# HW1. Quantum Mechanics for Nanoscienceand Nanotecnology. Solutions by ixp95504

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```									         HW 1. Quantum Mechanics for Nanoscience and
Nanotecnology. Solutions

January 29, 2004

1. (a)
N                     N                           N
1 − e−hν/kT           e−mhν/kT =             e−mhν/kT − e−hν/kT         e−mhν/kT
m=0                    m=0                        m=0
N                  N +1
=          e−mhν/kT −           e−mhν/kT = 1 − e−(N +1)hν/kT .                                (1)
m=0                 m=1

We now take the limit, N → ∞:

lim e−(N +1)hν/kT → 0.
N →∞

This is to say:
∞
1
e−mhν/kT =                      .                             (2)
m=0
1−   e−hν/kT

(b) Let’s write the energy of the system in a way that stresses its dependency on frequency: E =
E(ν). As mentioned in class, the mean energy of a system of oscillators with unrestricted
¯
values on energies, yields a constant value, per degree of freedom: E(ν) ∼ kT . This has
the problem that the energy density diverges when integrated over all frequency values.
¯
With this fact in mind, let us ﬁnd out how E(ν) looks once we use Planck’s hypotheses.
hν
First we make α ≡ kT , so
∞                                           ∞                                          ∞
hν           hν
¯
E(ν) ≡           nhνe−n kT     1 − e− kT     = αkT          ne−nα 1 − e−α = αkT 1 − e−α                 ne−nα .
n=0                                        n=0                                         n=0
(3)
Now, the hint given allows us to express
∞                        ∞
∂
ne−nα = −             e−nα .                                 (4)
n=0
∂α n=0

By (2), previous equation becomes:
∞
∂
ne−nα = −        (1 − e−α )−1 = (1 − e−α )−2 e−α .                          (5)
n=0
∂α

Finally,

¯                                                              1         hν
E(ν) = αkT 1 − e−α (1 − e−α )−2 e−α = αkT                          =     hν
.             (6)
eα − 1   Exp kT − 1

1
¯
Equation 6 coincides with the classical value (a.k.a. Rayleigh-Jeans relation) E(ν) for low
frequencies, where

hν         hν
Exp            −1∼      .
kT         kT
In next assignment you will verify this average doe not make the energy inside a cavity
diverge when integration is carried out over all frequencies. This was the ﬁrst occasion
where quantization of energies helped solving a statistical problem.

2

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