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Quantum Theory of Hydrogen quantum numbers (I will not finish this lecture 03-02-2005) “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 Picking up where we left off… we are “solving the hydrogen atom. We “separated” our 3D Schrödinger equation into 3 singlevariable differential equations… 6.3 Quantum Numbers We briefly saw two quantum numbers in the previous lecture: mℓ 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 mℓ and ℓ are. We find the first quantum number by solving the differential equation for . d2 2 + m = 0 2 d 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 jm solution = Ae . We will get the constant A by normalization. Now, because and +2 represent a single point in space, we must have Ae jm = Ae j m + 2 This happens only for mℓ = 0, 1, 2, 3, ... . For reasons which are not yet obvious, mℓ is called the magnetic quantum number. Our differential equation for is 1 d d sinθ + sinθ dθ dθ m2 =0 +1 2 sin θ It involves the term m2 . +1 2 sin θ 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 mℓ. ℓ is another quantum number, called the orbital quantum number, and the requirement on ℓ can be restated as mℓ = 0, 1, 2, 3, ..., ℓ. I’ll summarize OSE’s in a bit. Finally, the radial differential equation is 1 d 2 dR 2m e 2 + E r + 2 2 r dr dr 4ε0r +1 R=0 2 r It can be solved only for energies E which satisfy the same condition as we found on the energies for the Bohr atom: me 4 En = 322 ε0 2 E1 1 = 2 , n = 1, 2, 3... 2 n n n is called the principal quantum number. Here’s the differential equation for R again: 1 d 2 dR 2m e 2 + E r + 2 2 r dr dr 4ε0r +1 R=0 2 r 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). Summarizing our quantum numbers: n = 1, 2, 3, ... ℓ = 0, 1, 2, ..., (n-1) mℓ = 0, 1, 2, 3, ..., ℓ We summarize this section by noting that solutions to the Schrödinger equation for the hydrogen atom must be of the form = Rnℓ ℓmℓ mℓ , with conditions on the quantum numbers n, ℓ, and mℓ 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 wellknown, and Beiser tabulates some of them in Table 6.1. Note how Table 6.1 is set up. For n=1, the only allowed possibilities are ℓ=mℓ=0. For this case, Beiser lists the three solutions R, , and . For n=2, ℓ can be either 0 or 1. If ℓ=0 then mℓ=0. If ℓ=1 then mℓ=0 and mℓ=1 are allowed. The solutions for mℓ=1 are the same. Beiser tabulates the three solutions. 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 mℓ=-1. Then, according to table 6.1, 1 ± j = e 2 θ = 15 sin θ cos θ 2 and 4 r 2 - r / 3a0 R r = e . 2 3/2 81 30 a0 a0 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. 6.4 Principal Quantum Number Our result for the allowed values of the principal quantum number n and, and dependence of the electron energy En 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. 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: me 4 En = 322 ε0 2 E1 1 = 2 , n = 1, 2, 3... 2 n n It is true that any positive energy may lead to a solution to Schrödinger's equation… …but a positive energy means the electron is not bound, so we don't have a electron in the hydrogen atom. The only possible negative (bound electron) energies are those given by the equation above. 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. 1 d 2 dR 2m e 2 + E r + 2 2 r dr dr 4ε0r +1 R=0 2 r On the next slide, I’ll write it in an alternate form, which uses: E=K+V K = Kradial + Korbital V = - e2/ ( 4 0 r ) . The alternate form is: 1 d 2 dR 2m r + 2 K radial + K orbital 2 r dr dr +1 R=0 2 2mr 2 Ohhhhh nooooo… The above equation is supposed to have only r in it. But Korbital 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… 1 d 2 dR 2m r + 2 K radial + K orbital 2 r dr dr +1 R=0 2 2mr 2 …somehow make Korbital just “go away.” Let’s see… both Kradial and Korbital 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 Korbital "go away" from this equation is to have K orbital = 2 +1 2mr 2 . Some review from Physics 23. Orbital kinetic energy: K orbital Orbital angular momentum: 1 2 = m v orbital 2 L orbital = r m v orbital Just as we can write kinetic energy in terms of momentum, we can write orbital kinetic energy in terms of orbital angular momentum: K orbital L2 = 2 m r2 Combining our two equations for Korbital (the one on this slide, and the one on the previous slide), we find that… 2 K orbital = +1 2 m r2 L2 = 2 m r2 which gives us L= +1 . 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! L= The lowest possible nonzero L is 2 . Here’s a table from the good old days of spectroscopy, showing how we label angular momentum states: ℓ= 0 s 1 p 2 d 3 f 4 g 5 h 6 i Know how to use this table! How would you label an electron with n=3, ℓ=2? ℓ n It would be a 3d electron. Note that a 3d electron can have any one of several possible allowed mℓ'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? L= +1 . 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 mℓ gives the component of L in the direction of B: Lz = m . This is called “space quantization.” See your text. 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. 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: L= +1 Lz = m Because mℓ is at most equal to ℓ, then Lz=mℓħ 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 Lz 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 fullylabeled. Also, I used a circle instead of a semicircle due to the lack of a semicircle tool in Powerpoint. Here's why Lz 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. (The final 3 slides of lecture 21 are duplicated at the start of lecture 22.) 6.7 Electron Probability Density Recall that the volume element in spherical polar coordinates is dV = r 2 sin θ dr dθ d . Thus, the electron probability density in hydrogen is P(r,θ, ) dV = dV = RR r 2 sin θ dr dθ d * dV P(r,θ, ) dV = RR r 2 dr sin θ dθ d P(r,θ, ) dV = P(r) dr P(θ) dθ P() d P(r) dr = R*R r2 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: Probability = P r,θ, dV . Because is separable, and R, , and are orthonormal,* we can write the triple integral as three one-dimensional integrals. Probability r = Probability(θ) = Probability() = P r dr P(θ) dθ P() d *As one consequence of this, the R, , and in table 6.1 are independently normalized, and the product wave function is also normalized. 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. 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 <r>, 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)…