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) )( ∇+∇=∇ fggffg

(b) ×∇⋅=×⋅∇ − ⋅ ∇×baabba )()()(

Note: There was an error in the sign of the RHS in the original PS #5.

(c) −∇⋅=××∇ ⋅∇ + ∇ ⋅ − ∇ ⋅aabbaabba )()()()()( 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 ×= rBA as = ∇ × AB . 2

In order to answer this question we need to show that ∇ × = BA . Using the results above in 1c, we see that the curl of the vector potential is given by:

1 A ( ××∇=×∇ rB ) 2 . (1) 1 ⋅∇−⋅∇+∇⋅−∇⋅= rBBrrBBr ])()()()[( 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 ⋅∇+∇⋅−=×∇ BrrBA ])()([ . (2) 2 The first term on the RHS can be simplified as follows:

∂ ∂ ∂ [()( BBB ⋅++=∇⋅ () + + )]rzyxzyxrB zyx ∂ ∂ ∂zyx ∂ ∂ ∂ = [Bx + By + Bz ]r ∂x ∂y ∂z . (3) ∂ ∂ ∂ = [B + B + B ]( ++ zyx zyx ) x ∂x y ∂y z ∂z

BBB zyx =++= Bzyx

The second term can also be simplified as:

∂ ∂ ∂ =⋅∇ [()( + + () ++⋅ zyx )]BzyxzyxBr ∂ ∂ ∂zyx . (4) =++= 3]111[ BB

Substituting (3) and (4) into (2) yields the desired result, namely that

1 1 1 ])()([ ]3[ ]2[ ==+−=⋅∇+∇⋅−=×∇ BBBBBrrBA . (5) 2 2 2

(b) What is form of the vector potential for a magnetic field directed along the z axis?

If B = Bz )00( , i.e., the magnetic field is directed along the z axis, then the vector potential,

1 given by ×= rBA has the following form: 2

zyxzyx 1 1 1 1 rBA =×= BBB == 00 B +−= xByB yx ])()[( . (6) 2 2 zyx 2 z 2 z z zyx zyx

Therefore,

1 A (−= xByB )0 . (7) 2 zz

(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 rB ∇+×=∇+= 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 ∇ × = BA 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 ∂ ∂ ∂ ∂ ∂ttttt ∂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 ′ φφ −= (12) ∂t and since

∂f ∂ ∇ ′ φφ ∇−∇= φ −∇= ∇f )( (13) ∂ ∂tt

(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 ˆ A)( 2 ++∇−= ViH . (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 ( ) ×= rBrA (15) 2 will be modified to

1 1 1 1 ′( +≡ ′)() +×= ′)( ×+×= ′ )( ×+= rBrArBrBrrBrrArA ′ . (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 ˆ [ iH +∇−=Δ A′ 2 [])( A 2 ++∇−−+ ViV ])( 2 2 1 i +∇−= A′ 2 i +∇−− A 2 ])()[( 2 1 [( 2 ⋅∇−−∇= ′ − ii ′ +∇⋅ AAA ′2 () 2 ii +∇⋅−⋅∇−−∇− AAA 2 )] (17) 2 1 2[( i ⋅∇−= ′ + AA ′2 2() i +⋅∇−− AA 2 )] 2 1 i ⋅∇−= ′ )( +− ′ − AAAA 22 )( 2 which can be further manipulated to:

1 ˆ iH ⋅∇−=Δ ′ )( +− ′ − AAAA 22 )( 2 1 1 1 i (( ×+⋅∇−= ′ )) (( ×++− ′ − ArBAArBA 22 )) 2 2 2 1 1 1 i ×⋅∇−= ′)( 2 ×⋅++ ′)((( ×+ ′ − ArBrBAArB 22 )))( . (18) 2 2 4 1 1 1 i ×⋅∇−= ′)( ×⋅+ ′)( ×+ rBrBArB ′)( 2 2 2 8 1 1 1 i ×⋅∇−= ′)( ×⋅×+ ′)()( ×+ rBrBrBrB ′)( 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 Hˆ χχχ i ×⋅∇−=Δ ′)( ×⋅×+ ′)()( ×+ rBrBrBrB ′)( 2 χ 2 4 8 1 1 1 i χ ×⋅∇−= rB ′)( χχ ×⋅×+ rBrB ′)()( χχ ×+ rB ′)( 2 χ 2 4 8 1 1 1 i ×⋅∇−= rB ′)( ×⋅×+ rBrB ′)()( χχχχχχχχχ ×+ rB ′)( 2 χ (19) 2 4 8 1 1 1 p ×⋅= ′)( ×⋅×+ ′)()( ×+ rBrBrBrB ′)( 2 2 4 8 1 ×= rB ′)( 2 8 in which I have assumed a spherically symmetric (about the origin) electron density which implies that 0 == rp . 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 ×⋅− ryB )( . 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 χχ exp[ ⋅−= ′ rrB )]( ⊗≡× δχ (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 ⋅ ′×rrB )( as follows (using (16)): 2

1 1 ⋅ ′ )( ×⋅=× ′ ⋅= ′ − AArrBrrrB )()( (21) 2 2 which allows us to rewrite the magnetic field dependent function in the London orbital (20) as δ i ⋅−= ′ − AAr )](exp[ . (22)

To evaluate the energy in this case, we need to consider:

=Δ TE ˆ′ − Tˆ = Tˆ′ − Tˆ χχχδχδχδχδχδχδ (23)

1 where Tˆ is the kinetic energy operator given by (14) as ˆ iT +∇−= A)( 2 and Tˆ′ is the kinetic 2

1 energy operator at the displaced geometry, i.e., ˆ′ iT +∇−= 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 +∇−= A′ χδχδ ii )()( +∇−∇−= A′ χδδχδχ . (24) i δχ −∇−= ′ )()( +− AAA ′ χδχδ i )( +∇−= A χδδχ δ i +∇−= A)( χ

1 1 Since ˆ′ iT +∇−= A′)( 2 +∇−= A′ ii +∇−⋅ A′)()( , we can use (24) twice to rewrite the first 2 2 term on the RHS of (23) as:

Tˆ′ = Tˆ χχχδχδ . (25)

Using (25), our expression for ΔE in (23) vanishes. As noted above, the use of London orbitals therefore leads to energies that are invariant to the gauge origin.

Q4. Response of the hydrogen molecule in a minimal basis set to an applied electric field.

(a) Evaluate the matrix element σ Hˆ ()1 σ ∗ for an electric field parallel to the molecular axis in terms of atomic orbital contributions, to allow you to determine whether the result is zero or non-zero.

Let’s begin this problem by aligning the hydrogen molecule along the z-axis with HA at –a and

)1( HB at +a. The fluctuation potential, Hˆ , for a molecule in an applied electric field is given as:

Hˆ )1( E⋅−= μˆ . (26)

Since the applied electric field is parallel to the molecular axis, it is orientated along the z-axis and the perturbed Hamiltonian in (26) is therefore:

ˆ )1( z ( += zzEH ˆˆ 21 ) . (27)

The task at hand is to evaluate a matrix element which includes the following MOs (assuming no overlap between 1sA and 1sB):

1 σ += ss )11( (28a) 2 BA and

1 σ * −= ss )11( . (28b) 2 BA

Using (27) and (28), the aforementioned matrix element can be evaluated as follows:

1 Hˆ )1( σσ * += 11 ˆ )1( −11 ssHss 2 BA BA 1 += 11 ˆ −11 sszEss BAzBA 2 (29) 1 = 1[ ˆ − 11 ˆ + 11 ˆ − 11 ˆ szsszsszsszsE ]1 2 AAz BA AB BB 1 )]()[( −=+−−= aEaaE 2 z z which is non-zero for an applied electric field aligned parallel to the molecular axis. Therefore, to first order, the σ and σ* will “mix” through the dipole operator.

(b) Repeat for an electric field perpendicular to the molecular axis

In this case, the perturbed Hamiltonian will have either/both x and y components, and we would need to consider matrix elements of the form: 1 ˆ 1sxs BA (30) and

1 ˆ 1sys BA (31) both of which would vanish for the system in question. Therefore, to first order, an applied electric field aligned perpendicular to the molecular axis would not cause the σ and σ* MOs to mix.

(c) Use first order perturbation theory to discuss the form of the occupied orbital in the presence of applied fields parallel and perpendicular to the molecular axis.

The first-order correction to the wavefunction from perturbation theory is given by the following generic formula:

Hˆ ΨΨ )0()1()0( )1( 0 j )0( 0 =Ψ ∑ )0()0( Ψj . (32) j≠0 0 − EE j

Since we are dealing with the hydrogen molecule in a minimal basis, the only excited determinant is given by population of the σ* MO. So (32) simplifies to

ˆ )1( )1( H σσ * 0 =Ψ )0( )0( σ * (33) − EE σσ * and the only remaining piece is the perturbed Hamiltonian. In the case of the parallel electric

field, we found that the matrix element needed in (33) was equal to − z aE , while vanishing for perpendicular electric fields. Therefore, the occupied MO for hydrogen in an applied electric field aligned parallel to the molecular axis will now mix in some σ* or anti-bonding character.

The amount of anti-bonding character mixed into the occupied MO will be based on the field strength, the internuclear separation, and the energy eigenvalue difference between σ and σ*:

)1( − z aE 0 =Ψ )0( )0( σ * . (34) − EE σσ * (d) Discuss qualitatively how your answer would change if a second s-type basis function was added to each atom. And what about adding a shell of p-type functions to each atom?

If a second set of s-type basis functions were added to each atom, then the results of 4a and 4b

)1( )1( would remain the same, i.e., 0 ≠Ψ 0 for the parallel case while 0 =Ψ 0 for the perpendicular case (since the second s-type functions are spherically symmetrical). However, if a shell of p-type basis functions were added to each atom, then the results would change. Only the

)1( pz orbitals, like the s orbitals, would add to Ψ0 for the parallel case, but the px and py orbitals

)1( would contribute to Ψ0 for the perpendicular case.

5. Electrical polarizability.

(a) Write the perturbation theory expression for the polarizability of a 1-electron system whose eigenstates are known.

Using the perturbed Hamiltonian given by (26), the second-order correction to the energy takes on the following form:

| Hˆ ΨΨ 2)0()1()0( || )0( E μ Ψ⋅−Ψ |2)0( )2( 0 j 0 j E0 = ∑ )0()0( = ∑ )0()0( . (35) j≠0 0 − EE j j≠0 0 − EE j

This term is equivalent to the third term in the Taylor series expansion of the energy, i.e.,

2 )0( ∂E 1 T ∂ E E)( EE E⋅+= E ⋅− 2 E +⋅ L (36) ∂E E=0 2 ∂E E=0 so the electronic polarizability (the matrix of second derivatives of the energy with respect to the electric field) is given by comparison of (35) and (36) as

2)0()0( | 0 μ ΨΨ j | α −= 2∑ )0()0( . (37) j≠0 0 − EE j

(b) Suppose an electron in a neutral 1-electron atom is modeled as a particle in a 1-d box of side length 1 Angstrom. Evaluate the electronic polarizability along the x axis.

Assuming a box length of l, the energy eigenvalues of the 1-d particle in a box are given as

hn 22 E = (38) n 8ml 2 with corresponding wavefunctions given by

1 ⎛ 2 ⎞ 2 ⎛ πxn ⎞ ψ n = ⎜ ⎟ sin⎜ ⎟ n = ,3,2,1 K (39) ⎝ l ⎠ ⎝ l ⎠

Plugging the particle in a box wavefunctions (39) into our expression for the electronic polarizability (37) yields the following infinite sum:

2 ⎡ xx ˆˆ 1221 xx ˆˆ 1331 xx ˆˆ ψψψψψψψψψψψψ 1441 ⎤ αxx −= 2Ex ⎢ + + +L⎥ . (40) ⎣ − EE 21 − EE 31 − EE 41 ⎦

Using trigonometric identities, the contribution from the nth wavefunction can be shown to depend on l4. Numerical solution of (40) for the electronic polarizability can be performed using a spreadsheet or your favorite mathematical software package.

(c) Suggest how the polarizability depends on the box length, either analytically, or by repeating part (b) for a box twice as long (it is an algebraic dependence).

As mentioned above, the contribution from the nth wavefunction can be shown to depend on l4.