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236 7 QUANTUM THEORY
�e quantity in the parentheses is the commutator [p̂x , x̂].�e commutator of
the position and momentum operators is given by [7C.16–259], [x̂ , p̂x] = iħ,
therefore [p̂x , x̂] = −iħ
[â, â†]ψ = i[p̂x , x̂]ψ = i(−iħ)ψ = ħψ
Hence [â, â†] = ħ .
7D Translational motion
Answers to discussion questions
D7D.2 �e hamiltonian for a particle in a two- or three-dimensional box is separa-
ble into a sum of terms each of which depends on just one of the variables
x, y or z. �erefore the solutions to the Schrödinger equation are products
of separate functions of each of these variables, Ψ(x , y, z) = ψ(x)ψ(y)ψ(z).
In the hamiltonian the terms in the three variables are all of the same form,
so the wavefunctions are likewise of the same form. Similarly, the boundary
conditions along each direction are the same, so the quantization which this
imposes is the same. As a result the wavefunctions are simply a product of the
wavefunctions that would be found by solving the Schrödinger equation along
each direction separately.
�e energy levels of two- and three-dimensional boxes can be degenerate, that
is there are di�erent wavefunctions which have the same energy.�is is usually
a consequence of the box having a certain kind of symmetry, for example in a
two-dimensional box in which the two sides are the same length the levels with
quantum numbers n1 = l , n2 = m and n1 = m, n2 = l are degenerate.
Solutions to exercises
E7D.1(b) �e linear momentum of a free electron is given by
p = ħk = (1.0546 × 10−34 J s) × (5 × 109 m−1) = 5 × 10−25 kg m s−1
where 1 J = 1 kgm2 s−2 is used; note that 1 nm−1 = 1 × 109 m−1. �e kinetic
energy is given by [7D.2–261]
Ek =
(ħk)2
2m
= ((1.0546 × 10−34 J s) × (5 × 109 m−1))2
2 × (9.1094 × 10−31 kg)
= 2 × 10−19 J
E7D.2(b) �e electron is travelling in the positive x direction hence the momentum and
the velocity υ are both positive.�e momentum of this particle is given by
p = mυ = (1.0 × 10−3 kg) × (+10 ms−1) = +0.01 kgms−1