Page 227 - Principles and Applications of NanoMEMS Physics
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A. QUANTUM MECHANICS PRIMER                                   217



               E     = = ¦ ω  §   1  ·
                             ¨ +n
                                   ¸ .                                                                  (A.12)
                 MLC        n
                         n   ©    2 ¹
             However, if the field contains no phonons (n=0), the energy is not zero, but
             is given by,
               E     =  1  = ¦ ω .                                                                            (A.13)
                 MLC          n
                       2  n
             This, n=0, state is called the vacuum state, and the corresponding energy, is
             called zero-point energy. Notice that, since  = ,0n  , 1  , 2  3 ! ∞ , the zero-

             point vacuum energy is, in principle, infinite! In practice, however, various
             factors, such as, dielectric constant  cutoff, preclude  it  from  becoming
             infinity, although still very large.
                It we imagine the free-space  in  which  a  z-directed,  x-polarized
             electromagnetic  wave propagates as being  divided into  cubes  of  volume
             V =  L , then, the solution to its associated electric field wave equation,
                  3
                        1 ∂  2 E
               ∇ 2 E =        x  ,                                                                           (A.14)
                    x    2   2
                       c    t ∂
             may be obtained by separation of variables as,

               E  ( ) =tz,  ¦  a  q  ( ) ( ) zft  ,                                                             (A.15)
                 x          n  n   n
                         n
             where,    subject   to     the    spatial   boundary    conditions
             E  x  ( =z  , 0 t ) = E  x  ( = Lz  ,t ) 0= , one obtains,

               ª  d  2  2  º
               «   2  +  k n »  f  n () z = 0 →  f  n () z =  sin ( zk  n  ),                                    (A.16)
               ¬ dz      ¼
             where,

                               1
               k  = nπ / L  ,n  = ,  , 2  3 ! ∞ ,                                                      (A.17)
                n
             and,

               ª d  2   2  º
               «   2  + ω n  » q n  () =t  0  → q n () =t  cos (ω t  −  ) φ ,                               (A.18)
                                                 n
               ¬ dt      ¼
             where ω  =  ck . Writing the electric field solution as
                    n     n
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