uantization of Constrained Systems

A Thesis Presented to The Division of Mathematics and Natural Sciences Reed College

In Partial Fulfillment of the Requirements for the Degree Bachelor of Arts

Rory M. Donovan May 2004

Approved for the Division (Physics)

Darrell F. Schroeter

Acknowledgments I had a wonderful time working on my thesis, and I’d like to thank the entire Reed physics department for making it such an enjoyable process. I would particularly like to thank my advisor, Darrell Schroeter, without whose interest in my topic and incredible pedagogical ability none of this document would have been even remotely possible. Finally, to my friends and family, and especially my loving parents: thank you for your support throughout this process.

Table of Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Constraint Formalism . . . . . . . . . . . . . . . . . . 1.1 Lagrangian Formalism . . . . . . . . . . . . . . . 1.1.1 The Usual Constraint Condition . . . . 1.1.2 The Conservative Constraint Condition 1.2 Hamiltonian Formalism . . . . . . . . . . . . . . 1.2.1 The Usual Constraint . . . . . . . . . . . 1.2.2 The Conservative Constraint . . . . . . . 1.2.3 The Dirac Bracket . . . . . . . . . . . . 1.3 Example: The Spherical Shell . . . . . . . . . . .

1

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3 4 4 7 9 9 13 16 18

2 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

3 An Example: The Torus . . . . . . . . . . 3.1 Gaussian Normal Coordinates . . . 3.2 Setting up the Schrödinger Equation 3.3 Numerical Solutions . . . . . . . . . 3.3.1 Pictures of Wave Functions .

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29 29 31 35 35

Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

A Computation of the Dirac Bracket . . . . . . . . . . . . . . . . . . . . . . .

45

B Mathematica Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53

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List of Figures 1

A Bead on a Hoop . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.1

Nice Coordinates for a Hoop . . . . . . . . . . . . . . . . . . . . . . .

3

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

G.N.C. Coordinatization . . . . . . . Toroidal Geometry . . . . . . . . . . . Effective Potential Wells . . . . . . . . p = 1.5, m = 0 Even Wave Functions p = 1.5, m = 0 Odd Wave Functions p = 1.5, m = 1 Even Wave Functions p = 1.5, m = 1 Odd Wave Functions p = 1.1, m = 0 Even Wave Functions p = 1.1, m = 0 Odd Wave Functions Wave Functions With p as a Parameter

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Abstract The goal of this thesis was to learn about the quantization of constrained systems, specifically systems of single particles constrained to curved surfaces. To find a quantum theory for a system, the usual approach is to first find an appropriate classical theory; to that end the classical theory of constrained systems was investigated. From the Lagrangian formalism for constrained particles, which is well understood and widely employed, a classical Hamiltonian theory was found. Quantizing the classical Hamiltonian to find a quantum Hamiltonian operator was attempted, and achieved with a modicum of success. This quantum Hamiltonian was then used to investigate the influence of the curvature of a surface upon a wave function of a particle constrained to that surface. An in-depth example was worked out for a particle constrained to the surface of a torus, and numerical solutions to the Schrödinger wave equation were found for certain states of the system.

Introduction Despite almost a century of literature on the subject, obtaining a workable quantum theory for even a well understood classical system is not always a straightforward endeavor. The usual approach to quantizing a system requires that you first develop a Hamiltonian theory for the classical system. With a classical Hamiltonian in hand, you then apply the canonical substitutions, and expect (hope) that you will find yourself with a quantum Hamiltonian that accurately characterizes the system you are interested in.

Figure 1: A constrained system in two dimensions, where a particle is confined to move on the one-dimensional “surface” of a hoop.

Unfortunately, in attempting to use this prescription to find a quantum theory for a constrained system, one is balked at the first step. The Lagrangian formalism is the most natural framework in which to set up a constrained classical system, so the Hamiltonian formalism is not generally encountered in such problems. To get a quantum theory, though, we need a Hamiltonian formulation of the classical system, so it must somehow be extracted or extrapolated from the Lagrangian. Dirac provides a general procedure for doing so [1], but it is a process that requires a substantial amount of work; finding a suitable Hamiltonian by Dirac’s prescription comprises a large portion of this thesis. Even with a classical Hamiltonian in hand, though, a usable quantum Hamiltonian does not emerge very readily. The problem of operator ordering rears its ugly head and introduces some ambiguity into the quantum Hamiltonian. Resolving this ambiguity

2

CHAPTER 0. INTRODUCTION

is the second major hurdle that must be cleared on the route to a workable quantum theory, and it remains in my thesis as a problem that is only partially resolved. Assuming that the operator ordering is acceptably resolved so that we arrive at a workable quantum Hamiltonian, we are still left with the problem of solving the Schrödinger equation on the surface of constraint. This task comprises the third portion of my thesis, where I resort to numerical solutions for the wave functions of a particle confined to move on the surface of a torus.

Chapter 1 Constraint Formalism The problem of finding the classical dynamics of a particle constrained to move on a surface is ideally suited to the formalism of Lagrangian mechanics. The elegant method with which to approach the problem is to free yourself of the “Cartesian tyranny” that the Newtonian formalism foists upon us [2], and choose a set of generalized coordinates that has one coordinate everywhere normal to the surface of constraint. In this manner, the normal coordinate does not appear in the Lagrangian, and is thus a constant of the motion. We are then left to solve n − 1 differential equations that involve the n − 1 “meaningful” coordinates that acquire dynamics.

q₀ q₁

Figure 1.1: Nicely chosen generalized coordinates for a confined system. While generalized coordinates are an excellent tool for getting a feel for the symmetries of a system and picking out dynamical coordinates of interest, this approach is not for us. Eventually we will want to quantize the system, and generalized coordinates are not very well suited to quantum mechanics, since a general procedure for quantizing a system in anything but Cartesian coordinates does not exist. Even if we could Legendre transform to a Hamiltonian for the system, if we’re not working in Cartesian coordinates we would be unable to extract a quantum theory from the classical one.

CHAPTER 1. CONSTRAINT FORMALISM

4

This situation is unfortunate, but not disastrous. There is another method for enforcing a constraint on the motion of a particle, that of undetermined Lagrange multipliers. Given a functional description of the surface of constraint such that the function is numerically equal to zero, the function is inserted into the Lagrangian by means of a Lagrange multiplier, and n + 1 equations of motion are taken. This somewhat unnatural method requires us to work in a coordinate space that is of greater dimensionality than the number of “meaningful” coordinates, but gives us the freedom to choose any coordinate system we want in which to work. If we “choose” to work in Cartesian coordinates, we will at least have a hope of eventually finding a quantum theory for the system.

1.1 1.1.1

Lagrangian Formalism The Usual Constraint Condition

In an n dimensional space free of any external potentials, we are given a surface of constraint defined by the equation f (x) = 0 .

(1.1)

For example, if we were working in three dimensions and wanted the constraint surface to be a spherical shell of radius R, we would have f (x) =

p x2 + y2 + z2 − R .

(1.2)

Inserting the constraint by means of a Lagrange multiplier, we get a Lagrangian for the system: ˙ , L = L(ξ, ξ) (1.3) where ξ = (ξ1 , · · · , ξn+1 ) is a generalized coordinate vector, ξ = (x1 , · · · , xn , λ) ,

(1.4)

x = (x1 , · · · , xn ) are the Cartesian coordinates of the n-dimensional physical space in which we are working, and ξ˙ is the total time derivative of the generalized coordinate ξ,

˙ . ξ˙ = (x˙ 1 , · · · , x˙ n , λ)

(1.5)

Since we assume there to be no potential, the Lagrangian for the system is ˙ = 1 mx˙ a x˙ a − λ f (x) . L(ξ, ξ) 2

(1.6)

In general, I will use the index a for the x’s, and have it run from 1 to n, and use the index i for the ξ’s, and have it run from 1 to n + 1. I am also employing a version of the summation convention, where any two indices that are repeated in a product are summed over. The Euler-Lagrange equations d dt



∂L ∂ξ˙ i

 =

∂L , ∂ξi

(1.7)

1.1. LAGRANGIAN FORMALISM

5

are the conditions that must be satisfied to ensure that the classical action of the system is minimal, and thus are the equations we need to solve to get the equations of motion. For the Cartesian coordinates we have d dt



∂L ∂x˙ a



∂L ∂x˙ a



∂L , ∂xa

(1.8)


d mx˙ a = mx¨ a . dt

(1.9)

=

the LHS of which is d dt



=

The RHS gives
∂L ∂ = − λ f (x) ∂xa ∂xa = −λ f,a = −λna .

(1.10a) (1.10b) (1.10c)

Here f,a is shorthand for the partial derivative of f (x) with respect to xa , and na , defined as equal to f,a , is the a component of n, a vector normal to the surface. In fact, if we were to pick a suitable function, n will be the unit normal vector, which is much nicer to work with, so we will assume that such a suitable functional form for f (x) is always chosen (if it wasn’t, we could just normalize n, but we might as well cleverly chose f (x) and do the normalization work at that point, rather than later on). Setting the two sides of Equation 1.8 equal gives a differential equation for the motion of the particle, mx¨ a = −λna .

(1.11)

The Euler-Lagrange equation for λ, d dt



∂L ˙ ∂λ

 =

∂L ∂λ

(1.12)

just give us back the constraint condition 0 = − f (x) .

(1.13)

To solve the differential equation for x(t), we must solve explicitly for λ, or otherwise eliminate it from the equations of motion. While, it might seem unmotivated1 , it is 1 What

we’re doing is imposing a sort of “consistency condition” on the constraint, that it and all of its time derivatives vanish. This idea will be central in the Hamiltonian formalism, and will merit further exploration therein.

CHAPTER 1. CONSTRAINT FORMALISM

6 certainly true that f (x) = 0

=⇒ =⇒ =⇒ =⇒

d2 f (x) = 0 dt2  d ∂ f (x) dxa =0 dt ∂xa dt
d na x˙ a = 0 dt ˙ b x˙ b = 0 na x¨ a + n a

c b

=⇒

na x¨ + nb,c x˙ x˙ = 0

=⇒

na x¨ a = −nb,c x˙ b x˙ c .

(1.14a) (1.14b) (1.14c) (1.14d) (1.14e) (1.14f)

It might look a little odd, but there’s nothing mysterious going on in the second to last step above, it’s just the chain rule2 applied to n˙ b . If we multiply the equation of motion Equation 1.11 through by a factor of na , we get mx¨ a na = −λna na ,

(1.15)

but na is the unit normal vector, so that na na = 1 ,

(1.16)

mx¨ a na = −λ .

(1.17)

and we have Save for a factor of m, the LHS of this equation is the same as the LHS of Equation 1.14f, and setting the two equal we obtain a solution for λ in terms of the coordinates and velocities: λ = mna,b x˙ a x˙ b . (1.18) We’re finally ready to eliminate λ from the equation of motion by substituting the above expression for λ into Equation 1.11:
m x¨ a = − mnb,c x˙ b x˙ c na .

(1.19)

The m’s cancel, and the equation of motion for the system is x¨ a = −nb,c x˙ b x˙ c na .

(1.20)

We will derive this equation by four other methods in Section 1.1.2, Section 1.2.1, Section 1.2.2 and Section 1.2.3. The equation will be used as a point of comparison, to check if the other formalisms we will work with are “trustworthy” in the sense that they yield the same equations of motion for the system. 2 That

is, n˙ b =

˙b dn dt

=

∂nb dxc ∂xc dt

= nb,c x˙ c .

1.1. LAGRANGIAN FORMALISM

1.1.2

7

The Conservative Constraint Condition

It is worth noting3 that the usual constraint condition f (x) = 0

(1.21)

is not unique in its characterization of the constraint. For instance, instead of requiring that f vanish, we could require that the (total) time derivative of f vanish: f˙ (x) = 0 .

(1.22)

Applying the chain rule clarifies the motivation for this constraint condition, which we will call the conservative constraint condition: d f (x) f˙ (x) =

dt ∂ f (x) dxa = ∂xa dt = na x˙ a .

(1.23a) (1.23b) (1.23c)

So whereas the usual constraint condition fixes the motion of the particle by rather bluntly stating that the position vector of the particle must be an element of the surface submanifold, the conservative constraint condition is somewhat more subtle in its phrasing: it states that the velocity of the particle in the direction normal to the surface4 is equal to zero. So if we supply the appropriate initial conditions, namely that the particle starts out on the surface, the conservative constraint condition will keep it there. Using the same f (x) as before, we can write down a Lagrangian that characterizes the system in terms of the new constraint condition: ˙ = 1 mx˙ a x˙ a − λ f˙ (x) L(ξ, ξ) 2 = 12 mx˙ a x˙ a − λna x˙ a ,

(1.24a) (1.24b)

and use the same basic procedure to find the equations of motion. The Euler-Lagrange equations for the Cartesian coordinates are d dt



∂L ∂x˙ a

 =

∂L , ∂xa

(1.25)

so the LHS is d dt



∂L ∂x˙ a

 =


d mx˙ a − λna dt

˙ a, = mx¨ a − λna,b x˙ b − λn 3 The

(1.26a) (1.26b)

reason it is “worth noting” the non-uniqueness of the constraint formulation will become apparent when we attempt to quantize the system, and operator ordering becomes an issue. 4 One will recall that a function’s gradient is everywhere perpendicular the function’s surfaces of constant value, such as our constraint surface.

CHAPTER 1. CONSTRAINT FORMALISM

8 and the RHS is


∂L ∂ = a − λnb x˙ b a ∂x ∂x = −λnb,a x˙ b .

By equality of cross-partials, the λ terms cancel, and we are left with ˙ a. mx¨ a = λn

(1.27a) (1.27b) (1.28)

The Euler-Lagrange equation for λ, d dt



∂L ˙ ∂λ

 =

∂L , ∂λ

(1.29)

gives us back the constraint condition: 0 = na x˙ a = f˙ (x) .

(1.30a) (1.30b)

From one point of view, there’s no percentage in actually substituting to get rid of λ˙ , because we have equations of motion that are of exactly the same form as in the case of the usual constraint (compare Equations 1.28 and 1.30 with Equations 1.11 and 1.13). There is one small difference: where we previously had a − λ we now have a λ˙ in the equations of motion. If we believed that both of the constraint conditions were equivalent, that is they both yield the same equations of motion, we could deduce that the two equations of motion, Equation 1.28 and Equation 1.11, tell us that in fact ˙ conservative = −λusual , and that all the legwork for finding the explicit (λ˙ free) equation of λ motion for the conservative constraint is already done. But as of now, you might not be convinced of this, so let’s check. We can get rid of λ˙ by taking advantage of the same trick we used to get rid of λ when we employed the “usual” constraint condition, but this time it suffices to require that only the first time derivative of the constraint vanishes: f˙ (x) = 0

=⇒

d ˙ f (x) = 0 dt
d na x˙ a = 0 dt na x¨ a + na,b x˙ a x˙ b = 0

=⇒

na x¨ a = −na,b x˙ a x˙ b .

=⇒ =⇒

(1.31a) (1.31b) (1.31c) (1.31d)

If we multiply Equation 1.28 through by na and divide through by m, again noting that na na = 1, we can set the LHS of the resultant equation equal to the RHS of 1.31d to get 1 ˙ λ = −na,b x˙ a x˙ b . (1.32) m If we solve this equation for λ˙ and substitute it into Equation 1.28, we get a λ˙ -free version of the equations of motion, x¨ a = −nb,c x˙ b x˙ c na , (1.33) exactly the same equations of motion we got with the usual constraint procedure.

1.2. HAMILTONIAN FORMALISM

1.2

9

Hamiltonian Formalism

In order to deal with constraints in quantum mechanics, we must work in the Hamiltonian picture, and use that formalism to attack the same problem. Working from the Lagrangians we wrote down in Equations 1.6 and 1.24, we’d like to Legendre transform to the canonical Hamiltonian H = pi ξ˙ i − L , (1.34) where the conjugate momenta are defined by pi =

∂L . ∂ξ˙ i

(1.35)

˙ to H(ξ, ξ˙ , p) by Legendre’s prescription, What this entails is transforming from L(ξ, ξ) and then functionally inverting the ξ˙ i ’s in terms of the ξi ’s and pi ’s: ˙ , p)) , H(ξ, p) = pi ξ˙ i (ξ, p) − L(ξ, ξ(ξ

(1.36)

to eliminate them from the Hamiltonian, but said functional inversion is by no means guaranteed to be possible. While the constraint procedure (“ordinary” or “conservative”), and the resultant Lagrangian, will dictate the specifics of the transformation, it is possible to investigate the theory for very general constraints [1, 3]. We have no real use for the full theoretical machinery produced by such a general investigation, and the notation gets cumbersome very quickly. I think it will be more illuminating to look at the two cases separately.

1.2.1

The Usual Constraint

For the usual constraint condition, we have as our Lagrangian ˙ = 1 mx˙ a x˙ a − λ f (x) , L(ξ, ξ) 2

(1.37)

and we would like to find ˙ , p)) H(ξ, p) = pi ξ˙ i (ξ, p) − L(ξ, ξ(ξ
˙ , p) − = pa x˙ a (ξ, p) + pλ λ(ξ

1 ˙ a (ξ, p) x˙ a (ξ, p) − 2 mx


λ f (x) .

(1.38a) (1.38b)

The conjugate momenta for the Cartesian coordinates are ∂L ∂x˙ a = mx˙ a ,

pa =

(1.39a) (1.39b)

and the momentum conjugate to the Lagrange multiplier is ∂L ˙ ∂λ

(1.40a)

= 0.

(1.40b)

pλ =

CHAPTER 1. CONSTRAINT FORMALISM

10

The inversion of the x˙ a ’s in terms of the positions and momenta is clearly possible, we have from Equation 1.39b simply x˙ a =

1 m

pa .

(1.41)

We can’t, however, find a functional inversion for λ˙ ; Equation 1.40b leaves us wanting in that regard. Actually, we should have been expecting this, since there’s no λ˙ anywhere in the Lagrangian! So the Legendre transform, as it stands, is no good, since it is not invertible. It turns out that the problem is that the Lagrangian is “singular,” in the sense that there are a number of Hamiltonians that will Legendre transform to the original Lagrangian, so trying to transform to the Hamiltonian will not work. A Hamiltonian that will work is H 7−→ H? = H + uφ(ξ, p) ,

(1.42)

where u is a multiplier (like λ) and φ(ξ, p) is a function equal to zero (like f (x)), inserted into the Hamiltonian with the intention of keeping track of our non-invertible relation by means of a constraint. In our case, it’s pλ and λ˙ that are posing the inversion problem, so from Equation 1.40b we formulate our constraint as φ(ξ, p) = pλ ≈ 0.

(1.43a) (1.43b)

The ≈ symbol in Equation 1.43b is used to indicate that φ is only “weakly equal” to zero. That is, it should be included in the Hamiltonian, but can only be set equal to zero after taking brackets/finding equations of motion. Substituting the invertible relations, and inserting φ to take care of the non-invertible one, our (tentative) new Hamiltonian is H? (ξ, p, u) = 21m pa pa + λ f (x) + upλ . (1.44) To check if this is a good Hamiltonian to use (i.e. if we can Legendre transform from it to our original Lagrangian), lets try the inversion, L = pi ξ˙ i − H? .

(1.45a)

The velocities are given by ∂H? ∂pi ∂H ∂φ = +u ∂pi ∂pi ∂H ∂pλ +u = ∂pi ∂pi

ξ˙ i =

(1.46a) (1.46b) (1.46c)

which yields x˙ a =

1 m

pa

(1.47)

1.2. HAMILTONIAN FORMALISM and

11

˙ = u. λ

(1.48)

So we have our Lagrangian
˙ λ − L = x˙ a pa + λp

1 2m pa pa


+ λ f (x) + upλ ,

(1.49)

into with we can substitute these relations for pa and u, to get
˙ λ − L = mx˙ a x˙ a + λp

1 ˙ a x˙ a 2 mx


˙ λ . + λ f (x) + λp

(1.50)

Remarkably, the non-invertible pλ terms cancel, and we get ˙ = 1 mx˙ a x˙ a − λ f (x) , L(ξ, ξ) 2

(1.51)

our original Lagrangian. What we’ve done, in effect, has been to treat the non-invertible velocity, λ˙ , not as a velocity, but rather as a multiplier (renaming it u) for it’s erstwhile conjugate momentum, pλ , which we now treat as a constraint. We’ve arrived at a Hamiltonian that we should be confident in using, but it should only be trusted to the extent that it gives us the same equations of motion that the Lagrangian does. To that end, lets check what Hamilton’s equations of motion give us. For our Hamiltonian (1.52) H? = 21m pa pa + λ f (x) + upλ , we have ∂H? ∂ξi ∂H ∂φ =− i −u i ∂ξ ∂ξ ∂pλ ∂H =− i −u i ∂ξ ∂ξ ∂H =− i, ∂ξ

p˙ i = −

(1.53a) (1.53b) (1.53c) (1.53d)

where the last term was dropped since the ξ’s and p’s all vary independently. This equation tells us that ∂H ∂xa = −λna ,

p˙ a = −

(1.54a) (1.54b)

and ∂H ∂λ = − f (x)

p˙ λ = −

(1.55a)

≈ 0.

(1.55b) (1.55c)

CHAPTER 1. CONSTRAINT FORMALISM

12

Hamilton’s other canonical equation tells us that ∂H? ∂pi ∂H ∂φ = +u ∂pi ∂pi ∂H ∂pλ = +u , ∂pi ∂pi

ξ˙ i =

(1.56a) (1.56b) (1.56c)

from which we get x˙ a =

1 m

pa

(1.57)

and ˙ = u. λ

(1.58)

To solve the equations of motion, we need to get rid of the λ appearing in Equation 1.54b. To do this, we’ll impose the consistency condition that all of the time derivatives of the constraint pλ must vanish. Taking Poisson brackets with H? , we have   p¨ λ = p˙ λ , H?   = − f (x), H?

∂ − f (x) ∂H? ∂H? ∂ − f (x) − = ∂ξi ∂pi ∂ξi ∂pi 1 = − m na pa ≈ 0,

(1.59a) (1.59b) (1.59c) (1.59d) (1.59e)

which doesn’t fix λ, so we need to take another time derivative:   ... pλ = p¨ λ , H?   1 na pa , H? = −m

1 1 ∂ −m na pa ∂H? ∂H? ∂ − m na pa = − ∂ξi ∂pi ∂ξi ∂pi



1 1 1 = − m nb,a pb m pa − λna − m na 1

= − m2 na,b pa pb +

1 m

λ

≈ 0,

(1.60a) (1.60b) (1.60c) (1.60d) (1.60e) (1.60f)

which does fix λ for us. Evidently λ=

1 m

na,b pa pb ,

(1.61)

which is consistent with Equation 1.18. We can plug this expression into the equation of motion, Equation 1.54b, to eliminate λ, and we get 1 p˙ a = − m na nb,c pb pc .

(1.62)

1.2. HAMILTONIAN FORMALISM

13

Since we know from Equation 1.57 that pa = mx˙ a ,

(1.63)

we can get rid of the p’s in favor of x˙ ’s, and we find that x¨ a = −nb,c x˙ b x˙ c na .

(1.64)

These are the same equations of motion we arrived at in Equations 1.33 and 1.20 using the Lagrangian formalism, which is enough to convince me that H? is a valid Hamiltonian to use5 .

1.2.2

The Conservative Constraint

If we want to use the conservative constraint in the quantum theory, we need to find a Hamiltonian form for it as well. The same basic procedure used for the usual constraint will be employed here to find a Hamiltonian that yields valid equations of motion under Hamilton’s canonical equations. Our Lagrangian for the conservative constraint is L = 21 mx˙ a x˙ a − λ f˙ (x) =

1 ˙ a x˙ a 2 mx

b

− λnb x˙ .

(1.65a) (1.65b)

Following the precedent set in Section 1.2.1, we expect our Hamiltonian to be given by
H? = ξ˙ i pi + uφ − L .

The momenta are obtained from pi =

∂L , ∂x˙ i

(1.66) (1.67)

giving ∂L ∂x˙ a = mx˙ a − λna

pa =

(1.68a) (1.68b)

and ∂L ˙ ∂λ

(1.69a)

= 0.

(1.69b)

pλ =

Again, λ˙ is not invertible in terms of the coordinates and momenta, so we need to make its conjugate momenta a constraint, φ = pλ , 5 The

(1.70)

argument I used to find a Hamiltonian should make any good mathematician uneasy (to say the least). For a more rigorous derivation I refer those who are interested in such things to my main references, Dirac [1] and Henneaux [3].

CHAPTER 1. CONSTRAINT FORMALISM

14

inserted into the Hamiltonian by means of a multiplier u. Using Equation 1.68b we can invert the xa ’s in terms of the coordinates and momenta to find x˙ a =

1 m


pa + λna ,

(1.71)

and substitute this expression, along with Equation 1.70, into the Hamiltonian,
H? = x˙ a pa + upλ −

1 ˙ a x˙ a 2 mx


− λnb x˙ b ,

(1.72)

to get H? =

1 m



2
1 1 + λnb m pa + λna pa + upλ − 21 m m pa + λna pb + λnb .

(1.73)

This expression is a mess, but after performing the sum in the last term and expanding the squared term, it cleans up nicely and we find that H? =

1 2m

pa + λna


2

+ upλ .

(1.74)

We want to check to make sure the equations of motion that this Hamiltonian yields agree with those obtained using the Lagrangian method in Section 1.1.2. First, the p˙ ’s are given by p˙ i = −

∂H? , ∂ξi

(1.75)

so that we have ∂H? ∂xa
1 pb + λnb λnb,a = −m

p˙ a = −

(1.76a) (1.76b)

and ∂H? ∂λ
1 = − m nb pb + λ

p˙ λ = −

(1.77a)

≈ 0.

(1.77b) (1.77c)

Equation 1.77b is required to equal zero by consistency of the constraint, and would let us solve for λ, if we so desired. Now for the ξ˙ ’s, ξ˙ i =

∂H? , ∂pi

(1.78)

and we find that ∂H? ∂pa
1 =m pa + λna ,

x˙ a =

(1.79a) (1.79b)

1.2. HAMILTONIAN FORMALISM

15

and ∂H? ∂pλ = u.

˙= λ

(1.80a) (1.80b)

Equation 1.76b looks like the equation we want to solve, but to check it against the equations from the Lagrangian analysis, we want a differential equation in the coordinates and velocities, so we need to do some work to get rid of the momenta. From Equation 1.79b we know that pa = mx˙ a − λna .

(1.81)

Taking the time derivative of the above equation, we get ˙ a ˙ a + λn p˙ a = mx¨ a − λn




˙ a. = mx¨ a − λna,b x˙ b − λn

(1.82a) (1.82b)

We also know, from substituting Equation 1.79b into Equation 1.76b, that p˙ a = −λnb,a x˙ b .

(1.83)

Setting equal the RHS’s of Equation 1.82b and Equation 1.83, we get ˙ a = −λnb,a x˙ b . mx¨ a − λna,b x˙ b − λn

(1.84)

By the equality of cross partials the λ terms cancel, and we find that ˙ a. mx¨ a = λn

(1.85)

We still need to determine λ˙ ; to do this we’ll take another time derivative of the constraint, and require that it equal zero.   p¨ λ = p˙ λ , H? ∂p˙ λ ∂H? ∂H? ∂p˙ λ = − ∂ξi ∂pi ∂ξi ∂pi

= − m12 nb,a pb pa + λna −

(1.86a) (1.86b) 1 m

u+

1

m2


na pb + λnb nb,a

(1.86c)

This expression is messy, but it cleans up nicely if one notes that na na,b is always zero6 . So we have p¨ λ = − m12 na,b pa pb − ≈ 0, 6n

a na,b

=

1 2

na na


,b

=

1 2




1

,b

= 0 [4].

1 m

u

(1.86d) (1.86e)

CHAPTER 1. CONSTRAINT FORMALISM

16

and since Equation 1.80b tells us that u = λ˙ , we can solve the above equation for λ˙ , and find ˙ = − 1 na,b pa pb λ m 1

a

= − m mx˙ − λna




b


mx˙ − λnb na,b

= −mna,b x˙ a x˙ b .

(1.87a) (1.87b) (1.87c)

The λ terms in the second line dropped out because we again make use of the fact that ˙ into Equation 1.85, we get na na,b is identically zero. Substituting this expression for λ x¨ a = −nb,c x˙ b x˙ c na

(1.88)

as our final equations of motion. These are the same equations as we obtained in Equations 1.20, Equation 1.33, and Equation 1.64.

1.2.3

The Dirac Bracket

You might have noticed that by the time we actually found the equations of motion in the last section, we had employed four constraints: f (x), pλ , p˙ λ , p¨ λ , were all required to be weakly equal to zero. The usual constraint was no better, we managed to work ... our way up to pλ as a constraint before we were done. Needless to say, this is a rather cumbersome way of managing a constrained system. In our case it was only a couple of auxiliary “consistency constraints” that we needed to employ, but I think you could imagine a case where the number was much greater, and the ensuing equation solving mess much worse, than it was here. Instead of juggling the constraints around until after the Poisson brackets are taken, and only then setting them equal to zero, Dirac defined a “new Poisson bracket,” (later referred to as the “Dirac bracket” by others), which incorporates all of the constraints as strong equations [1]. The definition of the Dirac bracket in terms of the Poisson bracket is given by         A, B D = A, B P − A, φk P Ekl φl , B P , (1.89) where Ekl is defined such that   Ekl φl , φm P = δkm .

(1.90)

As Dirac points out, these new brackets have many of the familiar properties of Poisson brackets. They are antisymmetric, linear in both terms, and obey the bracket product rule and the Jacobi Identity. Moreover, these brackets will yield the same equations of motion as the Poisson brackets, since they are weakly equal to them:         g, H? D = g, H? P − g, φk P Ekl φl , H? P (1.91a)   ≈ g, H? P . (1.91b)   ˙ l are all weakly equal to zero, as they are This follows since the terms in φl , H? P = φ

time derivatives of constraints. This is more a requirement the brackets need to pass to be of any use (after all, they wouldn’t be any good if they didn’t yield the same equations

1.2. HAMILTONIAN FORMALISM

17

of motion as the Poisson brackets), rather than a reason to use them. What makes the Dirac Bracket interesting is that the Dirac Bracket of any function G(x, p) with any of the φ’s, say φn , is strongly zero:         G, φn D = G, φn P − G, φk P Ekl φl , φn P     = G, φn P − G, φk P δkn     = G, φn P − G, φn P = 0,

(1.92a) (1.92b) (1.92c) (1.92d)

where in the first step I just used the definition of E, from Equation 1.90. So since the Dirac bracket of any function of x and p with any of the φ’s is zero, we might as well set the constraints in the Hamiltonian strongly equal to zero, and use the Dirac bracket to get the equations of motion from this simplified Hamiltonian, HD =

1 2m pa pa .

(1.93)

It turns out, and is shown Appendix A, that for the usual constraint, f (x) = 0, the fundamental Dirac brackets are  a b x ,x D = 0  
pa , pb D = na,c nb − nb,c na pc  a  x , pb D = δab − na nb .

(1.94a) (1.94b) (1.94c)

From these relations we can find the equations of motion from the Hamiltonian HD .   x˙ a = xa , HD D   = xa , 21m pb pb D   = 21m xa , pb pb D      = 21m xa , pb D pb + pb xa , pb D   1 =m pb xa , pb D
1 =m pb δab − na nb =

1 m

pa .

(1.95a) (1.95b) (1.95c) (1.95d) (1.95e) (1.95f) (1.95g)

In the last step I used φ3 = p¨ λ = − m1 nb pb (defined in Appendix A) as a strong equation to set pb nb equal to zero, killing off the second term in the parentheses. For the other equation of motion, we get   p˙ a = pa , HD D   = pa , 21m pb pb D   = 21m pa , pb pb D      = 21m pa , pb D pb + pb pa , pb D   1 =m pb pa , pb D
1 =m pb na,c nb − nb,c na pc 1 = −m nb,c pb pc na ,

(1.96a) (1.96b) (1.96c) (1.96d) (1.96e) (1.96f) (1.96g)

CHAPTER 1. CONSTRAINT FORMALISM

18

where, again, I’ve employed φ3 as a strong equation in the last step to eliminate the first term in the parentheses. Equation 1.96g should look rather familiar; if we use Equation 1.95g to get rid of the p’s in favor of x˙ ’s, we find that x¨ a = −nb,c x˙ b x˙ c na ,

(1.97)

the same equation of motion that resulted from the previous approaches. I think that this computation is pretty good evidence that finding the equations of motion from the Hamiltonian is much easier if we use the Dirac bracket instead of the Poisson bracket. However, as Appendix A will attest, actually computing the E matrix and finding the fundamental Dirac brackets can be something of pain. The chore is a mindless one though, more of a prescribed ritual to carry out than an equation solving problem to puzzle through, as was the case when using the Poisson bracket. In our case, the “puzzle” wasn’t terribly hard to solve, but that was because only four constraints were necessary to use. Having to deal with even twice that number would be a bit overwhelming to me, and there could easily arise a situation in which many times more constraints than that were necessary.

1.3

Example: The Spherical Shell

At this point, we’ve found the equations of motion for a particle confined to a surface using five different methods, and I think it’s high time to work out an example to get a better feel for the applicability of the formalism. As a surface of constraint, the ndimensional spherical shell suggests itself as a nice example, both for the familiarity one might have with the solutions in the two- or three-dimensional case, and for the obvious symmetry such a system will yield in the equations of motion for any n. The initial legwork will be done for the n-dimensional case, and then we can specialize to the circle or spherical shell case when suitable. We note that xa xa = R 2 (1.98) gives an implicit description of an n-sphere with radius R. An equivalent functional characterization of the spherical shell we might write down would be f (x) =

√

xa xa − R

≈ 0,

(1.99a) (1.99b)

and indeed this is the constraint we will want to use for working through the problem [3]. The general result was the same for all five derivations, so obviously this example is applicable to all of the approaches. Lets unravel the general result for the equations of motion, x¨ a = −nb,c x˙ b x˙ c na .

(1.100)

The na term is the a component of n, a vector normal to the sphere, and can be obtained by differentiating √ f (x) = xa xa − R (1.101)

1.3. EXAMPLE: THE SPHERICAL SHELL

19

with respect to xa . This gives ∂ f (x) ∂xa  ∂ √ b b = a x x −R ∂x
1 1 ∂ xb xb = √ a 2 xb xb ∂x

na =

=√

1 xb xb

xa .

(1.102a) (1.102b) (1.102c) (1.102d)

You may note that from Equation 1.102d, n is in fact the unit normal vector, which is a result of a carefully selected functional description of the surface. For the slightly more complicated term in Equation 1.100 we have nb,c

∂ = c ∂x



xb



√ xd xd   1 ∂xb 1 b ∂ √ =√ +x ∂xc xd xd ∂xc xd xd 
−3/2 c  1 =√ 2x δbc + xb − 12 xd xd xd xd   xb xc 1 δbc − d d . =√ x x xd xd

(1.103a) (1.103b) (1.103c) (1.103d)

Now seems like the right time to fix the dimensionality of the problem. Let’s look at the n = 2 case (a particle confined to move on a ring), where we can write x = (x, y). Writing out explicitly the summation that the repeated indices imply, we get b c

"

nb,c x˙ x˙ = p

1 x2 + y2

   xy x2 2 x˙ + − 2 x˙ y˙ 1− 2 x + y2 x + y2     # yx y2 + − 2 y˙ x˙ + 1 − 2 y˙ 2 x + y2 x + y2


2 xy˙ − x˙ y =
3/2 . x2 + y2

(1.104a)

(1.104b)

Now we can plug this expression along with Equation 1.102d into Equation 1.100, to get the equations of motion:
2 xy˙ − x˙ y (x¨ , y¨ ) = − p
(x, y) x2 + y2 x2 + y2 3/2  2 xy˙ − x˙ y =− (x, y) . x2 + y2

1

(1.105a) (1.105b)

CHAPTER 1. CONSTRAINT FORMALISM

20

I don’t know if I’d actually want to solve these equations (at least not in Cartesian coordinates), but that’s not the point. In principle we’re done with the analysis of the system, what’s left is an exercise in solving differential equations. It might be nice to see them solved, though, and if we switch to polar coordinates, the process isn’t so bad. We note that the polar angle φ is given by φ = arctan

y x

,

(1.106)

so y d arctan dt x   y  dx  ∂  y  dy ∂ = arctan + arctan ∂x x dt ∂y x dt −y x = 2 x˙ + 2 y˙ x + y2 x + y2 xy˙ − x˙ y . = 2 x + y2

˙ = φ

(1.107a) (1.107b) (1.107c) (1.107d)

˙ into our equations of motion, to get We can substitute this expression for φ ˙ 2 (x, y) . (x¨ , y¨ ) = −φ

(1.108)

We now write x and y in terms of r and φ, and find that x = r cos φ

=⇒ =⇒


˙ x˙ = − r sin φ φ

2 ¨ − r cos φ φ ˙ x¨ = − r sin φ φ

(1.109a) (1.109b)


˙ y˙ = r cos φ φ

2 ¨ − r sin φ φ ˙ . y¨ = r cos φ φ

(1.110a) (1.110b)

and y = r sin φ

=⇒ =⇒

Inserting these expressions into the x and y components of Equation 1.108, we get

2
¨ − r cos φ φ ˙ = −φ ˙ 2 r cos φ − r sin φ φ

and or simplifying, and



2
¨ − r sin φ φ ˙ = −φ ˙ 2 r sin φ , r cos φ φ

(1.111a) (1.111b)


¨ =0 r cos φ φ

(1.112a)


¨ =0 r sin φ φ

(1.112b)

Squaring and adding these equations gives us
2 ¨ = 0, r2 cos2 φ + sin2 φ φ

(1.113)

1.3. EXAMPLE: THE SPHERICAL SHELL or simply

21

¨2 = 0. r2 φ

(1.114)

For a circle of fixed radius, r is constant, so ¨2 = 0 r2 φ

=⇒

¨2 = 0 φ

=⇒

¨ =0 φ

=⇒

˙ = const . φ

(1.115a) (1.115b) (1.115c)

This result should be familiar to anyone who has taken freshmen physics. It states that when there are no other external forces, the angular velocity of a particle moving in a circle is constant. For such a simple example, the formalism developed here is pretty drastic overkill, but it’s reassuring to see that a somewhat intimidatingly abstract formalism reproduces a familiar result in the appropriate context.

Chapter 2 Quantization Unfortunately, deriving a quantum theory from the classical theory for constrained particles is not a completely straightforward procedure, because the non-commutativity of the quantum mechanical operators presents a problem. Take for example the classical Hamiltonian obtained by following the conservative constraint procedure, H= =

1 2m 1 2m


pa δab − na nb pb


pa pa − pa na nb pb .

(2.1a) (2.1b)

Here I have plugged in for λ in Equation 1.74, and am assuming the use of Dirac brackets, so that the constraints are all set strongly equal to zero and thus drop out of the Hamiltonian. I have also dropped the subscript D from H, from now on I will no longer write the star/subscript D, so an unadorned H is taken to mean a Hamiltonian that gives acceptable equations of motion when the appropriate brackets are applied. In Equation 2.1b there is only one possible way of ordering the variables in the term pa pa , so even if we were concerned with ordering (not that we should be, in the classical theory), the ordering is completely unambiguous. Not so for the second term, pa na nb pb , where in general if we have four variables multiplied together there are 4 · 3 · 2 · 1 = 24 ways of ordering those variables. In the classical theory, it doesn’t matter at all what ordering we pick, since all the operators commute with each other we can push them around at our discretion. When we quantize the system, though, and the variables become non-commuting operators, different orderings of the operators will yield different results. We begin the attempt at a quantum theory by writing down, from the classical case for the conservative constraint, the Dirac brackets between the x’s and p’s:  a b x ,x D = 0   pa , pb D = 0  a  x , pb D = δab .

(2.2a) (2.2b) (2.2c)

These relations are taken from [4], but they could be found by the same procedure I used to compute the brackets for the usual constraint in Appendix A. Since we have a classical Hamiltonian for the system, all we have to do to quantize it is follow the standard

CHAPTER 2. QUANTIZATION

24

procedure: send the variables to operators, and introduce commutation relations given by ih times the Dirac brackets. That is, xa 7−→ xˆ a

(2.3a) (2.3b)

pa 7−→ pˆ a ,

and   A, B D 7−→

  ˆ , Bˆ . A ih 1

(2.3c)

I’m not going to use the result immediately, but it might be worth noting right now that the commutation relations between the xˆ ’s and pˆ ’s, from which, in general, we try to derive expressions for xˆ and pˆ , are exactly the same as for an unconfined particle in flat space1 , so we can use the result that xˆ a = xa h ∂ . pˆ a = i ∂xa

(2.4a) (2.4b)

Following this procedure, the quantum Hamiltonian we get is ˆ = H

1 ˆ a pˆ a 2m p

−

1 2m

ˆ a nˆ b pˆ b : , : pˆ a n

(2.5)

where : : means the most general hermitian linear combination of operator orderings that reduces to the classical result in the classical limit. The first term in the Hamiltonian is clearly the usual kinetic energy term, while the second term is, in some manner or other, related to the kinetic energy of the particle normal to the surface2 . In any event, it’s useful to use some shorthand for the terms, and ˆ = Tˆ − Tˆn H

(2.6)

makes some sort of sense. Since the nˆ ’s (functions only of the xˆ ’s) commute amongst themselves, as do the pˆ ’s, the 24 possible orderings for Tˆn reduce to only eight distinct quantum operators: ˆ 0 = pˆ a nˆ a nˆ b pˆ b , A

(2.7a)

ˆ 1 = nˆ a pˆ a pˆ b nˆ b , A ˆ 2 = nˆ a pˆ a nˆ b pˆ b A ˆ 3 = pˆ a pˆ b nˆ a nˆ b A ˆ 4 = nˆ a pˆ b nˆ b pˆ a A 1 This

(2.7b) and

ˆ 2 = pˆ b nˆ b pˆ a nˆ a , A

(2.7c)

and

ˆ 3 = nˆ b nˆ a pˆ b pˆ a , A

(2.7d)

and

ˆ = pˆ a nˆ b pˆ b nˆ a , A 4

(2.7e)

†

†

†

is one of the reasons I chose to work with the conservative constraint procedure; for the usual constraint procedure, it turns out that the commutation relations are different (and a bit more complicated), so it is not so easy to guess at representations of xˆ and pˆ in that case. 2 Classically, p n = n p = p , so we have (again, only classically) T ≡ 1 p2 = 1 p n n p a a a a n n 2m n 2m a a b b

25 ˆ 0 and A ˆ 1 are hermitian, so their hermitian conjugates are equal to them(note that A selves). Right now we have eight ambiguous parameters in the Hamiltonian (the eight coefficients on the A’s), and the plan is to reduce these ambiguous parameters to as few as possible. This derivation closely follows the work of Ikegami et al. in [4]. Since Tˆn needs to be hermitian in order to be an observable, define ˆ k+ = A ˆk +A ˆ †k , A ˆ k− = i A ˆk − ˆ A

A†k

(2.8a)


,

(2.8b)

ˆ k ’s (k = 2, 3, 4). We can then write the “hermitized versions” of the A X
1 ˆ 0 + c1 A ˆ1 + ˆ k+ + c0k A ˆ k− c0 A ck A Tˆn = 2m k=2 4

!

,

(2.9)

where the c’s are all real, which is certainly hermitian, since it is the sum of hermitian ˆ 0 from both sides we get operators. Rearranging Equation 2.9 and subtracting A ˆ 0 = c0 − 1 A ˆ 0 + c1 A ˆ1 + 2mTˆn − A


4 X


ˆ k+ + c0k A ˆ k− . ck A

(2.10)

k=2

ˆ k+ = 2A ˆ k and A ˆ k− = 0, so the correspondence principle tells us In the classical limit, A that c0 + c1 +

4 X

2ck = 1 .

(2.11)

k=2

We can use the constraint placed on the coefficients by the correspondence principle in
Equation 2.11 to substitute in for c0 − 1 , and get ˆ 0 = c0 − 1 A ˆ 0 + c1 A ˆ1 + 2mTˆn − A


4 X

ˆ k+ + c0k A ˆ k− ck A




(2.12a)

k=2

  4 4 X X
ˆ 0 + c1 A ˆ1 + ˆ k+ + c0k A ˆ k− = − c1 + 2ck A ck A k=2

k=2


ˆ1 −A ˆ0 + = c1 A

4 X

4
X ˆ k+ − 2A ˆ0 + ˆ k− ck A c0k A

k=2


ˆ1 −A ˆ0 + = c1 A

4 X


ˆ1 + = c1 δA

k=2

(2.12c)

k=2

ck



4

 X ˆk −A ˆ0 + A ˆ †k − A ˆ0 + ˆ k− (2.12d) A c0k A

k=2

4 X

(2.12b)


ˆ k + δA ˆ †k + i ck δA

k=2

4 X


ˆk −A ˆ †k , c0k A

(2.12e)

k=2

ˆj = A ˆj −A ˆ 0 . The terms in Equation 2.12e aren’t too difficult to compute, and where δA

CHAPTER 2. QUANTIZATION

26 it turns out that ˆ 1 = h2 Kˆ ,0 δA

(2.13a)

ˆ 2 = δA ˆ 3 = ih pˆ a nˆ a Kˆ δA

(2.13b)

ˆ 2 = δA ˆ 3 = −ih Kˆ nˆ a pˆ a δA

(2.13c)

ˆ 4 = δA ˆ =0 δA 4

(2.13d)

†

† †

and  ˆ2 −A ˆ †2 = A ˆ3 −A ˆ †3 = ih pˆ a , nˆ a Kˆ A

(2.13e)

ˆ4 −A ˆ = 0. A 4

(2.13f)

†

In the above expressions, K ≡ −na,a ,

(2.14a) (2.14b)

K,0 ≡ −na nb,b,a , 

and , is the anticommutator, given by  ˆ , Bˆ = A ˆ Bˆ + Bˆ A ˆ. A

(2.15)

When we plug these values into Equation 2.12e, we get
 



ˆ 0 = c1 h2 Kˆ ,0 + c2 + c3 ih pˆ a , nˆ a Kˆ 2mTˆn − A


   ˆ ˆ a K . (2.16a) + i c2 + c3 ih pˆ a , n 0

0

Since pˆ a , nˆ a Kˆ = ih Kˆ 2 − Kˆ ,0 , we can substitute this in, 








ˆ 0 = c1 h2 Kˆ ,0 − h2 c2 + c3 Kˆ 2 − Kˆ ,0 − h c02 + c03 pˆ a , nˆ a Kˆ , 2mTˆn − A








(2.16b)

and group some terms, to get  2   ˆ 0 = h γKˆ ,0 − ηKˆ 2 − hζ pˆ a , nˆ a Kˆ , 2mTˆn − A 4

(2.16c)

where γ = 4 c1 + c2 + c3 , η = 4 c2 + c3 , and ζ = c02 + c03 . We can now write Tˆn as





  h2  ˆ 1 ˆ ˆ 2 − hζ pˆ a , nˆ a Kˆ A0 + γK,0 − ηK . Tˆn = 2m 4 



(2.17)

It can be shown that ζ = 0, by requiring that ˆ

ˆ = 0, f ,H 

(2.18)

  d f = f ,H D = 0. dt

(2.19)

since of course classically our constraint is

27 So we have    ˆ = fˆ , Tˆ − Tˆn f ,H     = fˆ , Tˆ − fˆ , Tˆn

ˆ

= 0.

(2.20a) (2.20b) (2.20c)

For the first term, we get    f , Tˆ = fˆ , 21m pˆ a pˆ a     = 1 fˆ , pˆ a pˆ a + pˆ a fˆ , pˆ a

ˆ

2m ih

ˆ a pˆ a + pˆ a nˆ a n  h ˆ a , pˆ a , n = 2im =




2m

(2.21a) (2.21b) (2.21c) (2.21d)

  where in the second step I’ve used the result3 that fˆ , pˆ a = ihnˆ a . For the second term, we can use the fact that fˆ is a function only of xˆ to note that it will commute with the γ and η terms in Equation 2.17, so we have 
   ˆ 0 − hζ pˆ a , nˆ a Kˆ f , Tˆn = fˆ , 21m A      ˆ 0 − hζ fˆ , pˆ a , nˆ a Kˆ = 21m fˆ , A .

ˆ

(2.22a) (2.22b)

The term we want to look at in Equation 2.22b is the first one, ˆ

   ˆ 0 = fˆ , pˆ a nˆ a nˆ b pˆ b f ,A     ˆ a nˆ b pˆ b + pˆ a nˆ a fˆ , nˆ b pˆ b = fˆ , pˆ a n

(2.23a) (2.23b)

 ˆ

  ˆ   ˆ  ˆ ˆ a + pˆ a f , nˆ a nˆ b pˆ b + pˆ a nˆ a f , nˆ b pˆ b + nˆ b f , pˆ b = f , pˆ a n (2.23c)

ˆ a nˆ a nˆ b pˆ b + pˆ a nˆ a ihnˆ b nˆ b (2.23d) = ihn
ˆ b pˆ b + pˆ a nˆ a = ih n (2.23e)  ˆ a , pˆ a , = ih n (2.23f)   where in the third step I used the previously remarked upon result that fˆ , pˆ a = ihnˆ a , and in the fourth step I used the fact that nˆ a nˆ a = 1. Substituting this result back into Equation 2.22b, we get ˆ

f , Tˆn = 

1 2m

     ˆ a , pˆ a − hζ fˆ , pˆ a , nˆ a Kˆ ih n .

(2.24)

      Equation 2.20 tells us that fˆ , Tˆ = fˆ , Tˆn , so we can set our expression for fˆ , Tˆ from Equation 2.21d equal to the RHS of Equation 2.24, to get ih 2m 3 This

[5].

 ˆ a , pˆ a = n

1 2m

     ˆ a , pˆ a − hζ fˆ , pˆ a , nˆ a Kˆ ih n .

(2.25)

ˆ xˆ ), pˆ a = Q ˆ ,a , for any “nice” function Q( ˆ xˆ ). See follows from the more general result that Q( 



CHAPTER 2. QUANTIZATION

28

The LHS cancels with the first term in the parentheses, and we find that    ˆ a Kˆ = 0 . hζ fˆ , pˆ a , n

(2.26)

Since fˆ does not commute with pˆ a , the commutator in the above equation is not trivial, in fact      ˆ  f , pˆ a , nˆ a Kˆ = fˆ , pˆ a nˆ a Kˆ + fˆ , nˆ a Kˆ pˆ a         ˆ a Kˆ + pˆ a fˆ , nˆ a Kˆ + fˆ , nˆ a Kˆ pˆ a + nˆ a Kˆ fˆ , pˆ a = fˆ , pˆ a n

ˆ a nˆ a Kˆ + nˆ a Kˆ ihnˆ a = ihn

ˆ. = 2ihK

(2.27a) (2.27b) (2.27c) (2.27d)

ˆ is in general nonzero, so to make the whole LHS of Equation 2.26 go to zero, we K require that ζ = 0.

(2.28)

This result allows us to eliminate yet another parameter from Equation 2.17, so that we now have   2  1 h 2 ˆ ,0 − ηKˆ ˆ + Tˆn = γK . (2.29) A 2m 0 4 This is as good as it’s going to get in terms of determining the unknown parameters in our expression for Tˆn , so it’s time to plug it back in to the expression for the Hamiltonian, ˆ = Tˆ − Tˆn H

(2.30a) h

ˆ0 + A



i

ˆ ,0 − ηKˆ 2 γK h  i 2 ˆ a nˆ b pˆ b − h4 γKˆ ,0 − ηKˆ 2 = 21m pˆ a pˆ a − pˆ a n

2 ˆ a nˆ b pˆ b − 8hm γKˆ ,0 − ηKˆ 2 . = 21m pˆ a δˆ ab − n =

1 ˆ a pˆ a 2m p

−

1 2m

h2

4

(2.30b) (2.30c) (2.30d)

When we want to actually solve this equation, we would almost certainly need to fix the parameters γ and η to specific values. This is done in the next chapter, but here we should at least note that regardless of the specific values of the constants (unless they are both zero), confining a particle to a surface does indeed have a noticeable effect on the particle’s wave function: a term of the same order in h as the kinetic energy operator has appeared as an effective potential.

Chapter 3 An Example: The Torus 3.1

Gaussian Normal Coordinates

aŌ

aō

aŎ

Figure 3.1: Given an n− 1 dimensional surface embedded in n dimensional space, it is possible to recoordinatize the space, and find a set of orthonormal coordinates so that one coordinate, q0 , is always (locally) normal to the surface. The surface is then given by a constant value of the normal coordinate q0 . In Cartesian coordinates, a torus is described by the solutions to the equation z2 +

p 2 x2 + y2 − a = b2 ,

(3.1)

where b is the major and a the minor radius of the torus (see Fig. 3.1). Even though it is not essential to the analysis, shifting to a nicer coordinate system is very helpful in many regards. I will adopt the coordinates qµ = (ρ, θ, φ), where ρ ∈ [0, ∞) and θ, φ ∈ [0, 2π), such that r ρ=

z2 +

p

x2 + y2 − a

z p x2 + y2 − a y φ = arctan . x θ = arctan

2

(3.2a)

!

(3.2b) (3.2c)

CHAPTER 3. AN EXAMPLE: THE TORUS

30

Here ρ is the normal coordinate, and setting ρ = b defines the surface of a torus. Figure 3.1 is an attempt at illustrating the relevant geometry of the system. Also note that θ is not measured as usual from the z axis, but rather from the x-y plane (in the same manner as latitude is measured on a globe), so as to take advantage of the symmetry of the torus through the x-y plane.

j 

  

i

Q

R

Figure 3.2: A “slice” of the torus in the plane x = 0 (or alternatively, φ = 0). Rotate the figure about the z axis for the full effect.

The relationships corresponding to Equation 3.2c, giving x = (x, y, z) in terms of the qµ ’s are
x = ρ cos θ + a cos φ
y = ρ cos θ + a sin φ

(3.3a) (3.3b) (3.3c)

z = ρ sin θ .

From the relationships between the coordinates, we can determine the vielbeins: 

cos θ cos φ − ρ sin θ cos φ ∂x  cos θ sin φ − ρ sin θ sin φ eaµ ≡ = ∂qµ sin θ ρ cos θ a


 − a + ρ cos θ
sin φ a + ρ cos θ cos φ

0

(3.4)

3.2. SETTING UP THE SCHRÖDINGER EQUATION and

31





cos θ cos φ cos θ sin φ ∂qµ  1 1 µ − sin θ cos φ − sin θ sin φ = ρ e a≡ ρ a ∂x

−

sin φ ρ cos θ+a

sin θ 1 cos θ . ρ 0

cos φ ρ cos θ+a

(3.5)

From these we compute the metrics: 1 0 a a  gµν ≡ e µ e ν = 0 ρ2 0 0

0 , 0
2 a + ρ cos θ



and



1 0 µν µ ν g ≡ e ae a =  0



0



0 0


1 2 ρ

1 a+ρ cos θ

0

(3.6)


2

 .

(3.7)

Some curvature terms are now obtainable from the metric, where the roman indices indicate that we have dropped the zeroth row and column (corresponding to the normal coordinate) out of the metric since it doesn’t factor into the computation: kij ≡

− 12 gij,0

 =



0
, − a + ρ cos θ cos θ

−ρ

0 "

ki j ≡ gil klj =

1

−




#

0

ρ

0

cos θ a+ρ cos θ

−

(3.8)


.

(3.9)

The curvature scalars of interest, defined in Equation 2.14, are therefore 

i

K≡ki=−

and i

j

ρ

+

1 ρ+

 2

K,0 ≡ k j k i =

3.2

1

1 ρ

 +



(3.10)

a

cos θ

2

1 ρ+

a

.

(3.11)

cos θ

Setting up the Schrödinger Equation

We want to solve the time independent Schrödinger equation Hψ = Eψ

(3.12)

for a particle constrained to the surface of the torus. I have dropped the hat’s from the operators, since from now on we will be working only with the operators, and there should be no ambiguity as to what each symbol is referring to. For a particle of mass µ, Dirac’s prescription with the conservative constraint given by f˙ (x) = 0 , (3.13)

CHAPTER 3. AN EXAMPLE: THE TORUS

32

which we followed in Chapter 2, and the confining approach [6], which we have not examined at all, both yield an effective potential V due to the curvature (see Equation 2.30d): V=−


h2 γK,0 − ηK2 . 8µ

(3.14)

In the confining approach, there is no ambiguity as to what values to assign γ and η, while for Dirac’s prescription the ambiguity remains unless we make an assumption. The necessary assumption is that Tn is not just suggestive shorthand for a bunch of operators, but is actually made up of the normal component of the momentum: Tn ≡ 21 p2n ,

(3.15)

pn ≡ : pa na : .

(3.16)

where This requirement is based on the somewhat dubious claim that Tn is an “indecomposable fundamental physical quantity” [4], and I take it with a healthy bit of skepticism. It does, however, manage to bring Dirac’s prescription into accord with the the results from the confining approach, and by greatly reducing the operator ordering ambiguity in the Hamiltonian, lets us set γ = 2 and η = 1. Before writing out the Schrödinger equation, I am going to switch to Gaussian normal coordinates to tidy up the expression for the kinetic energy.  ∂
∂  ˆ a nˆ b pˆ b = a δab − na nb pˆ a δˆ ab − n ∂x ∂xb ∂ ∂ ∂ ∂ = a a − a na nb b ∂x ∂x ∂x ∂x 1 1 = g− 2 ∂i gij g 2 ∂j

(3.17a) (3.17b) (3.17c) (3.17d)

≡ ∆Σ ,

the Laplacian on the surface, written in normal coordinates, where the roman indices indicate that zeroth coordinate is not being summed over. So the kinetic energy operator becomes 2 (3.18) T = − 2hµ ∆Σ . and for an otherwise free particle confined to our surface, the Schrödinger equation then reads  
h2 h2 2 − ∆Σ − 2K,0 − K ψ = Eψ . (3.19) 2µ 8µ For our toroidal coordinates, 1

1

∆Σ = g− 2 ∂i gij g 2 ∂j " =

1

ρ2 a ρ

=

(3.20a)

+ cos θ

1 b2

a b

+ cos θ





∂θ

∂θ 



a b

a ρ


 + cos θ ∂θ +


+ cos θ ∂θ



a ρ

1
∂2 + cos θ φ

!#

1 2 +
2 ∂φ . a 2 b b + cos θ

(3.20b) ρ=b

(3.20c)

3.2. SETTING UP THE SCHRÖDINGER EQUATION

33

Plugging in for ∆Σ and the K terms, the Schrödinger equation becomes " −

1

b2

a b



a b






#

1 2 +
2 ∂φ ψ a 2 b b + cos θ


∂θ + cos θ ∂θ + cos θ " 2 !  2 #  2  1 1 1 1 2µE 1 − + − + ψ = 2 ψ , (3.21) 2 a a h 4 b b + cos θ b b + cos θ

which is a mess. The curvature term immediately simplifies, to give  2

1

2

+

b

2 !

1



b+

 −

a

cos θ

1 b

+

2

1 b+

 =

a

cos θ

1 b

−

2

1 b+

a

,

(3.22)

cos θ

and so Equation 3.21 reads " −

1 b2

a b

+ cos θ


∂θ



a b




+ cos θ ∂θ



#

1 2 +
2 ∂φ ψ a 2 b b + cos θ 1 1 1 − − 4 b b + cosaθ 

2 ψ=

2µE h2

ψ . (3.23)

The wave function ψ is a function of θ and φ, and we can use separation of variables, ψ(θ, φ) = Θ(θ)Φ(φ) ,

(3.24)

to find separable solutions. For the φ component the solution is the same as for spherical coordinates, Φ(φ) = eimφ , m = 0, ±1, ±2, . . . (3.25) Substituting this into Equation 3.23, and taking all the partials, we get a second order differential equation for Θ(θ), 

a b

   00 0 a + cos(θ) + cos(θ) Θ (θ) − sin(θ)Θ (θ) b  2  2 a 2µEb2  a 2 + + + cos(θ) − m Θ(θ) = 0 . (3.26) b 4b2 h2

If we let the ratio of the width of the torus to its “fatness”, ab , equal a new parameter p, p≡

and set n≡

a , b

2µEb2 h2

(3.27)

,

(3.28)

we can write i
h
p + cos(θ) p + cos(θ) Θ 00 (θ) − sin(θ) Θ 0 (θ) h
2 i + 14 p2 − m2 + n p + cos(θ) Θ(θ) = 0 . (3.29)

34

CHAPTER 3. AN EXAMPLE: THE TORUS

In this equation, there are three terms we can tune to find various solutions for Θ(θ). Of the three, p is actually dictated by the shape of the torus, so if we were given a system to solve, we couldn’t really touch it. However, since I get to propose the system, as well as try to solve it, we can change the value of p; looking at solutions for different values of p will hopefully give us a glimpse of how much the curvature affects the eigenstates of the system. The other two terms, m and n, are the “quantum numbers”. m is in fact what one usually thinks of as the azimuthal quantum number, since it takes on integer values, but n, even though it is not an integer, is also a quantization condition. That is, for given values of p and m, there will only be solutions of Equation 3.29 for certain values of n. Unfortunately, this equation is rather too difficult to solve analytically, so I used Mathematica to find some numerical solutions.

3.3. NUMERICAL SOLUTIONS

3.3

35

Numerical Solutions

We can find numerical solutions to the theta equation (and discover acceptable values of n) if we specify the values of the other parameters. p = 32 is a nice value for p. It gives us a torus with a hole that’s half as wide as the ring is thick. For this value of p, we can look at numerical solutions for m = 0, and for as many excited states as we care to examine. I used Mathematica to “shoot” for solutions1 , requiring the values of Θ(θ) and Θ 0 (θ) to match themselves at 0 and 2π, and got results accurate enough for my qualitative interests. In the pictures that follow, I’ve graphed the (unnormalized) θ wave function along √ side a plot of g |Θ(θ)|2 . I’ve done this because the plots of the θ wave function are rather deceptive. They all seem to show the particle having a preference for bunching up around the inner ring of the torus (θ = π). In fact it is the second plot which gives an honest account of the particle’s preferred position, since it takes into account the fact that the √ differential area element of the torus is a function of the coordinates: dA = g dθdφ, where g was previously defined as the determinant of the (covariant) metric.

3.3.1

Pictures of Wave Functions

As mentioned above, the p = 1.5 torus is a nice example of a typical torus, so I begin by plotting a number of numerical solutions for the wave functions. I have also plotted some of the low energy states for a p = 1.1 torus, for comparative purposes. The effective potential due to the curvature of the surface becomes more pronounced as p gets closer to 1, where the width of the hole goes to zero, so I finish by plotting a series of wave functions that all have the same number of nodes, and are all solved with the azimuthal quantum number, m, set to 0. By changing p, we should be able to observe the direct effects of the curvature of the surface upon the probability distribution of the particle. Π  2

Π

p = 1.5

3Π  2

2Π

Π  2

Π

p = 1.1

3Π  2

2Π

Π  2

Π

3Π  2

2Π

p = 1.01

Figure 3.3: Plots of the effective potentials for three values of p, as functions of θ. Note that the plots do not have the same vertical scale; the well gets deeper as p gets closer to 1. It is helpful, when viewing the wave functions, to keep in mind the effective potential due to the curvature that is acting on the particle. It is a potential well, centered at θ = π (the inner ring of the torus), which gets narrower and deeper as p approaches 1. 1 See

Appendix B.

CHAPTER 3. AN EXAMPLE: THE TORUS

36 Some p = 1.5 Wave Functions

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

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Π

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Θn (θ)

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

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3Π €€€€€€€€ 2

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3Π €€€€€€€€ 2

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3Π €€€€€€€€ 2

2Π

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Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

2Π

√

g |Θn (θ)|2

Figure 3.4: Approximations of the first five (unnormalized) even wave functions for m = 0, listed in descending order. Here n = −0.59, 0.44, 3.48, 8.48, 15.49.

3.3. NUMERICAL SOLUTIONS

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π

Π

Π

Π

Π

Θn (θ)

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

37

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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Π

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

√ g |Θn (θ)|2

Figure 3.5: Approximations of the first five (unnormalized) odd wave functions for m = 0, listed in descending order. Here n = 0.57, 3.50, 8.49, 15.48, 24.48.

CHAPTER 3. AN EXAMPLE: THE TORUS

38

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π

Π

Π

Π

Π

Θn (θ)

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

Π €€€€ 2

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

2Π

√

g |Θn (θ)|2

Figure 3.6: Approximations of the first five (unnormalized) even wave functions for m = 1, listed in descending order. Here n = 0.20, 1.95, 4.76, 9.58, 16.57.

3.3. NUMERICAL SOLUTIONS

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

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Π

Π

Π

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Θn (θ)

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3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

39

2Π

Π €€€€ 2

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

√ g |Θn (θ)|2

Figure 3.7: Approximations of the first five (unnormalized) odd wave functions for m = 1, listed in descending order. Here n = 1.21, 4.47, 9.55, 16.56, 25.56.

CHAPTER 3. AN EXAMPLE: THE TORUS

40 Some p = 1.1 Wave Functions

Π  2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π €€€€ 2

Π

Π

Π

Π

Π

Θn (θ)

3Π  2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

2Π

Π  2

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Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

√

g |Θn (θ)|2

Figure 3.8: Approximations of the first four (unnormalized) even excited state wave functions for p = 1.1, m = 0, listed in descending order. Here n = −0.10, 2.01, 6.55, 13.32.

3.3. NUMERICAL SOLUTIONS

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Π €€€€ 2

Π €€€€ 2

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Θn (θ)

41

3Π €€€€€€€€ 2

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3Π €€€€€€€€ 2

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

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3Π €€€€€€€€ 2

2Π

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Π

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

√ g |Θn (θ)|2

Figure 3.9: Approximations of the first four (unormalized) odd excited state wave functions for p = 1.1, m = 0, listed in descending order. Here n = −0.11, 1.85, 6.51, 13.31.

CHAPTER 3. AN EXAMPLE: THE TORUS

42 p as a Parameter

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3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

2Π

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3Π €€€€€€€€ 2

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2Π

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2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

2Π

2Π

2Π

2Π

2Π

√ g |Θn (θ)|2

Figure 3.10: Plotted are odd wave functions with 24 nodes and for which m = 0. From the top down, p changes from 1.5, to 1.1, 1.05, 1.01, 1.005, and to 1.001 at the bottom. As p gets closer to 1, the curvature and its effect become more pronounced.

Concluding Remarks The astute reader will notice that while I developed the classical theory of constrained systems on parallel tracks, exploring the formalism for both of the constraints f and f˙ , this double course was abandoned at Chapter 2, when I quantized the Hamiltonian for only one constraint. The absence of the other quantum theory was due mainly to a lack of time, which is unfortunate because there are many interesting comparisons to make between the two processes. The result for the constraint f = 0, not included here, is also particularly puzzling because while the classical Hamiltonian formulations for the two different constraints yield the same equations of motion, the quantum Hamiltonians one arrives at by Dirac’s procedure are different [4]. In fact they cannot be brought into accord at all without abandoning the correspondence between the f˙ procedure and the confining approach. It is unfortunate that I was unable to solve the torus example exactly, but there was no real reason to expect that I would be able to. Exactly solvable systems in quantum mechanics are few and far between; introducing the extra curvature term into the Schrödinger equation probably made the already bleak prospect of finding an exact solution even worse. The general inability to find the exact eigenfunctions for a Hamiltonian makes me speculate about the attempting to apply some sort of approximation technique to deal with the curvature term in the Hamiltonian. Variational principals and perturbation theory both seem like decent candidates, and perhaps the WKB approximation might be able to treat some of the higher excited states. I haven’t given much thought to this, other than that it might be an interesting direction to branch off into given more time, and that it might even be amenable to becoming a future senior thesis topic.

Appendix A Computation of the Dirac Bracket The Dirac Bracket is defined as         A, B D = A, B P − A, φk P Ekl φl , B P ,

(A.1)

  ∆kl = φk , φl P .

(A.2)

and we define For the case of the usual constraint, f (x) = 0, we had a total of four constraints we needed to use to determine u and λ. φ1 = pλ φ2 = p˙ λ = − f (x) 1 φ3 = p¨ λ = − m na pa ...

φ4 = pλ =

1 m

λ−

1

m2

na,b pa pb .

(A.3a) (A.3b) (A.3c) (A.3d)

The Poisson brackets are antisymmetric, so we only need to calculate six of the sixteen entries in ∆ to know the whole thing. The necessary entries to calculate are   ∆12 = φ1 , φ2 P   = pλ , − f (x) P =−

(A.4a) (A.4b)

∂pλ ∂ f (x) ∂ f (x) ∂pλ + ∂ξi ∂pi ∂ξi ∂pi

(A.4c)

= 0,

(A.4d)

  ∆13 = φ1 , φ3 P   1 = pλ , − m na pa P

(A.5a) (A.5b)

∂pλ ∂ =− i ∂ξ = 0,

1


n p ∂ a a m + ∂pi

1


n p ∂pλ a a m i ∂ξ ∂pi

(A.5c) (A.5d)

APPENDIX A. COMPUTATION OF THE DIRAC BRACKET

46

  ∆14 = φ1 , φ4 P  1 λ− = pλ , m ∂pλ ∂ = ∂ξi 1 , = −m

1

na,b pa pb

m2

1

λ− m

(A.6a) (A.6b)

 P

1


∂ na,b pa pb − ∂pi

m2

1

λ− m

1


na,b pa pb ∂pλ ∂ξi ∂pi

m2

(A.6c) (A.6d)

  ∆23 = φ2 , φ3 P   1 = − f (x), − m pa na P
1 ∂ pa na ∂ f (x) ∂ m − = ∂ξi ∂pi
1 = nb m δab na = =   ∆24 = φ2 , φ4 P  1 = − f (x), m λ− 1

1 m

(A.7a) (A.7b) 1


pa na ∂ f (x) ∂ξi ∂pi

m

(A.7d) (A.7e) (A.7f)

na na

1

,

m

1

m2

1

na,b pa pb

(A.8a) (A.8b)


P

∂ ∂ f (x) ∂ m λ − m2 na,b pa pb − = i ∂ξ ∂pi
!
∂ pa pb 1 = − nc − m2 na,b ∂pc   ∂pb ∂pa 1 = m2 nc na,b pa + pb ∂pc ∂pc
1 = m2 nc na,b pa δbc + pb δac
= m12 nb na,b pa + na na,b pb

(A.7c)

1 m

λ−

= 0,

1


na,b pa pb ∂ f (x) ∂ξi ∂pi

m2

(A.8c) (A.8d) (A.8e) (A.8f) (A.8g) (A.8h)

where in the last step I used the identity na na,b = 0.   ∆34 = φ3 , φ4 P (A.9a)  1  1 = − m pa na , m λ − m12 nb,c pb pc P (A.9b) !



1 1 ∂ pa na ∂ λ − m nb,c pb pc ∂ λ− m nb,c pb pc ∂ pa na 1 + (A.9c) = 2 − i m ∂ξ ∂pi ∂ξi ∂pi
!  !
1
∂ pb pc 1 ∂p a 1 = 2 pa na,d n − m nb,c,d pb pc na (A.9d) m b,c m ∂pd ∂pd 

 = m13 pa na,d nb,c pb δcd + pc δbd − nb,c,d pb pc na δad (A.9e)   (A.9f) = m13 pa na,d nb,d pb + pa na,d nd,c pc − nb,c,d nd pb pc   = m13 2pa nd,a nd,b pb − nb,c,d nd pb pc (A.9g) ≡ .

(A.9h)

47 So we have, for the usual constraint, 0  0 ∆kl =   0 

1 m

0 0 −

0

−

1 1

 , − 

0

0



0

m

m

1 m

(A.10)

0



From which we can find E, the matrix inverse of ∆, 

0

− m2 

0

 m2  Ekl =   0

0

−m

m

−m

0

0 0

m



0  . 0  0

(A.11)

We also know that  a b x ,x P   pa , pb P  a  x , pb P   λ, xa P   λ, pa P

=0 =0 = δab =0 = 0.

(A.12a) (A.12b) (A.12c) (A.12d) (A.12e)

The only remaining ingredients for the Dirac brackets are the Poisson Brackets of the form  a  x , φk P   pa , φk P   λ, φk P ,

(A.13a) (A.13b) (A.13c)

 a    x , φ1 P = xa , pλ P ∂xa ∂pλ ∂pλ ∂xa = − ∂ξi ∂pi ∂ξi ∂pi = 0,

(A.14a)

of which there are twelve to compute:

    pa , φ1 P = pa , pλ P ∂pa ∂pλ ∂pλ ∂pa − = ∂ξi ∂pi ∂ξi ∂pi = 0,     λ, φ1 P = λ, pλ P ∂λ ∂pλ ∂pλ ∂λ = i − ∂ξ ∂pi ∂ξi ∂pi = 1,

(A.14b) (A.14c) (A.15a) (A.15b) (A.15c) (A.16a) (A.16b) (A.16c)

APPENDIX A. COMPUTATION OF THE DIRAC BRACKET

48

 a    x , φ2 P = xa , − f (x) P

(A.17a)

a

=−

a

∂x ∂ f (x) ∂ f (x) ∂x + ∂ξi ∂pi ∂ξi ∂pi

(A.17b)

= 0,

(A.17c)

    pa , φ2 P = pa , − f (x) P

(A.18a)

=−

∂pa ∂ f (x) ∂ f (x) ∂pa + ∂ξi ∂pi ∂ξi ∂pi

(A.18b)

= na ,

(A.18c)

    λ, φ2 P = λ, − f (x) P

(A.19a)

=−

∂λ ∂ f (x) ∂ f (x) ∂λ + ∂ξi ∂pi ∂ξi ∂pi

(A.19b)

= 0,

(A.19c)

 a    1 x , φ3 P = xa , − m pb nb P 1

(A.20a)


∂ ∂xa ∂ m pb nb + i ∂ξ ∂pi ∂p b 1 = −m nb ∂pa 1 = −m nb δab =−

1




pb nb ∂xa ∂ξi ∂pi

m

(A.20b) (A.20c) (A.20d) (A.20e)

1

= − m na ,     1 pb nb P λ, φ3 P = λ, − m =−

∂λ ∂ ∂ξi

1

(A.21a)


pb nb ∂ + ∂pi

m

1




pb nb ∂λ ∂ξi ∂pi

m

(A.21b)

= 0,

(A.21c)

    1 pa , φ3 P = pa , − m pb nb P

(A.22a)

1 pb nb ∂ ∂pa ∂ m + i ∂ξ ∂pi 1 = m pb nb,a ,




=−

 a   1 x , φ4 P = xa , m λ− ∂xa ∂ ∂ξi

1

1

m2

m

1

m2

= −2 m12 na,b pb ,

pb nb ∂pa ∂ξi ∂pi

m

(A.22b) (A.22c)



(A.23a)

P




nb,c pb pc ∂ − ∂pi
∂ pb pc 1 = − m2 nb,c ∂pa
1 = − m2 nb,c pb δca + pc δba
= − m12 nb,a pb + na,c pc =

λ−

nb,c pb pc

1




1 m

λ−

1




nb,c pb pc ∂xa ∂ξi ∂pi

m2

(A.23b) (A.23c) (A.23d) (A.23e) (A.23f)

49    1 λ, φ4 P = λ, m λ− =

∂λ ∂ ∂ξi

1 m

1

nb,c pb pc

m2



(A.24a)

P

1

λ−




nb,c pb pc ∂ − ∂pi

m2

1 m

λ−

1




nb,c pb pc ∂λ ∂ξi ∂pi

m2

= 0,

(A.24b) (A.24c)

   1 pa , φ4 P = pa , m λ− 1

1

m2

nb,c pb pc



(A.25a)

P

1

1




∂ ∂pa ∂ m λ − m2 nb,c pb pc − i ∂ξ ∂pi 1 = m2 nb,c,a pb pc , =

m

λ−

1




nb,c pb pc ∂pa ∂ξi ∂pi

m2

(A.25b) (A.25c)

To summarize:  a  x , φ1 P  a  x , φ2 P  a  x , φ3 P  a  x , φ4 P

=0 =0 1 = −m na

= −2 m12 na,b pb ,

  λ, φ1 P   λ, φ2 P   λ, φ3 P   λ, φ4 P   pa , φ1 P   pa , φ2 P   pa , φ3 P   pa , φ4 P

=1 =0 =0 = 0,

=0 = na = =

1 m

1

na,b pb

m2

nb,c,a pb pc .

(A.26a) (A.26b) (A.26c) (A.26d) (A.27a) (A.27b) (A.27c) (A.27d) (A.28a) (A.28b) (A.28c) (A.28d)

Now we finally have all the ingredients for computing the Dirac brackets of interest,  a b       x , x D = xa , xb P − xa , φk P Ekl φl , xb P =0−0 = 0,         pa , pb D = pa , pb P − pa , φk P Ekl φl , pb P          = 0 − pa , φ2 P E23 φ3 , pb P + pa , φ3 P E32 φ2 , pb P 





1 1 nb,c pc + m na,d pd m − nb = − na − m − m = −na nb,c pc + nb na,d pd
= nb na,c − na nb,c pc ,

(A.29a) (A.29b) (A.29c) (A.30a) (A.30b) (A.30c) (A.30d) (A.30e)

APPENDIX A. COMPUTATION OF THE DIRAC BRACKET

50

 a        x , pb D = xa , pb P − xa , φk P Ekl φl , pb P     = δab − xa , φ3 P E32 φ2 , pb P


1 na m − nb = δab − − m = δab − na nb ,  a        x , λ D = xa , λ P − xa , φk P Ekl φl , λ P      = 0 − xa , φ4 P E41 φ1 , λ P


= − − 2 m12 na,b pb − m − 1 1 = 2m na,b pb ,

        pa , λ D = pa , λ P − pa , φk P Ekl φl , λ P          = 0 − pa , φ2 P E21 φ1 , λ P + pa , φ4 P E41 φ1 , λ P 





 = − na m2  − 1 + m12 nb,c,a pb pc − m − 1 
 1 = m2 na m13 2pa nd,a nd,b pb − nb,c,d nd pb pc − m nb,c,a pb pc  
1 =m 2na,c nb,c pa pb − nd nb,c,d pb pc na − na,b,c pb pc .

(A.31a) (A.31b) (A.31c) (A.31d) (A.32a) (A.32b) (A.32c) (A.32d) (A.33a) (A.33b) (A.33c) (A.33d) (A.33e)

To summarize:  a b x ,x D   pa , pb D  a  x , pb D  a  x ,λ D   pa , λ D

=0


= nb na,c − na nb,c pc = δab − na nb 1 = 2m na,b pb  
1 2na,c nb,c pa pb − nd nb,c,d pb pc na − na,b,c pb pc . =m

(A.34a) (A.34b) (A.34c) (A.34d) (A.34e)

Appendix B Mathematica Code I used Mathematica to “shoot” for solutions, requiring the values of Θ(θ) and Θ 0 (θ) to match themselves at 0 and 2π, and obtained some graphical results that were accurate enough for my largely qualitative interests. Below, I’ve given a sample of code that looks for odd solutions to the θ equation for p = 1.5, m = 0. It is set to look for acceptable values of n between 0 and 50, and when it finds one, to output a plot of Θ(θ), as well as √ a plot of g |Θ(θ)|2 , both on the interval [0, 2π]. acc = 1 * 10-4 ; p = 1.5; m = 1; DoA p2 i 2 y j z sol = NDSolveA9j -m + ÅÅÅÅÅÅ Å + n Hp + Cos@qDL2 z z Q@qD + j 4 k { Hp + Cos@qDL H-Sin@qD Q£ @qD + Hp + Cos@qDL Q££ @qDL == 0, Q@0D ã 0, Q£ @0D ã 1=, Q, 8q, 0, 2 p
p 3p Q@qD ê. sol, 8q, 0, 2 p<, Ticks Ø 990, ÅÅÅÅ , p, ÅÅÅÅÅÅÅÅ , 2 p=, 80<=E && 2 2 PlotAHp + Cos@qDL Abs@Q@qDD2 ê. sol, 8q, 0, 2 p<, p 3p Ticks Ø 990, ÅÅÅÅ , p, ÅÅÅÅÅÅÅÅ , 2 p=, 80<=E E, 2 2

8n, 0, 50, 0.001
Bibliography [1] P. A. M. Dirac, Lectures on Quantum Mechanics (Yeshiva University, New York, 1964). [2] N. Wheeler, Classical Mechanics: Fundamentals (Reed College Library, Portland, OR, 1982). [3] M. Henneaux and C. Teitelboim, Quantization of Gauge Systems (Princeton University Press, Princeton, NJ, 1992). [4] M. Ikegami, Y. Nagaoka, S. Takagi, and T. Tanzawa, Prog. Theoret. Phys. 88, 241 (1992). [5] D. J. Griffiths, Introduction to Quantum Mechanics (Prentice Hall, Inc., Upper Saddle River, NJ, 1995). [6] R. C. T. da Costa, Phys. Rev. A 23, 1981 (1982). [7] N. Ogawa, K. Fujii, and A. Kobushukin, Prog. Theoret. Phys. 83, 894 (1990). [8] C. Lanczos, The Variational Principals of Mechanics (University of Toronto Press, Toronto, 1966). [9] H. Goldstein, Classical Mechanics (Adison-Wesley, Reading, MA, 1965). [10] H. Aoki, M. Koshino, D. Takeda, and H. Morise, arxiv.org/cond-mat/0306729 (2003).