### B.3.a. Number Representation For One State

Consider the occupation of an 1-particle state |α .

Let | n α  be the n α  -particle state in which |α  is occupied n α  times.  Obviously, n α  must be a non-negative integer.  With the understanding that we are dealing only with the occupation of |α , we can suppress the subscript a for the sake of clarity.

For convenience, we shall assume the set {   |n   }  to be orthonormal, i.e.,

n|m= δ nm                               (1)

With  denoting the hermitian conjugate, the number n ˆ = a + a , creation a + = ( a )  and annihilation a= ( a + )  operators are defined by

n ˆ |n=n|n                               (2)

a + |n= c n+ | n+1                     (3)

and

a + |n= c n | n1                     (4)

where c n±  are constants.  Since n0 , we have

a|0=0            Þ            c 0 =0                      (5)

Taking the hermitian conjugates of (2,3) gives

n|a= n+1 | c n+ *                        (3a)

and

n| a + = n1 | c n *                      (4a)

Thus, (3a)´(3) gives

n|a a + |n= n+1 | c n+ * c n+ | n+1   =   | c n+ | 2

=n|a   c n+ | n+1 =n| c n+ c n+1, |n= c n+ c n+1,

= n+1 | c n+ * a +   |n=n| c n+1, * c n+ * |n= c n+1, * c n+ *

Þ            c n+ * = c n+1,                         (6)

Similarly, (4a)´(4) gives

n| a + a|n= n1 | c n * c n | n1   =   | c n | 2

=n| n ˆ |n=n

Without loss of generality, we can set all c n±  to be real and non-negative.  Hence,

c n = n                                   (7)

and, from (6),

c n+ = c n+1, = n+1          (8)

Thus, (3,4) become

a + |n= n+1 | n+1                (9)

and

a|n= n | n1                       (10)

Now, from (9), we have

a a + |n= n+1   a  | n+1   =( n+1 )|n

so that

[ a,   a + ] |n=( a a + a + a )|n   =|n          for all n

Hence, we have the commutator relation

[ a, a + ] =1                               (11)

Note that the results (9-11) are obtained under the implicit assumption that there is no upper limit to n.  Hence, it is applicable to bosons.  Conversely, we can say that bosonic behavior will be observed if we demand the commutator relation (11).

For fermions, each state can be occupied at most once., i.e., n can only take on two values, n=0,1 .  Hence,

( a ) 2 = ( a + ) 2 =0                                               (12)

n ˆ |n= a + a|n=n|n

a|0=0                            a|1= c |0

a + |0= c + |1                     a + |1=0

Þ

1| a + a|1=0| c * c |0= | c | 2

=1| n ˆ |1=1

0|a a + |0=1| c + * c + |1= | c + | 2

=0|a   c + |1=0| c + c |0= c + c

=1| c + * a +   |0=0| c * c + * |0= c * c + *

Keeping all c n±  real and non-negative, we have

c + = c =1

so that

a|0=0                            a|1=|0

a + |0=|1                         a + |1=0                           (13)

Now,

( a a + + a + a   )  |0=a a + |0=a  |1=|0

( a a + + a + a   )  |1= a + a|1=  |1

i.e.,

( a a + + a + a   )  |n=|n               for all allowable n.

Þ            [ a, a + ] + =a a + + a + a=1                           (14)