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QM Postulates

QM Postulates. 1. Y = function for particle/system Y ( x,y,z,t ). Operators ( Ĥ, x, p x , etc .) e.g. p x = - i ħ /2m d 2 (Y)/dx 2. ^. ^. Measurements & eigenfunctions e.g. Ĥ Y = E Y. 4. Expectation Values e.g. < p x > = ∫ Y * p x Y d t.

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QM Postulates

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  1. QM Postulates 1. Y= function for particle/system Y(x,y,z,t) Operators ( Ĥ, x, px, etc.) e.g. px = -iħ/2m d2(Y)/dx2 ^ ^ Measurements & eigenfunctions e.g. ĤY = EY 4. Expectation Values e.g. <px> = ∫ Y*pxYdt Time-dependent Schrödinger Equation ĤY = iħ d(Y)/dt 6. Spin (Pauli Exclusion Principle)

  2. A h a C B o The Heisenberg Uncertainty Principle The more certain you know the position of a particle, the less certain you know its momentum – the act of measuring the position will change its momentum. This uncertainty is intrinsic to Quantum Theory and is not a function of limitations in experimental design or instrumentation. sinA = o/h tanA = o/a cosA = a/h

  3. x screen detects photons y l w = slit width = Dx Dpx = 2px p w/2 px q py l/2 sin q = l/2 ÷ w/2 = l/w sin q = px/p & px = p sinq Dx • Dpx = 2plsub in l = h/p …. w 2psinq (let sin q = l/w) Dx • Dpx = 2h

  4. The Heisenberg Uncertainty Principle from slit exptdx dpx (Dpx) (Dx) = 2h General Rule (Dpx) (Dx) ≥ ħ/2 Try problem # 14

  5. V = ∞ V = ∞ Particle in a 1D box The Standing Wave V = 0 What is y? V • Ĥy= Ey -ħ2/2m {d2(y)/dx2} + V(y) y = 0 y = 0 d(sin cx)/dx = c coscx d(coscx)/dx = -c sin cx d(ecx)/dx = cecx y2 = 0 x a 0

  6. d2{Asin(cx)}/dx2 = A d2{sin(cx)}/dx2 1st derivative = c cos(cx) 2nd derivative = -c2 sin(cx) -c2 (Asin(cx)) = (-2mE/ħ2) (Asin(cx)) -c2 = (-2mE/ħ2) En = c2ħ2/(2m) = n2h2/(8ma2) • Ĥy= Ey Particle in a 1D box The Standing Wave y= A sin cx + B coscx V cos 0 = 1 …. B = 1 y= A sin cx y= 0 at x = a only if c = np/a E = n2h2/(8ma2) x a 0

  7. y= A sin cx = A sin (npx/a) Particle in a 1D box The Standing Wave E = n2h2/(8ma2) 0ay2 dx = 1 A20a sin2cx dx = 1 What is A?  sin2 cx dx = x/2 – {sin(2cx)}/(4c) A2 ( a/2 – 0 – 0 + 0) = 1 A2 = 2/a & A = (2/a)1/2 • = A sin cx = (2/a)1/2 sin (npx/a) • where a = length of box and n = integer from 1 → ∞

  8. V = ∞ V = ∞ Particle in a 1D box The Standing Wave V = 0 What is <x>? y= A sin cx <x> = a/2 V y = (2/a)1/2 sin (npx/a) What is <px>? • Ĥy= Ey <px> = 0 E = n2h2/(8ma2) y2 = 0 x a 0

  9. n = 2 n = 3 n = 1 yy2 n = 4

  10. 10.37 Probability that an electron in a 1D box would be between 0.495a and 0.505a (i.e. near the center) for n = 1→4. Probability = ∫0ay2dx y = (2/a)1/2 sin (npx/a) Probability ~ y2dx = (2/a) sin2(npx/a) • 0.01a where a = 0.5 Probability ~ 0.02 sin2(½np) • 0.01a where a = 0.5

  11. c b a Particle in a 3D box Y (x,y,z) = X(x) • Y(y) • Z(z) y = (2/a)1/2sin(nxpx/a)  (2/b)1/2sin(nypy/b)  (2/c)1/2 sin(nzpz/c) E = (h2/8m) (nx2/a2 + ny2/b2 + nz2/c2) Degeneracy ― multiple states of the same system have the same energy. Symmetry causes degeneracy Quantum numbers ― ynx,ny,nz ― y111, y112, y121, y211, etc. If a = b = c then E = (h2/8ma2) (nx2 + ny2 + nz2) and E(y112,) = E(y121) = E(y211)

  12. Approximation Methods ― Variation Used when you cannot find a function y, such that Ĥy = Ey It can be proven from QM that <E> > E ― therefore best f gives lowest <E> y = (2/a)1/2 sin (npx/a) f = (30/a5)1/2 x(a – x) Look for a function fthat is similar to anticipated y and for which ∫ f Ĥ fdt = <E> ~ E (obtained from experimental results on system).

  13. Approximation Methods ― Perturbation Ĥ = Ĥ0+ Ĥ' Ĥ0y= E0ysolve for E0 Ĥ'y E'yso E' =  yĤ'ydx For a particle in a box with a position dependent potential energy term, e.g. V =kx, this method can be applied. E ≈ E0 + E' = n2h2/(8ma2) + ka/2 In this perturbation model we are applying the correction to the energy of the ground state (n=1) only. In addition we are assuming that yis unchanged. Therefore calculating the <x> and <px> would not be affected. However, it is likely that the assumption that y is the same is not valid. (see example 12.9 on page 389)

  14. y = Cekx + De-kx Particle in Free Motion k (wavenumber)  E E = k2ħ2/2m •  (V-E) kħ = {2m(V-E)}½ ← L → • = Aeikx + Be-ikx • = 2A coskx y = A´eikx + 0 • e-ikx e = E/V tunneling If V >> E (kL >> 1) … T = 16e(1 – e)e-2kL

  15. 10.37 Probability that an electron in a 1D box would be between 0.495a and 0.505a (i.e. near the center) for n = 1→4. Probability = ∫0ay2dx y = (2/a)1/2 sin (npx/a) Probability ~ y2dx = (2/a) sin2(npx/a) • 0.01a where a = 0.5 Probability ~ 0.02 sin2(½np) • 0.01a where a = 0.5

  16. 2. Cesium has a work function (f in the equation for the photoelectric effect) of 2.14 eV. What is the minimum wavelength needed to emit an electron from Cs? Cs – work function = 2.14 eV = 3.43 x 10-19J E = hc/l and l = hc/E = 5.80 x 10-7 m = 580 nm What is the speed of the electron emitted when light of wavelength … E = E = hc/l ½mv2 = E - f and v = (2(E-f)/m)½ i) 550 nm 3.61 x 10-19 J 2.01 x 105 m/s ii) 450 nm 4.42 x 10-19 J 4.66 x 105 m/s iii) 350 nm 5.68 x 10-19 J 7.03 x 105 m/s is shined on Cs metal in a vacuum?

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