Quantum Mechanics: Schrödinger Equation and Uncertainty Principle - Prof. Leo Radzihovsky, Assignments of Quantum Mechanics

Download Quantum Mechanics: Schrödinger Equation and Uncertainty Principle - Prof. Leo Radzihovsky and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - 1: HW 2 Solutions Leo Radzihovsky Paul Martens September 22, 2007 1 Problem 1 Showing that the Schrödinger’s equation contains Hamilton-Jacobi equation (in the h̄ → 0 limit) and the continuity equation. We start with the Schrödinger equation − h̄ 2 2m ∇2Ψ + VΨ = ıh̄∂tΨ (1) we suppose Ψ = |Ψ| e ıSh̄ . We’ll need the derivatives ∇Ψ = (∇ |Ψ|) eı Sh̄ + |Ψ| ı h̄ ∇Se ıSh̄ (2) ∇2Ψ = ∇2 |Ψ| e ıSh̄ + 2ı h̄ ∇ |Ψ|∇Se ıSh̄ + ı h̄ |Ψ|∇2Se ıSh̄ − 1 h̄2 |Ψ| (∇S)2 e ıSh̄ (3) ıh̄∂tΨ = ıh̄∂t |Ψ| e ıS h̄ − |Ψ| ∂tSe ıS h̄ (4) Using the above we can write (1) − h̄ 2 2m ∇2 |Ψ|− ıh̄ m ∇ |Ψ|·∇S− ıh̄ 2m |Ψ|∇2S+ 1 2m |Ψ| (∇S)2+V |Ψ| = ıh̄∂t |Ψ|−|Ψ| ∂tS (5) both real and imaginary parts of this equation must be satisfied, so we can look at these parts independently. 1.1 real: − h̄ 2 2m ∇2 |Ψ|+ |Ψ| 2m (∇S)2 + V |Ψ| = − |Ψ| ∂tS (6) In the classical limit h̄→ 0 this becomes the H-J equation. 1 2m (∇S)2 + V = −∂tS (7) 1 1.2 Im: − h̄ m ∇ |Ψ| · ∇S − h̄ 2m |Ψ|∇2S = h̄∂t |Ψ| (8) to get the continuity equation we multiply both sides by 2 |Ψ|, and note that ∂ |Ψ|2 = 2 |Ψ| ∂ |Ψ|. ∂t |Ψ|2 +∇ · ( |Ψ|2 ∇S m ) = 0 (9) from the above equation we can identify P = |Ψ|2 as the probability density, and ~J = |Ψ|2︸︷︷︸ density ~∇S m︸︷︷︸ velocity as the probability current. 2 Problem 2 Show that Û = e −ı h̄ Ĥt satisfies the Schrödinger equation. 2.1 generally ıh̄∂tÛ = ıh̄ ( −ı h̄ ) Ĥe− ı h̄ Ĥt = ĤÛ (10) so U satisfies the Schrödinger equation. ıh̄∂tÛ = ĤÛ (11) 2.2 An explicit case For a free particle U (x, x′; t) = √ m 2πıh̄t eı m(x−x′)2 2h̄t (12) therefore ıh̄∂tU = √ m 2πıh̄ ıh̄ [ − t − 32 2 − ıt− 12 m (x− x ′)2 2h̄t2 ] eı m(x−x′)2 2h̄t and − h̄ 2 2m ∂2xU = − h̄2 2m √ m 2πıh̄ t− 1 2 ∂x ( ım h̄t (x− x′) eı m(x−x′)2 2h̄t ) = − h̄ 2 2m √ m 2πıh̄ t− 1 2 ( ım h̄t − (m h̄t )2 (x− x′)2 ) eı m(x−x′)2 2h̄t = √ m 2πıh̄ ıh̄ [ − t − 32 2 − ıt− 12 m (x− x ′)2 2h̄t2 ] eı m(x−x′)2 2h̄t Therefore our example for U is a solution to the Schrödinger equation 2 3.3 finding the normalized Lx eigenvectors We did this above they were |Lx = 0〉 = √ 2 2  10 −1  |Lx = ±1〉 = 1 2  ±1√2 ±1  (29) 3.4 given that we are in the Lz = 1 what is P (Lx = 0,±1) We have |Lx〉 in the Lz basis so we can basically read the probabilities off. PLz=1→Lx=±1 = |〈Lx = ±1|Lz = 1〉| 2 = 1 4 (30) PLz=1→Lx=0 = |〈Lx = 0|Lz = 1〉| 2 = 1 2 (31) 3.5 collapse of wavefunctions we started with the wavefunction |Ψ〉 = 1 2  11√ 2  (32) we find that L2z = 1, so that means the only Lz = ±1 is possible. So our state after is. |Ψafter〉 = ( 1 2 |Lz = 1〉+ √ 2 2 |Lz = −1〉 ) ︸ ︷︷ ︸ Lz 6=0 1√( 1 2 )2 + (√22 )2 (33) |Ψafter〉 = 1√ 3 |Lz = 1〉+ √ 2 3 |Lz = −1〉 (34) The probability that we get L2z = 1 as an outcome can be read off the state. 1 4︸︷︷︸ Lz=1 + 1 2︸︷︷︸ Lz=−1 = 3 4 (35) The probabilities of an Lz measurement afterwords can be read off the final state. P (Lz = 1) = 1 3 (36) P (Lz = 0) = 0 (37) P (Lz = −1) = 2 3 (38) 5 3.6 Probabilities to states Suppose we have a state |Ψ〉 = C+ |Lz = 1〉+ C0 |Lz = 0〉+ C− |Lz = −1〉 (39) we know that |C+|2 = 1 4 (40) |C0|2 = 1 2 (41) |C−|2 = 1 4 (42) The most general solutions to these are Cn = √ Pne ıδn (43) These phases are physical and can be measured by considering Lx = 0. P (Lx = 0) = |〈Lx = 0|Ψ〉|2 = ∣∣∣∣eıδ12 〈Lx = 0|Lz = 1〉+ eıδ2√2 〈Lx = 0|Lz = 0〉+ eıδ32 〈Lx = 0|Lz = −1〉 ∣∣∣∣2 = ∣∣∣∣∣12eıδ1 √ 2 2 + √ 2 2 eıδ20 + 1 2 eıδ3 ( − √ 2 2 )∣∣∣∣∣ 2 = ∣∣∣∣ 123/2 (eıδ1 − eıδ3) ∣∣∣∣2 = 1 2 sin2 ( δ1 − δ3 2 ) (44) 4 Problem 4 ~L = ~r × ~p = ijkripj (45) 4.1 Poisson bracket {Li, Lj} = {iklxkpl, jmnxmpn} = ilkjmn {xlpk, xmpn} = ilkjmn ∂ (xlpk)∂xq ∂ (xmpn)∂pq︸ ︷︷ ︸ δlqδnqpkxm − ∂ (xlpk) ∂pq ∂ (xmpn) ∂xq︸ ︷︷ ︸ δkqδmqxlpn  = ilkjmlpkxm − ilkjknxlpn = pjxi − pmxmδij − pixj + δijxnpn {Li, Lj} = xipj − xjpi = ijkLk (46) 6 4.2 commutation relations[ L̂i, L̂j ] = [x̂lp̂k, x̂mp̂n] ilkjmn = x̂l [p̂k, x̂mp̂n]︸ ︷︷ ︸ −ıh̄δkmp̂n + [x̂l, x̂mp̂n]︸ ︷︷ ︸ ıh̄δlnx̂m p̂k  ilkjmn = ıh̄ (x̂mp̂kinkjmn − x̂lp̂nilkjkn) = ıh̄ [x̂mp̂k (δkjδim − δknδij)− x̂lp̂n (δinδlj − δijδln)] = ıh̄ (x̂ip̂j − x̂j p̂i)[ L̂i, L̂j ] = ıh̄ijkLk (47) 5 Problem 5 Finding the energy eigenstates of a bouncing ball 5.1 Bohr-Sommerfeld quantization Bohr-Sommerfeld condition is ∮ pdz = nh̄ (48) Classically the energy of a ball is E = p2 2m +mgz ⇒ p = √ 2m (E −mgz) (49) Using this form for our momentum 2 √ 2m ∫ E mg 0 dz √ E − gmz = 2√mg 2 3 ( E mg )3/2 (50) so 4 3 √ 2m E 3/2 n mg = nh̄ (51) En = ( 9 32 mg2n2h̄2 )1/3 (52) we can use this energy to come up with an approximate maximum height for each state. zn ≈ En mg ⇒ zn ≈ ( 9n2h̄2 32m2g ) 1 3 (53) 7 8.3 Time evolution of wavefunction where p0 is the momentum of the particle. Mathematically we could have just multiplied by eıφ, but the choice here make physical sense. To see this we look at the time evolution of the state. Ψ (x, t) = ( 1 π∆2 ) 1 4 1√ 2πıa2 ∫ dx′ eı (x−x′)2 2a2︸ ︷︷ ︸ U e− (x′)2 2∆2 +ı p0x ′ h̄ (69) where a(t) (70) which we’ll call the diffusion length. This is a Gaussian integral, which can be done by “completing the square.” (This one comes up alot you should know how to do it). We use that∫ dx′eıhx ′− 12 Ax ′2 = √ 2π A e− 1 2 A −1h2 (71) Ψ (x, t) = 1 π1/4 1√ ∆ + ıa2∆ eF (72) where F = − ( x− p0a 2 h̄ )2 2 (∆2 + ıa2) + ı p0a 2 h̄ − ıp 2 0a 2 2h̄2 (73) substituting in for a2(t) = h̄tm . Ψ = √ 1− ı h̄tm/∆2 (π∆2)1/4 ( 1 + h̄ 2t2 m2 /∆ 4 )1/4 e− (x− p0tm ) 2 2∆2(1+ h̄2t2 m2 /∆4) e ı 2 h̄t/m ∆4+h̄2t2/m2 (x− p0 m t) 2 +ı p0x h̄ −ı p20t 2mh̄ (74) 8.4 The Probability density So the Probability density is P (x, t) = |Ψ(x, t)|2 = 1√ (π∆2) ( 1 + h̄ 2t2 m2 /∆ 4 )e− ( x− p0t m ) 2 ∆2(1+ h̄2t2 m2 /∆4) (75) 10 8.5 Expectation values 8.5.1 x The expectation value of x(t) gives a nice result. 〈t|x|t〉 = ( 1 π1/2∆(t) )∫ dx [( x− p0t m ) + p0t m ] e − (x− p0t m ) 2 ∆(t)2 (76) where ∆(t) = ∆ √ 1 + h̄2t2 m2 /∆4 (77) Writing x = x− p0tm + p0t m make this integral trivial. 〈x(t)〉 = p0t m (78) 8.5.2 x2 To calculate the variance we’ll need to know 〈 x2 〉 = √ 1 π∆(t)2 ∫ dx(x− x0 + x0)2e − (x−x0) 2 ∆(t)2 (79) = x0(t)2 + 1√ π∆(t)2 ∫ dyy2e − y 2 ∆(t)2 (80) here we call y = x− x0(t). This leads to 〈 x2 〉 = x20(t) + ∆(t)2 2 (81) 8.5.3 p we can calculate the momentum 〈p(t)〉 = 1√ π∆2(t) ∫ dxe − (x−x0(t)) 2 2(∆2−ıa2) e−ı p0x h̄ +ı p20t 2mh̄ −ıh̄∂x ( e − (x−x0(t)) 2 2(∆2+ıa2) eı p0x h̄ −ı p20t 2mh̄ ) (82) The derivative above requires a product rule. One term has the form ∫ dxxe−ax 2 = 0. So 〈p(t)〉 = p0√ π∆(t)2 ∫ dxe − (x−x0(t)) 2 ∆2(t) (83) 〈t|p|t〉 = p0 (84) 11 8.5.4 p2 we would also like to know〈 p(t)2 〉 = − h̄ 2√ π∆2(t) ∫ dxe − (x−x0(t)) 2 2(∆2−ıa2) e−ı p0x h̄ +ı p20t 2mh̄ ∂2x ( e − (x−x0(t)) 2 2(∆2+ıa2) eı p0x h̄ −ı p20t 2mh̄ ) = − h̄ 2√ π∆2(t) ∫ dxe − (x−x0(t)) 2 2(∆2−ıa2) e−ı p0x h̄ +ı p20t 2mh̄ ∂x [( − x− x0 ∆2 + ıa2 + ı p0 h̄ ) e − (x−x0(t)) 2 2(∆2+ıa2) eı p0x h̄ −ı p20t 2mh̄ ] = − h̄ 2√ π∆2(t) ∫ dxe − (x−x0(t)) 2 2(∆2−ıa2) e−ı p0x h̄ +ı p20t 2mh̄ [ − 1 ∆2 + ıa2 + ( x− x0 ∆2 + ıa2 )2 − p 2 0 h̄2 − 2ıp0 (x− x0) h̄ (∆2 + ıa2) ] e − (x−x0(t)) 2 2(∆2+ıa2) eı p0x h̄ −ı p20t 2mh̄ = 1√ π∆2(t) ∫ dx [ h̄2 ∆2 + ıa2 − ( h̄ ∆2 + ıa2 )2 (x− x0)2 + p20+ ı 2p0h̄ (x− x0) ∆2 + ıa2︸ ︷︷ ︸R =0 by sym  e∆2(x−x0)2∆4+a4 = p20 + h̄2 ∆2 + ıa ( 1− 1 ∆2 + ıa2 ∆4 + a4 2∆2 ) = p20 + h̄2 ∆2 + ıa ( 1 + ∆2 − ıa2 2∆2 ) ︸ ︷︷ ︸ (∆2+ıa2)/(2∆2) 〈 t|p2|t 〉 = p20 + h̄2 2∆2 (85) the RMS uncertainty on each p and x is δprms = √ 〈p2〉 − 〈p〉 = h̄√ 2∆ (86) δxrms = ∆(t)√ 2 = √ 2 2 ∆ √ 1 + h̄2t2 m2 /∆4 (87) 12