Anone who can contemplate quantum mechanics without getting di hasn t understood it. --Niels Bohr Lecture 7, p
Phsics Colloquium TODAY! Quantum Optomechanics Prof. Markus Aspelmeer, U. Vienna Massive mechanical objects are now becoming available as new sstems for quantum science. Recent eperiments, including laser-cooling of micro- and nanomechanical resonators into their quantum ground state of motion, provide the primar building blocks for full quantum optical control of mechanics, i.e., quantum optomechanics. This new frontier opens fascinating perspectives both for various applications and for unique tests of the foundations of quantum theor, for eample table-top eperiments eploring the interface between quantum phsics and gravit. 4 pm, 4 Loomis Special (Optional) Lecture Quantum Information One of the most modern applications of QM quantum computing quantum communication crptograph, teleportation quantum metrolog Prof. Kwiat will give a special 24-level lecture on this topic Sunda, Feb. 24 3 pm, 4 Loomis Attendance is optional, but encouraged. Lecture 7, p 2
Overview of the Course Up to now: General properties and equations of quantum mechanics Time-independent Schrodinger s Equation (SEQ) and eigenstates. Time-dependent SEQ, superposition of eigenstates, time dependence. Collapse of the wave function Tunneling This week: 3 dimensions, angular momentum, electron spin, H atom Eclusion principle, periodic table of atoms Net week: Molecules and solids, consequences of Q. M., Schrodinger s cat Metals, insulators, semiconductors, superconductors, lasers,.. Final Eam: Monda, Mar. 4 Homework 6: Due Saturda (March 2), 8 am Lecture 7, p 3
Lecture 7: Atomic States, Angular Momentum & Selection Rules Lecture 7, p 4
Toda Schrödinger s Equation for the Hdrogen Atom Radial wave functions Angular wave functions Angular Momentum Quantiation of L and L 2 Atomic Transitions Selection rules Lecture 7, p 5
Summar of S-states of H-atom The s-states (l=0, m=0) of the Coulomb potential have no angular dependence. In general: but: ( r,, ) = R ( r ) Y (, ) ψ θ φ θ φ ψ nlm nl lm ( r, θ, φ ) R ( r ) n00 n0 because Y 00 (θ,φ) is a constant. S-state wave functions are sphericall smmetric. Some s-state wave functions (radial part): ψ 20 (r,θ,φ) 2 : 5 R 0 R 20 R 30 0 r 4a 0 0 0a 0 r 0 5a r 0 http://www.falstad.com/qmatom/ Lecture 7, p 6
Total Wave Function of the H-atom We will now consider non-ero values of the other two quantum numbers: l and m. θ ( r,, ) = R ( r ) Y (, ) ψ θ φ θ φ nlm nl lm r n principal (n ) l orbital (0 l < n-) m magnetic (-l m +l) * φ The Y lm (θ,φ) are known as spherical harmonics. The are related to the angular momentum of the electron. * The constraints on l and m come from the boundar conditions one must impose on the solutions to the Schrodinger equation. We ll discuss them briefl. Lecture 7, p 7
Quantied Angular Momentum Linear momentum depends on the wavelength (k=2π/λ): p = ħk where ψ ( ) e Angular momentum depends on the tangential component of the momentum. Therefore L depends on the wavelength as one moves around a circle in the - plane. Therefore, a state with L has a similar form: im L = mħ where ψ ( r ) Y ( θ, φ) e φ L Z An important boundar condition: An integer number of wavelengths must fit around the circle. Otherwise, the wave function is not single-valued. ik lm We re ignoring R(r) for now. Re(ψ) Reminder: φ e imφ = cos(mφ) + i sin(mφ) This implies that and m = 0, ±, ±2, ±3, L = 0, ±ħ, ±2ħ, ±3ħ, Angular momentum is quantied!! http://www.falstad.com/qmatom/ Lecture 7, p 8
The l Quantum Number The quantum number m reflects the component of angular momentum about a given ais. L = mħ where m = 0, ±, ± 2,... In the angular wave function ψ lm (θ,φ) the quantum number l tells us the total angular momentum L. L 2 = L 2 + L 2 + L 2 is also quantied. The possible values of L 2 are: 2 2 L = l l + ħ l = ( ) where 0,, 2,... Wave functions can be eigenstates of both L 2 and L Z. For sphericall smmetric potentials, like H-atom, the can also be eigenstates of E. Such states are called orbitals. Summar of quantum numbers for the H-atom orbitals: Principal quantum number: n =, 2, 3,. Orbital quantum number: l = 0,, 2,, n- Orbital magnetic quantum number: m = -l, -(l-), 0, (l-), l Lecture 7, p 9
Angular Momentum & Uncertaint Principle ( l ) 2 2 L = l( l + ) ħ not ħ 2 Note that. Also, we describe angular momentum using onl two numbers, l and m. Q: Wh can t we specif all three components (e.g., L =(0,0,l) so that L 2 = l 2? A: The uncertaint principle doesn t allow us to know that both L = 0 and L = 0 unless L = 0 also. Proof b contradiction: Assume L =(0,0,l). L = r p, so if L points along the -ais, both r and p lie in the - plane. This means that = 0 and p = 0, violating the uncertaint principle. Thus, L must have a nonero L or L, making L 2 somewhat larger. We can t specif all three components of the angular momentum vector. This logic onl works for L 0. L = (0,0,0) is allowed. It s the s-state. All phsical quantities are subject to uncertaint relations, not just position and momentum. Lecture 7, p 0
Classical Picture of L-Quantiation e.g., l = 2 L = l( l 2 + ) ħ = 2( 2 + ) ħ = 6ħ 2 L L = 6 ħ +2ħ +ħ 0 -ħ -2ħ L = r p Lecture 7, p
The Angular Wave Function, Y lm (θ,ϕ) The angular wave function ma be written: Y lm (θ,φ) = P(θ)e imφ where P(θ) are polnomial functions of cos(θ) and sin(θ). To get some feeling for these angular distributions, we make polar plots of the θ-dependent part of Y lm (θ, φ) (i.e., P(θ)): l = 0 Y 0,0 = 4π l = Y,0 cosθ Y, ± sinθ θ 0 θ θ Length of the dashed arrow is the magnitude of Y lm as a function of θ. Lecture 7, p 2
The Angular Wave Function, Y lm (θ,ϕ) l = 2 θ 2,0 2 ( θ ) Y 3cos 2 θ Y 2, ± sinθ cosθ 0 Y 2, ± 2 2 sin θ θ + - Lecture 7, p 3
Probabilit Densit of Electrons Let s look at the angular momentum states of the hdrogen atom. Probabilit densit = Probabilit per unit volume = ψ nlm 2 R nl 2 Y lm 2. The densit of dots plotted below is proportional to ψ nlm 2. s state 2s state 2p states n,l,m =,0,0 2,0,0 2,,{0,±} Lecture 7, p 4
Act. Suppose the electron is in the l=, m= state. In what direction(s) (at what θ), is the electron most likel to be found? Y, ± sinθ a. θ = 0 (north pole) b. θ = 45 c. θ = 90 (equator) Lecture 7, p 5
Solution. Suppose the electron is in the l=, m= state. In what direction(s) (at what θ), is the electron most likel to be found? Y, ± sinθ a. θ = 0 (north pole) b. θ = 45 sinθ is maimum at θ = 90. c. θ = 90 (equator) l = 0 θ l = 0 θ θ Length of the dashed arrow is the magnitude of Y lm as a function of θ. Y 0,0 = 4π Y,0 cosθ Y, ± sinθ Lecture 7, p 6
Clindrical Smmetr Wh do none of the graphs displa φ-dependence? (The all have clindrical smmetr.) For a given m, the φ dependence of ψ is e imφ. When we square it to find the probabilit, e imφ e -imφ =. In order to see φ dependence, we need a superposition of different m s. For eample, consider the superposition: (l =, m = +) & (l =, m = -). This will have an aimuthal wave function: e iφ + e -iφ cos φ, i.e., lobes along the -ais: Similar arguments eplain how to create the usual d lobes, from l =2, m = ±2 superpositions: - + - + See Supplement for more info. Lecture 7, p 7
Wh are these distributions important? The govern the bonding and chemistr of atoms. In particular, the determine the angles at which different atoms bond: the structure of molecules & solids. Historical Labeling of Orbitals Notation from 9 th centur Angular momentum quantum # spectroscop l = 0 s sharp l = l = 2 p principle d diffuse l = 3 f fundamental Lecture 7, p 8
Act 2 How does the angular part of the wave function depend on the principal quantum number, n? a. The number of lobes increases as n increases. b. As n increases, the wave function becomes more concentrated in the plane. c. No dependence. Lecture 7, p 9
Solution How does the angular part of the wave function depend on the principal quantum number, n? a. The number of lobes increases as n increases. b. As n increases, the wave function becomes more concentrated in the plane. c. No dependence. The principal quantum number describes the radial motion, not the angular motion. R nl (r) depends on n, but Y lm (θ,φ) does not. Lecture 7, p 20
Effect of l on Radial Wave Functions R n,l n= n=2 n=3 : l < n (Total energ must alwas be larger than rotational part.) l = 0 l = 0 l = l = 0 l = 2: a. For fied l, the number of radial nodes increases with n. b. For fied n, the number of radial nodes decreases with l. (E = T rad + T rot + U(r), i.e., radial KE decreases as rotational KE increases ). l = 2 3: # radial nodes = (n-) - l. 4: ψ(r=0) = 0 for l 0 Do ou understand wh? (i.e., a phsics eplanation) The energ eigenvalues do not depend at all on l. E n = -3.6 ev/n 2 This is onl true for the Coulomb potential. Lecture 7, p 2
Hdrogen Atom States: Summar Ke Points: n: principal quantum # l: orbital quantum # m l :orbital magnetic quantum # 0-5 n=3 n=2 Energ depends onl on n 2 κe 3.6 ev E n = = 2a n n 0 2 2 E -0 For a given n, there are n possible angular momentum states: l = 0,,..., n- For a given l, there are 2l + possible -components: m l = -l, -(l -), 0 (l -), l -5 n= Therefore, a level with quantum number n has n 2 degenerate states. Lecture 7, p 22
Hdrogen Atom States: Summar 0-5 n=3 n=2 (n,l,m l ) (3,0,0) (3,,-), (3,,0), (3,,) (3,2,-2), (3,2,-), (3,2,0), (3,2,), (3,2,2) (2,0,0) (2,,-), (2,,0), (2,,) E -0-5 n= (,0,0) Lecture 7, p 23
Transitions in the Hdrogen Atom Consider the three lowest energ levels of the hdrogen atom. What wavelengths of light will be emitted when the electron jumps from one state to another? Lecture 7, p 24
Transitions in the Hdrogen Atom Consider the three lowest energ levels of the hdrogen atom. What wavelengths of light will be emitted when the electron jumps from one state to another? Solution: E 2 = 0.2 ev E 3 = 2. ev E 32 =.9 ev E = -3.6 ev/n 2, so E = -3.6 ev, E 2 = -3.4 ev, and E 3 = -.5 ev. There are three jumps to consider, 2-to-, 3-to-, and 3-to-2. The photon carries awa the energ that the electron loses. λ = h/p = hc/e λ 2 = 22 nm λ 3 = 02 nm λ 32 = 653 nm hc = 240 evnm Two wavelengths are in the ultraviolet. The 3-to-2 transition gives a visible (red) photon. Lecture 7, p 25
Optical Transitions between Atomic Levels Consider the n = and 2 levels of hdrogen: E n = 2 n = E c f = = h λ hc 240 ev nm λ = = E E The atom can make transitions b emitting (n: 2 ) or absorbing (n: 2) a photon. In general, the time-dependent solution of the SEQ in the time-dependent EM field shows the wave function oscillating between the two eigenstates of the energ (that is, the were eigenstates before the field showed up!). Not all transitions are possible. For eample, one must conserve angular momentum (and the photon has l = ). Stationar States: s 2s 2p www.falstad.com/qmatomrad s ± 2s s ± 2p photon Superpositions: No electricdipole moment Oscillating electric-dipole couples to photons Each photon carries ћ of angular momentum. Forbidden transition l = 0 Allowed transition l = ± Lecture 7, p 26
Allowed Transitions for H (You will observe some of these transitions in Lab 4.) n s p d f g l = 0 2 3 4 Energ (ev) 4 3 0.00-0.85 -.5 Selection Rule for electricdipole (photon has l = ) transitions: 2-3.40 l = ± m = 0, ± Forbidden transition l = 0 E n = 3.6 ev 2 n NOTE: It is possible (but unlikel) for the photon to have l or for more than one photon to be involved. This permits other l and m. -3.6 ev Lecture 7, p 27
Net Time Electron spin and the Stern-Gerlach eperiment Net Week Multi-electron Atoms Covalent bonds Electron energ bands in Solids QM in everda life Lecture 7, p 28
Supplement: Superposition and Chemical Bonding Chemical bonds are stronger when the bonding electrons in each atom lie on the side near the other atom. This happens as a result of superposition. A state with definite (l,m) is smmetrical, but a superposition does not have to be. The eample here is called an sp hbrid : Y 0,0 = 4π Y,0 cos θ Y 00 + Y 0 θ θ + 0 - Lecture 7, p 29
Supplement: Chemistr Notation From chemistr ou ma be familiar with states like d, etc. How do these relate to our Y lm? d means l=2. stands for a particular superposition of different m s. d = (Y 22 +Y 2-2 )/ 2. The probabilit distribution is shown here: Which set of states is right? It depends on the problem ou want to solve. In a strong magnetic field the m states are (approimatel) the energ eigenstates, because the magnetic moment determines the energ. In a crstalline environment, states like ma be better, because the interaction with nearb atoms dominates the energ. Lecture 7, p 30