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244 7 QUANTUM THEORY
(d) �e wavefunction is an eigenfunction of the hamiltonian which, because
the potential is zero inside the box, is just the operator for kinetic en-
ergy.�is operator can be written p̂2x/2m, so the wavefunction is also an
eigenfunction of p̂2x with eigenvalue 2m times the energy eigenvalue. In
this case, the eigenvalue of p̂2x is 2m × n2h2/8mL2 = n2h2/4L2. Because
the wavefunction is an eigenfunction of p̂2x , the expectation value is equal
to the eigenvalue: ⟨p2x⟩ = n2h2/4L2 .
P7D.8 (a) Because the particle is moving back and forth with constant speed its
probability density within the box must be uniform at P0.�e total prob-
ability must be 1, so it follows that ∫
L
0 P0 dx = 1. Evaluating the integral
give P0L = 1, hence P0 = 1/L, as required.
(b) �e average value of any quantity is the value of that quantity in an in-
�nitesimal interval at an arbitrary position, multiplied by the probability
of being in that interval and then summed (integrated) over all possible
positions. If the probability density is P(x), the required average is ⟨xn⟩ =
∫
L
0 x
nP(x)dx.�e integral is only over the range 0 to L because outside
this region the probability density is zero.
(c)
⟨x⟩ = ∫
L
0
xP(x)dx = (1/L)∫
L
0
x dx
= (1/L) (x2/2)∣L0 = (1/L)(L2/2 − 0) = L/2
⟨x2⟩ = ∫
L
0
x2P(x)dx = (1/L)∫
L
0
x2 dx
= (1/L) (x3/3)∣L0 = (1/L)(L3/3 − 0) = L2/3
(d) ⟨x⟩ is identical with the quantum result for all values of n. �e quan-
tum result for ⟨x2⟩ is L2(1/3 − 1/2n2π2). For large n the second term
in parentheses becomes negligible compared to the �rst, in which case
⟨x2⟩ = L2/3, which is the classical result. �e correspondence principle
is satis�ed.
P7D.10 (a) Outside the box, L/2