like a dumbbell wearing a hula hoop. Four of the seven new f orbitals are rotated versions of the same six lobe structure, looking a little like a circle of six kidney beans. Two of the f orbitals are rotated versions of an eight lobe structure, and finally we have a lone m equals 0 f orbital. This process of adding a new family of orbitals with each higher energy level continues on in principle without end. However, for our purposes, these four families of orbitals, s, p, d and f are sufficient. Now in terms of reproducing the Bohr models explanation of the hydrogen spectrum, all that matters are the energy levels. The single quantum number n. Those values naturally arise in the Schrodinger's solution. If explaining the Reidberg formula was our only interest, we could forget about the rest of the details. But naturally we might be curious about all these different types of orbitals. Is there any significance to the fact that higher energy levels have a greater number of distinct solutions? And how about the specific shapes of all these orbitals? Do these mean anything physically? Or are they just some abstract mathematical result? Let's first just look at the number of orbitals for each energy level. The first energy level has a single s orbital. The second energy level has a single s and three p orbitals for a total of four independent states. The third energy level adds five d orbitals for a total of nine and the fourth energy level adds seven f orbitals for a total of sixteen. There are one, four, nine, and sixteen different electron states for the first, second, third, and fourth energy levels. Does this sequence, one, four, nine, sixteen correspond to anything in the physical world? Well let's multiply those numbers by two. We'll get to the significance of that later. But this gives us the sequence of numbers two, eight, eighteen, and thirty-two. Now, does that sequence show up anywhere? The similarities and differences of the chemical properties of the elements are represented by the periodic table. As the number of electrons per atom increases from one for hydrogen up to more than one hundred, the periodic table displays groups of atoms arranged in columns. Each atom in a group shares important chemical properties. Each period ends on the right with one of the so-called noble gases, helium, neon, etc., which are practically chemically inert. There are two elements in the first period, hydrogen and helium, so there's our two. The second period contains eight elements, lithium through neon. There's our eight. The third period also contains eight elements. The fourth period contains eighteen elements, as does the fifth period. There's our eighteen. Finally, the sixth and seventh periods contain thirty-two elements. So the sequence two, eight, eighteen, and thirty-two that corresponds to twice the number of orbitals for the different energy levels of hydrogen corresponds to the structure of the periodic table. That certainly seems significant. And indeed, the structure of the periodic table is well understood in terms of our SPD and F orbitals.