Quantum Dynamics of Coin Tossing

IASc-INSA-NASI Summer ResearchFellowship 2013

Project Report

Quantum dynamics of coin-tossing

Mohit Pandey 1

under the guidance of

Dr.Arul Lakshminarayan 2

1IISER, Kolkata Email: [email protected] Institute of Technology, Madras Email: [email protected]

1

Acknowledgement

I would be like to acknowledge Professor Lakshminarayan’s great insightand experience in giving me the problem, which was accessible to me withmy present level of education and yet challenging at each and every step. Itprovided me an opportunity to think afresh about the things learnt beforeand also to learn a lot of new stuff. Further, I would like to thank him forthe freedom and flexibility he granted me while working on the problem.

There are many warm memories for me to carry back to home. One-to-one discussions which I had with him invariably gave me an easier way to dothe calculation, which I would have been struggling to complete. After thediscussions, often I went back with a clearer head and thinking why I didn’tthink it before! Moreover,the thursday chalk-and-talk presentations werea good excuse for shrugging off the laziness, besides being a great learningexperience. It was educational to hear other students of the group explainingtheir different problems and how they were attacking.

I would like to thank Department of Physics, IIT-M for the classes onwide-array of topics-experimental and theoretical. It exposed me to bothcutting-edge research being undertaken and chinks in my armor of Physicseducation. Of course, Professor V.Balakrishnan’s classes deserve a specialmention.

To conclude, I owe to Prof. Lakshminarayan the most important lessonlearnt during the project. You put in efforts in getting to the solution afterworking through some equations and then you think you are done. ButPhysics doesn’t end here. Now it’s important to think why did you get thissolution, does it seem reasonable to you, how can you check if you got itright(apart from just rechecking your calculation). In short, to think whythings are the way they are. Now that’s called Physics!

2

Abstract

The basic question is how random coin-tossing is. It has been attacked usingquantum mechanics. Classically, it has been studied in [1]. An attempt hasbeen made to build upon it. The problem becomes much richer with thetypical gear shift expected from classical to quantum region. In quantumdomain, we would be studying wave-packet of the coin to understand itstime-evolution and dependence of evolution on initial conditions.

3

Contents

1 Classical tossing of a coin 51.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 The preimages of Head . . . . . . . . . . . . . . . . . . . . . . 71.3 Probability of heads . . . . . . . . . . . . . . . . . . . . . . . 91.4 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Quantum tossing of a coin 102.1 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.1.1 Coherent state . . . . . . . . . . . . . . . . . . . . . . 102.1.2 Initial state . . . . . . . . . . . . . . . . . . . . . . . . 112.1.3 Evolution . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.2 Initial state . . . . . . . . . . . . . . . . . . . . . . . . 142.2.3 Evolution . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.4 Quantum Probability of head . . . . . . . . . . . . . . 17

Appendix A Detailed Calculations 20A.1 CBH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20A.2 exponential constants . . . . . . . . . . . . . . . . . . . . . . 21A.3 wave-function in position basis . . . . . . . . . . . . . . . . . 21A.4 Expectation of absolute of sin φ . . . . . . . . . . . . . . . . . 28A.5 Quantum Probability of heads . . . . . . . . . . . . . . . . . . 30A.6 Mean angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4

Chapter 1

Classical tossing of a coin

1.1 Introduction

Here we would be reproducing the results of [1] and answering the questionsraised in the paper which are:

Why is the outcome of a coin-toss considered to be random, even thoughit is uniquely determined by the laws of physics and the initial conditions?If it is random, then we should not be able to predict anything.Then whyis there a definite probability associated with each outcome, regardless ofhow the coin is tossed? Finally, if there is definite probability for each out-come(which we from our experience, it does exist) how it can be calculated?

In our attempt to answer these questions, we write the Hamiltonian ofa coin, with mass m and moment of interia I, tossed up in air as

H =p2x

2m+p2φ

2I+mgx (1.1)

where px & pφ being linear and angular momentum, respectively. Theaxes,with y-axis on ground, are shown in the following figure.

The coin-tosses are usually restricted to the height where gravitationalpotential can be approximated as mgx. Thanks to the symmetry of the coin,x and y components of total angular momentum cancel to give the resultantpointing in z-direction. As a result, effectively its angular momentum issame as that of a stick of same moment of inertia rotating around an axisperpendicular to the plane on which it’s placed.

5

Figure 1.1: Coordinate system for coin-tossing

Now, let’s get down to business. Using Hamilton’s equations, forH(x, px, pφ)

∂H∂x

=− px (1.2)

∂H∂px

=x (1.3)

∂H∂φ

=− pφ (1.4)

∂H∂pφ

=φ (1.5)

On solving above equations, we get nicely separated linear(equation 1.6 )and angular equations of motion (equation 1.7).

x(t)=− g (1.6)

φ(t)=0 (1.7)

With initial data as

x(0) = a, x(0) = u (1.8)

φ(0) = 0, φ(0) = ω (1.9)

we get

x(t) = a+ ut− gt2

2(1.10)

φ(t) = ω (1.11)

Here we have assumed that we toss with the head facing up. We shall assumethat whichever side of the coin is up at the time t0 > 0 when it hits the floor

6

remains up. Floor is like sand where it gets stuck. Now when will the coinend its motion with head up? It will do so if at the time t0

2nπ − π

2< φ(t0) < 2nπ +

π

2(1.12)

And how do we find the time t0 ? It is smallest positive root of:

x(t0)− a| sin(φ(t0))| = 0 (1.13)

1.2 The preimages of Head

The set of all pairs of u, ω of nonnegative values for which the coin ends upheads,we shall call the pre-image of heads in the u, ω plane, and we shalldenote H( and pre-image of tails by T) First, we consider end points of theintervals in (1.12), which are given by

φ(t0) = 2nπ ± π

2(1.14)

Using this, we get t0 =2u

g. As a result,

ω =(

2nπ ± π

2

) 1

2u/g, n = 0, 1, 2, 3... (1.15)

We see that there are two branches of solution – ω+ and ω−. In first quad-rant of u-ω plane, ω− starts from n=1, while ω+ from n=0. Thus, in firstquadrant of u-ω plane, we would arrange firstly, ω+ with n=0 followed byω− with n=1, then ω+ with n=1 followed by ω− with n=2 and so on.

7

Figure 1.2: Keller plot of u/g versus ω

The region closest to u/g−ω axes correspond to H followed by the stripof T. The strips in figure(1.2) are alternately H and T.

Let’s try to understand the plot. We concentrate on the strip closestto axes. For points close to origin, it is obvious that we get a head. Thestate of the coin stayed roughly the same as what it started throughout itsevolution. So, there is no surprise we get a head. Further, for points closeto u/g-axis, the angular velocity ω is close to 0, then the coin again ends upwith head as ω was not high enough to change its face facing up. For largeru (and Tflight larger), ω has to be smaller and smaller for one to get a head.Lastly, for points close to ω axis, at first sight it’s surprising that one getsa head at all if one thinks that ω is quite high as compared to u/g, we caneither land up with head or tail. But it is not so. Since, u/g tends to be0, so goes the time of flight. Thus, the coin gets no time to change its’ facealthough ω is quite high. That is why the strip gets narrower as the ω getshigher. Thus, again we land up with head.

Other strips can be understood on similar lines. Also as value of u andω increases, it’s noticed that area of alternate strips keeps on decreasing.Thus, for a slight change for initial conditions, you can either land up withhead or tail.

Moreover, on putting in the values of u and ω of the typical coin tosses,we do find that they lie in the far upper right region of the above figure.

8

1.3 Probability of heads

Let’s take u and ω to be random variables with continuous probability den-sity p(u, ω) with non-zero value only where u, ω > 0. Thus, probabilitythat one gets a head would be given by

PH =

∫ ∫Hp(u, ω)dudω (1.16)

Keller showed that

PH = limu→∞ω→∞

∫ ∫Hp(u, ω)dudω =

1

2(1.17)

In this way, he arrives at the empirical result that probability for getting ahead is same as that of getting a tail.

1.4 Result

The crux of Keller’s work lies in the u− ω plot which is an important toolfor thinking about the problem. Also, we find that coin-tossing is not thatrandom as often told. In some cases, you can exactly tell the outcome of asingle coin-toss.

9

Chapter 2

Quantum tossing of a coin

Using Dirac’s recipe of going from classical to quantum-mechanical dynamicsby replacing observables with corresponding operators, we get Hamiltonianof a quantum toss as

H =p2

2mi+mggx +

L2z

2I(2.1)

where mi and mg denote inertial and gravitational masses, respectively.The basic problem is to find the evolution of the state ket |ψ(t)〉, which

describes the dynamics of coin, when started with an initial ket |ψ(0)〉.Since, Hamiltonian of the system is independent of time, we can write

|Ψ(t)〉 = exp (−iHt/~) |Ψ(0)〉= exp (−iHxt/~) |Ψx(0)〉 ⊗ exp (−iHφt/~) |Ψφ(0)〉 ∵ [Hx,Hφ] = 0

where ⊗ stands for direct product between x and φ space.As a result, the problem has been broken down into two separate parts-

translation and rotation of the coin.

2.1 Translation

To find out the evolution of wave-function, we need the initial state. Beforethat, let’s take a detour to study coherent states.

2.1.1 Coherent state

A coherent state is a state which is an eigen-state of annihilation/loweringoperator a, i.e.

a|α〉 = α|α〉 (2.2)

where

a =

√miω

2~

(x +

ip

miω

)(2.3)

10

After this, let’s study the translation in position and momentum basiswhich would be found useful later.

Position boost

〈x|e−ibp/~|x0, p0〉 = 〈x− b|x0, p0〉

=1

π1/4√σ

exp

[ip0(x− b)

~− (x− (x0 + b))2

2σ2

]= exp

[− ip0b

~

]〈x|x0 + b, p0〉 (2.4)

Thus, position gets a boost by momentum operator, which is the translationgenerator.

e−ibp

~ |x0, p0〉 = e−ip0b

~ |x0 + b, p0〉 (2.5)

Momentum boost

〈x|e−iax/~|x0, p0〉 = e−iax/~〈x|x0, p0〉

=1

π1/4√σ

exp

(−iax~

)exp

[ip0(x)

~− (x− x0)2

2σ2

]= 〈x|x0, p0 − a〉 (2.6)

Thus, momentum gets a boost by position operator, which is the generatorof translation in momentum space.

e−iax

~ |x0, p0〉 = |x0, p0 − a〉 (2.7)

2.1.2 Initial state

Keeping into mind, Heisenberg’s uncertainty relationship, the initial stateshould be chosen so that lower limit of product of uncertainties of positionand momentum is reached. This can be achieved by using coherent state.Let the initial state be chosen as follows

〈x|x0, p0〉 =,1

π1/4√σ

exp

[ip0x

~− (x− x0)2

2σ2

], σ2 =

~miω

(2.8)

It can also be represented in momentum basis as

〈p|x0, p0〉 =

√σ√

2

~√π

exp

[ix0

~(p0 − p)−

(p− p0)2σ2

~2

], σ2 =

~miω

(2.9)

11

Now, the initial state chosen is an eigenfunction of a with eigen-equationgiven by

a

∣∣∣∣x0 + ip0

miω

⟩=

(x0 + i

p0

miω

) ∣∣∣∣x0 + ip0

miω

⟩(2.10)

We can also label the state as an ordered pair (x0, p0). This can be thoughtof as a circular gaussian state with miω = 1 to give same uncertainty inposition and momentum. Now, let’s attack the problem.

2.1.3 Evolution

Let’s study the evolution of translation part of state ket dropping the x-labelfrom kets.

|Ψ(t)〉 = exp (−iHxt/~) |Ψ(0)〉

= exp

(− ip2t

2mi~− itmggx

~

)|Ψ(0)〉

= exp(−iαp2 − iβx

)|Ψ(0)〉 (2.11)

where

α =t

2mi~, β =

mggt

~(2.12)

To nicely separate out the operators involved in the exponential, we usingBaker Campbell Hausdorff(BCH) formula [3] for two operators X and Y:

log(eXeY

)= X + Y +

1

2[X,Y ] +

1

12[X, [X,Y ]]− 1

12[Y, [X,Y ]] + · · ·

|Ψ(t)〉 = exp(−iαp2 − iβx

)|x0, p0〉

= exp

(2

3iαβ2~2 − iαp2

)exp(−iβx) exp(−i~αβp)|x0, p0〉, refer to Appendix

= exp

(2

3iαβ2~2 − iαβp0~− iαp2

)|x0 + αβ~2, p0 − β~〉

= exp

(2

3iαβ2~2 − iαβp0~

)∫dp exp(−iαp2)〈p|x0 + αβ~2, p0 − β~〉|p〉

(2.13)

Thus,

〈p|Ψ(t)〉 = exp

(2

3iαβ2~2 − iαβp0~− iαp2

)〈p|x0 + αβ~2, p0 − β~〉

=

√σ

~√πe−

i3αβ

2~2e−ix0~ (p− (p0 − β~))e−i~αβp ×

e−iαp2exp−(p+ β~− p0)2σ2

2~2(2.14)

12

Putting the values of α and β

〈p|Ψ(t)〉 =

√σ

~√π

exp i(mi/~){−(mg/mi)2 1

6g2t3} − ix0

~(p− (p0 −mggt))×

exp−i ~(mg/mi)12gt

2p− i tp2

2mi~ −(p− (p0 −mggt))

2σ2

2~2

(2.15)

But the task in not yet finished. After some tedious but straightforwardalgebra (for details, refer to Appendix), we get wave-function in positionbasis as

〈x|Ψ(t)〉

= N exp

− i6m2gg

2t3

mi~−tx0( p0mi

− mg

migt)

2σst−

(x− x0 −

mggt2

2mi

)2

4σst

×

exp

i(p0 −mggt)

(x− p0t

2mi

)σ

2~st

(2.16)

where

st =σ

2

(1 +

it~miσ2

), σ2 =

~miω

(2.17)

and

N =

√2

(πσ2)1/4

√√√√√1 +

(t~

4miσ2

)2

+ 1 + i

√√√√√1 +

(t~

4miσ2

)2

− 1

(2.18)

2.2 Rotation

2.2.1 Basis

We would be using the eigen-kets of φ and Lz as basis-{|φ〉}&{|m〉}. φ ∈[−π, π] has been chosen so that the interval is symmetric about φ = 0. Also

〈φ|m〉 =1√2πeimφ, m ∈ Z (2.19)

13

m is restricted to integral values because of periodicity of angle basis.

〈φ+ 2π|m〉 = 〈φ|m〉 (2.20)

As a result, {|m〉} basis set is discrete while {|φ〉} continuous, which isevident as ∑

m

|m〉〈m| =∫ π

−πdφ|φ〉〈φ| = I (2.21)

2.2.2 Initial state

Unlike translation part, Heisenberg’s uncertainty relationship for angle φand angular momentum Lz(= pφ) doesn’t hold true.[2]

∆φ∆Lz � ~/2 (2.22)

All eigen-states of Lz violate it. In these states,

∆ Lz = 0,∆φ =π√3

(2.23)

Why don’t we use one of these as our initial state? This state is quiteeconomical with the product of uncertainties in Lz and φ being zero. Butit’s time-evolution is too drab. If the initial state is an angular momentumeigen-state, then the expectation value of angular momentum would remainconstant because Hφ is only dependent on Lz. The evolved state at time twould differ only by a phase-factor. This is not suitable for the problem athand. For coin-tossing, expectation value of angular momentum needs tovary as a function of time.

Instead, taking cue from the translation part, we start with a guassianstate localized around φ = 0

〈φ|Ψ(0)〉 =1

(πσ2)1/4e

iL0φ

~ e−φ2

2σ2 , σ � π (2.24)

The expectation values of initial angle and angular momentum being 0 andL0. The condition on σ is necessary so that approximately

〈φ+ 2π|Ψ(0)〉 ' 〈φ|Ψ(0)〉 (2.25)

14

In angular momentum basis

〈m|Ψ(0)〉

=

∫ π

−πdφ〈m|φ〉〈φ|Ψ(0)〉

=1

(πσ2)1/4

∫ π

−πdφ〈m|φ〉e

iL0φ

~ e−φ2

2σ2

=1√

2π × (πσ2)1/4

∫ π

−πdφe−imφe

iL0φ

~ e−φ2

2σ2

' 1√2π × (πσ2)1/4

∫ ∞−∞

dφe−imφe

iL0φ

~ e−φ2

2σ2

=

(σ√π

)1/2

e−(m−L0/~)2σ2(2.26)

Thus, we find the value of 〈m|Ψ(0)〉 approximately. But wait for asecond! Why don’t we start with a state whose value is exactly equal to theabove value? Since, the Hamiltonian is just dependent on Lz, in m-basis,the state would be simply evolving by a phase factor. Thus, it’s better totake an initial state to be exactly given by (2.26) but then the initial statein φ basis would be now approximately given by the form (2.24). We neednot find its exact form in φ basis. Thus,

〈m|Ψ(0)〉 = Ne−(m−L0/~)2σ2(2.27)

where N is a normalization constant. As a bargain, this makes the period-icity exact because

〈φ|Ψ(0)〉 =∑m

eimφ√2π〈m|Ψ(0)〉 (2.28)

At the end, we have come up with an initial state with necessary ingredientswith N given by

N2 =1∑m=∞

m=−∞ e−2(m−L0/~)2σ2 (2.29)

This can be computed numerically. There is no dynamical role of N.

2.2.3 Evolution

z-subscript and φ superscript from Lz and |Ψφ〉, respectively can be droppedsafely for a while.

15

Figure 2.1: | sinφ(t)|

2.2.4 Quantum Probability of head

Since, the above plot doesn’t throw much light, let’s find the quantum prob-ability of a head. If the angle φ ∈ [−π/2, π/2] at time t, then we would saythat the coin has head facing up. The probability that one gets a head if onestart with head face(with regards to coordinate axes shown in figure(1.1))is given by

PH(t) =

∫ π/2

−π/2dφ|Ψ(φ, t)|2

=1

2+N2

π

∑m−m′=odd

exp[− i(m2 −m′2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]sin(m−m′)π/2

m−m′(2.35)

Again the details can be found on Appendix. On plotting it, we get figure(2.2.4).

17

Figure 2.2: PH(t)

The units have been chosen so that time-period comes to be unity in theplot. We find that PH oscillates around classical value of half, which is whatwe expected in a way. But the kink at half-time period is not understoodas of now.

18

Conclusion

This concludes the work that I have done on the problem as of now. TheKeller’s work [1] was revisited first so that to get a grip of things in clas-sical domain. Then quantum version of the problem was attacked. Wave-functions of coin, for its both translation and rotation, were obtained. Fur-ther, the quantum probabilty PH(t) and expectation of absolute of sin φ asbeen computed. The PH(t) oscillates around the classical value half.

The problem remains unsolved and a lot of work needs to be done. Weneed to understand the oscillations and other features in both PH(t) and〈| sinφ|〉 plots. Understanding them is crucial. Moreover, an equivalentquantum plot of Keller’s u-ω is the key thing which remains evasive. Itwould tell us the dependence of the outcome of the quantum coin tossingon the initial conditions.

In short, the ground has been set and one needs to capitalize on that tounravel the physics of quantum coin tossing.

19

Appendix A

Detailed Calculations

Here the detailed calculations are given for the expressions shown in thereport

A.1 CBH

We use the following trick

eA+B = eA(e−AeA+B

)= eAeB−

12

[A,B]+ 112

[A,[A,B]]+ 112

[A+B,[A,A+B]]··· (A.1)

Let’s apply this with

A = −iαp2,B = −iβx (A.2)

Let’s first calculate

eiαp2e−iαp

2 − iβx

= e−iβx−12(2i~αβp) + 1

12 [−iαp2, 2i~αβp] + 112 [−iαp2 − iβx, 2i~αβp] · · ·

= e−iβx− i~αβp + 0 + 112 [−iβx, 2i~αβp]

= e−iβx− i~αβp + 16 iαβ

2~2(A.3)

Now, it’s easy to see that

e−iαp2 − iβx

= e−iαp2 (eiαp

2e−iαp

2 − iβx)

= e16 iαβ

2~2e−iαp

2e−iβx− i~αβp (A.4)

Again

A = −iβx,B = −i~αβp (A.5)

20

e−iβx− i~αβp

= e−iβx(eiβxe−iβx− i~αβp

)= e−iβxe−i~αβp−

12 [−iβx,−i~αβp]

= e−iβxe−i~αβp + 12 i~αβ (A.6)

Stitching them together, we get

e−iαp2 − iβx = e

23 iαβ

2~2e−iαp

2e−iβxe−i~αβp (A.7)

A.2 exponential constants

For r =√a+ ib, δ = tan−1(b/a)

e−iδ/2√r

=1√r× (cos(δ/2)− i sin(δ/2))

=1√r×

(√1 + cos δ

2− i√

1− cos δ

2

)=

1√2r×(√

1 + cos δ − i√

1− cos δ)

=1√2r×(√

1 + a/r − i√

1− a/r)

=1

r√

2×(√r + a− i

√r − a

)=

1

r√

2× r + a+ r − a(√

r + a+ i√r − a

)=

√2(√

r + a+ i√r − a

) (A.8)

A.3 wave-function in position basis

〈x|Ψ(t)〉

= exp

(2

3iαβ2~2

)exp(−iαβp0~)

∫dpe−iαp

2〈p|x0 + αβ~2, p0 − β~〉〈x|p〉

= e23iαβ2~2e−iαβp0~I (A.9)

21

I =

∫dpe−iαp

2〈p|x0 + αβ~2, p0 − β~〉〈x|p〉

=

√σ

2π~2√π

∫dp exp

[−iαp2 + ipx/~+ i(x0 + αβ~2)(p0 − β~)/~

]×

exp

[−i(x0 + αβ~2)p/~− (p− (p0 − β~))2σ2

2~2

]=

√σ

2π~2√πei(x0+αβ~2)(p0−β~)/~

∫dp exp

[−iαp2 + ipx/~

]×

exp

[−i(x0 + αβ~2)p/~− (p− (p0 − β~))2σ2

2~2

]=

√σ

2π~2√πei(x0+αβ~2)(p0−β~)/~I1 (A.10)

I1 =

∫dp exp

[−iαp2 + ipx/~− i(x0 + αβ~2)p/~− (p− (p0 − β~))2σ2

2~2

]I1 = exp

[−(p0 − β~)2 σ

2

2~2

] ∫dp exp

[−iαp2 + ip

(x− x0

~− αβ~

)−p

2σ2

2~2+p(p0 − β~)σ2

~2

]I1 = exp

[−(p0 − β~)2 σ

2

2~2

]I2 (A.11)

I2 =

∫dp exp

[−(iα+

σ2

2~2

)p2 + ip

(x− x0

~− αβ~

)+p(p0 − β~)σ2

~2

]=

∫dp exp

[−(iα+

σ2

2~2

){p2 + pµ

}]=

∫dp exp

[−(iα+

σ2

2~2

){(p− µ/2)2 − µ2

4

}]= exp

[(iα+

σ2

2~2

)µ2

4

] ∫dp exp

[−(iα+

σ2

2~2

){(p− µ/2)2

}]= exp

[(iα+

σ2

2~2

)µ2

4

]I3 (A.12)

where

µ =

i

(x− x0

~− αβ~

)+

(p0 − β~)σ2

~2

σ2

2~2+ iα

(A.13)

22

and

I3 =

∫dp exp

[−(iα+

σ2

2~2

){(p− µ/2)2

}]=

√π

re−iδ/2 (A.14)

where

r =

√σ4

4~4+ α2, δ = tan−1(2α~2/σ2) (A.15)

From equation(A.8), we have

e−iδ/2√r

=

√2(√

r + a+ i√r − a

) , a =σ2

2~2(A.16)

Thus, keeping constants together, we have√σ

2π~2√π

√π

re−iδ/2

=

√σ

2π~2√π

√π

√2(√

r + a+ i√r − a

)=

√σ

~2√π

1(√r + a+ i

√r − a

)=

√σ

~2√π

1(√σ4

4~4+ α2 +

σ2

2~2

)1/2

+ i

(√σ4

4~4+ α2 − σ2

2~2

)1/2

=

√σ

~2√π

1

σ

~√

2

(√1 +α2

σ4/4~4+ 1

)1/2

+ i

(√1 +

α2

σ4/4~4− 1

)1/2

=

√2

σ√π

1(√1 +α2

σ4/4~4+ 1

)1/2

+ i

(√1 +

α2

σ4/4~4− 1

)1/2

= N (A.17)

23

Let’s simplify the expression

〈x|Ψx(t)〉

= N exp

[2

3iαβ2~2 − iαβp0~+ i(x0 + αβ~2)(p0 − β~)/~

]×

exp

[−(p0 − β~)2 σ

2

2~2+

(iα+

σ2

2~2

)µ2

4+

]= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~− (p0 − β~)2 σ

2

2~2

]×

exp

[+

(iα+

σ2

2~2

)µ2

4

]= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~− (p0 − β~)2 σ

2

2~2

]×

exp

1

4

(iα+

σ2

2~2

) i(x− x0

~− αβ~

)− (p0 − β~)σ2

~2

σ2

2~2+ iα

2

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~− (p0 − β~)2 σ

2

2~2

]×

exp

{i

(x− x0

~− αβ~

)+

(p0 − β~)σ2

~2

}2

4

(σ2

2~2+ iα

)

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~− (p0 − β~)2 σ

2

2~2

]×

exp

−(x− x0

~− αβ~

)2

+(p0 − β~)2σ4

~4

4

(σ2

2~2+ iα

)×

exp

2i(p0 − β~)σ2

~2

(x− x0

~− αβ~

)4

(σ2

2~2+ iα

)

24

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~

]×

exp

−(x− x0

~− αβ~

)2

− 2iα(p0 − β~)2σ2

~2

4

(σ2

2~2+ iα

)×

exp

2i(p0 − β~)σ2

~2

(x− x0

~− αβ~

)4

(σ2

2~2+ iα

)

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~

]×

exp

−(x− x0

~− αβ~

)2

− 2iα(p0 − β~)2σ2

~2

4

(σ2

2~2+ iα

)×

exp

2i(p0 − β~)σ2

~2

(x− x0

~

)4

(σ2

2~2+ iα

)×

exp

−2i(p0 − β~)σ2

~2αβ~

4

(σ2

2~2+ iα

)

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~

]×

exp

−(x− x0

~− αβ~

)2

4

(σ2

2~2+ iα

)×

exp

2i(p0 − β~)σ2

~2

(x− x0

~

)4

(σ2

2~2+ iα

)×

exp

−2i(p0 − β~)σ2

~2α(p0 − β~+ β~)

4

(σ2

2~2+ iα

)

(A.18)

25

= N exp

[− i

3αβ2~2 + ix0(p0 − β~)/~

]×

exp

−(x− x0

~− αβ~

)2

4

(σ2

2~2+ iα

)×

exp

i(p0 − β~)σ2

~2

(x− x0

~− αp0

)2

(σ2

2~2+ iα

)

= N exp

− i3αβ2~2 − αx0(p0 − β~)/~(σ2

2~2+ iα

)×

exp

−(x− x0

~− αβ~

)2

4

(σ2

2~2+ iα

)×

exp

i(p0 − β~)σ2

~2

(x~− αp0

)2

(σ2

2~2+ iα

)

= N exp

− i3αβ2~2 − ~αx0(p0 − β~)

~2

(σ2

2~2+ iα

)×

exp

−(x− x0 − αβ~2

)24~2

(σ2

2~2+ iα

)×

exp

i(p0 − β~) (x− αp0~)σ2

2~3

(σ2

2~2+ iα

)

(A.19)

26

= N exp

− i3αβ2~2 − ~αx0(p0 − β~)

σ2

(1

2+ i

α~2

σ2

)×

exp

−(x− x0 − αβ~2

)24σ2

(1

2+ i

α~2

σ2

)×

exp

i(p0 − β~) (x− αp0~)σ2

2~σ2

(1

2+ i

α~2

σ2

)

= N exp

[− i

3αβ2~2 − ~αx0(p0 − β~)

σst

]×

exp

[−(x− x0 − αβ~2

)24σst

]×

exp

[i(p0 − β~) (x− αp0~)σ2

2~σst

](A.20)

where

st = σ

(1

2+ i

α~2

σ2

)(A.21)

Putting the value of α, β

〈x|Ψ(t)〉

= N exp

− i6m2gg

2t3

mi~−tx0( p0mi

− mg

migt)

2σst−

(x− x0 −

mggt2

2mi

)2

4σst

×

exp

i(p0 −mggt)

(x− p0t

2mi

)σ

2~st

(A.22)

where

st =σ

2

(1 +

it~miσ2

), σ2 =

~miω

(A.23)

27

and

N =

√2

(πσ2)1/4

√√√√√1 +

(t~

4miσ2

)2

+ 1 + i

√√√√√1 +

(t~

4miσ2

)2

− 1

(A.24)

A.4 Expectation of absolute of sin φ

for equation (1.13)

〈| sinφ|〉Ψ(t)

= 〈Ψ(t)|(| sinφ|)|Ψ(t)〉=

∑m,m′

〈Ψ(t)|m〉〈m|(| sinφ|)|m′〉〈m′|Ψ(t)〉

=∑m,m′

〈Ψ(t)|m〉〈m′|Ψ(t)〉∫ ∫

dφdφ′〈m|φ〉〈φ(| sinφ|)|φ′〉|〈φ′|m′〉

=∑m,m′

〈Ψ(t)|m〉〈m′|Ψ(t)〉∫ π

−πdφ| sinφ|〈m|φ〉〈φ|m′〉

=∑m,m′

〈Ψ(t)|m〉〈m′|Ψ(t)〉I

28

where

I =

∫ π

−πdφ| sinφ|〈m|φ〉〈φ|m〉

=1

2π

[∫ π

−πdφ| sinφ|ei(m′−m)φ

]=

1

2π

[−∫ 0

−πdφ sinφeikφ +

∫ π

0dφ sinφeikφ

], k = m′ −m

=1

2π × 2i

[−∫ 0

−πdφ(eiφ − e−iφ)eikφ +

∫ π

0dφ(eiφ − e−iφ)eikφ

]=

1

2π × 2i

[∫ 0

−πdφ(−e(k+1)iφ + e(k−1)iφ) +

∫ π

0dφ(e(k+1)iφ − e(k−1)iφ)

]

=1

2π × 2i2

− e(k+1)iφ

k + 1

∣∣∣∣∣0

−π

+e(k−1)iφ

k − 1

∣∣∣∣∣0

−π

+e(k+1)iφ

k + 1

∣∣∣∣∣π

0

− e(k−1)iφ

k − 1

∣∣∣∣∣π

0

=

1

2π × 2i2

[−1− (−1)k+1

k + 1+

1− (−1)k−1

k − 1+

(−1)k+1 − 1

k + 1− (−1)k−1 − 1

k − 1

]=

1

2π × 2i2

[−2

1− (−1)k+1

k + 1+ 2

1− (−1)k−1

k − 1

]=

1− (−1)k+1

2π

[1

k + 1− 1

k − 1

]=

1 + (−1)k

2π

[−2

k2 − 1

]=

1 + (−1)k

π(1− k2), k = m′ −m (A.25)

What about the limiting value, i.e. k → ±1 ? We cannot use the aboveexpression while taking a limit as k has been assumed to be an integer. Ondoing a separate calculation for k = ± 1(or the whole calculation can berepeated keeping into mind k ∈ R and then taking a limit), the expectationvalue for such values of k comes out to be zero.

29

A.5 Quantum Probability of heads

for equation(2.35)

PH(t) =

∫ π/2

−π/2dφ|Ψ(φ, t)|2

=

∫ π/2

−π/2dφ〈Ψ(t)|φ〉〈φ|Ψ(t)〉

=

∫ π/2

−π/2dφ

( ∞∑m=−∞

N√2π

exp[−imφ+im2t~

2I− (m− L0/~)2σ2]

)×( ∞∑

m′=−∞

N√2π

exp[im′φ− im′2t~2I

− (m′ − L0/~)2σ2]

)

=N2

2π

∞∑m,m′=−∞

exp[− i(m′2 −m′2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]

∫ π/2

−π/2dφei(m

′−m)φ

=N2

2π

∞∑m,m′=−∞

exp[− i(m′2 −m2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]2 sin(m′ −m)π/2

m′ −m

=N2

2π

∑m−m′=odd

exp[− i(m′2 −m2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]2 sin(m′ −m)π/2

m′ −m+N2

2π

∑m=m′

π exp[−2(m− L0/~)2σ2]

=1

2+N2

2π

∑m′−m=odd

exp[− i(m′2 −m2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]2 sin(m′ −m)π/2

m′ −m(A.26)

30

A.6 Mean angle

〈φ(t)〉 =

∫ π

−πdφφ|Ψ(φ, t)|2

=

∫ π

−πdφφ〈Ψ(t)|φ〉〈φ|Ψ(t)〉

=

∫ π/2

−π/2dφφ

( ∞∑m=−∞

N√2π

exp[−imφ+im2t~

2I− (m− L0/~)2σ2]

)×( ∞∑

m′=−∞

N√2π

exp[im′φ− im′2t~2I

− (m′ − L0/~)2σ2]

)

=N2

2π

∞∑m,m′=−∞

exp[− i(m′2 −m′2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]

∫ π

−πdφφei(m

′−m)φ

=N2

2π

∑m−m′ 6=0

exp[− i(m′2 −m2)t~

2I− (m− L0/~)2σ2]×

exp[−(m′ − L0/~)2σ2]2i

(m′ −m)2(sin(m′ −m)π − π(m′ −m) cos(m′ −m)π)

=N2

2π

∑m′−m>0

exp[−(m′ − L0/~)2σ2 − (m− L0/~)2σ2]×

−2iπ

(m′ −m)cos(m′ −m)π × (A.27)[

exp

{− i(m

′2 −m2)t~2I

}− exp

{− i(m

′2 −m2)t~2I

}]=

N2

2π

∑m′−m>0

exp[−(m′ − L0/~)2σ2 − (m− L0/~)2σ2]×

−2iπ

(m′ −m)cos(m′ −m)π × 2i sin

(m′2 −m2)t~2I

exp[−(m′ − L0/~)2σ2]2 sin(m′ −m)π/2

m′ −m(A.28)

=N2

2π

∑m′−m>0

exp[−(m′ − L0/~)2σ2 − (m− L0/~)2σ2]×

4π

(m′ −m)cos(m′ −m)π × sin

(m′2 −m2)t~2I

(A.29)

31

Bibliography

[1] Joseph B Keller. The probability of heads. The American MathematicalMonthly, 93(3):191–197, 1986.

[2] Asher Peres. Quantum theory: concepts and methods, volume 72.Springer, 1995.

[3] Wikipedia. Baker–campbell–hausdorff formula — wikipedia, the freeencyclopedia. 2014. [Online; accessed 15-July-2013].

32

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