PROBLEM SET #5 SOLUTIONS by Robert A. DiStasio Jr. Q1. Vector calculus refresher. Prove the following vector calculus identities by explicit expansion and rearrangement, or from simpler identities. (a) ∇()fg = f ∇ g + g ∇ f (b) ∇ ⋅()()()a × b = b ⋅ ∇× a− a⋅ ∇× b Note: There was an error in the sign of the RHS in the original PS #5. (c) ∇ ×()()()()()a × b = b ⋅∇ a − a⋅∇ b+ ∇ ⋅ b a− ∇ ⋅ a b (d) ∇ ⋅( ∇ ×b ) = 0 (e) ∇ ×( ∇f ) = 0 Please see any text on vector calculus for the proof of these vector identities. If you have any questions on these vector identities or the basic machinery of vector calculus, please feel free to see me during office hours. Q2. Nuts and bolts of molecules and their interaction with magnetic fields. (a) For a constant magnetic field B, explicitly verify that it can be represented by the 1 vector potential A= B × r as BA= ∇ × . 2 In order to answer this question we need to show that ∇ × AB= . Using the results above in 1c, we see that the curl of the vector potential is given by: 1 ∇ ×A = ∇ ×( B × r) 2 . (1) 1 =[(r ⋅∇ ) B − ( B ⋅∇+∇⋅ ) r ( r ) B −∇⋅ ( B ) r ] 2 Since B is a constant magnetic field, then B is not a function of r or t and therefore the first and last terms vanish (derivatives wrt position). We are now left with considering 1 ∇ ×A =[()()] − B ⋅∇ r + ∇ ⋅ r B . (2) 2 The first term on the RHS can be simplified as follows: ∂ ∂ ∂ (B⋅∇ ) r = [(BBB x + y + z)( ⋅ x+ y+ z)] r x y z ∂x∂ y∂ z ∂ ∂ ∂ = [Bx + By + Bz ]r ∂x ∂y ∂z . (3) ∂ ∂ ∂ = [B + B + B ](xx+ y y + z z) x ∂x y ∂y z ∂z =BBBxx + y y + z z = B The second term can also be simplified as: ∂ ∂ ∂ (∇ ⋅r ) B = [( x+ y+ z)(⋅x x + y y + z z)] B ∂x∂ y∂ z . (4) =[1 + 1 + 1]BB = 3 Substituting (3) and (4) into (2) yields the desired result, namely that 1 1 1 ∇×=−⋅∇+∇⋅A[()()] B r r B =−+[ B 3 B ] =[2 B ] = B . (5) 2 2 2 (b) What is form of the vector potential for a magnetic field directed along the z axis? If B = (0 0Bz ) , i.e., the magnetic field is directed along the z axis, then the vector potential, 1 given by A= B × r has the following form: 2 x y z x y z 1 1 1 1 A= B × r = BBB == 0 0 B =[( −B y )x + ( B x ) y ] . (6) 2 2 x y z 2 z 2 z z x y z x y z Therefore, 1 A =( − B y B x 0) . (7) 2 z z (c) Show that the addition of the gradient of a scalar potential f to this vector potential reproduces the same magnetic field. If we were to add the gradient of the scalar potential, ∇f , to A yielding a new vector potential, ~ A , then ~ 1 AA= + ∇f =B × r + ∇f . (8) 2 To show that this modified vector potential reproduces the same magnetic field, we need to ~ prove that ∇ ×AA = ∇ × , since we have already proven that ∇ × AB= in 2a. Considering the difference between the curl of the original vector potential and the curl of the modified vector potential, we have ~ ~ (∇×AAAA)()( −∇× =∇× −) =∇×∇ (f ) = 0 (9) in which I have used the vector identity given above in 1e (i.e., that the curl of a gradient is zero). Therefore, adding the gradient of the scalar potential to the original vector potential reproduces the same magnetic field. (d) The electric field is derived from a scalar potential φ , and the vector potential A ∂A according to E = −∇φ − . How must the scalar potential be modified to reproduce the ∂t same electric field when the vector potential is modified as in (c)? Note: This question refers to the work done in 2c not 2b. ~ From 2c above, the modified vector potential is given above by (8) as AA= + ∇f . The electric field is given in terms of the original vector potential as: ∂A E = −∇φ − . (10) ∂t In order to maintain the same electric field using the modified vector potential, we must have ~ EE= . Considering the difference between these two quantities, ~ ~ ~ ∂A ∂AAAA∂ ∂ ∂ ∂ ∂ EE− =( −∇φ − )()− −∇φ − = − = − ()()A + ∇f = − ∇f (11) ∂t ∂t∂ t∂ t∂ t∂ t ∂t we see that the time derivative of the gradient of the added scalar potential is the only term that would cause a difference in the electric field. Therefore, we could modify the scalar potential in (10) via ∂f φ′ = φ − (12) ∂t and since ∂f ∂ ∇φ′ = ∇ φ − ∇ = ∇φ − ()∇f (13) ∂t∂ t (assuming that f is a well-behaved function and hence the time and position partial derivatives commute) this will allow us to reproduce the same electric field using the modified vector potential in (8). Q3. The effect of translations on the energies of atomic orbitals in magnetic fields. (a) An electron is in an atomic orbital χ , centered at the origin, and is subject to a uniform magnetic field B. Let the energy of the atom at this point be E. Evaluate the energy when the atom is displaced a distance y from the origin. From class, we know that the semi-classical Hamiltonian for a one-electron system in a magnetic field is: 1 Hˆ =() − i ∇ +A 2 + V . (14) 2 The kinetic energy term, not V, will be affected by displacement of the electron through the vector potential. Denoting a displacement vector as r′ , the original vector potential 1 A( r) = B × r (15) 2 will be modified to 1 1 1 1 ArArr′( )()≡ + ′ = Brr ×() + ′ = Br × + BrAr × ′ =() + Br × ′ . (16) 2 2 2 2 If we want to find the energy of the displaced atom relative to the energy of the atom at the origin, E, we can evaluate the expectation value of ΔHˆ . The form of ΔHˆ is given by (using (14), (15), and (16)) 1 1 ΔHˆ =[ ()][ − i ∇ + A′ 2 +V −()] − i ∇ +A 2 + V 2 2 1 =[( −i ∇ + A′ )2 − ( −i ∇ + A )2 ] 2 1 =[( −∇2 −i ∇ ⋅ AAA′ − i ′⋅∇ + ′2 )(− −∇2 −i ∇ ⋅AAA − i ⋅∇ + 2 )] (17) 2 1 =[( − 2i ∇ ⋅ AA′ + ′2 )− ( − 2i ∇ ⋅AA + 2 )] 2 1 = −i ∇ ⋅()AAAA′ − + ()′2− 2 2 which can be further manipulated to: 1 ΔHˆ = − i ∇ ⋅()AAAA′ − + ()′2− 2 2 1 1 1 = −i ∇ ⋅((A + B × r′))− A +(( A + B × r′))2− A 2 2 2 2 1 1 1 = −i ∇ ⋅()B × r′ +(( A2 + A ⋅ ( B × r′ ) +())) B × r′ 2− A 2 . (18) 2 2 4 1 1 1 = −i ∇ ⋅()B × r′ + A ⋅() B × r′ +() B × r′ 2 2 2 8 1 1 1 = −i ∇ ⋅()B × r′ +()() B × r ⋅ B × r′ +() B × r′ 2 2 4 8 To evaluate the expectation value of ΔHˆ we sandwich it between the atomic orbital χ and integrate (assuming this atomic orbital is normalized): 1 1 1 χΔHˆ χ = χ −i ∇ ⋅()B × r′ +()() B × r ⋅ B × r′ +() B × r′ 2 χ 2 4 8 1 1 1 = −i χ ∇ ⋅()B × r′ χ+ χ ()()B × r ⋅ B × r′ χ+ χ ()B × r′ 2 χ 2 4 8 1 1 1 =χ −i ∇ χ ⋅ χ()B × r′ χ+ χ()()B × r χ ⋅ χ B × r′ χ+ χ ()B × r′ 2 χ (19) 2 4 8 1 1 1 =p ⋅()B × r′ +()() B × r ⋅ B × r′ +() B × r′ 2 2 4 8 1 =()B × r′ 2 8 in which I have assumed a spherically symmetric (about the origin) electron density which implies that p=0 = r . Therefore, the energy is not invariant with respect to the gauge origin (which the vector potential depends on). This will happen during calculations that use a finite (incomplete) basis unless you use London orbitals for instance (see 3b). (b) Repeat your evaluation of the energy of the atom after displacement when the ⎡ e ⎤ atom is described by χ exp −i B ⋅() y × r . This transformation defines a “London ⎣⎢ 2h ⎦⎥ orbital”. Discuss briefly the advantages of this type of orbital for use in evaluating the magnetic properties of molecules. Note: The London orbital requires a negative sign within the exponential. In atomic units, the London orbital takes on the following form: i χ= χ exp[ −B ⋅( r′× r )] ≡χ ⊗ δ (20) LO 2 where I have used δ to symbolize the part of the London orbital that depends on the magnetic 1 field. Using one of the triple product rules, we can rewrite B⋅() r′× r as follows (using (16)): 2 1 1 B⋅() r′× r = r ⋅()() B × r′ = r ⋅ A′ − A (21) 2 2 which allows us to rewrite the magnetic field dependent function in the London orbital (20) as δ =exp[ −ir ⋅ ( A′ − A )] . (22) To evaluate the energy in this case, we need to consider: ΔET = χδˆ′ χδ− χδTˆ χδ= χδTˆ′ χδ− χTˆ χ (23) 1 where Tˆ is the kinetic energy operator given by (14) as Tˆ =() − i ∇ + A 2 and Tˆ′ is the kinetic 2 1 energy operator at the displaced geometry, i.e., Tˆ′ =() − i ∇ + A′ 2 . To evaluate the first term on 2 the RHS of (23), we will first consider the action of the operator (−i∇ + A′) on the ket χδ : ()()−i ∇ + A′ χδ = −i ∇ + A′ χδ = −i ∇χδ + A′ χδ = −i()() ∇χ δ − i χ ∇ δ + A′ χδ . (24) = −i ()() ∇χ δ − AAA′ −χδ + ′ χδ = −i () ∇χ δ + A χδ =δ () −i ∇ + A χ 1 1 Since Tˆ′ =() − i ∇ + A′ 2 =()() −i ∇ + A′ ⋅ − i ∇ + A′ , we can use (24) twice to rewrite the first 2 2 term on the RHS of (23) as: χδTˆ′ χδ= χTˆ χ . (25) Using (25), our expression for ΔE in (23) vanishes.