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.




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