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852_2010hw3_Solutions

# 852_2010hw3_Solutions - PHYS852 Quantum Mechanics II Spring...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 3 Topics covered: Unitary transformations, translation, rotation, vector operators 1. [25]Symmetry: A quantum system is said to posses a ‘symmetry’ if the Hamiltonian operator, H , is invariant under the associated transformation. In other words, if H ′ = H , where H ′ := U † HU . (a) [5] Show that H ′ = H is equivalent to [H, U ] = 0 Start with U † HU = H Hit from the left with U and use U U † = I to get HU = U H . Put both terms on the l.h.s. to get HU − U H = 0 or equivalently [H, U ] = 0. (b) [5] Any hermitian operator can be used to generate a unitary operator via U = e−iGφ , where G† = G is the ‘generator’ of the symmetry transformation, and φ is a free parameter. Show that [H, G] = 0 is necessary and suﬃcient for H to be symmetric under U . If [H, G] = 0 then it follows that [H, f (G)] = 0 for any single-variable function f (x). As U is of this form, it follows that [H, U ] = 0 so that H is symmetric with respect to the U . (c) [5] Show that when [H, G] = 0, the probability distribution over the eigenvalues of G does not change in time. In QM this means that G is a ‘constant of motion’. Must a QM constant of motion have a well-deﬁned value? Since [H, G] = 0 it follows that simultaneous eigenstates of H and G exist. We can label them |n, g so that H |n, g = En |n, g and G|n, g = g|n, g . The most general state is then |ψ (t) = n g cn,g (t)|n, g . From Schr¨dinger’s equation we ﬁnd that o d n, g|ψ (t) dt i = − n, g|H |ψ (t) d cn,g (t) = dt = −iωn cn,g (1) so that cn,g (t) = e−iωn t cn,g (0), which gives |ψ (t) = n g cn,m (0)e−iωn t |n, g (2) The projector onto the subspace with eigenvalue g is I (g) = n |n, g n, g|, so that the probability for the system to be in an eigenstate of G with eigenvalue g is P (g, t)) = = ψ (t)|I (g)|ψ (t) n,n′ ,n′′ g ′ ,g ′′ c∗ ′ ,g′ (0)eiωn′ t cn′′ ,g′′ (0)e−iωn′′ t n′ , g′ |n, g n, g|n′′ , g′′ n c∗ ′ ,g′ (0)eiωn′ t cn′′ ,g′′ (0)e−iωn′′ t δn′ ,n δg′ ,g δn′′ ,n δg′′ ,g n = n,n′ ,n′′ g ′ ,g ′′ = n |cn,g (0)|2 (3) which we see is independent of time. (d) [5] If a system possesses ‘translational symmetry’ what operator is a constant of motion? i The translation operator is UT (d) = e− dP so that P is the generator of translation. Thus in a system with translational symmetry, momentum will be conserved. 1 (e) [5] Consider a particle described by the Hamiltonian H= P2 + V (X ). 2M (4) What operator is the generator of translation? Show that H has translational symmetry only if V (x) = V0 . For H to possess translational symmetry required [H, P ] = 0. This then requires [V (X ), P ] = d 0. We know that [V (X ), P ] = i V ′ (X ), so translational symmetry requires dx V (x) = 0, or equivalently V (x) = V0 . 2 2. [25] Consider a system described by the Hamiltonian H= P2 1 + M ω 2 X 2 + M gX, 2M 2 (5) where g has units of acceleration. † † (a) [5] Show that UT (d)XUT (d) = X + d and UT (d)P UT (d) = P . We have † X ′ = UT (d)XUT (d) = † dxUT (d)|x x x|UT (d) = dx|x − d x x − d| = dx|x (x + d) x| = dx|x x x| + d dx|x x| (6) = X +d We know that P ′ = P because [UT (d), P ] = 0, given that UT (d) is a function of P . † (b) [5] Solve for d and E0 such that H ′ := UT (d)HUT (d) satisﬁes H ′ = E0 + P2 1 + M ω2X 2 2M 2 (7) We start from † H ′ = UT (d)HUT (d) † † UT (d)P UT (d)UT (d)P UT (d) 1 † † † + M ω 2 UT (d)XUT (d)UT (d)XUT (d) + M gUT (d)XUT (d) 2M 2 1 P ′2 2 + M ω 2 X ′ + M gX ′ = 2M 2 P2 1 = + M ω 2 (X + d)2 + M G(X + d) 2M 2 1 1 P2 (8) + M ω 2 X 2 + (M ω 2 d + M G)X + M ω 2 d2 + M Gd = 2M 2 2 Therefore the linear term will cancel for G d=− 2 (9) ω giving = H′ = = 1 P2 + M ω2X 2 + 2M 2 P2 1 + M ω2X 2 − 2M 2 so we see that E0 = − 3 1 M G2 M G2 − 2 ω2 ω2 2 MG 2ω 2 M G2 2ω 2 (10) (11) ′ (c) [5] Let |φ′ , n = 0, 1, 2, . . . be the nth eigenstate of H ′ , with corresponding eigenvalue En . What n ′ are En and φ′ (x) = x|φ′ ? n n We have ′ H ′ |φ′ = En |φ′ (12) n n Because H ′ is just an SHO plus a constant, we know that ′ En = ω n + 1 2 − M G2 2ω 2 (13) As a constant shift in the zero-point energy doesn’t change the shape of the wavefunction, we have 2 1 1 Hn (x/λ) e− 2 (x/λ) φ′ (x) = √ (14) n n n!λ π2 where λ = Mω . (d) [5] Show that |φn := UT (d)|φ′ is an eigenstate of H with eigenvalue En . What is the relationn ′ ship between En and En ? H |φn = HUT (d)|φ′ n † = UT (d)UT (d)HUT (d)|φ′ n = UT (d)H ′ |φ′ n ′ = UT (d)En |φ′ n ′ = En UT (d)|φ′ n ′ = En |φn (15) ′ with the deﬁnition H |φn = En |φn we see that En = En . (e) [5] What is the relationship between φn (x) := x|φn and φ′ (x)? What is φn (x)? n φn (x) = = = x|φn x|UT (d)|φ′ n x − d|φ′ n = φ′ (x − d) n this gives φn (x) = √ 1 π 2n n!λ (16) Hn (x + G/ω 2 )/λ e− 2 ((x+G/ω 1 4 2 )/λ 2 ) (17) 3. [10] Show explicitly that the momentum operator of a particle P is a vector operator with respect to rotation. Show that the operator P 2 = P · P is invariant under rotation about any axis (hint: chose a coordinate system where the axis of rotation is the z-axis). To show that P is a vector, we can consider an inﬁnitesimal rotation about the z axis. ′ Px = e = i i ǫLz Px e− ǫLz i i 1 + ǫ(XPy − Y Px ) Px 1 − ǫ(ZPy − Y Px ) i = Px + ǫ [XPy − Y Px , Px ] (18) = Px − ǫPy ′ Py = e i ǫLz Py e− i ǫLz i i 1 + ǫ(XPy − Y Px ) Py 1 − ǫ(ZPy − Y Px ) = i = Py + ǫ [XPy − Y Px , Py ] = Py + ǫPx ′ Pz = e = i ǫLz (19) i Pz e− ǫLz i i 1 + ǫ(XPy − Y Px ) Pz 1 − ǫ(ZPy − Y Px ) i = Pz + ǫ [XPy − Y Px , Pz ] = Pz so that (20) ′ 1 −ǫ 0 Px Px ′ Py = ǫ 1 0 Py ′ 001 Pz Pz for a ﬁnite rotation this becomes ′ Px cos(θ ) − sin(θ ) 0 Px ′ Py = sin(θ ) cos(θ ) 0 Py ′ Pz 0 0 1 Pz (21) (22) which is the same as P ′ = M (θ )P P2 ′ 2 2 ′ ′ ′ = Px + Py + Pz (23) 2 2 = (cos θPx − sin θPy )2 + (cos θPy + sin θPx )2 + Pz 2 2 2 = cos2 θPx − cos θ sin θ (Px Py + Py Px ) + sin2 θPy + cos2 θPy + cos θ sin θ (Py Px + Px Py ) 2 2 + sin2 θPx + Pz 2 2 2 = (cos2 θ + sin2 θ )(Px + Py ) + Pz 2 2 2 = Px + Py + Pz = P2 (24) 5 4. [40/35] Consider an inﬁnitesimal rotation about an arbitrary axis, described by the unitary operator UR (ǫ) = e− i L·ǫ i i i = 1 − L1 ǫ1 − L2 ǫ2 − L3 ǫ3 . (25) where L = j Lj ej and ǫ = j ǫj ej , with {e1 , e2 , e3 } being a right-handed set of orthogonal unit vectors. Using this notation, the angular momentum components are given by Lj = k,ℓ ǫj,k,ℓRk Pℓ , with ǫj,k,ℓ being the totally antisymmetric Levi-Cevita tensor, any index repeated 0; ǫjkℓ = 1; cyclic permutations of {j, k, ℓ} = {1, 2, 3} . (26) −1; cyclic permutations of {j, k, ℓ} = {3, 2, 1} The components of R and P satisfy the commutation relation [Rj , Pk ] = i δj,k . † ′ (a) [10] Evaluate Rj = UR (ǫ)Rj UR (ǫ) for each component of the position operator R = j Rj ej , and use this to deduce the 3 × 3 matrix, M (ǫ) that rotates an ordinary vector by the inﬁnitesimal angle ǫ. In the inﬁnitesimal limit, we have UR (ǫ) = 1 − i j ǫj Lj , so that ′ Rj = 1+ i ǫk Lk Rj 1+ i k = Rj + i ǫℓ Lℓ ℓ ǫk [Lk , Rj ] k = Rj + i ǫk ǫkℓm [Rℓ Pm , Rj ] k ℓm = Rj + i k ℓm = Rj + ǫk ǫkℓmRℓ (−i ) δm,j ǫk ǫkℓj Rℓ kℓ = δj,ℓ + ℓ ǫk ǫkℓj Rℓ (27) k This tells us that Mjk = δjk + ǫℓkj ǫℓ ℓ = δjk − so that ǫjkℓ ǫℓ ℓ 1 − ǫ3 ǫ2 1 − ǫ1 M (ǫ) = ǫ3 − ǫ2 ǫ1 1 6 (28) (29) (b) [5] Show that M (−ǫ) = M T (ǫ), then show that M T (ǫ) = M −1 (ǫ) by showing that M T (ǫ)M (ǫ) = I. Mjk (−ǫ) = δjk + ǫjkℓ ǫℓ ℓ = δkj − ǫkjℓ ǫℓ ℓ = Mkj (ǫ) T = Mjk (ǫ) M T (ǫ)M (ǫ) jk (30) T Mjℓ (ǫ)Mℓk (ǫ) = ℓ = δjℓ + m ℓ = ℓ δℓk − ǫjℓm ǫm δjℓ δℓk − = δjk − δjℓ ǫℓkn ǫn + ℓn ǫℓkn ǫn n ǫjℓm ǫm δℓk ℓm ǫjkn ǫn + n ǫjkm ǫm m (31) = δjk (c) [5] Now consider a ﬁnite rotation by δ = j δj ej , M N (δ/N ). described by the 3 × 3 matrix M (δ ). Clearly we must have M (δ ) = Take the limit as N → ∞, and use your result to part (a) to show that we can put M (δ ) into the form: M (δ ) = lim 1− N →∞ 1 Λ(δ ) N N = e−Λ(δ) where Λ(δ ) is a 3 × 3 antisymmetric matrix, whose components are given by Λj,k (δ ) = M (δ ) = M δ /N N (32) ℓ ǫj,k,ℓ δℓ . (33) where Mjk (δ/N ) = δjk − ǫjkℓ ℓ 1 = δjk − N = δjk − 7 δℓ N ǫjkℓ δℓ ℓ 1 Λjk (δ ) N (34) (d) [5] Show that the eigenvalues of Λ(δ ) are ω0 = 0, and ω± = ±iδ, where δ = |δ |. The eigenvalues of Λ(δ ) are solutions to −ω δ3 −δ2 det −δ3 −ω δ1 δ2 −δ1 −ω (35) =0 which gives the characteristic equation 2 2 2 −ω 3 − ωδ1 + δ1 δ2 δ3 − ωδ3 − δ1 δ2 δ3 − ωδ2 = 0 (36) ω ω 2 + δ2 = 0 (37) which simpliﬁes to so that the solutions are ω0 = 0, and ω± = ±iδ. (e) Show that the eigenvectors of Λ(δ ) are u0 = u± = δ δ 2 (δ1 δ2 ± iδδ3 )e1 + (δ2 − δ2 )e2 + (δ2 δ3 ∓ iδδ1 )e3 2 2δ2 (δ2 − δ2 ) Λ(δ )u0 − ω0 u0 j = (38) (39) Λjk (δ )u0,k k = ǫjkℓδℓ u0,k kℓ 1 δ = = ǫjkℓ δℓ δk kℓ δ×δ j =0 Λ(δ )u± − ω± u± j = k = kℓ = (40) (Λjk − ω± δjk ) u±,k ǫjkℓ u±,k δℓ ∓ iδu±,j u± × δ ∓ iδu± j =0 (41) where we have used u± × δ = e1 (u±,2 δ3 − δ2 u±,3 ) + e2 (u±,3 δ1 − δ3 u±,1 ) + e3 (u±,1 δ2 − δ1 u±,2 ) 2 2 2 2 = e1 δ2 δ3 − δ2 δ3 − δ2 δ3 ± iδδ1 δ2 + e2 δ1 δ2 δ3 ∓ iδδ1 − δ1 δ2 δ3 ∓ iδδ3 2 2 + e3 δ1 δ2 ± iδδ2 δ3 − δ1 δ2 + δ2 δ1 2 2 = ±iδ e1 (δ1 δ2 ± iδδ3 ) + e2 −δ1 − δ3 + e3 (δ2 δ3 ∓ iδδ1 ) = ±iδu± 8 (42) (f) [5] Based on your result to part (e), show that = u0 (u0 · V ) + u− eiδ (u+ · V ) + u+ e−iδ (u− · V ) M (δ )V (43) where V is an arbitrary vector. We start from M (δ ) = e−Λ(δ ) (44) Λ(δ ) = |u0 ω0 u0 | + |u+ ω+ u+ | + |u− ω− u− | (45) M (δ ) = |u0 e−ω0 u0 | + |u+ e−ω+ u+ | + |u− e−ω− u− | (46) + u− eiδ u∗ · V − (47) + u− eiδ u+ · V (48) Using Dirac notation, we have so that switching back to standard notation, this gives M (δ )V = u0 u∗ · V 0 + u+ e−iδ u∗ · V + noting that u∗ = u0 , and u∗ = u∓ , this gives ± 0 M (δ )V = u0 u0 · V + u+ e−iδ u− · V (g) [5+5 bonus] Based on your results to parts (e) and (f), show that † V ′ = UR (δ )V UR (δ ) = M (δ )V = δ (δ · V ) δ(δ · V ) δ×V sin(δ) + V− cos(δ) + 2 2 δ δ δ (49) V ′ = M (δ )V = δ δ·V δ2 + cos(δ) u+ u− · V − i sin(δ) u+ u− · V u+ · V = u− · V = u+ u− · V + u− u+ · V 1 + u − u+ · V − u − u+ · V (50) 2 (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 (51) 2 (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 (52) 2 2δ2 (δ2 − δ2 ) 2 2δ2 (δ2 − δ2 ) = 2 (δ1 δ2 + iδδ3 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 (δ1 δ2 − iδδ3 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 22 2 δ1 δ2 δ3 − δ1 δ2 δ3 (δ1 δ2 + δ2 δ3 ) δ1 δ2 = V3 V1 − 2 V2 + 2 2 2 δ δ2 (δ1 + δ3 ) δ2 (δ2 − δ2 ) 2 22 2 2 δ1 δ2 δ1 δ3 δ2 (δ1 + δ3 ) − δ1 (δ1 + δ3 ) V1 − 2 V2 − 2 V3 = 2 2 δ δ δ2 (δ1 + δ3 ) δ1 (53) = V1 − 2 δ · V δ + 9 u+ u− · V − u− u+ · V = 1 − = = = u+ u− · V + u − u+ · V u+ u− · V − u − u+ · V u+ u− · V + u− u+ · V 2 2 3 2 (δ1 δ2 + iδδ3 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 (δ1 δ2 − iδδ3 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 2 δ3 (δ2 − δ2 ) δ2 δ2 + δ2 δ3 i V2 + i 1 2 2 2 V3 δ(δ2 − δ2 ) δ(δ − δ2 ) δ3 V2 δ2 V3 −i +i δ δ i (54) δ×V δ 1 2 2 (δ2 − δ2 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 2 (δ2 − δ2 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 − 2 2δ2 (δ2 − δ2 ) 2 δ2 − δ2 δ2 δ3 δ1 δ2 V2 − 2 V3 = − 2 V1 + δ δ2 δ δ2 (55) = V2 − 2 δ · V δ =− 2 2 (δ2 − δ2 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 2 (δ2 − δ2 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 + 2 2δ2 (δ2 − δ2 ) δ1 V3 δ3 V1 −i =i δ δ i (56) = δ×V δ 2 =− = 2 (δ1 δ2 − iδδ3 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 (δ1 δ2 + iδδ3 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 2 2δ2 (δ2 − δ2 ) δ3 (57) = V3 − 2 δ · V δ + u+ u− · V − u− u+ · V 3 = − = 2 (δ1 δ2 − iδδ3 ) (δ1 δ2 − iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 + iδδ1 )V3 2 2δ2 (δ2 − δ2 ) 2 (δ1 δ2 + iδδ3 ) (δ1 δ2 + iδδ3 )V1 + (δ2 − δ2 )V2 + (δ2 δ3 − iδδ1 )V3 2 2δ2 (δ2 − δ2 ) i (58) δ×V δ 3 10 Putting these pieces together gives V′ = δ δ·V δ2 + V − δ δ·V δ2 11 cos(δ) + δ × V sin(δ) δ (59) ...
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