CHM 671
Homework set # 2
Due: Thursday, September 7th
1) Read Chapter 1 in the 4th edition Atkins & Friedman's Molecular Quantum
Mechanics book.
2) Do problems 1.1, 1.3, 1.4, 1.7, 1.9, 1.11 in the book.
1.1 (a)
(f + g)dx =
1/2
(b) (f + g)
= f
f dx +
1/2
+g
1/2
gdx; linear.
; nonlinear.
(c) f (x + a) + g(x + a) = f (x + a) + g(x + a); linear.
(d) f (−x) + g(−x) = f (−x) + g(−x); linear.
Exercise: Repeat the exercise for (a) differentiation, (b) exponentiation.
c P.W. Atkins and R.S Friedman.
1
1.3 (a) (d/dx)eax = aeax ; eax is an eigenfunction,
eigenvalue a.
ax2
ax2
ax2
ax2
(d/dx)e
;e
= 2axe
= 2a xe
not an e.f.
(d/dx)x = 1; x not an e.f.
(d/dx)x2 = 2x; x2 not an e.f.
(d/dx)(ax + b) = a; ax + b not an e.f.
(d/dx) sin x = cos x; sin x not an e.f.
(b) (d2 /dx2 )eax = a2 eax ; eax is an eigenfunction, eigenvalue a2 .
2
2
2
2
(d2 /dx2 )eax = 2aeax + 4a2 x2 eax ; eax not an e.f.
(d2 /dx2 )x = 0 = 0x; x is an e.f.; e.v. is 0.
(d2 /dx2 )x2 = 2; x2 not an e.f.
(d2 /dx2 )(ax + b) = 0 = 0(ax + b); ax + b is an e.f.; e.v. is 0.
(d2 /dx2 )sin x = − sin x; sin x is an e.f.; e.v. is −1.
2
Exercise: Find the operator of which eax is an eigenfunction. Find the
eigenfunction of the operator ‘multiplication by x2 ’.
c P.W. Atkins and R.S Friedman.
1
1.4 Use the correspondence in Section 1.5.
(a)
T = p2 /2m = −(h̄2 /2m)(d2 /dx2 ) in one dimension.
T = p2 /2m = −(h̄2 /2m) (∂ 2 /∂x2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )
= −(h̄2 /2m)∇2 in three dimensions.
(b) 1/x −→ multiplication by(1/x)
(c) µ = ei r i −→ multiplication by ei r i
i
i
(d)
lz = xpy − ypx = (h̄/i){x(∂/∂y) − y(∂/∂x)}
= (h̄/i)(∂/∂φ) for x = r cos φ, y = r sin φ
(e) δx2 = x2 − x2 −→ multiplication by {x2 − x2 }
δp2 = p2 − p2 −→ {−h̄2 (∂ 2 /∂x2 ) − p2 }
Exercise: Devise operators for 1/r, xpx , and eαx .
c P.W. Atkins and R.S Friedman.
1
1.7 (a)
2
ψa∗ T ψb dτ
= −(h̄ /2m)
∞
ψa∗ (d2 /dx2 )ψb dx
−∞
ψa∗ (d2 /dx2 )ψb dx
ψa∗ (d/dx)(dψb /dx)dx
= ψa∗ (dψb /dx) − (dψa∗ /dx)(dψb /dx)dx
udv = uv − vdu
∗
∗
= ψa (dψb /dx) − (dψa /dx)ψb + (d2 ψa∗ /dx2 )ψb dx
=
′
At each limit ψ ∗ ψ ′ and ψ ∗ ψ disappear; therefore
∞
2
2
2 ∗
∗
ψa T ψb dτ = −(h̄ /2m)
(d ψa /dx )ψb dx = (T ψa∗ )ψb dτ
−∞
But T ∗ = T ; hence
tian.
ψa∗ T ψb dτ = (T ψa )∗ ψb dτ , and so T is hermi-
(b)
ψa∗ lz ψb dτ = (h̄/i)
2π
ψa∗ (∂ψb /∂φ)dφ
0
2π
∗
∗
= (h̄/i) ψa ψb − (∂ψa /∂φ)ψb dφ 0
2π
(∂ψa∗ /∂φ)ψb dφ =
= −(h̄/i)
0
= (lz ψa )∗ ψb dτ
2π
0
and so lz is hermitian.
Exercise: Confirm that px , lz2 , and lz3 are hermitian.
c P.W. Atkins and R.S Friedman.
1
(lz ψa )∗ ψb dφ [lz∗ = −lz ]
1.9 (a)
d px ∝ sin(πx/L) sin(πx/L)
dx
∝ sin(πx/L)|cos(πx/L) = 0
(b)
p2x = 2mT = 2mE [V = 0] [see eqn 2.30]
2 h
= 2m
[n = 1] = h2 /4L2
8mL2
Alternatively, integrate explicitly.
Exercise: Evaluate (a) p3x , (b)p4x .
c P.W. Atkins and R.S Friedman.
1
1.11 Use the relations [x, px ] = ih̄, [y, py ] = ih̄, [z, pz ] = ih̄, all others zero.
(a) [x, y] = 0 [basic relation]
(b) [px , py ] = 0 [basic relation]
(c) [x, px ] = ih̄ [basic relation]
(d)
[x2 , px ] = x2 px − px x2 = x2 px − px xx = x2 px − xpx x + [x, px ]x
= x2 px − x2 px + x[x, px ] + [x, px ]x = xih̄ + ih̄x = 2ih̄x
(e)
[xn , px ] = xn px − px xn = xn px − xpx xn−1 + [x, px ]xn−1
= xn px − x2 px xn−2 + x[x, px ]xn−2 + [x, px ]xn−1
= xn px − xn px + xn−1 [x, px ] + xn−2 [x, px ]x
+ . . . + x[x, px ]xn−2 + [x, px ]xn−1
= nih̄xn−1
Exercise: Evaluate [x, p2x ], [x, pnx ], [x2 , p2x ].
c P.W. Atkins and R.S Friedman.
1