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Quantum Theory of Hydrogen shrödinger's equation for hydrogen separation of variables “A facility for quotations covers the absence of original thought.”— Dorothy L. Sayers

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Chapter 6 Quantum Theory of the Hydrogen Atom 6.1 Schrödinger's Equation for the Hydrogen Atom We have discovered a "new" theory—quantum mechanics as exemplified by Schrödinger's equation. We tested it on three simple model systems in Chapter 5. We ought to test it on something a bit more “realistic,” like an atom. What’s the simplest atom you can think of? “A physicist is an atom’s way of knowing about atoms.”—biochemist George Wald

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The hydrogen atom is the simplest physical system containing interaction potentials (i.e., not just an isolated particle). Simple: one proton, one electron, and the electrostatic (Coulomb) potential that holds them together. The potential energy in this case is just (the attractive potential between charges of +e and –e, separated by a distance r). This is a stationary state potential (no time dependence). We could just plug it in to Schrödinger’s equation to get

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The potential looks quite simple, but it is a function of r, not x or (xyz). What can we do about that? Lots of luck solving that! (To say nothing of the fact that we shouldn’t be using the 1D Schrödinger equation for a 3D problem.) We need to let the symmetry of the problem dictate our mathematical approach.

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The spherically symmetric potential “tells” us to use spherical polar coordinates! http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html

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In spherical polar coordinates, r is the length of the radius vector from the origin to a point (xyz) is the angle between the radius vector and the +z axis and is the angle between the projection of the radius vector onto the xy plane and the +x axis point (xyz)

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The equations on the previous slide tell us how to express (r ) in terms of (xyz). We can also express (xyz) in terms of (r ): Now we can re-write the 3D Schrödinger equation in three dimensions, and in spherical polar coordinates, as

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If we plug in our potential V and multiply both sides by r 2 sin 2 , we get This looks nasty! We had better have a little talk about this equation! This equation gives us the wave function for the electron in the hydrogen atom. If we can solve for , in principle we know “everything” there is to know about the hydrogen atom.

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When we solved Schrödinger's equation in one dimension, we found that one quantum number was necessary to describe our systems. At the end of this section Beiser tells what the quantum numbers for the hydrogen atom are, and gives their possible values. I’ll skip that for now, because until we see where they come from and what they mean, they aren't of much use to us. For example, in the Bohr atom, the electron moves in an orbit, but we need only one parameter to specify its position in the fixed orbit, so we only need one quantum number. Here, in three dimensions and with three boundary conditions, we will find that we need three quantum numbers to describe our electron.

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Another comment: we are really solving Schrödinger's equation for the electron in a hydrogen atom, aren't we. Nevertheless, we talk about solving the "hydrogen atom," because our solution will provide us with much of what we need to know about hydrogen. 6.2 Separation of Variables We now “solve” the hydrogen atom. Here are some math activities, arranged from “fun” to “ugh:” solving linear algebraic equations solving coupled algebraic equations (e.g. xy together) solving linear differential equations solving coupled differential equations (e.g. derivatives mixed together)

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We have a coupled linear differential equation to solve. Maybe if we are clever, like we were with the tunneling calculation, we can make the problem easier. A big improvement would be to uncouple the variables. Stated more mathematically, when we have an equation like the one above, we like to see if we can "separate" the variables; i.e., "split" the equation into different parts, with only one variable in each part. Our problem will be much simplified IF we can write I’ve never known you (as a class) to shy away from leaps of logic, so how about if we assume that and see where it leads us?

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Let’s assume and see where it takes us. If our assumption works, in the orderly world of mathematics we know it must have been right. With this assumption, the partial derivatives in Schrödinger's equation become The partial derivatives become full derivatives because R, , and depend on r, , and only.

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To separate variables, plug = R into Schrödinger's equation and divide by R . The result is We have separated out the variable! The term is a function of only. Let's put it over on the right hand side of the equation. This gives us…

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This equation has the form f is a function of r and only, and g is a function of only. “How can this be? The RHS has only in it (but no r and ), and the LHS has only r and in it (but no ).” And LHS=RHS? “And yet you’re telling me LHS = RHS. I repeat, how can this be?” Yup, you heard right!

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Only one way! “Are you telling me everything is just a constant?” Absolutely not! It’s just that the particular combination of terms on the LHS happens to add up to a constant, which is the same as the constant given by the particular combination of terms on the RHS. This is really good. We've taken the one nasty equation in r , and separated it into two equations, one in r , and the other in only. Do you think maybe we can separate the r part…

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It turns out (although we won't do the math in this course) that the “constant” must be the square of an integer. If not, our differential equations have no solution. Thus, we can write the RHS of this equation as Where did this m ℓ come from? It’s an integer. We just “happened” to give it that “name.” Kind of hard to see, but that’s a lowercase script ℓ

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The LHS of our big Schrödinger equation also must equal m ℓ 2. If we set the LHS equal to m ℓ 2, divide by sin 2 , and rearrange, we get Once again we have separated variables. The LHS is a function of r only, and the RHS is a function of only. Again, the only way to satisfy this equation is for LHS=a constant=RHS. Solution of the resulting differential equations will result in restrictions on this constant. In this case, the constant must equal an integer times the next larger integer: ℓ ( ℓ +1).

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We have taken our initial differential equation and split into 3. Here are the pieces, rewritten slightly: “You sure went to a lot of trouble, and all you’ve done is replaced one differential equation by three.” But in the “one” the variables were coupled, and in the “three” the variables are separated. Huge improvement!

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The first attempt to LOG into ARPANET (precursor to the Internet), fall 1969 ( http://www.netvalley.com/intval1.html ): http://www.netvalley.com/intval1.html

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