Quantum Mechanics for Mathematicians

Taking the derivative, the Lie algebra representation is given by

π′n(X)f=

d

dt

πn(etX)f|t=0=

d

dt

f(etX·z 1 ,etX·z 2 )|t=0

whereX∈gl(2,C) is any 2 by 2 complex matrix. By the chain rule, for
(
z 1 (t)
z 2 (t)

)

= (etX)T

(

z 1

z 2

)

=etX

T

(

z 1

z 2

)

this is

πn′(X)f=

(

∂f

∂z 1

,

∂f

∂z 2

)(

d

dt

etX

T

(

z 1

z 2

))

|t=0

=

∂f

∂z 1

(X 11 z 1 +X 21 z 2 ) +

∂f

∂z 2

(X 12 z 1 +X 22 z 2 )

where theXjkare the components of the matrixX.
Computing what happens forXj=−iσ 2 j(a basis ofsu(2), we get

(πn′(X 3 )f)(z 1 ,z 2 ) =−

i

2

(

∂f

∂z 1

z 1 −

∂f

∂z 2

z 2

)

so

π′n(X 3 ) =−

i

2

(

z 1

∂

∂z 1

−z 2

∂

∂z 2

)

and similarly

πn′(X 1 ) =−

i

2

(

z 1

∂

∂z 2

+z 2

∂

∂z 1

)

, π′n(X 2 ) =

1

2

(

z 2

∂

∂z 1

−z 1

∂

∂z 2

)

Thezn 1 −kzk 2 are eigenvectors forS 3 =iX 3 with eigenvalue^12 (n− 2 k) since

π′n(S 3 )zn 1 −kzk 2 =

1

2

((n−k)zn 1 −kzk 2 −kz 1 n−kz 2 k) =

1

2

(n− 2 k)z 1 n−kz 2 k

z 1 nwill be an explicit highest weight vector for the representation (πn,Vn).
An important thing to note here is that the formulas we have found forπ′n
are not in terms of matrices. Instead we have seen that when we construct our
representations using functions onC^2 , for anyX∈gl(2,C)),πn′(X) is given by
a differential operator. These differential operators are independent ofn, with
the same operatorπ′(X) on all theVn. This is because the original definition
of the representation
(π(g)f)(x) =f(g−^1 ·x)

is on the full infinite dimensional space of polynomials onC^2. While this space
is infinite dimensional, issues of analysis don’t really come into play here, since
polynomial functions are essentially an algebraic construction.
Restricting the differential operatorsπ′(X) toVn, the homogeneous poly-
nomials of degreen, they become linear operators on a finite dimensional space.