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Getting used to quantum weirdness
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Going from 1D to 2D to 3D
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Multiple Dimensions (Separation of Variables)
(-ħ22/2m + U)f = Ef If U(x,y) = Ux(x) + Uy(y) Then f(x,y) = X(x).Y(y) and E = Ex + Ey [Solve two 1-D problems]
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Multiple Dimensions (Separation of Variables)
Nx = 100, Ny=100, N = 10,000 10,000 eigenvalues But we can also solve two 1d problems involving two sets of 100 x 100 matrices But this only gives us 200 eigenvalues! Where are the rest?
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Multiple Dimensions (Separation of Variables)
Can mix and match any 100 x eigenvalues with any 100 y eigenvalues 1 1 2 2 3 3 100 100
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2-D Box fpq = 4/L2 sin(ppx/L).sin(qpy/L) Epq = ħ2p2(p2+q2)/2mL2 (p, q = 1, 2, 3, …)
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fpq = 4/L2 sin(ppx/L).sin(qpy/L)
2-D Box f22 fpq = 4/L2 sin(ppx/L).sin(qpy/L) Epq = ħ2p2(p2+q2)/2mL2 (p, q = 1, 2, 3, …) 8E0 f21 f12 5E0 2E0 f11
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Square 2-D Box f22 8E0 f21 f12 5E0 This has nodes along x and y
(or x+y, x-y etc) 2E0 f11
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Circular 2-D Box 8E0 5E0 2E0
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Compare
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Circular 2D box
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What about a circular 2D box?
Angular orthogonality m Radial orthogonality This has nodes along r and q Angular nodes .. Orbitals l
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Where do these nodes come from?
2F = (-2mE/ħ2)F 1/r∂/∂r(r∂F/∂r) + 1/r2∂2F/∂j2 F = Rnm(r)Qm(j) ∂2Q/∂j2 = -m2Q, solution e±imj m integer
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Radial Solution for free electron
2F = (-2mE/ħ2)F r2∂2R/∂r2 + r∂R/∂r + (r2/l2- m2 ) R = 0 = ħ/√2mE is the de Broglie wavelength Oscillatory radial solutions Rnm(r) = Jm(r/ln) Bessel Functions
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Quantization Fnm = Jm(rxnm/a)eimj Rnm(a) = Jm(a/ln) = 0
a/ln = xnm, nth root of Jm En = ħ2xnm2/2ma2 Fnm = Jm(rxnm/a)eimj
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Radial Solution for Coulomb electron
2F = (-2m(E-U)/ħ2)F r2∂2R/∂r2 + r∂R/∂r + (r2/l2 + r/a0 - m2) R = 0 Different boundary condition R(∞) = 0
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Radial Solution for Coulomb electron
r2∂2R/∂r2 + r∂R/∂r + (r2/l2 + r/a0 - m2) R = 0 Near r 0, get R = rm Near r ∞, get R = eir/l = e-r/|l| Set R = rme-r/|l| f(r) rf’’+ f’[(2m+1)-2r/l] + f’[1/a0-(2m+1)/l] = 0 Solutions are associated Laguerre polynomials This equation has an interpretation we’ll come back to later
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Radial Solution for Coulomb electron
Try f(r) = Spcprp cp/cp-1 = [(2p+2m-1)/l-1/a0]/(p+2m) For series to be finite, need (2p+2m-1)/l-1/a0 = 0 E = -ħ2/2ml2 = -(ħ2/2ma02) x 1/(2p+2m-1)2 = 4E0/(2n-1)2 = E0/(n-1/2)2 Twice the energy of 3D Hydrogen !!
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