Lecture 16: 3D Potentials and the Hydrogen Atom. 1 = π. r = a 0. P(r) ( ) h E. E n. Lecture 16, p 2

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1 It was almost as incredible as if ou fired a 15-inch shell at a piece of tissue paper, and it came back to hit ou! --E. Rutherford (on the discover of the nucleus) ecture 16, p 1

2 ecture 16, p ecture 16: 3D Potentials and the Hdrogen Atom 5 P(r) 0 4a r r = a 0 z ( ) 8 z n n n n n n m h E z + + = ) ( ) ( ) ( ),, ( z z ϕ ϕ ϕ ψ = o a / r 3 o e a 1 ) r ( = π ψ 136 n ev. E n =

3 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, H atom Angular momentum, electron spin Net week: Eclusion principle, periodic table of atoms, molecules Solids, Metals, insulators, semiconductors Consequences of Q. M., Schrodinger s cat, superconductors, lasers,.. Final Eam: Monda, Oct. 14 Homework 6: Due Saturda (Oct. 1), 8 am ecture 16, p 3

4 Toda 3-Dimensional Potential Well: Product Wave Functions Degenerac Schrödinger s Equation for the Hdrogen Atom: Semi-quantitative picture from uncertaint principle Ground state solution Sphericall-smmetric ecited states ( s-states ) ecture 16, p 4

5 Quantum Particles in 3D Potentials A real (D) quantum dot So far, we have considered quantum particles bound in one-dimensional potentials. This situation can be applicable to certain phsical sstems but it lacks some of the features of most real 3D quantum sstems, such as atoms and artificial structures. One consequence of confining a quantum particle in two or three dimensions is degenerac -- the eistence of several quantum states at the same energ. To illustrate this important point in a simple sstem, let s etend our favorite potential - the infinite square well - to three dimensions. ecture 16, p 5

6 Particle in a 3D Bo (1) The etension of the Schrödinger Equation (SEQ) to 3D is straightforward in Cartesian (,,z) coordinates: ħ ψ ψ ψ + + U( z,, ) ψ Eψ + = m z 1 m Kinetic energ term: ( p + p + pz ) where ψ ψ (,, z) et s solve this SEQ for the particle in a 3D cubical bo: z U(,,z) = outside bo, or or z < 0 0 inside bo outside bo, or or z > This U(,,z) can be separated : U(,,z) = U() + U() + U(z) U = if an of the three terms =. ecture 16, p 6

7 Particle in a 3D Bo () Whenever U(,,z) can be written as the sum of functions of the individual coordinates, we can write some wave functions as products of functions of the individual coordinates: (see the supplementar slides) ψ (,, z) = f ( ) g( ) h( z) For the 3D square well, each function is simpl the solution to the 1D square well problem: nπ h n En fn ( ) = N sin = m Similarl for and z. D wave functions: 0 n sin π n sin π (n,n ) = (1,1) (n,n ) = (1,) 0 Each function contributes to the energ. The total energ is the sum: Etotal = E + E + E z 0 (n,n ) = (,1) 0 (n,n ) = (,) ecture 16, p 7

8 Particle in a 3D Bo (3) The energ eigenstates and energ values in a 3D cubical bo are: nπ nπ nzπ ψ = N sin sin sin z E h = + + 8m ( n n n ) n n n z z where n,n, and n z can each have values 1,,3,. z This problem illustrates two important points: Three quantum numbers (n,n,n z ) are needed to identif the state of this three-dimensional sstem. That is true for ever 3D sstem. More than one state can have the same energ: Degenerac. Degenerac reflects an underling smmetr in the problem. 3 equivalent directions, because it s a cube, not a rectangle. ecture 16, p 8

9 Cubical Bo Eercise z Consider a 3D cubic bo: Show energies and label (n,n,n z ) for the first 11 states of the particle in the 3D bo, and write the degenerac, D, for each allowed energ. Define E o = h /8m. E (n,n n z ) Degenerac 6E o (,1,1) (1,,1) (1,1,) D=3 3E o (1,1,1) D=1 ecture 16, p 9

10 Solution z Consider a 3D cubic bo: Show energies and label (n,n,n z ) for the first 11 states of the particle in the 3D bo, and write the degenerac, D, for each allowed energ. Define E o = h /8m. E (n,n n z ) Degenerac 1E o D=1 (,,) 11E o (3,1,1) (1,3,1) (1,1,3) D=3 9E o (,,1) (,1,) (1,,) D=3 E h = + + 8m ( n n n ) n n n z z n,n,n z = 1,,3,... 6E o (,1,1) (1,,1) (1,1,) D=3 3E o (1,1,1) D=1 ecture 16, p 10

11 Act 1 For a cubical bo, we just saw that the 5 th energ level is at 1 E 0, with a degenerac of 1 and quantum numbers (,,). 1. What is the energ of the net energ level? a. 13E 0 b. 14E 0 c. 15E 0. What is the degenerac of this energ level? a. b. 4 c. 6 ecture 16, p 11

12 Solution For a cubical bo, we just saw that the 5 th energ level is at 1 E 0, with a degenerac of 1 and quantum numbers (,,). 1. What is the energ of the net energ level? a. 13E 0 b. 14E 0 c. 15E 0 E 1,,3 = E 0 ( ) = 14 E 0. What is the degenerac of this energ level? a. b. 4 c. 6 ecture 16, p 1

13 Solution For a cubical bo, we just saw that the 5 th energ level is at 1 E 0, with a degenerac of 1 and quantum numbers (,,). 1. What is the energ of the net energ level? a. 13E 0 b. 14E 0 c. 15E 0 E 1,,3 = E 0 ( ) = 14 E 0. What is the degenerac of this energ level? a. b. 4 c. 6 An ordering of the three numbers will give the same energ. Because the are all different (distinguishable), the answer is 3! = 6. Question: Is it possible to have D > 6? Hint: Consider E = 6E 0. ecture 16, p 13

14 Non-cubic Bo Consider a non-cubic bo: z The bo is stretched along the -direction. What will happen to the energ levels? Define E o = h /8m 1 1 E 1 > 1 11E o 9E o 6E o 3E o ecture 16, p 14

15 Consider a non-cubic bo: Solution z The bo is stretched along the -direction. What will happen to the energ levels? Define E o = h /8m 1 1 E h E = E n + n + n ( ) ( ) nnnz 0 z 8m Smaller than E 0 1 > 1 11E o 9E o 6E o 3E o The others are left for ou. (,1,1) (1,1,) D= (1,,1) D=1 (1,1,1) D=1 1: The smmetr is broken for, so the 3-fold degenerac is lowered. A -fold degenerac remains, because and z are still smmetric. : There is an overall lowering of energies due to decreased confinement along. ecture 16, p 15

16 Consider a particle in a D well, with = =. 1. Compare the energies of the (,), (1,3), and (3,1) states? a. E (,) > E (1,3) = E (3,1) b. E (,) = E (1,3) = E (3,1) c. E (,) < E (1,3) = E (3,1) Act. If we squeeze the bo in the -direction (i.e., < ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) ecture 16, p 16

17 Solution Consider a particle in a D well, with = =. 1. Compare the energies of the (,), (1,3), and (3,1) states? a. E (,) > E (1,3) = E (3,1) b. E (,) = E (1,3) = E (3,1) c. E (,) < E (1,3) = E (3,1) E (1,3) = E (3,1) = E 0 (1 + 3 ) = 10 E 0 E (,) = E 0 ( + ) = 8 E 0 E h 8m 0. If we squeeze the bo in the -direction (i.e., < ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) ecture 16, p 17

18 Solution Consider a particle in a D well, with = =. 1. Compare the energies of the (,), (1,3), and (3,1) states? a. E (,) > E (1,3) = E (3,1) b. E (,) = E (1,3) = E (3,1) c. E (,) < E (1,3) = E (3,1) E (1,3) = E (3,1) = E 0 (1 + 3 ) = 10 E 0 E (,) = E 0 ( + ) = 8 E 0 E h 8m 0. If we squeeze the bo in the -direction (i.e., < ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) Because <, for a given n, E 0 for motion is larger than E 0 for motion. The effect is larger for larger n. Therefore, E (3,1) > E (1,3). Eample: = ½, = 1: E (1,3) = 13 E (3,1) = 37 We sa the degenerac has been lifted. ecture 16, p 18

19 Another 3D Sstem: The Atom -electrons confined in Coulomb field of a nucleus Earl hints of the quantum nature of atoms: Discrete Emission and Absorption spectra When ecited in an electrical discharge, atoms emit radiation onl at discrete wavelengths Different emission spectra for different atoms Atomic hdrogen λ (nm) Geiger-Marsden (Rutherford) Eperiment (1911): Measured angular dependence of a particles (He ions) scattered from gold foil. Mostl scattering at small angles supported the plum pudding model. But Occasional scatterings at large angles Something massive in there! α v Au θ Conclusion: Most of atomic mass is concentrated in a small region of the atom a nucleus!

20 Rutherford Eperiment

21 Atoms: Classical Planetar Model (An earl model of the atom) Classical picture: negativel charged objects (electrons) orbit positivel charged nucleus due to Coulomb force. F -e There is a BIG PROBEM with this: +Ze As the electron moves in its circular orbit, it is ACCEERATING. As ou learned in Phsics 1, accelerating charges radiate electromagnetic energ. Consequentl, an electron would continuousl lose energ and spiral into the nucleus in about 10-9 sec. The planetar model doesn t lead to stable atoms.

22 Hdrogen Atom - Qualitative Wh doesn t the electron collapse into the nucleus, where its potential energ is lowest? We must balance two effects: As the electron moves closer to the nucleus, its potential energ decreases (more negative): U κe = r However, as it becomes more and more confined, its kinetic energ increases: p ħ r KE ħ mr Therefore, the total energ is: ħ κe E = KE + PE mr r E has a minimum at: r ħ mκe = a nm The Bohr radius of the H atom. At this radius, E 4 mκ e = 13.6 ev ħ The ground state energ of the hdrogen atom. Heisenberg s uncertaint principle prevents the atom s collapse. One factor of e or e comes from the proton charge, and one from the electron. ecture 16, p

23 Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the effective Bohr radius a 0,He with the usual Bohr radius for hdrogen, a 0 : a. a 0,He > a 0 b. a 0,He = a 0 c. a 0,He < a 0 Act 3 r ħ mκe = a nm The Bohr radius of the H atom.. What is the ratio of ground state energies E 0,He /E 0,H? a. E 0,He /E 0,H = 1 b. E 0,He /E 0,H = c. E 0,He /E 0,H = 4 ecture 16, p 3

24 Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the effective Bohr radius a 0,He with the usual Bohr radius for hdrogen, a 0 : a. a 0,He > a 0 b. a 0,He = a 0 c. a 0,He < a 0. What is the ratio of ground state energies E 0,He /E 0,H? a. E 0,He /E 0,H = 1 b. E 0,He /E 0,H = c. E 0,He /E 0,H = 4 Solution ook at how a 0 depends on the charge: ħ ħ a0 a0 a 0, He = mκ e mκ( e) e This should make sense: more charge stronger attraction electron sits closer to the nucleus ecture 16, p 4

25 Consider an electron around a nucleus that has two protons, (an ionized Helium atom). 1. Compare the effective Bohr radius a 0,He with the usual Bohr radius for hdrogen, a 0 : a. a 0,He > a 0 b. a 0,He = a 0 c. a 0,He < a 0. What is the ratio of ground state energies E 0,He /E 0,H? a. E 0,He /E 0,H = 1 b. E 0,He /E 0,H = c. E 0,He /E 0,H = 4 Solution ook at how a 0 depends on the charge: ħ ħ a0 a0 a 0, He = mκ e mκ( e) e This should make sense: more charge stronger attraction electron sits closer to the nucleus Clearl the electron will be more tightl bound, so E 0,He > E 0,H. How much more tightl? ook at E 0 : 4 κ κ ( ) E = m e E = m e e = 4E ħ ħ 0, H 0, He 0, H In general, for a hdrogenic atom (onl one electron) with Z protons: E = Z E 0, Z 0, H ecture 16, p 5

26 Net ectures Angular momentum atomic orbitals Spin Pauli Eclusion Principle ecture 16, p 6

27 Supplement: Separation of Variables (1) In the 3D bo, the SEQ is: ħ ψ ψ ψ ( U( ) + U( ) + U( z) ) ψ = Eψ m z NOTE: Partial derivatives. et s see if separation of variables works. Substitute this epression for ψ into the SEQ: ψ (,, z) = f ( ) g( ) h( z) ħ d f d g d h gh + fh + fg + ( U( ) + U( ) + U( z) ) fgh = Efgh m d d dz NOTE: Total derivatives. Divide b fgh: ħ 1d f 1d g 1d h ( U( ) + U( ) + U( z) ) = E m f gd hdz ecture 16, p 7

28 Supplement: Separation of Variables () Regroup: ħ 1d f ħ 1d g ħ 1d h ( ) ( ) ( ) + U + + U + + U z = E mf mgd mhdz A function of A function of A function of z We have three functions, each depending on a different variable, that must sum to a constant. Therefore, each function must be a constant: ħ 1d f + U( ) = E mf ħ 1d g + ( ) U = E mgd ħ 1d h + ( ) U z = E mhdz E + E + Ez = E z Each function, f(), g(), and h(z) satisfies its own 1D SEQ. ecture 16, p 8