A Non-standard Geometric Quantization of the

Harmonic Oscillator

Adrian P. C. Lim

Department of Mathematics, Cornell University

Email: al382@cornell.edu

Abstract

In the standard geometric quantization of the harmonic oscillator in (R2, dp ∧ dx),

using standard holomorphic coordinate z = x−ip induces a standard complex structure

which Kahler polarizes the pre-quantum line bundle. Hence, the Quantum Hilbert

space maybe identified with holomorphic functions which are L2 integrable with respect

to Gaussian measure, denoted by H. The corresponding quantized Hamiltonian is

then given by ~z ∂∂z

. We propose using a different almost complex structure J (under

some restrictions) compatible with dp∧ dx and construct a Quantum Hilbert space HJ

containing non-trivial L2 sections and a quantized Hamiltonian QJ . The main result

is that (HJ , QJ) is unitarily equivalent to (H, ~z ∂∂z

).

1 Introduction

Let (M, w) be a symplectic manifold of dimension 2n. Quantization of a sympectic manifold

requires a construction of a Hermitian complex line bundle B → M with a metric compatible

connection ∇ such that the curvature of ∇ is 1i~ω. Here, ~ is Planck’s constant divided

by 2π. This line bundle B exists if and only if 12π~w ∈ H2(M,Z). If the manifold is

simply connected, then up to equivalence, there is a unique line bundle (B) with a covariant

derivative ∇ such that the curvature is equal to 1i~ω.

1

The pre-quantum Hilbert space is then the space of sections of B which are square

integrable with respect to the Liouville volume form ωn. We can then define the pre-

quantum operator Q(f) of a function f : M → R as

Q(f) = −i~∇Xf+ Mf (1.1)

where Xf is the Hamiltonian vector field given by

df + ω(Xf , ·) = 0 (1.2)

and Mf is the multiplication operator.

It is generally agreed that the pre-quantum Hilbert space is too big. To ’trim’ down

the size of this Hilbert space, one chooses a polarization on the tangent space of TmM .

By complexification of the tangent bundle TMC, one considers a polarization of TMC by

picking out a subset of dimension n out of these 2n variables. Namely, we choose n directions

such that the sections in TMC are constant. This is Kahler polarization.

In more details, we define an almost complex structure J on TM compatible with ω

such that the triple (M, ω, J) becomes a Kahler manifold. Then our Quantum Hilbert

space, H is defined to be all complex polarized L2 sections in TMC such that it is constant

in the T (0,1)M directions. However, given such a section l, there is no reason for Q(f)l

to be back in the Hilbert space H, where Q(f) is as defined in Equation (1.1). Hence we

need to impose restrictions on the choice of J . It turns out that J needs to be constant

on the flow generated by the Hamiltonian vector field Xf in order for Q(f)l to be back

in H. Given a Kahler potential φ, this quantum Hilbert space is unitarily equivalent to

the space of holomorphic, square integrable with respect to e−φ/~ωn functions, denoted by

HL2(M, e−φ/~ωn). This unitary map is given by ke−φ/2~ ∈ H 7→ k ∈ HL2(M, e−φ/~ωn).

We then define an operator Q(f) acting on HL2(M, e−φ/~ωn), unitarily equivalent to Q(f)

via this map.

In this article, we consider a harmonic oscillator in R2 with the Hamiltonian H = 12 (x2 +

p2) where p is the conjugate momentum. In the ’standard’ geometric quantization, the

choice of Jstd is

Jstd =

0 1

−1 0

.

2

This Jstd is compatible with the canonical ω = dp ∧ dx and further more, is constant in

the flow of XH . We choose our holomorphic coordinates to be z = x − ip and our Kahler

potential to be H. Under this map (x, p) 7→ z = x − ip, we identify our Hilbert space

HL2std(R2, e−H/~ω), with Hstd, the space of holomorphic functions in C, square integrable

with respect to Gaussian measure. Under this map, the operator Qstd(H) corresponding to

Qstd(H), is given by Qstd(H) = ~z ∂∂z . (See Section 3.)

However, there are many complex structures J one can define on TM which are compat-

ible with dp∧dx and constant on the flow of XH , different from Jstd. By choosing a J (under

some restrictions) and Kahler polarize as described earlier, we obtained a ’new’ Hilbert space

HL2(R2, e−φ/~ω) with an operator Q(H). We now would like to compare these 2 Hilbert

spaces (Hstd, Qstd(H)) and (HL2(R2, e−φ/~ω), Q(H)) and see if these 2 spaces are unitarily

equivalent to each other. There is no reason why these 2 should be unitarily equivalent.

However, it turns out that they are. This is the content of this article.

This article is organized as follows. We begin by describing Kahler polarization in detail

in the next section. The reader familiar with the Kahler polarization of the harmonic

oscillator may skip directly to Section 4.

After the theoretical setup, we will work out the example of the harmonic oscillator in

R2, using the standard almost complex structure Jstd =

0 1

−1 0

. This is a standard

Kahler polarization of the harmonic oscillator in R2.

Following the standard description of the harmonic oscillator, we will describe Kahler

polarization using a different complex structure J . We will then show that this Quan-

tum Hilbert space (HL2(R2, e−φ/~ω), Q(H)) contains non-trivial complex polarized sections

which are L2 integrable and that it is unitarily equivalent to the Quantum Hilbert space

(Hstd, Qstd(H)) obtained using Jstd.

2 Kahler Polarization

We begin with the definition of a Kahler manifold.

3

Definition 2.1 A Kahler manifold is a triple, (M,ω, J) where M is a complex manifold

with almost complex structure J associated to the complex structure on M , ω is a symplectic

form on M such that

1. J∗ω = ω, (i.e. ω(Jv, Jw) = ω(v, w) for all v, w ∈ TmM and m ∈ M), and

2. v → ω(Jv, v) is a positive definite quadratic form.

Let (M, ω, J) be a Kahler manifold. The form ω and the almost complex structure J

will be extended to TMC by complex linearity.

Notation 2.2 We will denote the space of holomorphic functions on M by H(M).

Definition 2.3 We say a vector field Z is of type (1,0) if Z = X − iJX, for some vector

field X. A section k ∈ Γ(B) is said to be complex polarized if

∇Zk = 0

for all vector fields Z of type (1, 0), where B is a Hermitian complex line bundle.

Remark 2.4 Note that f ∈ H(M) iff Zf = 0 for all type (1, 0) vector fields Z.

Definition 2.5 The Quantum Hilbert space is

H = k ∈ K : k is complex polarized

where K = L2(B,ωn).

We would like to make the prequantum operators Q(f) = −i~∇Xf+Mf act on H. There

is no canonical way to do this since typically (ignoring domain questions) Q(f)H " H. The

following theorem gives a sufficient condition on f so that Q(f)H ⊂ H.

Theorem 2.6 Suppose that (M, ω, J) is a Kahler manifold and B is a Hermitian complex

line bundle over M with a metric – covariant derivative ∇ such that Curvature(∇) = 1i~ω.

If f ∈ C∞(M) is such that LXfJ = 0 where L denotes Lie derivative, then Q(f)H ⊂ H.

4

Proof. If k ∈ H (i.e. ∇Zk = 0 for all Z ∈ T (1,0)M), we must show that ∇Z (Q(f)k) = 0

for all type (1, 0) vector fields Z. Now

∇Z (Q(f)k) = ∇Z

((−i~∇Xf+ f

)k)

= −i~∇Z∇Xfk +∇Z (fk)

= −i~[∇Z ,∇Xf]k +

(Zf

)k

= −i~ω(Z, Xf )

i~k − i~∇[Z,Xf ]k + df

(Z

)k

= −df(Z)k − i~∇[Z,Xf ]k + df(Z

)k

= −i~∇[Z,Xf ]k.

So to finish the proof, we need only realize that [Z, Xf ] is of type (0, 1). To verify this

remark, write Z = W + iJW for some real vector field W, then

−[Z,Xf ] = LXfZ = LXf

(W + iJW )

= LXfW + i

(LXf

J)W + iJLXf

W

= LXfW + iJLXf

W ∈ T (0,1).

Because of this theorem, if we want to quantize the Hamiltonian system with Hamiltonian

f, then we should choose J such that LXfJ = 0. In this case, we will be able to quantize f

in an unambiguous way. Since

LXfJ =

d

dt

∣∣∣0e−tXf∗ JetXf e

tXf∗ ,

where etXf is the flow of Xf , it follows that

JetXf (m) = etXf∗ Jme

−tXf∗ ∈ End(TetXf (m)M

). (2.1)

So Equation (2.1) shows that J is determined on each point of the orbits of etXf once it is

known at one point. Thus if we can find a slice (S) to the orbits Xf , then J is determined

by its values on S.

Remark 2.7 In general, Equation (2.1) does not extend to give us an unambiguously defined

J on M . For example, if the orbits of etXf are periodic, then there will be several values of t

such that etXf (xo) = x, where x0 is the point in S which is in the same orbit as x. However,

JetXf (xo) may be different for these values of t. To get an unambiguously defined J , the

value of J at x0 must commute with eTXf∗ , where T is the period of the orbit of etXf (x0).

5

Proposition 2.8 Suppose that S is a slice to the orbits Xf and J is given on S such that

J2 = −1, J∗ω = ω on S, and ω(v, Jv) > 0 for all v ∈ TmM \ 0 and m ∈ S. Then J

defined on each orbit of etXf by Equation (2.1) has these properties for all m ∈ S.

Proof. By Equation (2.1), for m ∈ S and v, w ∈ TetXf (m)M,

J2etXf (m)

=(etXf∗ Jme

−tXf∗)2

= etXf∗ J2

me−tXf∗ = e

tXf∗ (−1) e−tXf∗ = −1,

ω(JetXf (m)v, JetXf (m)w) = ω(etXf∗ Jme−tXf∗ v, e

tXf∗ Jme−tXf∗ w)

= ω(Jme−tXf∗ v, Jme

−tXf∗ w) = ω(e−tXf∗ v, e−tXf∗ w)

= ω(v, w),

and

ω(v, JetXf (m)v) = ω(v, etXf∗ Jme

−tXf∗ v) = ω(e−tXf∗ v, Jme−tXf∗ v) > 0,

where we have made repeated use of the fact that(etXf

)∗ω = ω for all t. This is because

LXfω = dω(Xf , ·) + iXf

dω = −ddf = 0

and hence

0 = LXfω =

d

dt

∣∣∣0

(etXf

)∗ω.

Definition 2.9 Let (M,ω, J) be a Kahler manifold. A real valued function φ is called a

Kahler potential if

i∂∂φ = ω.

Theorem 2.10 Suppose that (M, ω, J) is a Kahler manifold and that there exists a globally

defined Kahler potential φ. Then

θ = − Im ∂φ = −(∂φ− ∂φ

)

2i(2.2)

is a symplectic potential. That is ω = dθ.

6

Proof. The proof is the following computation:

dθ = (∂ + ∂)θ = − (∂ + ∂)(∂φ− ∂φ

)

2i

= −(∂∂φ− ∂∂φ

)

2i= i∂∂φ = ω.

Theorem 2.11 Suppose that (M, ω, J) is a Kahler manifold and that there exists a globally

defined Kahler potential φ. Let θ be as in Equation (2.2) and B = M × C be the Hermitian

line bundle with covariant derivative ∇ = d + 1i~θ where θ is described as above. Then the

Quantum Hilbert space H may be described as

H = ke−φ/2~ ∈ K : k ∈ H(M) ∼= HL2(M, e−φ/~ωn).

Let f ∈ C∞(M), then for any k ∈ H(M),

Q(H)k := eφ/2~Q(f)(ke−φ/2~

)= −i~Xfk + (f + iZfφ) k

= −i~Zfk + (f + iZfφ) k, (2.3)

where Zf = 12 (Xf − iJXf ) . If LXf

J = 0, then Zf is a holomorphic vector field and

(f + iZfφ) is a holomorphic function.

Proof. The main point is to show that ∇Ze−φ/2~ = 0 for all Z ∈ T (1,0)M. This is easily

checked using

∇Ze−φ/2~ = (Z +1i~

θ(Z))e−φ/2~,

Ze−φ/2~ = − Zφ

2~e−φ/2~,

and from Equation (2.2),

1i~

θ(Z) = − 1i~

Im ∂φ(Z) = − 1i~

(∂φ− ∂φ

)(Z)

2i=

12~

Zφ.

This proves the theorem because of the product rule for ∇.

In more detail, if k ∈ H(M), then

∇Z

(ke−φ/2~

)= Zk · e−φ/2~ + k∇Ze−φ/2~ = 0 + 0 = 0.

7

Conversely if ke−φ/2~ ∈ H, then

0 = ∇Z

(ke−φ/2~

)= Zk · e−φ/2~ + k∇Ze−φ/2~ = Zk · e−φ/2~

which implies that Zk = 0 for all Z ∈ T (1,0)M, i.e. that k ∈ H(M). We may write

Xf = Zf + Zf . Because ∇Zf

(ke−φ/2~) = 0 and

θ(Zf ) = −(∂φ− ∂φ

)(Zf )

2i=

Zfφ

2i,

we find that

eφ/2~Q(f)(ke−φ/2~

)= eφ/2~ (−i~∇Xf

+ f) (

ke−φ/2~)

= eφ/2~ (−i~∇Zf+ f

) (ke−φ/2~

)

= −i~Zfk + fk − i~keφ/2~∇Zfe−φ/2~

= −i~Zfk + fk − i~keφ/2~(

Zf +1i~

θ(Zf ))

e−φ/2~

= −i~Zfk +(

f − Zfφ

2i+ i~

Zfφ

2~

)k

= −i~Zfk + (f + iZfφ) k.

Let zj be a set of local holomorphic coordinates on M . Now suppose that LXfJ = 0

and write X for Xf , Z for Zf , ∂j for ∂∂zj and Z =

∑j cj∂j . It is well known that LXJ = 0

is equivalent to requiring the coefficients cj to be holomorphic, i.e. that Z is a holomorphic

vector field. (See Appendix.) To show that (f + iZfφ) is holomorphic it suffices to prove

that W (f + iZfφ) = 0 for all type T (1,0)M vector fields W. Recall that ω = i∂∂φ and

∂∂φ(W , Z) = d∂φ(W , Z) = W ∂φ(Z)− Z∂φ(W )− ∂φ([W , Z])

= −ZWφ− [W , Z]φ = −WZφ,

wherein we have used LZJ = 0 to conclude that [W , Z] is still type T (0,1)M. (See the last

paragraph in the proof of Theorem 2.6.) Using this result,

W (f + iZfφ) = Wf + iWZfφ = Wf − i∂∂φ(W , Zf )

= df(W )− ω(W , Zf ) = df(W ) + ω(Zf , W ) = 0.

The last line follows from Equation (1.2).

8

The previous theorem tells that our Quantum Hilbert space H (the space of L2 complex

polarized sections) may be identified with the space of holomorphic functions, L2 integrable

with respect to the form e−φ/~ωn, denoted by HL2(M, e−φ/~ωn). The corresponding Hamil-

tonian operator Q(H) is given by Equation (2.3). Note that this operator Q(f) is unitarily

equivalent to Q(f).

3 Standard Kahler Polarization of the Harmonic Oscil-

lator in R2

We can now apply the theory developed in the previous section to the simple harmonic

oscillator. Let M = T ∗R ∼= R × R with the usual symplectic form ω = dp ∧ dx. Note

that a 2-dimensional symplectic vector space with any compatible almost complex structure

(satisfying (1) and (2) in Definition 2.1) is always a Kahler manifold. This is because on R2,

any almost complex structure J is integrable, and thus by Newlander-Nirenberg Theorem ,

(R2, J) is a complex manifold. (See [Newlander and Nirenberg(1957)].) Let H ∈ C∞(R2)

be the Hamiltonian of the harmonic oscillator, H given by H(x, p) = 12p2 + 1

2x2.

The Hamiltonian vector field of H is

XH = p∂x − x∂p,

where ∂x := ∂∂x and ∂p := ∂

∂p . Hamilton’s equations of motion are

x(t) = Hp(x, p) = p(t),

p(t) = −Hx(x, p) = −x(t)

or equivalently

d

dt

x(t)

p(t)

= Ω

x(t)

p(t)

where

Ω :=

0 1

−1 0

. (3.1)

9

Newton’s equations of motion are

x(t) + x(t) = 0

from which we easily derive

etΩ =

cos t sin t

− sin t cos t

. (3.2)

After identifying TmM with M, Equation (2.1) becomes

JetΩm = etΩJme−tΩ. (3.3)

Hence we seek to find a Jm on the slice of orbits of etXH which has the properties in

Proposition 2.8.

Proposition 3.1 Suppose ω is the standard symplectic form on R2 and J be a 2×2 matrix

such that J2 = −I and J∗ω = ω1. Then J is of the form

J =

a b

− (1 + a2

)b−1 −a

(3.4)

and J is positive, i.e. and ω(Jv, v) > 0 for all v 6= 0, iff b > 0. If we replace b by(1 + a2

)1/2b

in this last expression we arrive at an alternate form for J, namely

J =

a(1 + a2

)1/2b

− (1 + a2

) 12 b−1 −a

.

Proof. Let J0 =

0 1

−1 0

and J =

a b

c d

so that ω (v, w) := v · J0v The condition

J∗ω = ω implies

Jv · J0Jw = v · J0w for all v, w ∈ R2

which implies that J trJ0J = J0 where

J trJ0J =

a c

b d

0 1

−1 0

a b

c d

=

0 ad− bc

−ad + bc 0

.

1Notice that this condition implies that ω(Jv, w) is a symmetric form. Indeed,

ω(Jv, w) = −ω(Jv, JJw) = −ω(v, Jw) = ω(Jw, v).

10

So this condition is equivalent to det J = 1 which is also clear from the exterior algebra

construction of the determinant.

The condition that J2 = −1 becomes

a b

c d

a b

c d

=

bc + a2 ab + bd

ac + cd bc + d2

=

−1

0

0

−1

and hence that

b (a + d) = 0 = c (a + d) and bc + a2 = −1 = bc + d2.

It follows from these equations that a2 = d2 or a = εd with ε ∈ ±1 . If ε = 1 we then have

ba = 0 = ca

in which case either a = 0 and hence c = −1/b and hence

J =

0 b

−b−1 0

or a 6= 0 and hence b = c = 0 and in which case bc + a2 = −1 does not have a solution. If

ε = −1, then we must still satisfy bc + a2 = −1 or bc = −1− a2 or c = − 1+a2

b , i.e.

J =

a b

− (1 + a2

)b−1 −a

.

Finally we have

ω(Jv, v) = Jv · J0v = −J0Jv · v

and

−J0J = −

0 1

−1 0

a b

− (1 + a2

)b−1 −a

=

1b

(a2 + 1

)a

a b

which is positive if b > 0 since det (−J0J) = 1.

Let us consider the case when a = 0, b = 1. Then Equation (3.3) reduces to the standard

almost complex structure, Jstd :=

0 1

−1 0

. Hence ξ(x, p) = x − ip = z is a global

11

holomorphic chart ξ : R2 → C such that ∂z = 12 (∂x + i∂p) is a T (1,0)R2 vector field. Let the

Kahler potential φ = H = 12 (p2 + x2) = 1

2 |z|2, which follows from

i∂∂12(x2 + p2) =

i

8(∂2

x + ∂2p)(x2 + p2)dz ∧ dz

=i

2(dx− idp) ∧ (dx + idp)

= dp ∧ dx.

Note that the Kahler potential is never unique. We can always add the real part of a

holomorphic function to it. Similarly we can also add the real part of an anti-holomorphic

function. By Theorem 2.10, ω = dθ, where θ = 12 (pdx − xdp). In this case we may take

B = M × C and ∇ = d + 1i~θ. Then the covariant derivative ∇ is metric compatible and

Curvature(∇) = 1i~dθ. Since R2 is simply connected, this Hermitian line bundle with this

covariant derivative (B,∇) is unique.

By Theorem 2.11, ZH is a holomorphic vector field. With XH = p∂x−x∂p and z = x−ip,

ZH =12(XH − iJXH) =

12

(p∂x − x∂p + i(x∂x + p∂p)) = i(x− ip)12(∂x + i∂p) = iz∂z.

Our Kahler potential φ thus satisfies

H + iZHφ =12zz − z

12∂zzz = 0.

Therefore using Theorem 2.11, we may identify our Quantum Hilbert space, via the map

ξ : (x, p) → z = x−ip, with Hstd = HL2(C, e−(x2+p2)/~dp∧dx). Under this ’ξ’ identification,

our quantized Hamiltonian Qstd(H) is thus given by

Qstd(H) = −i~ · iz∂z = ~z∂z.

It is then a straight forward calculation to show that zn∞n=0 are the eigenfunctions of

Qstd(H). In fact, one can check that they are orthogonal and span Hstd. Let us summarize

these facts in a theorem.

Theorem 3.2 Let (R2, ω, Jstd) be our Kahler manifold with ω = dp ∧ dx and

Jstd =

0 1

−1 0

.

12

Let H = 12 (x2 + p2) be the Hamiltonian. Under the map ξ : (x, p) → z = x − ip, our

Quantum Hilbert space is identified with

Hstd = HL2(C, e−(x2+p2)/~dp ∧ dx).

The corresponding quantized Hamiltonian Qstd(H) is given by

Qstd(H) = ~z∂

∂z

and zn : n = 0, 1, 2, . . . are the eigenfunctions of Qstd(H) = ~z ∂∂z . Furthermore, they

form an orthogonal basis for Hstd.

4 A Non-standard Kahler Polarization

We now want to consider the quantization of the harmonic oscillator, using a different almost

complex structure J , given in Proposition 3.1. Let r2 = x2 + p2. For r ≥ 0, let a(r) and

b(r) be given smooth functions with some restrictions near 0. That is, a(r) → 0, b(r) → 1

as r → 0.

Now the orbits of etΩ are circles, from Equation (3.2). Choose the slice of orbits of etΩ

to be the positive real line. By Proposition 3.1, we should let

J(r,0) := J(x,p)

∣∣∣x=r,p=0

=

a(r) b(r)

− 1b(r)

(1 + a2(r)

) −a(r)

. (4.1)

Then use Equation (3.3) to define J on each orbit of etΩ. By Proposition 2.8, J defined in

this way is compatible with the standard symplectic form ω.

It is clear from Equation (3.2) that ’t’ takes values [0, 2π]. The period of each orbit of etΩ

is 2π and e2πΩ is just the identity map from R2 to R2. Hence we can use Equation (3.3) and

define J unambiguously on all of R2. Thus Equation (3.3) defines a J on all of R2 which is

compatible with the standard symplectic form. (See also Remark 2.7.) The assumptions on

a and b will guarantee that J is continuous on R2, but it may not be smooth at the origin.

We make an additional assumption that J is smooth at the origin.

13

Let

V =(

b , i− a

)e−tΩ

∂x

∂p

and

V =(

b ,−i− a

)e−tΩ

∂x

∂p

.

Note that here, ’t’ is not the usual angular variable, in the sense that etΩ is rotation in the

clockwise direction, as opposed to the usual angular variable in polar coordinates, where

positive means going in the anti-clockwise direction. It is easily checked that JV = iV and

JV = −iV .

As remarked earlier, (R2, ω, J) with a compatible almost complex structure J is always

a Kahler manifold. By the Uniformization Theorem (See [Farkas and Kra(1992)]), since R2

is simply connected and non-compact, (R2, ω, J) is conformally equivalent to either C or the

complex unit disc D. Let ζ be such a map ζ : R2 −→ C or D. We will write down a formula

for ζ in section 4.1. But for the time being, we don’t need the explicit formula. We will

assume that ζ(0) = 0. Then there exists a non-zero function ψ such that

ζ−1∗

∂

∂w= ψV,

ζ−1∗

∂

∂w= ψV

where w is a linear holomorphic coordinate in C or D.

We will continue to label points on R2 with (x, p). The uniformization map ζ maps

(x, p) ∈ R2 7→ w = ζ(x, p) ∈ C or D.

Notation 4.1 We will in future write ∂∂w as ∂w. Similarly write ∂

∂w as ∂w. In linear

coordinates, we will write w = α − iβ. This is to be consistent with our z = x − ip.

Therefore,

∂w =12(∂α + i∂β), ∂w =

12(∂α − i∂β),

where we write

∂α :=∂

∂α, ∂β :=

∂

∂β.

14

4.1 Uniformization Map

By definition,

J(r,t) := JetΩ(r,0) = etΩJ(r,0)e−tΩ.

It is convenient to use this chart (r, t), where x = r cos t, p = −r sin t. Hence we can write

∂x and ∂p as

∂x = cos t∂r − sin t

r∂t, ∂p = − sin t∂r − cos t

r∂t

where ∂r := ∂∂r and ∂t := ∂

∂t . As a reminder, this ’t’ is rotation in the clockwise direction.

Thus in this chart (r, t),

V = etΩ

b

−i− a

·

∂x

∂p

=(

b −i− a

)

cos t − sin t

sin t cos t

cos t − sin tr

− sin t − cos tr

∂r

∂t

= b∂r +i + a

r∂t.

Now, we seek holomorphic functions f(r, t) such that

V f = 0

or (b∂r +

i + a

r∂t

)f(r, t) = 0. (4.2)

Lemma 4.2 Let iXH = i(p∂x − x∂p). Suppose f ∈ C∞(R2,C) such that −iXHf = λf ,

λ ∈ R \ 0. Then

f(r, t) = f(r, 0)eiλt.

Proof. Note that

−iXHf(r, t) = −i∂

∂s

∣∣∣s=0

f(r, t + s)

= λf(r, t).

This is the same as

−i∂

∂tf(r, t) = λf(r, t),

15

or∂

∂tf(r, t) = iλf(r, t).

So

f(r, t) = f(r, 0)eiλt.

With this lemma, we can choose f to be a product of a radial function and a function

dependent only on the angular variable. Let us now consider a function of the form f(r)eit,

which is an eigenfunction of −iXH with eigenvalue 1. Note that f is only dependent on the

radial distance r and that f might take complex values. If f(r)eit is holomorphic, it must

satisfy the Cauchy-Riemann Equations. The Cauchy-Riemann Equations (4.2) are

b(r)f ′(r)eit + ii + a(r)

rf(r)eit = 0

or

f ′(r) =1− ia(r)

rb(r)f(r). (4.3)

This differential equation is of separable type and the solution is given by

f(r) = f(1)ec(r), r ∈ (0,∞), (4.4)

where c(r) =∫ r

11−ia(r)

rb(r) dr. Since we are looking for non-trivial solutions, there is no loss of

generality in assuming f(1) = 1. However, since a, b are smooth and a(r) → 0, b(r) → 1 as

r → 0, f(r) → 0 as r → 0. Hence, we can extend the solution of Equation (4.3) to [0,∞),

given by

f(r) =

ec(r) r ∈ (0,∞);

0, r = 0.(4.5)

On closer examination of the second part of the formula, the magnitude of f , |f | is given

by exp(∫ r

11/rb(r) dr) and the ’angular’ part is given by exp(− ∫ r

1a(r)/(rb(r)) dr). Thus

one might think of b as controlling the rate at which |f | is increasing and a is controlling

the ’angular’ part. Also, the smaller b is, the greater the increase in the |f |.

It is not difficult to see that f maps the positive x-axis in R2 to a line in C which does not

cross itself. Since b > 0, |f | is always increasing and continuous and thus 0 ≤ r 7→ |f(r)| = ρ

is a one-to-one and onto map on [0,∞). In other words, the function f(·) maps r to one

16

and only one point on the circle with radius ρ = |f(r)|. Therefore, the map r 7→ f(r)

traces a path l in the complex plane. When we apply eit to f , we are in fact rotating this

path l in the counterclockwise direction and after a revolution of 2π, we return back to the

original path l. Thus f(r)eit is a one to one and onto map from R2 to C and furthermore,

is holomorphic, i.e. it satisfies the Cauchy-Riemann Equation (4.2). Henceforth, we can

choose our uniformization map ζ : R2 → C to be

ζ(r, t) = f(r)eit

where f(r) is defined in Equation (4.5).

4.2 Formula for Kahler potential

By definition of Kahler potential, we want to find a function φ such that

i∂w∂w(φ ζ−1 ξ)dw ∧ dw = (ζ−1 ξ)∗ω.

Note that w = α− iβ = ξ(α, β) in linear coordinates. Now,

∂w∂w =12(∂α + i∂β) · 1

2(∂α − i∂β) =

14∆

where ∆ = ∂2α + ∂2

β . Define φ := φ ζ−1 ξ. Hence

i(∂w∂wφ) dw ∧ dw = i∆φ

4dw ∧ dw = (ζ−1 ξ)∗(dp ∧ dx) =

∣∣∣∣∣∣∣pβ xβ

pα xα

∣∣∣∣∣∣∣dβ ∧ dα,

where x := x ζ−1 ξ and p := p ζ−1 ξ. One notices that the right hand side is just the

Jacobian matrix

Jβ,α(p, x) =

∣∣∣∣∣∣∣pβ xβ

pα xα

∣∣∣∣∣∣∣of ζ−1 ξ. But

idw ∧ dw = 2dβ ∧ dα.

Thus∆φ

4· 2dβ ∧ dα = Jβ,α(p, x)dβ ∧ dα.

Hence φ must satisfy the following partial differential equation

12∆φ = Jβ,α(p, x).

17

Let us change to coordinates (r, t) and (ρ, θ) in R2. To be consistent, ρ is the usual

radial coordinate and θ is measuring rotation in the clockwise direction. Now under the

uniformization map ζ : (r, t) → χ(r)eiη(r)+it, with

χ(r) = |ζ(r)| = |f(r)|

and

η(r) =∫ r

1

−ia(r)rb(r)

dr.

Note that χ is an increasing positive valued function, dependent on r only. Using ξ−1

to identify C with R2, as a map from R2 to R2, (r, t) 7→ (χ(r), η(r) + t). Let µ(ρ, θ) :=

θ − η χ−1(ρ). Inverting this map, ζ−1 ξ : (ρ, θ) → (χ−1(ρ), µ(ρ, θ)). So,

x(ρ, θ) = χ−1(ρ) cos µ(ρ, θ),

p(ρ, θ) = −χ−1(ρ) sin µ(ρ, θ).

In coordinates (ρ, θ),

∂α = cos θ∂ρ − sin θ

ρ∂θ, ∂β = − sin θ∂ρ − cos θ

ρ∂θ,

where ∂ρ := ∂∂ρ and ∂θ := ∂

∂θ . By a direct computation, we get

pβ = sin θ sin µdχ−1

dρ− χ−1 sin θ cosµ(η χ−1)ρ + cos θ

cosµ

ρχ−1,

xα = cos θ cos µdχ−1

dρ+ χ−1 cos θ sinµ(η χ−1)ρ + sin θ

sinµ

ρχ−1,

pα = − cos θ sin µdχ−1

dρ+ χ−1 cos θ cosµ(η χ−1)ρ + sin θ

cos µ

ρχ−1,

xβ = − sin θ cos µdχ−1

dρ− χ−1 sin θ sin µ(η χ−1)ρ + cos θ

sinµ

ρχ−1.

Finally, after some manipulation, we get∣∣∣∣∣∣∣pβ xβ

pα xα

∣∣∣∣∣∣∣=

χ−1

ρ

dχ−1

dρ.

Converting the Laplacian in terms of coordinates (ρ, θ), we now have to solve

12∆φ =

12

(∂2

∂ρ2+

1ρ

∂

∂ρ+

1ρ2

∂2

∂θ2

)φ =

χ−1

ρ

dχ−1

dρ. (4.6)

To motivate the solution of this partial differential equation, let us look at iZHφ+H. By

Theorem 2.11, iZHφ + H is holomorphic. Since we can add a real part of any holomorphic

18

function to the Kahler potential, we will assume that iZHφ + H is 0. Now for the moment,

let us further assume that φ is a radial function, independent of the angular coordinate. It

is a reasonable assumption since the RHS of Equation (4.6) is a radial function.

Now −iζ∗ZH is a vector field on C. Therefore,

−iζ∗ZHw = −iZH(w ζ) = −iXHζ = ζ.

The second equality follows from XH = ZH + ZH and ζ is holomorphic. Hence, the push

forward of −iZH under ζ is w∂w. Since iζ∗ZH(φ ξ−1) + H ζ−1 = 0, thus,

−w∂w(φ ξ−1) +12|χ−1|2 ξ−1 = 0,

w∂

∂w(φ ξ−1) =

12|χ−1|2 ξ−1.

Now in cartesian coordinates, w = α − iβ and ∂w = (∂α + i∂β)/2. Note that α = ρ cos θ,

β = −ρ sin θ, hence α− iβ = ρeiθ. Therefore,

(α− iβ)12(∂α + iβ)φ =

12|χ−1|2,

12ρeiθ

[cos θ∂ρ − sin θ

ρ∂θ + i

(− sin θ∂ρ − cos θ

ρ

)∂θ

]φ =

12|χ−1(ρ)|2,

ρeiθe−iθ d

dρφ = |χ−1|2,

ρdφ

dρ= |χ−1|2.

Solving for this differential equation, we get the following formula for φ,

φ(ρ) =∫ ρ

0

|χ−1|2(ρ)ρ

dρ. (4.7)

So now we have a likely candidate for φ, under the assumption that it is a radial function.

We now have to verify that it indeed is a Kahler potential, using Equation (4.6). Using the

formula as in Equation (4.7),

∂2

∂ρ2φ =

∂

∂ρ

( |χ−1|2ρ

)

=2ρ|χ−1| |dχ−1

dρ− 1

ρ2|χ−1|2, (4.8)

and1ρ

∂

∂ρφ =

|χ−1|2ρ2

. (4.9)

19

Adding up Equations (4.8) and (4.9),

12∆φ =

12

[∂2

∂ρ2φ +

1ρ

∂

∂ρφ

]=

χ−1

ρ

dχ−1

dρ.

This verifies that φ defined by Equation (4.7) is indeed a Kahler potential. Since r 7→ χ(r) =

ρ, we have the following theorem.

Theorem 4.3 The Kahler potential φ is a radial function on R2 and is given by

φ(r) =∫ χ(r)

0

|χ−1|2(ρ)ρ

dρ. (4.10)

With this choice, iZHφ + H = 0.

4.3 On the Unitary Equivalence

Define Q(H) as in Equation (1.1) with (B,∇) and Q(H) as described in Theorem 2.11. We

will show that (HL2(R2, e−φ/~ω), Q(f)) is unitarily equivalent to (Hstd, Qstd(H)).

Proposition 4.4 The holomorphic eigenfunctions of −iXH are given by f(r)neint for n =

0, 1, 2, . . ., where f is a radial function that solves

f ′(r) =(1− ia(r))

b(r)rf(r).

Proof. By Lemma 4.2, we know that all eigenfunctions of −iXH are in the form g(r)eiλt

for some constant λ and radial function g. However, since t is the angular coordinate with

values running from 0 to 2π and we are looking at holomorphic functions, λ has to be an

integer. The Cauchy-Riemann equation for an eigenfunction fn with integer eigenvalue n is

given by

f ′n(r) =n(1− ia(r))

b(r)rfn(r).

The only solution is given by fn, where f = f1. Now f(r) → 0 as r → 0, and thus for

negative values of n, fn(r) has a singularity point at the origin. Hence, n has to be non-

negative and the only possible holomorphic eigenfunctions are (f(r)eit)n for n = 0, 1, 2, . . ..

Corollary 4.5 The only holomorphic eigenfunctions of Q(f) are given by ζn : n = 0, 1, 2, . . ..

20

Proof. By definition,

ZH =12(XH − iJXH)

and hence

XH = ZH + ZH .

Thus, for any holomorphic function k ∈ H(C),

XH(k ζ) = ZH(k ζ).

Hence the only holomorphic eigenfunctions of −iZH are given by ζn : n = 0, 1, 2, . . ..With the choice of φ, iZHφ + H = 0 and therefore,

Q(H)ζn = −iZHζn = −iXHζn = ~nζn.

Theorem 4.6 Suppose for constants m, M , 0 < m ≤ b ≤ M . Then (HL2(R2, e−φ/~ω), Q(H))

is unitarily equivalent to (Hstd, Qstd(H)).

Proof. We will first show that ζn are L2 integrable with respect to e−φ/~ω for n =

0, 1, 2, . . .. Recall χ(r) = |ζ(r)| = f(r). Note that

dχ(r)dr

=1

rb(r)e∫ r1

1rb(r) dr.

By the substitution rule, ρ = χ(r) and thus dρ = χ′dr or dr = dρ/χ′. For n ≥ 0, we want

to compute

∫

R2|ζ|2ne−φ/~ ω = 2π

∫

R+χ2n(r)e−φ(r)/~rdr

= 2π

∫

R+ρ2n(χ−1(ρ))2e−φ(ρ)/~b(χ−1(ρ))e−

∫ χ−1(ρ)1

1rb(r) drdρ.

Since b is bounded below by m and bounded above by M , then

e∫ r1

1Mr dr ≤ χ(r) = e

∫ r1

1rb(r) dr ≤ e

∫ r1

1mr dr

and after integrating, we have

r1/M ≤ χ(r) ≤ r1/m.

21

This inequality says that χ is growing at least the 1/M -th power of r but at most 1/m-th

power of r. Thus when we take the inverse of χ−1, we have

ρm ≤ χ−1(ρ) ≤ ρM .

Using these bounds on χ−1, we have

φ(ρ) ≥∫ ρ

0

ρ2m

ρdρ ≥ 1

2mρ2m,

and ∫ χ−1(ρ)

1

1rb(r)

dr ≥∫ χ−1(ρ)

1

1Mr

dr =1M

log χ−1(ρ).

Thus

∫

R2|ζ|2ne−φ/~ ω ≤ 2π

∫

R+ρ2n(χ−1(ρ))2e−φ(ρ)/~M(χ−1(ρ))−1/Mdρ

≤ 2Mπ

∫

R+ρ2nρM(2−1/M)e−ρ2m/2m~dρ

= 2Mπ

∫

R+ρ2n+2M−1e−ρ2m/2m~dρ < ∞

for any n. Hence ζn, n = 0, 1, 2, . . . are L2-integrable with respect to e−φ/~ω.

Next, we would like to show that this set forms a basis in HL2(R2, e−φ/~ω). Recall that

HL2(R2, e−φ/~ω) is the space of holomorphic functions which are L2 integrable with respect

to e−φ/~ω. It suffices to show that if g ∈ HL2(R2, e−φ/~ω) and∫R2 gζme−φ/~ω = 0 for any

m, then g ≡ 0. However, any holomorphic function g ∈ H(R2) can be written as a Taylor

series g =∑∞

n=0 anζn (pointwise).

Let B(σ) be a disc centered at the orgin with radius σ. In coordinates (r, t), ζ(r, t) =

f(r)eit and hence

∫

B(σ)

gζme−φ/~ω =∞∑

n=0

an

∫ π

−π

ei(n−m)t dt

∫ σ

0

fn+m(r)e−φ(r)/~ rdr.

Note that if m 6= n,∫ π

−πei(n−m)t dt = 0 and thus if we let

pn := 2π

∫ ∞

0

f2n(r)e−φ(r)/~ rdr, γn(σ) :=2π

pn

∫ σ

0

f2n(r)e−φ(r)/~ rdr,

then ∫

B(σ)

gζme−φ/~ω = amγm(σ)pm.

22

Observe that 0 ≤ γn(σ) ≤ 1 and limσ→∞ γn(σ) = 1. Thus

0 =∫

R2gζme−φ/~ω = lim

σ→∞

∫

B(σ)

gζme−φ/~ω

= ampm limσ→∞

γm(σ) = ampm,

for any m. Therefore, am = 0 for all m and hence g ≡ 0. This shows that ζn : n =

0, 1, 2, . . . forms a basis for HL2(R2, e−φ/~ω). A similar argument will also show that ζn is

orthogonal to ζm if m 6= n. By the definition of pn, ζn/pn∞n=0 hence forms an orthonormal

basis and are normalized eigenfunctions of Q(H).

Now let qn∞n=0 be a sequence of positive numbers such that zn/qn are normalized

eigenfunctions of Qstd(H). Note that zn/qn∞n=0 is an orthonormal basis for Hstd. The

linear map Ψ : HL2(R2, e−φ/~ω) → Hstd sends for n = 0, 1, 2, . . .,

Ψ(

1pn

ζn

)=

1qn

zn.

Then Ψ is a unitary map and Qstd(H)Ψ = ΨQ(H).

Remark 4.7 We can impose growth conditions on b in the statement of Theorem 4.6 instead

of assuming boundedness to show that ζn is integrable.

5 Discussion

It is indeed surprising that 2 different Kahler polarizations yield unitarily equivalent quan-

tum spaces. Is there a reason why this is true for the harmonic oscillator?

I believe that the main reason is because of the rotational invariance of the classical

Hamiltonian. Because of the rotational invariance, the Hamiltonian vector field XH is just

∂/∂t where t is angular variable measured clockwise. Hence the flow of the Hamiltonian

vector field is just the circles. Given −iXH , each eigenfunction is of the form g(r)eiλt, where

(r, t) are coordinates on R2. For any λ ∈ R, its multiplicity will be infinite.

We chose a non-standard complex structure J on R2 such that it is ’constant’ along the

flow. Kahler polarization involves restricting ourselves to an invariant subspace of the pre-

quantum line bundle. The positivity of Kahler polarization somehow restricts the values of

23

the eigenvalues to be positive integers and the eigenspace corresponding to each non-negative

integer has multiplicity one.

The eigenfunctions of −iXH with eigenvalue 1 are of the form ζ(r, t) = g(r)eit. There

are many choices of g, which give this same eigenvalue. Each J will pick a particular choice

of g and we chose such a ζ to be our uniformization map. As such, the quantum space for

the harmonic oscillator is determined solely by ζ, since the quantum space is spanned by

ζn∞n=0. In this sense, each J we pick determines ζ, which in turn determines the quantum

space. Furthermore, because ζ(r, t) = g(r)eit, and given a rotational invariant measure,

the set ζn∞n=0 forms an orthonormal basis (suitably normalized). By mapping ζ for a

particular choice of J to ζ of another choice J gives us the unitarily equivalence of the 2

respective quantum spaces. Of course, we need to verify these statements by calculations

but hopefully this gives some insight as to why the harmonic oscillator gives such special

results.

6 Acknowledgements

The author would like to thank Professor Bruce Driver, Professor Len Gross and Professor

Brian Hall for the invaluable discussions. Parts of this article were taken from Professor

Driver’s personal notes.

A Almost Complex Structure

Let M be a complex manifold with complex dimension n and z = x + iy be a holomorphic

coordinate chart. Then

∂zk =12(∂xk − i∂yk), ∂zk =

12(∂xk + i∂yk)

and thus

∂xk = ∂zk + ∂zk , ∂yk = −i(∂zk − ∂zk).

Let TM be the tangent space to M as a real manifold. Then there exists a unique complex

structure J on TM with

J∂zk = i∂zk , J∂zk = −i∂zk .

24

Thus

J∂xk = −∂yk , J∂yk = ∂xk .

Definition A.1 Let M be a complex manifold. A smooth section Z ∈ Γ∞(TMC) is said to

be a holomorphic vector field if Z ∈ Γ∞(TM (1,0)) and Zf is holomorphic on any open set

U ⊂ M where f is holomorphic.

Theorem A.2 Assume that M is a complex manifold with the induced almost complex

structure J . Then the following are equivalent.

1. LXJ ≡ 0.

2. [X, JY ] = J [X, Y ] for all smooth sections Y ∈ Γ∞(TM).

3. (X − iJX) is a holomorphic vector field.

Proof. Let Y ∈ Γ∞(TM), then

[X, JY ] = LX(JY ) = (LXJ)Y + JLXY = (LXJ)Y + J [X,Y ],

and hence

(LXJ)Y = J [X,Y ]− [X, JY ]. (A.1)

This implies the equivalence of (1) and (2).

Write X = a∂x + b∂y and Y = u∂x + v∂y, so that JY = −v∂x + u∂y. Any W ∈ TM (1,0)

can be written as W = 12 (X − iJX). If f is holomorphic, then Wf = 0, i.e. iJXf = Xf or

−iJXf = Xf . Thus

Wz =12(Xz − iJXz) =

12(Xz + Xz) = Xz = a + ib.

Hence12(X − iJX) = F∂z

where F := a + ib.

25

Let us compute (LXJ)Y using Equation (A.1).

(LXJ)Y = JXu∂x + Xv∂y − (Y a∂x + Y b∂y) − −Xv∂x + Xu∂y − (JY a∂x + JY b∂y)

= −Xv∂x + Xu∂y − (−Y b∂x + Y a∂y) − −Xv∂x + Xu∂y − (JY a∂x + JY b∂y)

= (JY a + Y b)∂x + (−Y a + JY b)∂y

= (JY a + Y b) + i(−Y a + JY b)∂z + (JY a + Y b)− i(−Y a + JY b)∂z

= (JY − iY )F∂z + (JY + iY )F ∂z

= (JY − iY )F∂z + (JY − iY )F∂z.

Thus, LXJ = 0 iff (JY − iY )F = 0 iff F is holomorphic. This shows the equivalence of (2)

and (3).

References

[Farkas and Kra(1992)] H. M. Farkas and I. Kra, Riemann surfaces, volume 71 of Graduate

Texts in Mathematics, Springer-Verlag, New York, second edition, 1992.

[Folland(1995)] Gerald B. Folland, Introduction to partial differential equations, Princeton

University Press, Princeton, NJ, second edition, 1995.

[Fritzsche and Grauert(2002)] Klaus Fritzsche and Hans Grauert, From holomorphic func-

tions to complex manifolds, volume 213 of Graduate Texts in Mathematics, Springer-

Verlag, New York, 2002.

[Newlander and Nirenberg(1957)] A. Newlander and L. Nirenberg, Complex analytic coor-

dinates in almost complex manifolds, Ann. of Math. (2), 65:391–404, 1957.

[Woodhouse(1992)] N. M. J. Woodhouse, Geometric quantization, Oxford Mathematical

Monographs. The Clarendon Press Oxford University Press, New York, second edition,

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