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SEMICONDUCTOR LASERS

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SEMICONDUCTOR LASERS Powered By Docstoc
					         SEMICONDUCTOR LASERS

Semiconductor lasers are similar
conceptually to gas and solid state
lasers, except for:
   scale length
   energy levels
SCLs are generally fabricated in
direct gap materials:
Energy




                    Eg(GaAs)
                      = 1.43 eV
                     ~ 867 nm
                      
                          k

                                      VI-1
AlGaInN
  ||




          VI-2
VI-3
           THRESHOLD GAIN

                       2  th L
             R1R 2 e              1

             1
So  th       ln( R1R 2 )
            2L

If L = 300 µm, R1 = R2 = 0.3, then

                  ~ 40 cm–1!
              th 




                                       VI-4
         BEAM DIVERGENCE


                      q ≈ /D ≈ 1/2
Active
Region
                  q11 ≈ /D ~ 1/30
         20-30 mm       ( = 1 mm)



          MODE SPACING

                   c
     cavity        
                      ~ 125 GHz
                  2nL

          For L = 300 µm


                                       VI-5
3. Short length leads to large c

              c       3  1010
 cavity ~       ~
            2nL 2  4  300  10  4
            11 1
 1.25  10 s


II. SEMICONDUCTOR
    FUNDAMENTALS

   A. Density of electrons/holes

   1. Free electrons are found in
      the conduction band.
      Free holes are found in the
      valence bands.

                                       VI-6
                                                          (E-Ec)½ ~ ke
                                                          Conduction band
                              Empty States
Electron Energy



                                (holes)
                                                    Fn (Quasi-Fermi level)
      e-Ec




                  Ec                              e’s (filled states)
                       Egap = Ec-Ev
                  Ev                            p’s (empty states)
Hole Energy
    Ev-e




                                                    Fp (Quasi-Fermi level)
                          e’s (filled states)
                                                          Valence band

                                                          (Ev-E)½ ~ kh




Fn = maximum energy of e in
     conduction band as T  0

Fp = maximum energy of p in
     valence band at T  0



                                                                         VI-7
     2. The density of
         p' s in valence  band is
        e' s in conduction 
                           
       
n    E c (E)f c (E) dE
           c            Probability of occupy
                              state at E
                                Fermi function
                   Density of available states at E
       
p    E   v
                v (E)1  f v (E)  dE
                             If a state is not
                             occupied by an
                             electron, it is
                             occupied by a hole.




                                                      VI-8
Conduction Band                        Valence Band
                       3

         1  2m* 
                       2
                          1
c(E) =        e (E  E ) 2
          2 2        c
        2   
                                             3

                                1  2m* 
                                             2
                                                         1
                  v(E) =        2 2 
                                        h (E  E) 2
                                                 v
                              2   
                 1
fc(E) =
        exp(E  Fn ) / kT   1
                                         1
                               exp (E  Fp ) / kT   1
                  fv(E) =

m* (GaAs) m*    e  0.067             m*n  0.48
(effective    m0                      m0
mass)
             m0 = 0.911 × 10–27 g




                                                       VI-9
                    ABSORPTION AND
                  EMISSION OF RADIATION


For intrinsic material:
Carrier Density




                    Fp   Ev   Ec   Fn     E




                                              VI-10
ABSORPTION
          E
          Fn
               e–
          Ec

          Ev
               holes
          Fp
                    –
                    e


Absorption occurs if h  Fn – Fp

Emission can occur over the energy
interval: Eg  h  Fn – Fp
             ~


                                    VI-11
               GAIN AND ABSORPTION

                         nc(E2) – nv(E1)
  Gain


                     gain
                    Eg Fn – Fp
  Absorption




                                    h
               Transparency
                    Absorption:
                    a  nv(E2)–nc(E1)

h  E 2  E1,
                *
E2  Ec  *
              mp
                    h  E g 
           m p  m*
                  e




                                             VI-12
Similarly,

                 *
               mn
E1 = Ev –              {h – Eg}
             mp  mn
              *    *

where
  h  Eg + (E2 – Ec) + (Ev – E1)

    GAIN IN SEMICONDUCTORS

Let’s assume that we have in some
way produced free electrons and
―free holes.‖




                                    VI-13
 The emission rate is:

R 21  B21    h   j ()
                          Density of
     Photon               States
     Density
        f c (E 2 )1  f v (E1 ) 
         Occupancy Probability


                                             
                      1                              1
  *
2m e ( E 2    Ec )   2      *
                           2m h ( E v    E1 )       2



                           *
                          mn
       (E2 – Ec) =         *
                               (Ev – E1)
                          me

The photon energy is h = E2 – E1,
Egap = Ec – Ev
                                             VI-14
h = Egap + (E2 – Ec) + (Ev – E1) or

           m*
E2 = Ec + * h      (h – EG)
         mh  me
        Energy of electron above Ec

           m*
E1 = Ev – * e    (h – EG)
         mh  me
          Energy of hole below Ev

For a given h, E2 is ―deeper‖ into
conduction band in GaAs since
m h  m e  Best to use np
  *     *                  +




                                   VI-15
III. GAIN IN SEMICONDUCTORS

  A. Let’s assume that we have in
     some way generated free
     electrons and ―free holes.‖
     1. Gain is ―automatically‖
        obtained.
     2. The question is which
        electrons can recombine
        with which holes.

  B. Joint density of states.
     1. An important parameter is
        the ―joint density of states‖
        — j.



                                   VI-16
   2. j expresses the total
      availability of states for the
      electron-hole pairs which
      recombine to produce h.
                       3

    j ()  2  2  h  E g  2
              h  2m r 
                       2         1

             4   
  #
  3         hj (h)   #
cm hz
                     cm3J


                          * *
                        memh
 mr = reduced mass =
                       me  mh
                        *     *




                                  VI-17
     C. 1. The gain coefficient is
           given by

d()
        ( )                          ng = group index
 dz
     [(Rate of Emission)–(Rate of Absorption)] (ng/c)
              R21                R12
                         3
                     #/cm -s
         = photon density #/cm                 3




So
                ng  1
        () =   (R21 – R12)
                c 
                




                                                        VI-18
       2. The rates of emission and
          absorption are

Emission                             Fermi Function
R 21  B21    h   j ()      [fc(E2)(1– fv(E2))]
          Photon
         Density                   e in       hole in
                                conduction    valence
 Einstein
Coefficient      Availability  Probability of
                  of States occupation of states

R12  B12    h   j ()       [fv(E1)(1– fc(E2))]
Absorption                        e in
                                valence        hole in
                                            conduction




                                                         VI-19
         3. Solving for gain coefficient
            (B21 = B12 since g2 = g1)

                      ng 
   ()  B21  h    j ()  f c (E 2 )  f v (E1 )
                      c 
                      
                  2
         A 21     0      j ()f c (E 2 )  f v (E1 )
                 8n g
                     2

                      1
f c (E 2 ) 
                     E 2  Fn ) 
             1  exp
                        kT     
                                       m*
                          E 2  E c  * h * (h  E g )
                                     me  mh
                     1
f v (E1 ) 
                    Fp  E1 ) 
                    kT 
            1  exp           
                              
                                            m*
                          E1  E v          e       (h  E g )
                                        m*
                                         e    m*
                                                h
      1                                 1
A 21  , n = recombination lifetime =
      n                               p
                                                            VI-20
       DERIVATION OF GAIN
        COEFFICIENT FOR
     SEMICONDUCTOR LASERS
                           ( cm 1 )
The gain coefficient  ()        is
given by
     d()
              ( )  ( )
       dz
                             n h
             (R 21  R12 ) 
                            c
where
   () = Laser energy density
                  3
           (W/cm )
   R21 = Rate of emission
          (photons/cm3-s)
   R12 = Rate of absorption
                         3
          (photons/cm -s)
   n = index of refraction
                                       VI-21
   c = speed of light

Solving for () we get
            n  h R  R 
    ()             21
             c   ()     12
           
The rates of emission and
absorption are
R21 = B21 · () · j() ·
        f c (E 2 )1  f v (E1 )
                                 EMISSION
R12 = B12 · () · j() ·
        f v (E1 )1  f c (E 2 )
                             ABSORPTION
where
B12 = B21 = Einstein coefficients
            (for g1 = g2)
                                       VI-22
j() = Joint density of states
                    3

 j ()  2  2  h  E g 2 #
           h  2m r 
                    2        1

         4                 cm3hz

E2 = Electron or hole energy in
       conduction band
E1 = Electron or hole energy in
       valence band
h = E2 – E1 = Photon energy
            m* 
E2 = Ec +  * h *  (h – Eg)
           m  m 
            e       h
            m* 
E1 = Ev –  * e *  (h – Eg)
           m  m 
            e       h
                  1
fc(E2) =
                 E 2  Fn 
         1  exp          
                 kT 
                                  VI-23
                 1
fv(E1) =
                 E1  Fp 
                 kT 
         1  exp         
                         
Note that:
a. fc(E2)(1 – fv(E1)) is the joint
   probability of finding an electron
   in the conduction band and hole
   in the valence band.
b. fv(E1)(1 – fc(E2)) is the joint
   probability of finding an electron
   in the valence band and a hole in
   the conduction band.




                                   VI-24
Solving for () we get

 ( )
        n  ()  f (E )  f (E )
 B21h  j          c  2     v 1
        c
          0
           2
 A 21            j ()  f c (E 2 )  f v (E1 )
         8n   2

                                              + ++
where we can approximate for n p
junctions

      1
A 21  , n = electron
      n
                          1
recombination lifetime =
                         p

                                                 VI-25
C. SIMPLE HOMOJUNCTION DIODE
   LASER
  1. A typical semiconductor laser
     consists of a forward biased
     pn junction in which an
     inversion is obtained.

  2. Assume we have a heavily
     doped n+p++ junction.
                        l

                    L

          –
      v       n+
          +                 d
               ++
              p




  3. The active region is roughly
     determined by the contact size
                                 VI-26
     and diffusion length of
                              ++
     minority carriers (e in p )

Volume = l · L · d
d  LDIF  (D)1/2
                             2
D = Diffusion coefficient (cm /s)
 = Recombination lifetime
    (s) of e’s

For electrons in GaAs,

                     1
Dn  15 cm /s, n 
           2
                    p
          –10
  2 × 10 cm /s,3

p = hole density

For a doping density of  1019 cm–3,
                                    VI-27
                   1
n 
     2  10 10 (cm3 /s) 1019 cm 3
             –10
    5 × 10 s,

so
                                 –10
d  LDIF  (15 (cm /s) · 5 × 10 s)
                    2                  1/2

          0.87 µm  1 µm

     4. The current density through
        junction is

               qn d A
            j
                     cm 2



                                       VI-28
D. CALCULATION OF GAIN
1. Specify kT, Egap, nindex, m* , m*
                              h    e
               +
2. Assume np junction; specify
   NA
                   2
3. Choose j(A/cm ), h > Egap
              m*
4. E2 – Ec = * h * (h – Eg),
            mh  me
                *
              me
   Ev – E1 = *      (h – Eg),
            me  mh
                  *

                      E v  Fp
5. p  NA. Compute
                         kT
   (do integral or see Table 11.2)
         1                       1
6. n                  1/2
             , d = (Dn) , A21 =
        p                      n
                                     VI-29
       qdn e        j  n
7. j =       , ne =
        n           qd
           Fn  E c
8. Compute          (see Table
             kT
   11.2)
9. E2 – Fn = (E2 – Ec) – (Fn – Ec) 
    fc(E2)
    E1 – Fp = (Ev – Ep) – (Fv – E1) 
    fv(E1)
10. j() = hj (h)
                             3
             h  2m r       2   1
            2  2  (h  E g ) 2
            4   
                     2
11. () = A21         0
                       2
                              j()
                 8n index
           f c (E 2 )  f v (E1 )
                                      VI-30
E. Example: GaAs, kT = 0.026 eV,
            ng = 3.58, Eg = 1.43
            eV, j = 40 kA/cm2,
                     (pulsed!)
 *
mh    0.55 m0, me
                 *
                      0.067 m0 ,
              19     –3
NA = 1.3 × 10 cm
                  *
                mh
1. E2 – Ec = *          (h – Eg)
             m h  m* e
                  0.55
           =                 (1.46 – 1.43)
             0.55  0.067
                          –2
           = 2.674 × 10 eV
                   *
                me
2. Ev – E1 = *          (h – Eg)
             me  mh  *

                 0.067
           =                 (1.46 – 1.43)
             0.067  0.55
                          –3
           = 3.258 × 10 eV
                                        VI-31
                                           Fn

                                 0.0533 eV
                                          E2
                                      0.027
                                          Ec
                  Eg = 1.43 eV
                                       Ev
                                      E1
          0.018                                 0.00326
     Fp




                                      –19            –3
3. For p  NA = 1.3 × 10                        cm ,
   from Table 11.2,
         1              1
4. n       
        p   1.3  1019  2  10 10
                  –10
      = 3.85 × 10 s
                  1/2
   d = (Dn)
     = (15 × 3.85 × 10–10)1/2
     = 0.76 µm
                                                          VI-32
       n e dq
5. j =        ,
        n
        j  n
   ne =
         dq
         40  103  3.85  10 10
      =
        0.76  10  4 1.6  10 19
                     18    –3
      = 1.267 × 10 cm

                     Fn  E c
6. From Table 11.2,           = 2.05,
                       kT
   Fn = Ec + 0.0533 eV

7. j() = hj(h)
                       3
            h  2m r  2     1/2
         = 2  2  (h – Eg)
           4   

                                      VI-33
          6.62  10  27
        
              4 2
                                         3
    2  0.911  10  27 (0.067  0.55) 2
                                       
    (1.054  10  27 ) 2 0.067  0.55 
                                       
        (1.6 × 10–12)1/2 h  E g eV
                                       1
                                       2


= 2.06 × 10 h  E g 
            5              1
                           2   s
                           eV cm3

8. E2 – Fn = (E2 – Ec) – (Fn – Ec)
                        –2
           = 2.674 × 10 – 0.0533
           = –0.02656 eV
E1 – Fp = (Ev – Fp) – (Ev – E1)
        = 1.794 × 10–2 – 3.258 × 10–3
        = 1.468 × 10–2 eV


                                      VI-34
                  1
fc(E2) =
                 E 2  Fn 
         1  exp
                    kT   
                1
       =                   = 0.7353
         1  exp(1.022)

                 1
fv(E1) =
                 E1  Fp 
                 kT 
         1  exp         
                         
                1
       =                 = 0.3625
         1  exp(0.565)

                 0
                  2
9. () = A21       2
                        j()
                8n g
                           1
  [fc(E2) – f1(E1)], A21 
                           n
                                      VI-35
        1           (8505  10 8 ) 2 cm 2
=                 
  3.85  10 10 s       8  3.582

   · 2.06 × 105 (1.46 – 1.43)1/2      s
                                     cm 3

   [0.7353 – 0.3625] = 776 cm–1




                                            VI-36
For ABSORPTION, the rate is:

R12 = B12 ·  · h · j()
           f v (E1 )1  f c (E 2 ) 

Setting B21 = B12 since g1 = g2, the
gain coefficient is found to be:

                n
() = B21 h j() ·
                c
          f c (E 2 )  f v (E1 )
Rewriting…
              2
() = A21           j()
            8n 2
         · f c (E 2 )  f v (E1 )
                          gain coefficient
                                           VI-37
where
                                        1
                        E 2  Fn 
   f c (E 2 )  1  exp
                           kT  
                                        1
                         Fp  E1 
   f v (E1 )  1  exp  kT    
                                 
and
                          1
         A 21 
                 recombinat ion
                     lifetime

Note: In these eqns., j() is the
―JOINT DENSITY OF STATES‖ that is a
measure of the availability of an
   +                   –     +
e-h pair (not just an e or h !) at
frequency .
                                             VI-38
                           3
            h  2m r      2    1
   j ()  2  2  (h  E g ) 2 ,
           4   
     1       1
 cm Hz


          m* m*
           n p
mr                  is the reduced mass
        *
       mn      mp
                 *
          –   +
for an e -h pair.




                                       VI-39
      DERIVATION OF GAIN
       COEFFICIENT FOR
    SEMICONDUCTOR LASERS
                             –1
The gain coefficient () (cm ) for
a semiconductor laser is given by
d()
         ( )  ( )
  dz
                        n  h
        (R 21  R12 ) 
                       c
where () = Laser energy
                    density (W/cm3)
        R21 = Rate of emission
                               3
                 (photons/cm -s)
        R12 = Rate of absorption
                               3
                 (photons/cm -s)
        n = Index of refraction
        c = Speed of light
                                  VI-40
Solving for g(n) we get

             n   h  ( R  R )
    ( )               21
              c    ( )     12
            

The rates of emission and
absorption are

R21 = B21 · () · j()
       f v (E 2 )1  f v (E1 )  Emission

R12 = B12 · () · j()
     f v (E1 )1  f c (E 2)  Absorption

where
B12 = B21 = Einstein coefficients
            (for g1 = g2)
                                          VI-41
j() = Joint density of states
                    3
          h  2m r 2         1   #
 j ()  2  2  (h  E g ) 2

         4                    3
                                cm hz

E2 = Electron or hole energy in
     conduction band
E1 = Electron or hole energy in
     valence band
h = E2 – E1 = Photon energy

             m* 
E 2  E c   * h * (h  E g )
            m  m 
             e     h
                  m* 
      E1  E v   * e * (h  E g )
                 m  m 
                  e   h

                                  VI-42
                      1
f c (E 2 ) 
                     E 2  Fn 
             1  exp
                        kT   
                                 1
             f v (E1 ) 
                                 E1  Fp 
                                 kT 
                         1  exp         
                                         
Note that:

a. fc(E2)(1 – fv(E1)) is the joint
   probability of finding an electron
   in the conduction band and a hole
   in the valence band.
b. fv(E1)(1 – fc(E2)) is the joint
   probability of finding an electron
   in the valence band and a hole in
   the conduction band.

                                        VI-43
Solving for () we get

()
          n   () · [f (E ) – f (E )]
= B21 h   j           c 2      v 1
         c
       2
= A21 0 2 j() · [fc(E2) – fv(E1)]
      8n

where we can approximate for n+p++
junctions

        1
A21  ,
        n
n = electron recombination lifetime
      1
   =
     p

                                      VI-44
1. Specify kT, Egap, nindex, h >
   Egap, m* , m*
          h    e
               + ++
2. Assume n p junction; specify
   NA
                      2
3. Choose j(A/cm )
                   *
                 mh
4. E2 – Ec =              (h – Eg),
               m*  m*
                h    e
                   *
                 me
   Ev – E1 =              (h – Eg)
               me  mh
                *    *

5. Assume p  NA. Compute
   E v  Fp
            [Do integral or see
      kT
   Table 11.2]

                                       VI-45
6. Obtain lifetime and A21. n 
    1              1/2       1
        , d = (Dn) , A21 =
   p                      n
7. Get electron density from
                        qdn e
   current density. j =       ,
                         n
          j  n
   ne =
           qd
           Fn  E c
8. Compute          (see Table
             kT
   11.2)
9. Get parameters required for
   Fermi functions.
   E2 – Fn = (E2 – Ec) – (Fn – Ec) 
   fc(E2)
   E1 – Fp = (Ev – Ep) – (Fv – E1) 
   fv(E1)
                                   VI-46
10. j() = hj (h)
                           3
             h  2m r     2     1
            2  2  (h  E g ) 2
            4   
                     2
11. () = A21         0
                       2
                              j()
                 8n index
           f c (E 2 )  f v (E1 )




                                      VI-47
       MODEL FOR A SIMPLE
  HOMOJUNCTION DIODE LASER
   A simple homojunction
semiconductor laser consists of a
forward biased pn junction in which
an inversion is obtained.
   Assume we have a heavily doped
n+p++ junction as for a GaAs device.
                         l

                    L

          –
      v       n+
          +                  d
               ++
              p




The active region is roughly
determined by the contact size and

                                  VI-48
diffusion length of minority carriers
       ++
(e in p )
          Volume = V = l · L · d
          d  LDIF  (D)1/2

where LDIF is the electron diffusion
      length
      D = Diffusion coefficient
            (cm2/s)
       = Recombination
            lifetime of e’s (s)

For electrons in GaAs,
                              1
        Dn  15 cm /s, n 
                    2
                             p
          2 × 10–10 cm3/s,
        p = hole density
                                   VI-49
For a doping density of  1019 cm–3,
                   1
n 
     2  10 10 (cm3 /s) 1019 cm 3
    5 × 10–10s,
so
d  LDIF  (15 (cm2/s) · 5 × 10–10s)1/2
           0.87 µm  1 µm

The current density through
junction is
              qn d A
           j
                     cm 2


The procedure for calculating gain
is then:


                                     VI-50
IV. LASER OSCILLATION
  A. With the gain computed in
      the previous section, laser
      power can be computed in the
      ―conventional‖ way.
      1. Although the laser is
         essentially homogeneously
         broadened, it does have
         inhomogeneous character
         due to cross relaxation
         between modes.
      2. Broad gain and short
         cavity result in finite
         number of discrete
         longitudinal modes.
  B. Laser Power
      1. Assumptions
         a. Iout() =
                                VI-51
  0  1I   1 ln( R R ) g ()
         sat  2       1 2 
 th                          
      b. Isat =
                h                 1
                           , s 
       stim spontaneous         A 21
                   A 21c 2
     c. stim              g( 0 ),
                  8n g  2
                       2

                       2
        g ( 0 ) 
                    total
         total = FWHM of ()
     d. Lasing occurs on
        discrete longitudinal
        cavity modes
                         c
         cavity 
                      2n g L

                                   VI-52
e. Although the laser is, in
    principle,
    homogeneous, there is
    inhomogeneous
    character, line
    narrowing, and cavity
    ―Q‖ considerations. For
    purposes of this
    example, assume a line
                                th
    shape function for the j
    cavity mode,
                     1
                   2 
 g j ( )                     ,
                               2
                      2   
            (   j )   
                          2 


                               VI-53
         20 GHz
             cavity
          
             5  10

C. Example
   1. GaAs, T = 300 K, NA = 1.3
      × 1019 cm–3, j = 7 kA/cm2,
      R1 = R2 = 0.3, Lcavity = 300
                      –1
      µm, a = 10 cm , lcontact =
      20 µm
      a. Threshold gain th  50
         cm–1
      b. Oscillation occurs on
         discrete cavity modes
      c. Total laser power 58
         mW


                               VI-54
2. Lcavity = 200 µms
               –1
   th  70 cm
   Fewer cavity modes
   Power  8.8 mW

3. Lcavity = 400 µms
               –1
   th  40 cm
   Power  0.136 W

4. NOTE: Total currents are
   large (0.1-0.3 AMP) and so
   these would most likely be
   pulsed devices, to lower
   threshold gain.




                           VI-55
V. HETEROJUNCTION LASERS
   A. Conventional lasers suffer
      from at least three ills.
      1. Injected carriers diffuse
         laterally, thereby reducing
         gain and increasing
         threshold current.
      2. Cavity properties are poor.
      3. Not tunable (fixed band
         gap).
   B. To some extent, all issues are
      addressed by heterostructure
      or heterojunction lasers.
      1. These are lasers which
         have layers of AlxGa1-xAs,
         GaAs, AlxGa1-xAs.


                                  VI-56
l
    n AlxGa1-xAs GaAs
                                    x ≈ 0.1-0.3
    p AlxGa1-xAs

        AlxGa1-xAs
          GaAs
                         Ec
                              Band gap increases
E                             with increasing Al.
                         Ev

                     l
                              Index of refraction
Index




                              decreases with
                              increasing Al.
                     0
          2. Lower Ec (higher Ev) tends
             to ―channel carriers‖ to a
             small region. This
             increases current density,
             increases gain.
                                               VI-57
AlxGa1-xAs GaAs AlxGa1-xAs
          e    e

       p              p

   3. Higher index of refraction
      tends to produce a
      waveguide structure (like a
      light-pipe or fiber).

AlxGa1-xAs GaAs AlxGa1-xAs


    Low n            Low n


            High n

                               VI-58
VII. Quantum Well Lasers
A. The use of ―quantum wells‖ has
   a more subtle effect. The
   physically narrow wells not only
   restrict current flow but also
   limit momenta of
   electrons/holes, and therefore
   increase density of states and
   gain.

   The electron DeBroglie
   wavelength is
             h      h
       D  
             p (2m*E)1/2
                               *
   For an electron in GaAs,   me   =
   0.067 m0, E = 0.1 eV
                                       VI-59
         6.63  10  27 erg  s
D 
     (2  0.067  0.911  10  27
      0.1  1.6  10 12 erg)1/2
   = 150 Å

   If the width of the quantum well
   d  , then the electron takes on
      ~
   wave behavior, and the
   continuum density of states does
   not apply.

B. The energy of an electron in the
   conduction band is quantized as

E2 = (l, k)
          l  
            2 2 2          2
                        2 k e
   = Ec +    * 2
                   +       *
          2m e L z       2m e
                                    VI-60
E1 = (n, k)
         n   2 2 2    2 k  h 2
  = Ev –    * 2
                  –         *
         2m h L z         2m h

   Assume energy level structure

Lx << DEBROGLIE

Lz, Ly >> D




                                       VI-61
                           (P)e
                  (Px)e
        Ec           E2
             EG           (P)h
        Ev
                    E1
                          Ly
                  (Px)h
                   Lx      z       y
                                    x

             ˆ
-Momentum in x is quantized : Px
-Momentum in y, z is ―continuum‖:
             ˆ ˆ
                      Py  Pz  P
                       2    2




                                   VI-62
From the ―particle in a box‖ of
width L quantum mechanics
problem, the quantized momentum
wave vectors are
        l
  kx =     , l = quantum number
       Lx
         mv 
    k =         continuum part
           
where P = momentum =  · k.
Since the total kinetic energy of a
particle is
       2     2     2
     P      Px    P
El           
     2m 2m 2m
     l 2 2 2 k 2  2
           2
               
      2mL x      2m      1 2
                           mv 
                         2
                                      VI-63
Then for electrons in conduction
band, holes in valence band

                          l         
                        2 2 2          2
E 2 l, k  e   E c 
                                   k e
                             * 2
                                    *
                          2m e L x 2m e

   
E1 n, k  p   
               n 2 2  2 k  2 
        Ev             
                                 p
               2m 2 L2     2m 2 
                   p x         p 


The object is to ultimately
maximize the density of states.



                                       VI-64
 ( ) 
              0
               2
     A 21                 j ()f c (E 2 )  f v (E1 )
            8n      2


           Lz (  100 Å)
                ~
Ga1-xAlxAs GaAs Ga1-xAlxAs

                                                              1
                                                  2 m*      2
                     E2 (2, k = 0)   k  e     2e
                                                         
                                                          
                     E2 (1, k = 0)                      
                                             E e  E 2 l, k   0
                                                                    1
Ec                                                                  2



     EG        w0         k is in/out
                           of paper
                                                              1
Ev                                                2 m*      2

                     E1 (1, k = 0)   k  h     2h
                                                         
                                                          
                                                        
                     E1 (2, k = 0)
                                             E1 l, k   0  E h 
                                                                        1
                                                                        2

                 x



Selection rules require n = l, (k)e =
(k)h = k
                                                                    VI-65
w = E2(l, k) – E1(l, k) = EG +
          2   2 l 22 
               k      
            *         2 
         2m r       Lz 

            * *
          mem h
 *
mr   
         me  mh
          *     *


C. The density of states in each
   band is also quantized.

               *
e(E) = 
               me
                     H E  E ( l, k ) 
           l 1 
                2

                 H = Heaviside function
                               1 x>0
                               0 x0
                                           VI-66
                  *
h(E) = 
                  mh
                        H  E ( l, k )  E 
            l 1 
                    2


                                  (E)
                                     Conventional
                            I3
                                  I2 > I1 (current)
                  I1    I2
Gain




       EG                                    w
                Fn – Fp

                                 (E)
                                        Quantum well

                                        I3
                       I2                    
Gain




             I1
       EG                                    w



                                                       VI-67
Scaling with Lx



Ec                                2 2
                  2 2    E 
           E                  2m* L2
                                    e x
EG              2m* L2
                    e x

      Lx
     w1                  w2

              w1 > w2
               1 < 2




                                     VI-68
        NONLINEAR OPTICS:
    HIGHER ORDER PROCESSES
I. LINEAR VS. NONLINEAR
PROCESSES
   A. The majority of laser
       interactions (stimulated
       emission, spontaneous
       emission, absorption) result
       from interactions with a
       single photon.
       1. Conservation of energy
          considerations require E2 –
          E1 = w
                                    E2
                       h      h

                                    E1

                                    VI-69
   2. There is exchange of a
       single quantum of energy.
B. It is possible, however, for
   the atom, molecule or crystal
   to simultaneously interact
   with multiple quanta of a
   single photon energy or
   different photon energies as
   long as energy and
   momentum are conserved. If
   we have two photons
   generating a third, this is
   called ―three-wave mixing.‖

 h1
 h2        Crystal         h3



                              VI-70
         1. Second Harmonic
            Generation
                               Either a real or a
                       E2      virtual energy level
     h2
h1
 = h2            h3 = 2(h1)
 = h
         h1
                       E1

         2. Sum Frequency
            Generation

                              E2
                 h2
     h1 ≠ h2              h3 = h1 + h2
                 h1
                              E1
                                                VI-71
   3. Difference Frequency
      Generation
                  E2
                h2
        h1       h3 = h1 – h2
h1 ≠ h2
                h3


C. It is possible to, in principle
   have n-photon processes. For
   example, third harmonic
   generation (4-wave mixing).
                          E2
               h3
h1 = h2               h4 = 3h
    = h3 = h h2
               h1
                          E1
                                VI-72
D. Typical Experiments
        1. Harmonic generation
            Crystal            n      1
Laser           , n                  1      31 = 3
                                        1


        2. Intracavity harmonic
           generation
                 R at , n
                      Laser   Crystal
                                              n

                                             T high at n
                                             R high at 



Nd:YAG: 1 = 1.064 µm
        2 = 532 nm
        3 = 355 nm
        4 = 266 nm

                                                     VI-73
           3. Mixing                    1
                                 1
                                 1    2    2  3

                          1

             1             2 = 21    2, 1        31
   Laser
                       Crystal               Crystal


                        Tunable Dye Laser
                           2              4 =
                                      3 2(21) – 3
            1    2 = 21            2, 3
   Laser
             Crystal                         Crystal
                                             or Gas

Nd:YAG:                                              2        3

       1 = 1.064 µm    
                          
                                        1
                                        2
                                                 2     2        4

       2 = 532 nm
Rh6G (dye): 3 = 640 nm
                 c
       4 =               = 168 nm
            2(21 )   3
            (tunable by varying 3)
                                                            VI-74
II. POLARIZATION AND
SUSCEPTIBILITY
     A. The interaction of photons
        with matter is through the
        susceptibility of the
        material.
        1. Wave equation         
                        E P 
          E  m0 e0
           2
                               
                      t  t t 

                        susceptibility
                       
 P  Polarization  e0 E
                     complex
                     (may be a tensor)

                2
               E      P
                        2
  E  m 0e 0 2  m 0 2
   2
               t      t
                                     VI-75
        n 2  2E                  2P
       2 2                  m0 2
        
        c t
                               t
                               
         Free space     Generation/absorption
         propagation    of radiation; dispersion


        2. The polarization is
           actually a function of
           higher orders of the
           electric field
           (perturbative analysis;
           valid only if I n  I n1 ).
         
   P  e0  n E n
            n 1

       e0 ( 1E   2 E 2  3E 3  )
                   
Source of                   multiwave
linear processes (absorption, emission)
                                           VI-76
  3. If the material through
     which the electric field
     propagates has a
     susceptibility with non-
     zero, nonlinear terms
     (n, n > 1), then it has
     the ability to interact
     with the electric field to
     produce radiation by
     higher order processes.
     This leads to harmonic
     generation and other
     effects.

                 k1          k3
                 k2
                             z
B. 3-Wave Mixing

                             VI-77
              1. Let’s assume that we
                 have 3-wave mixing.
w1 + w2 = w3 w2                w3
(energy conservation)
                      w1             1 E j0 (z)        2
                                                  k 
                                    E j0 z             
                  Ej 
                           1
                           2
                             E j0 expiw jt  k jz   cc
                             complex conjugate


              2. Each frequency
                 component must satisfy
                 the wave equation.

                                  2E j         2P
       2 E j  m 0e 0                     m0 2
                                    t 2       t

             ETOTAL = E1 + E2 + E3
                                                            VI-78
             3. The polarization is

P    e0 (1E total   2 E total )
                            2

                                                   m
                                   en ergy momentu
            31                                  
     e 01   E j0 expi(w jt  k jz)   cc 
             j12                               

             31
     e 0 2   E j0 exp i(w jt  k jz)   cc 
              j12
                3 1                                  
                E l 0 expi(wl t  k l z)   cc 
              l 12                                  

When we multiply out the terms for
the nonlinear polarization, we get
terms which correspond to each
frequency, as well as sums and
differences of frequencies. We can
                                                         VI-79
pick out those terms for w1, w2, and
w3.

For example for w1 = w3 – w2

P2 w1   
            e 0 2
               2
                  E 30E 20 expi(w3 w2 ) t
                          *


              (k 3  k 2 )z    cc

whereas for w3 = w1 + w2

          e 0 2
P2 w1         
             2
E10E 20 expi(w1  w2 )t
                  (k1  k 2 )z    cc
     Source term for w3 based on E1 and E2


                                             VI-80
            4. The right side of the
                   wave equation then
                   becomes (for w1)
             E1
              2
                            2
 m 0e 0 2  m 0 2 Pw1 
             t           t
                 1                             
 m 0e 0 (iw1 ) 2 E10 expi(w1t  k1z)   cc
                                              k12
                 2
                                             2
         2 e0
 m 0 (iw1 ) 1E10 expi(w1t  k1z)   cc
             2                              
 m 0 (iw1 )e 0 2
         2

     E 30 E 20 expiw1  (k 3  k z )z   cc

                                       w1 n1
                                        2 2
Note:    m 0e 0w1 (1  1 )  m 0e1w1  2  k1
                2                   2        2
                                        c
                   k1  m 0e1 2 w1
                                 1




           5. The Laplacian term to
              the wave equation is
                                                    VI-81
        2 1
 E j  2  E j0 exp i(w jt  k jz)   cc
  2
                                                
       z  2                                   
       1   2 E j0          E j0            
                   i 2k j        k j E j0 
                                       2
       2  t                 z              
                                             
        exp i(w jt  k jz)   cc


            E j0     2 E j0
Since    kj                 ,   then
             z       z  2



                  1 2                E j0 
          E j   k j E j0  i 2k j
            2
                  2                   z  
                    exp i(w jt  k jz)   cc


           6. So if we now group
              terms having the same
              frequency [exp(iwjt)] we
              get
                                                    VI-82
        dE10
  ik1        exp(ik1z)
         dz
             m 0w12
                   e 0 2 E 30 E* expi(k 3  k1 )z 
                                 20
                2

Divide by ik1exp(–ik1z)
                       1
dE10    iw1  m 0     2
                  e 0 2 E 30 E*
 dz      2  e1 
             
                                 20

            exp i(k 3  k 2  k1 )z 

                               1
          dE*        iw2  m 0 
                               2
            20           e  e 0 2 E10 E 30
                                           *
           dz         2  2
                       exp i(k1  k 3  k 2 )z 

                                             1
                     dE30    iw  m         2
                            3  0  e 0 2 E10 E 20
                      dz      2  e3 
                                   exp i(k1  k 2  k 3 )z 

                                                          VI-83
       7. Observations
          a. The growth or decay
             of each field depends
             on the other 2 fields.
             A field at E1 generates
             a field at E3.

          b. In order that the
             ―beating‖ between the
             waves does not
             average to zero, the
             wave vectors must
             sum to zero.

For E30, this means k1 + k2 – k3 = 0
             k3 = k1 + k2


                                  VI-84
 C. Second Harmonic
    Generation
  1. In second harmonic
      generation
            w2
                   w3          w1 = w2
            w1

w1 = w2, E10 = E20, 3 =   1
                           2
                               1,
         w3 = 2 w1
   2. Assume that we are
      perfectly phase matched.
        k3 = k1 + k2
   3. Assume E10 is real.
   E10 = E20 = E10  E*
                *
                      20

                                     VI-85
                         1
  dE10    iw1  m 0     2
             e  e 0 2 E 30E 20
                                *
   dz      2  1
                 1
dE 20 iw2  m 0 2                dE10
            e  e 0 2 E10E 30  dz
                             *
  dz      2  1
In order for these equations to hold,
E30 must be imaginary (that is, 90°
out of phase from E1). Defining
E  iE 30
  30
                  dE1 dE10 dE 20
E1  E10  E 20 ,           
                   dz     dz     dz
                     1
   dE1       m 0  e  E  E1 
                     2
        w1  0 2 30  
    dz       e1          2
                             1
         dE
           30   w3  m 0  2   E12

          dz          e  e 0 2 4
                  2  3
                                      VI-86
III. EXAMPLES

   A. Typical values of 2 are
        –10   –9
      10 -10 cm/V

To generate a second harmonic field
equal to the first harmonic field in a
distance l, we would need…




                                   VI-87
PHASE MATCHING

A. There is an additional
   requirement that the photon
   fields conserve momentum.

B. If we have sum mixing

e = h(3 – (1 + 2) = 0
                  
      k  k 3  k1  k 2   0
       2n
      k
           0
               vacuum



                                     VI-88
This is usually only a problem if
there is dispersion (i.e., n is a
function of ).

C. Example:
                 vacuum
              
Sum mixing of 1  4000 Å n1  3
                    2  3000 Å n 2  21.9
                    3  1714 Å n 3  2.8
                                       
            5 k1   1               5 k2
k1  4.71 10     cm k 2  6.06  10       cm 1
               k1                       k2
                                   
   ~ k 0 k1
                        ~ 1.29 k 0 k 2
                         
         k1                         k2


            1
3 =                = 1714 Å,         n = 2.8
       1  1 
         
        1   2
                                             VI-89
                                   
                 1 k 3            k3
k 3  10.2  10 cm ,
               5
                          2.17 k 0
                     k3             k3
                              
Phase matching requires k 3  k1  k 2
                        
            k1           k2
             q1         q2
                   
                   k3

k1 cos q1 + k2 cos q2 = k3
k1 sin q1 = k2 sin q2
           cos q1 + 1.29 cos q2 = 2.17
                   sin q1 = 1.29 sin q2
     q1 = 21.3°
     q2 = 16.4°
                                   VI-90
 
k 2 (
      = 300
            0 Å) Crystal
     2
 21.3°
                         
 16.4°                   k 3 (3 = 1714 Å)
  1 = 400 0 Å)
k 1(



IMPORTANT:

 is a function of the orientation of
the crystal! So in addition to phase
matching the crystal must be oriented
properly.




                                        VI-91

				
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