Quantum Theory of Hydrogen quantum numbers (I will not finish this lecture ) “When it comes to atoms, language can be used only as in poetry. The poet, too, is not nearly so concerned with describing facts as with creating images.”—Neils Bohr

6.3 Quantum Numbers We briefly saw two quantum numbers in the previous lecture: m ℓ and ℓ. They were “separation constants,” and you were told that the differential equations for hydrogen could be solved only if they took on integral values. In this section, we find out what the allowed values of m ℓ and ℓ are. Picking up where we left off… we are “solving the hydrogen atom. We “separated” our 3D Schrödinger equation into 3 single- variable differential equations…

We find the first quantum number by solving the differential equation for . That equation should look familiar to you; you've seen it a number of times before. It has solutions which are sines and cosines, or complex exponentials. We write the general solution We will get the constant A by normalization. Now, because  and  +2  represent a single point in space, we must have This happens only for m ℓ = 0,  1,  2,  3,...

For reasons which are not yet obvious, m ℓ is called the magnetic quantum number. Our differential equation for  is It involves the term It turns out that from differential equations that the equation for  can be solved only if ℓ is an integer greater than or equal to the absolute value of m ℓ.

It can be solved only for energies E which satisfy the same condition as we found on the energies for the Bohr atom: ℓ is another quantum number, called the orbital quantum number, and the requirement on ℓ can be restated as m ℓ = 0,  1,  2,  3,...,  ℓ. Finally, the radial differential equation is n is called the principal quantum number. I’ll summarize OSE’s in a bit.

Here’s the differential equation for R again: Note that the product ℓ ( ℓ +1) shows up in the equation for R, and n comes out of solving this equation. Another math requirement for valid solutions is that n  ( ℓ +1). We can express the requirement that n=1,2,3,… and n  (ℓ+1) as a condition on ℓ: ℓ = 0, 1, 2,..., (n-1). ℓ = 0, 1, 2,..., (n-1) n = 1, 2, 3,... m ℓ = 0,  1,  2,  3,...,  ℓ Summarizing our quantum numbers:

We summarize this section by noting that solutions to the Schrödinger equation for the hydrogen atom must be of the form  = R n ℓ  ℓ m ℓ  m ℓ, with conditions on the quantum numbers n, ℓ, and m ℓ as discussed above. We aren't going to go any further with our solutions to the Schrödinger equation, other than to note that they are well- known, and Beiser tabulates some of them in Table 6.1.

For n=2, ℓ can be either 0 or 1. If ℓ =0 then m ℓ =0. If ℓ =1 then m ℓ =0 and m ℓ =  1 are allowed. The solutions for m ℓ =  1 are the same. Beiser tabulates the three solutions. Note how Table 6.1 is set up. For n=1, the only allowed possibilities are ℓ =m ℓ =0. For this case, Beiser lists the three solutions R, , and . Here's an example. Suppose we have an electron with a principal quantum number n=3 (corresponding to the second excited state of the Bohr hydrogen atom) and orbital and magnetic quantum numbers ℓ =2 and m ℓ =-1. Then, according to table 6.1,

and The wave function  is the product of all three of those functions. I wouldn't care to calculate and plot the wave function by hand, but with Mathcad the problem is rather easy. With the wave functions in Table 6.1, you can calculate all sorts of fun stuff, like ground state energies (see the example on page 207), excited state energies, expectation values, probabilities, etc.

Our result for the allowed values of the principal quantum number n and, and dependence of the electron energy E n on n, turn out to be exactly the same as for the Bohr model. Is this just luck or was Bohr on to something deeper? In fact, it is not just luck. Both results depend on the wave nature of the electron. 6.4 Principal Quantum Number The Bohr model is, however, unable to provide additional details which the full quantum mechanical solution does.

Electron energies in the hydrogen atom are quantized, and they are negative numbers: It is true that any positive energy may lead to a solution to Schrödinger's equation… The only possible negative (bound electron) energies are those given by the equation above. …but a positive energy means the electron is not bound, so we don't have a electron in the hydrogen atom. None of this information is new; we have seen it all before.

6.5 Orbital Quantum Number I've already written down the differential equation for R(r) several times. Sigh… I guess I’d better repeat it again. On the next slide, I’ll write it in an alternate form, which uses: E = K + V K = K radial + K orbital V = - e 2 / ( 4   0 r ).

The alternate form is: Ohhhhh nooooo… The above equation is supposed to have only r in it. But K orbital depends on tangential velocity, so it seems to have angular dependence in it! I have two choices: throw out the last lecture and a half and start over, or…

…somehow make K orbital just “go away.” Let’s see… both K radial and K orbital are positive numbers, so they can’t somehow cancel each other out… …but if the radial equation is to really have only radial dependence in it, it clearly cannot have an orbital kinetic energy term. The only way to make K orbital "go away" from this equation is to have

Some review from Physics 23. Just as we can write kinetic energy in terms of momentum, we can write orbital kinetic energy in terms of orbital angular momentum: Combining our two equations for K orbital (the one on this slide, and the one on the previous slide), we find that… Orbital kinetic energy: Orbital angular momentum:

which gives us Because the orbital quantum number ℓ is quantized, the electron’s orbital angular momentum L (a vector) is also quantized. In fact, L is quantized in (non-integral) multiples of ħ. The lowest possible L is zero, when ℓ =0. Huh? Are you trying to tell me an orbiting electron in hydrogen can have zero angular momentum?

The ground-state hydrogen electron has zero angular momentum! It cannot be orbiting in any classical sense! The lowest possible nonzero L is Here’s a table from the good old days of spectroscopy, showing how we label angular momentum states: Know how to use this table! How would you label an electron with n=3, ℓ=2? It would be a 3d electron. n ℓ ℓ = spdfghi It would be a 3d electron. Note that a 3d electron can have any one of several possible allowed m ℓ 's.

See table 6.2, page 210 of Beiser for designation of atomic states. 6.6 Magnetic Quantum Number Adding to our house of cards, we have come up with an orbital quantum number (and found an orbiting electron that doesn’t “go around”). Regardless, an electron in a hydrogen atom is in an “orbit.” It has an angular momentum vector L (a vector). We found the magnitude of L above: What else does any vector have?

A direction, of course. What does the direction of L in an isolated hydrogen atom mean? Not much! (Actually, nothing, unless you specify a coordinate system.) But an orbiting electron is like a current in a loop, which gives rise to a magnetic field, and can interact with an external magnetic field. An external magnetic field therefore gives us a meaningful reference for specifying the direction of the electron orbital angular momentum vector in the hydrogen atom. By convention, we put our hydrogen atom's z-axis along the direction of the applied magnetic field B.

Then m ℓ gives the component of L in the direction of B: Huh? Where did this come from? Straight out of nowhere. Not from anything we've done here! Let’s just accept it as true. If you really want to see, come to my office some time and we’ll dig out my graduate text on quantum mechanics… OK, so now we know how to calculate a hydrogen atom electron’s total angular momentum, and it’s z-component, the part parallel to an applied magnetic field B. This is called “space quantization.” See your text.

Thinking back to Physics 24, if you had a loop of wire capable of rotating, and the loop carried a current, and was placed in an external magnetic field, what would the loop do? The current loop would experience a torque, and, in the absence of external forces, rotate until the torque became zero. In the process, the magnetic field due to the current loop would line up with the external magnetic field.

What does quantum mechanics say about the “same” scenario in the hydrogen atom? Or, in the language of the electron in hydrogen, can L ever be parallel to B? Answer: Because m ℓ is at most equal to ℓ, then L z =m ℓ ħ is always less than L. Huh? Does this mean that my current loop in the previous slide can never exactly “line up” its magnetic field with the external magnetic field? Quantum mechanics says “yes.” However, the quantum numbers will be so large that you will never see the difference between “almost” and “exactly” aligned.

Suppose we place a hydrogen atom in a magnetic field. Is L z always the maximum possible (i.e., does the hydrogen atom always try to "line up" with the field)? Answer: not necessarily. “Space quantization.” See figure 6.4. Understand this figure for test/quiz questions! The figure to the right is (intentionally) not fully- labeled. Also, I used a circle instead of a semicircle due to the lack of a semicircle tool in Powerpoint.

Here's why L z is always less than L. If L could point exactly along the z axis (magnetic field axis) then the electron orbit would lie exactly in the xy plane and the uncertainty in the z position coordinate would be zero. The momentum uncertainty in the z direction would then be infinite. This is intolerable for an electron in an atom. The tilt of L with respect to the z-axis lets us satisfy the uncertainty principle. October 24, 2003, go here next.here (The final 3 slides of lecture 21 are duplicated at the start of lecture 22.)

6.7 Electron Probability Density Thus, the electron probability density in hydrogen is Recall that the volume element in spherical polar coordinates is dV ** P(r) dr = R * R r 2 dr (the probability of finding the electron within infinitesimal dr centered at r)

We often wish to calculate the probability of finding the electron in some volume element in space: Because  is separable, and R, , and  are orthonormal,* we can write the triple integral as three one-dimensional integrals. *As one consequence of this, the R, , and  in table 6.1 are independently normalized, and the product wave function is also normalized.

Important note: in spherical polar coordinates, 0  r  , 0    , and 0    2 . It makes no sense to calculate probabilities outside these regions. When you calculate, the integral goes from 0 to , NOT from -  to  ! Some terminology and important notes (I discuss only the radial part, but it applies to the angular parts too)… I am using my own personal shorthand notation that Probability(r) means “the probability of finding the electron within some dr centered at r,” etc.