Atomic and Molecular Structure
(01:160:421/16:160:521)
Answers to Homework Problems, Week 4
7.1
<fm|Â|fn> = ∫ fm*(Âfn)dτ = <fm|Âfn>
7.2
<cfm|Â|fn> = ∫(cfm)*(Âfn)dτ = ∫c*fm*(Âfn)dτ = c*∫ fm*(Âfn)dτ
<fm|Â|cfn> = ∫fm*(Âcfn)dτ = ∫fm*(Âfn)dτ = c∫fm*(Âfn)dτ
These two expressions are equal if c* = c, i.e. the constant c is real.
7.14
<2p1|Lz|3p-1> = <2p1|Lz3p-1> = <2p1|(-1ħ) 3p-1> = (-ħ) <2p1| 3p-1> = 0
because two eigenfunctions of a Hermitian operator (Lz in this case) with different
eigenvalues (1ħ and –1ħ in this case) are orthogonal.
<3p0|Lz|3p0> = <3p0|Lz3p0> = <3p0|(0ħ) 3p0> = (0ħ) <3p0| 3p0> = (0ħ)(1) = 0
7.15
Harmonic oscillator: < fm |Ĥ| fn > = < fm |Ĥ fn > = < fm |Enfn > = En< fm |fn > =
Enδmn = En = (n + ½) hν if m = n; zero otherwise.
Particle in a 1D box: < fm |Ĥ| fn > = < fm |Ĥ fn > = < fm |Enfn > = En< fm |fn > = Enδmn =
En = n2h2/(8mL2) if m = n; zero otherwise.
7.21
Iff f(x,y,z) = f(-x,-y,-z), then f is even
Iff f(x,y,z) = -f(-x,-y,-z), then f is odd
If f(x,y,z) ≠ ±f(-x,-y,-z), then f is neither even nor odd
1
For the hydrogen-like atom, V = -Ze’2/(x2 + y2 + z2)1/2 = -Ze’2/r
which is (clearly!) an even function; (-x)2 = x2 etc.. The kinetic energy operator is (just
as clearly!) also an even function (∂2/∂x2 = ∂2/∂(-x)2 etc.), so the Hamiltonian for a
hydrogen-like atom is even. Hence the wavefunctions may be labeled by parity, since the
parity operator and the Hamiltonian commute.
(a) Ψ2s is an even function; no angular dependence whatsoever
(b) Ψ2px transforms as the x-coordinate and is hence an odd function
(c) Ψhybrid = (Ψ2s + Ψ2px) is neither even nor odd. Yet, it is an eigenfunction of H for
an H-like atom:
Ĥ Ψhybrid = Ĥ (Ψ2s + Ψ2px) = Ĥ Ψ2s + Ĥ Ψ2px = E2sΨ2s + E2pxΨ2px = E2Ψ2s +
E2Ψ2px = E2(Ψ2s + Ψ2px) = E2Ψhybrid
Remember: s orbitals and p orbitals of the same principal quantum number are
degenerate in an H-like atom.
7.23
<2s|x|2px> := (even)(odd)(odd) := (even)(even) = even; integral may be different
from zero
<2s|x2|2px> := (even)(even)(odd) := (even)(odd) = odd; integral is definitely zero
<2py|x|2px> := (odd)(odd)(odd) := (odd)(even) = odd; integral is definitely zero
7.27
When is <v2|x|v1> ≠ 0 ?????
The vibrational wavefunctions are even, when the vibrational quantum number is even (0,
2, 4, etc..); they are odd, when the vibrational quantum number is odd (1, 3, 5…etc.).
2
The operator x is odd….hence, in order for the integral <v2|x|v1> to possibly survive, v2
and v1 must differ in parity.
Or, almost the same thing stated in a slightly different way, if v2 and v1 have the same
parity, then the integral is definitely zero; if v2 and v1 have opposite parity, then the
integral may be different zero.
The vibrational wavefunctions Ψv(x) contain Hermite polynomials in x. If the quantum
number is v2(v1), then the power of the polynomial is v2(v1). The Hermite polynomials
form an orthogonal set, <vm|vn> = 0 unless vm = vn. Hence, in order for the integral
<v2|x|v1> to survive, the quantum numbers v2 and v1 can only differ by one unit (i.e. the
two Hermite polynomials in Ψv2(x) and Ψv1(x) can only differ by one power of x), i.e.,
∆v = v2 - v1 = ±1.
∆v = v2 - v1 = ±1 is the standard selection rule for infrared vibrational
spectroscopy.
7.31
Ψ = (6)-1/2 2p1 – i(2)-1/2 2p0 – (3)-1/2 3d1
If Lz is measured for an electron in the state Ψ, then the possible outcomes are 1ħ
from 2p1, 0ħ from 2p0, and 1ħ from 3d1. The probabilities of each outcome are
P(Lz = 1ħ) = (6)-1/2(6)-1/2 + [-(3)-1/2][-(3)-1/2] = 1/6 +1/3 = 1/2
P(Lz = 0ħ) = [-i(2)-1/2 ][i(2)-1/2] = 1/2
(Note: 1/6 + 1/3 +1/2 = 2/3 + 1/3 = 1….….)
<Lz>Ψ = (1/6)(1ħ) + (1/2)(0ħ) + (1/3)(1ħ) = ħ /2
3
7.32
Ψ = (6)-1/2 2p1 – i(2)-1/2 2p0 – (3)-1/2 3d1
If L2 is measured on an electron in the state Ψ, then the possible outcomes are 1(1+1)ħ2
= 2ħ2 from 2p1, 1(1+1)ħ2 = 2ħ2 from 2p0, and 2(2+1)ħ2 = 6ħ2 from 3d1.
P(L2 = 2ħ2) = (6)-1/2(6)-1/2 + [-i(2)-1/2 ][i(2)-1/2] = 1/6 +1/2 = 2/3
P(L2 = 6ħ2) = [-(3)-1/2][-(3)-1/2] = 1/3
(Note: 1/6 + 1/2 + 1/3)
< L2>Ψ = (1/6)(2ħ2) + (1/2)(2ħ2) + (1/3)(6ħ2) = 10ħ2/3
7.33
Ψ = (6)-1/2 2p1 – i(2)-1/2 2p0 – (3)-1/2 3d1
En = (-Z2/2n2) = (-1/2n2) in au for an H-atom
If E is measured on an electron in the state Ψ, then the possible outcomes are E2
from 2p1, E2 from 2p0, and E3 from 3d1. The probabilities of each outcome are
(6)-1/2(6)-1/2 = 1/6; [-i(2)-1/2 ][i(2)-1/2] = 1/2; and [-(3)-1/2][-(3)-1/2] = 1/3.
< H>Ψ = <E> = (1/6)E2 + (1/2)E2 + (1/3)E3
7.34 Done in 7.27
7.49
See “Answers to Selected Problems” in the back of the text, page 727
4
7.55
LzYℓ,m (θ,φ) = mћ Yℓ,m (θ,φ)
m = 0, ±1, ±2…±ℓ
So, if ℓ = 2, then m = -2, -1, 0, 1, or 2.
The matrix representation of Lz will hence be a (square) 5x5 matrix. All off-diagonal
elements will be zero. Why? Because we are setting up the matrix representation of Lz
using a basis set (Yℓ,m; l = 2; m = -2, -1, 0, 1, or 2), which consists of the eigenfunctions
of the Lz operator, viz.
<Yℓ,m| Lz |Yℓ,m’> = mћ δmm’
or, equivalently, as a matrix (Lz)ij = 0 i ≠ j
The diagonal elements of the Lz matrix will have the values
(Lz)11 = <Y2,-2| Lz |Y2,-2> = –2ћ, (Lz)22 = <Y2,-1| Lz |Y2,-1> = -1ћ,
(Lz)33 = <Y2,-0| Lz |Y2,0> = 0ћ, (Lz)44 = <Y2,1| Lz |Y2,1> = 1ћ, and
(Lz)55 = <Y2,2| Lz |Y2,2> = 2ћ
7.56
LzYℓ,m (θ,φ) = m ћ Yℓ,m (θ,φ)
m = 0, ±1, ±2…±ℓ
(∆Lz) 2 = <Lz2> - <Lz>2
(a) Ψ = 2pz = 2p0 , i.e. m = 0 for an electron in a 2pz orbital
(∆Lz)2Ψ=2pz = <Lz2>Ψ=2pz - <Lz>2Ψ=2pz = <2pz |Lz2|2pz > - <2pz |Lz|2pz >2 =
<2pz | m2 ћ2 (with m = 0)|2pz > - <2pz | m ћ (with m = 0) |2pz >2 = 0 – 0 = 0
so (∆Lz)Ψ=2pz = 0.
5
Actually, and more to the point, the uncertainty is zero because the evaluation is being
made over an eigenfunction (2pz) of the operator (Lz). The uncertainty is also zero
because of the m = 0 factor, but that is really just adding insult to injury. Thus, (∆Lz) = 0
also for Ψ = 2p1 , 2p-1, 3d2 , 3d1 etc.
(b) Ψ = 2px = (2)-1/2(2p1 + 2p-1), i.e. an electron in a 2px orbital does not have a specific
m value; 2px is not an eigenfunction of Lz. However, it is written as a linear combination
of functions (2p1, 2p-1), which are eigenfunctions of Lz
(∆Lz)2Ψ=2px = <Lz2>Ψ=2px - <Lz>2Ψ=2px = <2px |Lz2|2px > - <2px |Lz|2px >2
<Lz>Ψ=2px = <2px |Lz|2px > = Prob(Lz = 1ћ)x(1ћ) + Prob(Lz = -1ћ)x(-1ћ) =
(1/2)x(1ћ) + (1/2)x(-1ћ) = 0
<Lz2>Ψ=2px = <2px |Lz2|2px > = <2px |Lz2|(2)-1/2(2p1 + 2p-1) >
= <2px |Lz|(2)-1/2((1ћ)2p1 + (-1ћ)2p-1) > = <2px |(2)-1/2((1ћ)(1ћ)2p1 + (-1ћ)(-1ћ)2p-1) >
= (1ћ)2<2px |(2)-1/2(2p1 + 2p-1) > = (1ћ)2<2px |2px > = (1ћ)2
So that (∆Lz)2Ψ=2px = <Lz2>Ψ=2px - <Lz>2Ψ=2px = (1ћ)2 + 0 = (1ћ)2
and
(∆Lz)Ψ=2px = 1ћ
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7.63
(a) F Only in “favorable” cases – such as H independent of t – is it possible to factor
Ψ(q,t) into separate functions of t and of q
(b) T true statement ---the problem, in particular in QM, is that getting exact knowledge
of the present state of the system is normally not trivial. Then having to solve the
TimeDependent-SE afterwards is generally tough.
(c) F Only true if H is independent of time, since only then do we have HΨ = EΨ
(d) F Only true if the eigenfunctions have the same eigenvalue (degeneracy)
(e) F See Postulate 3
(f) T By definition
(g) F They may have some eigenfunctions in common, but they will not have all (i.e. the
complete set) in common.
(h) F If there are no degeneracies in either set of eigenvalues, then the statement is true.
When degeneracies are present, linear combinations of degenerate wave functions may
have to be made. They can always be formed in such a manner that if two Hermitian
operators commute, then a common set of eigenfunctions does exist.
(i) F The degeneracy issue again….if the two functions are part of a degeneracy, then
they may not be orthogonal. However, they can be made orthogonal (Gram-Schmidt
orthogonalization). If the two functions belong to different eigenvalues, then (of course)
they are always orthogonal automatically.
(j) F State functions are solutions to the TimeDependent-SE or the TimeIndependent-SE
(when H is independent of time)
(k) T Shown in Problem 3.41
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(l) F Wavefunctions have units, no matter whether they are normalized or not.
(m) F All eigenvalues of Hermitian operators will be real.
(n) T Both quantities, <fm|Â|fn> and <fm|fn> are definite integrals, so they must be
numbers
(o) T Follows from Postulate 3
(p) F The relation is valid as long as f and g are 'well behaved' functions.
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Exercise (5.9), p. 98: [px, Ĥ] = [px, (px2 + py2 + pz2)/2m + V(x,y,z)] =
[px, px2/2m] + [px, py2/2m] + [px, pz2/2m] + [px,V(x,y,z)] =
0
+
0
+
0
-iћ[∂/∂x,V(x,y,z)] =
-iћ {∂V(x,y,z)/∂x + V(x,y,z)∂/∂x - V(x,y,z)∂/∂x} = -iћ ∂V(x,y,z)/∂x
5.3 [x, px3] = [x, px px2] = px[x, px2] + [x, px]px2 = px(2ћ2∂/∂x) + iћ px2 =
(-iћ∂/∂x)(2ћ2∂/∂x) + iћ(-iћ∂/∂x)(-iћ∂/∂x) = -3iћ3∂2/∂x2 = (3ћ3/i) ∂2/∂x2
5.17 [Lx2, Ly] = Lx[Lx, Ly] + [Lx, Ly]Lx = Lx(iћLz) + (iћLz)Lx = (iћ){ LxLz + LzLx}
= (iћ)[ Lx,Lz]+
where the anti-commutator of two operators, A and B, is defined as [A,B]+ = AB + BA
5.25
L2Yℓ,m (θ,φ) = ℓ (ℓ +1) ћ 2Yℓ,m (θ,φ) ℓ = 0, 1, 2, 3, 4..
LzYℓ,m (θ,φ) = m ћ Yℓ,m (θ,φ)
m = 0, ±1, ±2…±ℓ
Lz2Yℓ,m (θ,φ) = m2 ћ2 Yℓ,m (θ,φ)
m = 0, ±1, ±2… ±ℓ
Lz3Yℓ,m (θ,φ) = m3 ћ3 Yℓ,m (θ,φ)
m = 0, ±1, ±2… ±ℓ
5.29 (a) The possible outcomes from measuring Lz are 2ћ, 1ћ, 0ћ (=0), -1ћ, -2ћ
(b) ℓ(ℓ +1) ћ 2 = 12 ћ 2 => ℓ = 3 => m = 3, 2, 1, 0, -1, -2, or –3.
The possible outcomes from measuring Lz are 3ћ, 2ћ, 1ћ, 0ћ (=0), -1ћ, -2ћ, -3ћ
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5.34 (a) T A particular ℓ value leads to (2ℓ + 1) degeneracy - all the possible m values
(b) F (operator)(eigenfunction) = (eigenvalue)(eigenfunction) does not mean
(operator) = (eigenvalue) !! Nonsense.
(c) T
(d) T
(e) T Eigenvalue = 0
(f) F If A and B do not commute, then it is impossible to find a complete set of functions
which are eigenfunctions of both operators
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