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**Quantum Theory of Hydrogen**

electron probability density radiative transitions Zeeman effect “Every Hun has value—if only to serve as a bad example.”—Attila the Hun

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From the last lecture...

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**Caution: here again is how to calculate the probability:**

comes from volume element dV Here is how to calculate the expectation value: comes from definition of <G> Raise your hand if you plan on forgetting the r2. Raise your hand if you plan on forgetting the r in <r>. You’re joking, right? Let’s use the convention that P(r) stands for the probability density function R2r2, and P(r1rr2) stands for the probability of finding the electron somewhere between r1 and r2 (i.e., the integral).

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We have been manipulating the radial part of the wave function, because that’s where all the “action” is. But hydrogen is a 3D object. As a start towards 3D visualization, let’s “spin” the 1s radial wave function about r=0. This is supposed to help you “see” how the wave function drops off in all directions. Imagine a 3D object with this shape (kind of like a certain chocolate candy).

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**But the probability density is the meaningful “thing.”**

Here’s the 1s probability density, “spun” about r=0. Actually, there is a small but finite probability of finding the 1s electron “inside” the nucleus. I suppose you want to see 3D models of orbitals. You have to take the radial probability density, “fold in” the angular probability densities (which for the 1s are uniform) and somehow display in 3D.

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**Here are a couple of representations of the 1s probability density.**

Because the 1s (or 2s, 3s, …) probability density is spherically symmetric, you usually see it represented as a sphere. The figure on the previous slide clearly shows that the sphere is “hollow” and the probability density varies with r. People who represent this with “3D” drawings try to use shading to represent relative probabilities. It is difficult to show the “hollowness” of the 1s probability density. Here are a couple of representations of the 1s probability density. HAtomPDFs.htm StuHome/cabell_f/Density.html

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Here are more images illustrating how probability densities are displayed. They are from Houghton Mifflin Online (http://hmchemdemo.clt.binghamton.edu/zumdahl/docs/chemistry/07atomstructure/library/0707.htm). According to this source, the figure on the left illustrates the electron probability distribution, and the figure on the right illustrates “the surface that contains 90% of the total electron probability (the size of the orbital, by definition).”

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The images on the previous slide still imply that the 1s electron probability distribution is something like a peach. A better analogy would be a tennis ball with a thick but “fuzzy” skin. Anyway, now that you have been exposed to some of the subtleties of displaying probability densities, you should be responsible enough to view the pages listed here. (broken, March 2005)

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**Section 6. 7 contains much of the “testable” material of chapter 6**

Section 6.7 contains much of the “testable” material of chapter 6. The earlier sections are important (especially quantum numbers and angular momentum) but many of the problems come from 6.7, so be sure to study it well. Important ideas (quantum mechanics works very well for describing the hydrogen atom, but we need to modify our classical thinking in several ways): We can only give probabilities for finding an electron at some set of coordinates. The electron doesn't orbit the nucleus in any conventional sense. You should think of the pictures shown here as representing fuzzy clouds. The clouds aren't electrons, but they show where electrons are likely to be found. The denser the "cloud," the more likely is the electron to be found. The electron doesn't orbit the nucleus in any conventional sense.

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**6.8 Radiative Transitions**

We are well on our way to understanding the hydrogen atom, as well as why Bohr’s model had some real physics behind it. We have found that electrons don’t orbit in the classical sense (like a planet). In fact, our electron wave functions are “stationary states.” That helps us understand why the electrons don’t radiate continuously, although we haven’t yet done the mathematics to show this. Now it’s time to do “that” mathematics.

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**First consider an electron in hydrogen in a state with principal quantum number n.**

Its energy is En, its wave function has a frequency fn=En/h, and its wave function is This wave function now includes time dependence. The expectation value of the electron's position is which Beiser shows is independent of time. There is nothing here which tells us the electron radiates energy, and in fact, it doesn't (the electron doesn't oscillate with time).

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When the electron is excited to another quantum state m, it doesn't radiate energy after it reaches its final state; that is, as long as it is in the state m. However, the electron does have to get from state n to m, and in the process it does radiate (or absorb) energy. We calculate the energy radiated (or absorbed) by considering the wave function which represents an electron which may exist in state m or n. The magnitudes of a and b tell us the relative probabilities of the two states. Kind of like Schrödinger’s cat. The magnitudes of a and b tell us the relative probabilities of the two states.

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It is straightforward, if slightly tedious, to calculate the expectation value <x> for such an electron. You could reproduce this calculation with some assistance on my part. However, I have never asked this on an exam. The only "trick" in the calculation is the fact that The above two equations come from a more thorough investigation of the mathematics of quantum mechanics. Here, we just have to accept them. With these “tricks” we can calculate <x>.

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**The result of the calculation is:**

where If the electron is in state n, a=1, b=0 and only the first integral contributes to <x>, just as we would expect. If the electron is in state m, a=0, b=1, and only the second integral contributes to <x>, just as we would expect. However, when the electron is in transition between states n and m, both a and b are nonzero, and all three integrals, including the third one, contribute to <x>.

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**The third integral corresponds to an electron "jumping" between quantum state n to quantum state m.**

The cosine term gives the frequency of radiation emitted by the electron oscillating ("jumping") between quantum states n and m. The important result here is that an electron radiates electromagnetic energy when it jumps between quantum states, and that the frequency of emitted radiation is

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We do have a problem here--we've mixed classical (accelerated electron emits radiation) and quantum theory. Strictly speaking, according to this interpretation and calculation, the electron should emit electric dipole radiation instead of a single photon. Beiser clears this up in the next section. Our ideas are OK; but a complete and correct treatment requires quantum electrodynamics, and there we find that single photons are, indeed, emitted.

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**6.9 Selection Rules One more time:**

For an electron to get from one quantum state to another, it must radiate/absorb energy. For this to happen, the integral must be nonzero. What if the integral is zero? If the integral is zero, the electron can never get between the two quantum states. There is no path connecting the states!

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Thus, calculation of the integral tells us which transitions are allowed (the integral is finite) or forbidden (the integral is zero). The wave functions for the hydrogen atom are known, and the integral can be evaluated for all possible transitions. The integral is nonzero when These are selection rules for the allowed transitions. In other words, the integral is nonzero only for these particular changes in ℓ and mℓ. Hey! Things are starting to fit together. When the integral is “big”, the probability of transition is large (the transition happens “fast”). When the integral is small, the probability of transition is small (the transition happens “slow”—a metastable state!).

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Note that ℓ = 1 implies a change in the angular momentum of the atom during an allowed transition. (Why?) Angular momentum must be conserved, so the photon must carry off angular momentum. Photon of linear momentum p carries off angular momentum of r x p. Angular momentum of electron participating in event must change (angular momentum is conserved). p r When mℓ = 0, there is no change in the angular momentum along the z axis. This means that the photon is emitted in the x-y plane. When mℓ = 1, the photon is emitted along the z axis. Study figure 6.13 (see next slide). Be able to explain parts of it to me on an exam or quiz!

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**Figure 6.13 shows allowed transitions in hydrogen.**

It is only concerned with the ℓ = 1 requirement. The arrows in the figure indicate the allowed transitions. Why is the transition indicated by the red arrow forbidden? Notice the energy scaling; “ground” is 0 eV instead of -13.6 eV (perfectly “legal”).

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**The subsection on quantum electrodynamics is interesting (virtual photons???) but is not testable.**

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**6.10 Zeeman Effect Good physics! Skip most! (sob) Why—not enough time!**

Here’s the idea: electrons with different n differ in energy. This section shows that if you put an atom in a magnetic field, states with the same n but different mℓ are split slightly in energy. This section shows that if you put an atom in a magnetic field, states with the same n but different mℓ are split slightly in energy. (Aha—that’s why mℓ is called the magnetic quantum number!) mℓ=-1 mℓ=0 mℓ=1 2p means n=0 ℓ=1, mℓ=0, 1.

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Spectral lines corresponding to a particular n (we studied those in chapter 4) are “split” into separate lines corresponding to different mℓ. The splitting is small but measurable. It is called the Zeeman effect, named after the scientist who discovered it in (Nobel Prize, 1902) Explanation of the Zeeman effect is a dramatic confirmation of the validity of quantum mechanics. None of the material in this section (including these two Powerpoint slides on it) will be tested or quizzed.

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**Homework problems for me to set up on the board:**

What are the angles between L and the z-axis for ℓ=2? Find the most probable value of r for a 3d electron in hydrogen. How much more likely is the electron in a ground-state hydrogen atom to be at the distance a0 from the nucleus than at the distance 2a0?

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**There are “zillions” of web sites about the hydrogen atom**

There are “zillions” of web sites about the hydrogen atom. Every physics teacher who does anything relating to modern physics and puts class material on the web seems to do the hydrogen atom. A search using Google for “hydrogen atom schrodinger equation” (but don’t put the words in quotes when you do the search) comes up with about 4300 sites. Many of them are much nicer looking than mine, but the basic concepts presented are (not surprisingly) virtually the same. The sites below are repeats of the sites on slide 15. Here are a few choice sites (search words in parentheses): (hydrogen atom electron probability density) Slightly different search: (hydrogen atom electron orbital) All of these links were “live” as of fall 2003, but some may now be “dead.”

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