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228 7 QUANTUM THEORY
Solutions to exercises
E7C.1(b) To construct the potential energy operator, replace the position x in the clas-
sical expression by the position operator x̂.�is operator is just multiplication
by x, x̂ = x×, therefore the potential energy operator is V̂ = De(1 − e−ax)2 .
E7C.2(b) A function ψ is an eigenfunction of an operator Ω̂ if Ω̂ψ = ωψ where ω is a
constant called the eigenvalue.
(i) (d2/dx2) cos(kx) = −k2 cos(kx). Hence cos kx is an eigenfunction of
the operator d2/dx2, with eigenvalue −k2 .
(ii) (d2/dx2)eikx = (d/dx)ikeikx = −k2eikx . Hence eikx is an eigenfunction
of the operator d2/dx2, with eigenvalue −k2 .
(iii) (d2/dx2)kx = (d/dx)k = 0 = 0 × kx. Hence kx is an eigenfunction of
the operator d2/dx2 with eigenvalue 0 .
(iv) (d2/dx2)e−ax
2
= (d/dx)(−2axe−ax
2
) = 2ae−ax
2
(2ax2 − 1). Hence this
function is not an eigenfunction of the operator d2/dx2.
E7C.3(b) Wavefunctions ψ1 and ψ2 are orthogonal if ∫ ψ∗1ψ2 dτ = 0, [7C.8–254]. Here
ψ1(x) = sin(4πx/L), ψ2(x) = sin(2πx/L), and the region is 0 ≤ x ≤ L. �e
integral is evaluated using Integral T.5
∫ ψ∗1ψ2dτ = ∫
L
0
sin(4πx/L) sin(2πx/L)dx
= [(L/4π) sin(2π) − (L/12π) sin(6π)] = 0
where sin(nπ) = 0 for integer n is used. �us, the two wavefunctions are
orthogonal.
E7C.4(b) Wavefunctions ψ1 and ψ2 are orthogonal if ∫ ψ∗1ψ2 dτ = 0, [7C.8–254]. Here
ψ1(x) = cos(3πx/L), ψ2(x) = cos(5πx/L), and the region is −L/2 ≤ x ≤ L/2.
�e integral is evaluated using Integral T.6
∫ ψ∗1ψ2 dτ = ∫
L/2
−L/2
cos(3πx/L) cos(5πx/L)dx
= (−L/4π) sin(−2πx/L) + (L/16π) sin(8πx/L)∣L/2−L/2
= [(−L/4π) sin(−π) + (L/16π) sin(4π)]
− [(−L/4π) sin(π) + (L/16π) sin(−4π)] = 0
where sin(nπ) = 0 for integer n is used. �us, the two wavefunctions are
orthogonal.