Lecture35.pptx

# Lecture35.pptx - 4/13/11 PHYS 360 Quantum Mechanics...

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Unformatted text preview: 4/13/11 PHYS 360 Quantum Mechanics Wed Apr 13, 2011 Lecture 35: Does understanding hydrogen help at all with helium? Exchange Forces For simplicity let’s work things out in just one dimension. We’ll consider a set of one ­parWcle wave funcWons that are orthogonal and normalized. If we have two dis\$nguishable parWcles the combined wave funcWon is: ψ x1 , x2 = ψ a ( x1 )ψ b ( x2 ) ( ) (You can think of these one ­parWcle states as eigenfuncWons of the inﬁnite square well, for example.) HW #10 due Monday, Apr 18: Ch 5, #5, 7, 9, 13, 14 Now consider indis\$nguishable parWcles…. 1 2 For symmetric wave funcWons: Case 1: Dis\$nguishable par\$cles ψ + x1 , x2 = ( ) 1 2 ⎡ψ a x1 ψ b x2 + ψ b x1 ψ a x2 ⎤ ⎣ ⎦ () () () () (x − x ) 1 2 2 d = x2 a + x2 b −2 x a x b For anWsymmetric wave funcWons: Case 2: Indis\$nguishable par\$cles ψ − x1 , x2 = ( ) 1 2 ⎡ψ a x1 ψ b ( x2 ) − ψ b x1 ψ a x2 ⎤ ⎣ ⎦ () () () (x − x ) 1 2 2 ± = x2 a + x2 b −2 x a x b 2 x 2 ab Now, for all three cases let’s calculate the expectaWon value of the square of the distance beteen the two parWcles: ...where x ab ≡ ∫ xψ a x ψ b x dx () * () (x − x ) 1 2 2 2 = x12 + x2 − 2 x1 x2 3 This last term is called an overlap integral. It is zero unless there is some overlap in the wave funcWons. 4 DisWnguishable (x − x ) 1 2 2 d = x2 a + x2 b −2 x a x b ≡ ( Δx ) 2 ab 2 d Symmetric AnWsymmetric (x − x ) 1 2 2 + = x2 a + x2 b −2 x a x b −2 x x b +2 x (x − x ) 1 2 2 − = x2 a + x2 b −2 x 2 ab a Symmetric ( Δx ) ( Δx ) 2 + = ( Δx ) ( Δx ) 2 d −2 x 2 ab Note: Even though electrons are fermions, the spaWal part can be either symmetric or anWsymmetric (because of the spin degree of freedom). AnWsymmetric 2 − = 2 d +2 x 2 ab 5 6 1 4/13/11 1.  Why is it hard to squeeze atoms closer together in solids? 2.  What keeps the Sun from condensing into a solid? 3.  What is a neutron star? 4.  How can a neutron star become a black hole? 1.  Why is it hard to squeeze atoms closer together in solids? Electrons are fermions, and Pauli exclusion prevents them from being “crowded.” (a.k.a. degeneracy pressure) 2.  What keeps the Sun from condensing into a solid? It’s hot. Excess kineWc energy, radiaWon pressure, etc. prevents atoms from being conﬁned to a small volume. 3.  What is a neutron star? If the Sun turned cold, then gravity would conﬁne a hydrogen atom to such a small volume that the electron would join the proton to make a neutron. But neutrons also are spin ­½ fermions, so Pauli exclusion prevents total collapse. 4.  How can a neutron star become a black hole? GravitaWonal pressure can be so great that moWons become relaWvisWc, and Pauli exclusion is overcome. Neutrons condense into black hole singularity. 7 8 Now, let’s consider atoms with more than one electron…. The Hamiltonian for an atom with Z electrons: 2 2 z⎧ e2 ⎪ − 2 ⎛ 1 ⎞ Ze ⎫ 1 ⎛ 1 ⎞ Z ⎪ H = ∑⎨ ∇j −⎜ ⎟ r ⎬ + 2 ⎜ 4πε ⎟ ∑ ⎝ 4πε 0 ⎠ j ⎭ ⎝ j =1 ⎩ 2 m ⎪ ⎪ 0 ⎠ j ≠ k r j − rk KineWc energy of each electron; no nuclear moWon Nuclear ­electron Coulomb potenWal energy Electron ­electron Coulomb repulsion 9 10 As usual our goal is to solve the Schrödinger equaWon: 5.2.1 Helium Helium has a nucleus with Z=2 plus two electrons: ⎧ 2 2 1 2e2 ⎫ ⎧ 2 2 1 2e2 ⎫ 1 e2 ⎪ ⎪⎪ ⎪ H = ⎨− ∇1 − ∇− ⎬ + ⎨− ⎬+ 4πε 0 r1 ⎭ ⎩ 2 m 2 4πε 0 r2 ⎭ 4πε 0 r1 − r2 ⎪ ⎪⎪ ⎪ ⎩ 2m Hψ = Eψ However, we require that the mulW ­electron wave funcWon, ψ r1 , r2 , ..., rz χ s1 , s2 , ..., sz ( )( ) …must be an\$ ­symmetric. That means, the overall sign of the wave funcWon must change if the coordinates (space & spin) of any two electrons are exchanged. This looks like the hydrogen atom, except with Z=2. So does this… But this interac\$on term is a big problem! So, let’s see what happens if we just drop the last term…. 11 12 2 4/13/11 Drop the interacWon term: ⎧ 1 2e ⎪ 2 H approx ≈ ⎨− ∇1 − 4πε 0 r1 ⎪ 2m ⎩ 2 2 ⎫⎧ 1 2e ⎪⎪ ∇2 − ⎬ + ⎨− 2 4πε 0 r2 ⎪ ⎪ 2m ⎭⎩ 2 2 ⎫ ⎪ ⎬ ⎪ ⎭ The “hydrogenic” (i.e. one electron atoms) energies depend on Z2, so going from hydrogen to helium is a factor of 4: E = 4 En + En ' The ground state is: ( ) Only electron #1 Only electron #2 The soluWons are products of hydrogenic wave funcWons: 8 −2 r +r / a ψ 0 r1 , r2 = ψ 100 ( r1 )ψ 100 ( r2 ) = 3 e ( 1 2 ) πa ( ) ψ r1 , r2 = ψ nlm ( r1 )ψ n ' l ' m ' ( r2 ) “Hydrogenic:” ( ) Z=1: ψ 100 (r,θ , φ ) = 1 π a3 1 e− r / a Z=2: ψ 100 (r,θ , φ ) = 13 ⎛ a⎞ π⎜ ⎟ ⎝ 2⎠ 3 e−2 r / a = 8 −2 r / a e π a3 14 NoWce that this is a symmetric wave funcWon: 8 −2 r +r / a ψ 0 r1 , r2 = ψ 100 ( r1 )ψ 100 ( r2 ) = 3 e ( 1 2 ) πa In this model the predicted ground state energy is: ( ) E0 = 4 −13.6 − 13.6 eV = −109 eV But the measured value is E0 = −78.975 eV ( ) = ψ 0 r2 , r1 ( ) This means the spin part must be anWsymmetric: ⎫ ⎧1 ψ 0 r1 , r2 χ ( s1 , s2 ) = ψ 0 r1 , r2 00 = ψ 0 r1 , r2 ⎨ ↑↓ − ↓↑ ⎬ ⎪ ⎪ ⎩2 ⎭ ( ) ( ) ( ) ( ) (How would you measure this?) s = 0 (singlet) Obviously, ignoring electron ­electron repulsion is a big problem, but at least it’s a start. Does the overall sign change if you exchange parWcles 1 & 2? 15 16 Put one electron in an excited state (& leave the other in the ground state): Parahelium: spin singlet Orthohelium: spin triplet ψ nlmψ 100 These have to be made symmetric or anWsymmetric: ψ ± (r1 , r2 ) = A [ψ nlm (r1 )ψ 100 (r2 ) ± ψ 100 (r1 )ψ nlm (r2 )] …and these must be matched with anWsymmetric or symmetric spin funcWons: ⎧ 11 = ↑↑ ⎪ 1 ⎪ ↑↓ + ↓↑ ⎨ 10 = 2 ⎪ ⎪ 1 − 1 = ↓↓ ⎩ ⎫ ⎪ ⎪ ⎬ s = 1 (triplet ) ⎪ ⎪ ⎭ Parahelium has the symmetric spaWal wave funcWon. That means the two electrons are a bit closer together on average. This increases their Coulomb repulsion, causing them to have higher energies (less negaWve). ( ) ⎧ ⎫ 1 ↑↓ − ↓↑ ⎬ s = 0 ( singlet ) ⎨ 00 = ⎪ ⎪ 2 ⎩ ⎭ ( ) 17 18 3 ...
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