BACHELOR THESIS

for the Degree ofBachelor of Science (BSc)

at theUniversity of Vienna

Quantum Information and Entropy

submitted by

Sridhar BulusuMEng

supervised by

Univ.Prof.Dr Reinhold Bertlmann

Contents

1 Acknowledgements 3

2 Shannon and von Neumann Entropy 4

3 Entropy Inequalities for Finite Dimensional Hilbert Spaces 4

4 Problem with the Peierls-Bogylobov-Inequality 8

5 Dirac’s opinion on the derivative of the logarithm 9

6 Complex Vector Operations 106.1 Multiplication of Complex Numbers . . . . . . . . . . . . . . . . . . . . . 106.2 Comparison of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 11

7 Separability - Entanglement - Negentropy - Non-Locality 137.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

8 The Problem of Time 218.1 Principle of Least Action in Classical Mechanics . . . . . . . . . . . . . . 218.2 Relativistic Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

9 Clock Time and Entropy 239.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239.2 Inaccessibility of coordinate time . . . . . . . . . . . . . . . . . . . . . . . 239.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

A Proposing an Extension of Einstein’s Energy-Mass equivalence 29

B Using the Definition of Euler Mascheroni Constant to find∫∞

0 1/xdx 30

References 31

2

1 Acknowledgements

I would like to specially thank the following people for their support and inspiration

Univ.Prof.Dr. Reinhold Bertlmann- My supervisor for his faithful guidance.

Dr. Atchyuta Rama Chandra Murty Bulusu and Uma Bulusu- My parents for their lifelong support.

BSc. Johanna Matausch- Deriving the integral using the definition of the Euler-Mascheroni Constant would not

have been possible without her.

Mag. Renate Till, Mag. Martin Walter and Mag. Andrea Stettinger- My high school Physics, Mathematics and English teachers for their inspiration.

Srinath Bulusu- For helping me create the graphics for the geometry of the states.

LaTex- For providing excellent free software for writing documents.

3

2 Shannon and von Neumann Entropy

In 1948 Claude Shannon published his well known paper ”A Mathematical Theory ofCommunication” [1]. The main theme of this paper was to find under what conditionsa message sent by a transmitter can be recovered by a receiver. In doing so he defineda quantity called Entropy used as a measure of Information. The problem of recoveringa message is similar to the idea in quantum mechanics of measurement: How to gainknowledge of the system by performing a measurement test. In this sense the quantumsystem represents an analogy to the message to be transmitted in Communication Theoryand the measurement outcome to the received message. Shannon derived a formula forthe Entropy [1]:

H =∑i

pilog(pi) (1)

In 1932 John von Neumann extended the idea of Entropy to quantum systems by in-troducing a density matrix in ”Mathematical Foundations of Quantum Mechanics” [2].Previously a system could be in a superposition of pure states: |Ψ〉 =

∑i |ψi〉 But with

the help of a density matrix we could create a mixture of pure states by adding projectorsof each pure states |ψi〉 with a probability λi such that

∑i λi = 1 and λi > 0:

ρ =∑i

λi |ψi〉 〈ψi| (2)

The density operator therefore has a few interesting properties which are easy to prove[3] and are listed here:

ρ† = ρ Hermitian

ρ2 = ρ Idempotent if pure

Tr [ρ] = 1 sum of the eigenvalues have to be 1

Tr[Aρ] = E[A] Procedure for calculating the Expected Value of the Observable A

Tr[ρ2] < 1 If it is in a mixed state

ρ ≥ 0 All eigenvalues (probabilities) are greater or equal to 0

Thus since density matrices are the direct analog to the probability Von Neumann definedthe Entropy of a quantum system described by the density matrix ρ as:

S = −Tr [ρ logρ] (3)

3 Entropy Inequalities for Finite Dimensional Hilbert Spaces

In this section we will restate some of the Lemmas of the paper ”Entropy Inequalities”by Araki and Lieb [5] and show some of the proofs.

4

Definition 3.1 If ρXY is a density matrix in the Hilbert Space HX ⊗ HY . Then thepartial trace over the Hilbert Space HY is ρX

TrY

[ρXY ] = ρX (4)

We will use throughout the following notation.

Definition 3.2 If ρXY denotes the density matrix on HX ⊗ HY then we denote theEntropy associated with ρXY by SXY .

One very important theorem in the Araki-Lieb Paper [5] is the following:

Theorem 3.3 let ρXY Z be a density matrix on HX ⊗ HY ⊗ HZ . Then the followinginequality holds.

SXY Z ≤ SXY + SY Z + log

[TrY

[ρ2Y

]]≤ SXY + SY Z (5)

Furthermore if [ρY ⊗ IZ , ρY Z ] = 0 holds then:

SXY Z + SY ≤ SXY + SY Z (6)

To prove this Theorem we need Klein’s, Peierls-Bogolybov and Golden Thompson in-equalities, of which the first two we will prove and the last one we will only state.

Lemma 3.4 Klein Inequality [5][10]For convex f and Hermitian A and B the following inequality holds

Tr[f(A)− f(B)− (A−B)f ′(B)

]≥ 0 (7)

Proof of Lemma 3.4 Rearrange such that:

Tr[f(A)− f(B)] ≥ Tr[(A−B)f ′(B)] (8)

Let C = A−B and t ∈ [0, 1] Then we can define a function

φ(t) = Tr[f(B + tC)] (9)

φ(1) = Tr[f(A)]

φ(0) = Tr[f(B)](10)

Differentiating φ(t) wrt. to t yields:

d

dtφ(t)|t=0 =

d

dtTr[f(B + tC)]|t=0 = Tr[(A−B)f ′(B)] (11)

Therefore:

φ(1)− φ(0) ≥ limt→0

φ(t)− φ(0)

t→ 0 (12)

because the RHS is monotonically decreasing with t.

5

Lemma 3.5 Peierls-Bogolybov inequality [5]If E and F are Hermitian, Tr[eE ] = 1 and f ≡ Tr[FeE ] then

Tr[eE+F

]≥ ef (13)

Proof of Lemma 3.5Using Lemma 3.4 and setting f(x) = ex A = E + F, B = E + f1.Rearranging

Tr[eE+F

]≥ Tr

[(F − fI)eE+fI + eE+f1

](14)

Noting that[E,1] = 0→ eE+f1 = eEef1 = eEef (15)

Using the definition f ≡ Tr[FeE ]

RHS = Tr[FeEef ]− Tr[feEef

]+ Tr

[eE+fI

]= Tr[FeE ]ef − f Tr

[eE]ef + Tr

[eE+f1

]= ef

(16)

Therefore:Tr[eE+F

]≥ ef (17)

Lemma 3.6 Golden-Thompson Inequality [5]If A and B are Hermitian then

Tr[eA+B] ≤ Tr[eAeB] (18)

Proof of Theorem 3.3 [5]Given a positive definite operator

(ρXY Z = eRXY Z

)∈ (H1 ⊗H2 ⊗H3)

We define ρXY = eRXY , RXY ≡ RXY ⊗IZ . Moreover we will use Tr to denote the Traceover all Hilbert Spaces. Partial Traces will be specially denoted.

∆ = SXY Z − SXY − SY Z = Tr [ρXY Z (−RXY Z +RXY +RY Z)] (19)

Choose F = −RXY Z +RXY +RY Z , E = ρXY Z and ∆ = f

∆ = SXY Z − SXY − SY Z = Tr[eRXY Z (−RXY Z +RXY +RY Z)

](20)

Using Lemma 3.5:

e∆ ≤ Tr[eRXY Z−RXY Z+RXY +RY Z

]= Tr

[eRXY +RY Z

](21)

And finally using Lemma 3.6:

e∆ ≤ Tr[eRXY eRY Z

]= Tr [ρXY ρY Z ] (22)

6

Since ρXY ≡ ρXY ⊗IZ and ρY Z ≡ IX ⊗ ρY Z . We can first trace out the Z and then theX and we are then left with

e∆ = TrY

[ρ2Y

]≤ 1 equal if pure (23)

This shows that:

SXY Z − SXY − SY Z ≤ log(TrY

[ρ2Y

])≤ 0 (24)

or rearranging:

SXY Z ≤ log(TrY

[ρ2Y

])+ SXY + SY Z ≤ SXY + SY Z (25)

Which is exactly the first part of Theorem 3.3.For the second part consider that [ρY , ρY Z ] = 0 Then we can apply the same procedureas above:

∆ = SXY Z + SY − SXY − SY Z (26)

e∆ = Tr[eRXY +RY Z−RY

]≤ Tr

[eRXY eRY Z−RY

]= Tr

[eRXY eRY ZeRY

](27)

= Tr[ρXY ρY Zρ

−1Y

]= 1 (28)

Which yields the desired result, since we can first Trace out over HZ then over HX andthen we are left with Tr[ρ2

Y ρ−1Y ] = 1:

SXY Z + SY ≤ SXY + SY Z (29)

Corollary 3.7 If ρXY is a density matrix on HX ⊗HY then

SXY ≤ SX + SY (30)

Proof of Corollary 3.7 Using the second part of Theorem 3.3, interchaning Y ↔ Zand letting Z be one dimensional shows this result.

Definition 3.8 A density matrix ρ is pure if it is a projection operator onto a one-dimensional subspace, i.e. ρ |µ〉 = 〈λ|µ〉 |λ〉 for |λ〉 being fixed and normed.

Lemma 3.9 If ρXY is a pure state density matrix on HX ⊗ HY and let f(.) be a realvalued function with f(0) = 0. Then:

Tr [f(ρX)] = Tr [f(ρY )] (31)

Proof of Lemma 3.9Applying Definition 3.8

ρXY |µ〉 = 〈λ|µ〉 |λ〉 and |λ〉 =∑i

pi |λXi〉 ⊗ |λY i〉 (32)

7

|λXi〉 and |λY i〉 can be chosen as orthogonoal. Now letting Pν,i = |λν,i〉 〈λν,i| be theProjection onto the one dimensional subspace of Hν then:

ρν =∑i

p2iPν,i =

∑i

p2i |λν,i〉 〈λν,i| (33)

Therefore ρX and ρY have the same eigenvalues with the same multiplicities. In partic-ular it can be shown then that SX = SY .

Lemma 3.10 [5] Let ρX be a density matrix on HX . Then there exists a Hilbert SpaceHY and a pure state density matrix ρXY on HX ⊗HY such that

ρX = TrY

[ρXY ] (34)

Proof of Lemma 3.10 Let ρX =∑iλiPi be the spectral decomposition with λi ≥ 0

and Pi |µ〉 = |λi〉 〈λi|µ〉 where |λi〉 form an orthonormal basis. Assume that dim (HY ) ≥dim (HX) and that it has the orthonormal basis |δi〉. Now let ρXY be the projectionoperator on the one dimensional subspace of HX ⊗HY containing the vector∑iλ1/2 |λi〉 ⊗ |δi〉. Then

TrY

[ρXY ] = ρX (35)

Lemmas 3.9 and 3.10 can be used to prove more Entropy inequalites. However theinequalities derived so far are sufficient for our purpose.

4 Problem with the Peierls-Bogylobov-Inequality

In the Peierl’s Bogylobov inequality in Lemma 3.5 we make one assumption: Namelythat Tr[eE ] = 1 where E is Hermitian. If we assume that E is Hermitian then we cando a spectral decomposition.

E =∑i

λi |λi〉 〈λi| (36)

If we take the Exponential of this then we can write it as:

Tr

[∑i

eλi |λi〉 〈λi|

]= 1 (37)

because of linearity of the Trace:∑i

eλiTr [|λi〉 〈λi|] = 1 (38)

Since the trace is independent of basis we can use the Basis of E:∑i

eλi∑j

〈λj |λi〉︸ ︷︷ ︸δij

〈λi|λj〉︸ ︷︷ ︸δij

= 1 (39)

8

∑i

eλi∑j

〈λj |λi〉︸ ︷︷ ︸δij

〈λi|λj〉︸ ︷︷ ︸δij

= 1 (40)

∑i

eλi∑j

δijδij︸ ︷︷ ︸δii

=∑i

eλiδii =∑i

eλi = 1 (41)

For real λi this can only be satisfied if there is only one eigenvalue λ = 0. So E = 0.However in the Proof of Theorem 1.3 we chose E = ρXY Z to prove the inequality given.But then ρXY Z would be not be a density matrix any more since its trace is not 1. So theonly other possibility would be if the eigenvalues could be complex or negative, in whichcase the density matrix would not really be hermitian or a density matrix respectively.

5 Dirac’s opinion on the derivative of the logarithm

The problem that the density matrix might have complex eigenvalues could be relatedto the fact that the derivative of the logarithm does, using a special argument presentedby Dirac [13], contain an imaginary part as well. We are used to:

d

dx[log(x)] =

1

x

∣∣∣∣x ∈ R+\{0} (42)

However Dirac in Principles of Quantum Mechanics [13] states that this equation needsexamination in the neighbourhood of the origin. So let us consider the integral:∫ a

−a

1

xdx = lim

ε→0

∫ −ε−a

1

xdx+

∫ a

ε

1

xdx = 0 (43)

Because 1/x is odd.Now let us do the following:

d

dxlog(x) =

1

x(44)

To check whether this is true we can integrate this function:∫ a

−a

d

dxlog(x) =

∫ a

−a

1

x(45)

The RHS vanishes as we just showed. So now we need to evaluate the LHS.

LHS =

∫ a

−a

d

dxlog(x) = log(a)− log(−a) = log(−1) (46)

However this term is complex.

log(−1) ≡ a→ ea = −1→ a = i (1 + 2n)π

∣∣∣∣ let us say thatn = 0 (47)

9

So we can conclude that we can extend the definition of the derivative of the logarithm:

d

dxlog(x) =

1

x− iπδ(x) (48)

We can check this by integrating again:∫ a

−a

d

dxlog(x)dx =

∫ a

−a

1

x− iπδ(x)dx (49)

log(−1) = −iπ∣∣∣∣exp(.) (50)

elog(−1) = e−iπ (51)

−1 = e−iπ (52)

Which is correct. Finally we note that an imaginary eigenvalue might have somethingto do with oscillations. Further reserach for justification is necessary.

6 Complex Vector Operations

Since we have shown that complex numbers might play a crucial role in the Entropyand because of the fact that during the mathematical algorithm for finding conditionson the Entropy for 2-qubit states we found that the computer sometimes printed errorsthat complex comparisons are not allowed. Therefore it might be useful for the futureto define complex number comparisons. [14]

6.1 Multiplication of Complex Numbers

Let us undertake the multiplication of complex numbers and develop a Matrix-vectorformalism for dealing with complex numbers:

zw = (a+ ib)(c+ id) = ac− bd+ i(ad+ bc) (53)

So this can be written in matrix form in two different waves:

zw =

[a −bb a

]×[cd

]=

[ac− bdad+ bc

]wz =

[c −dd c

]×[ab

]=

[ac− bdad+ bc

] (54)

So every complex numbers vector representation has a multiplication-matrix represent-ation. And the multiplication of complex numbers can be represented by a matrixmultiplication which is real-antisymmetric. Interestingly the determinant of the matrixmultiplication representation of a complex numebr is nothing but the modulus of thecomplex number. How can we define an orthogonal to a complex vector? For general

10

two dimensional real vectors all we need to do is switch the elements and multiply oneby −1. We will do exactly the same for complex numbers.

z =

[ab

]z⊥ =

[±b∓a

](55)

So we can define a matrix which does this operation, call it the complex orthogonalisationmatrix CO.

z⊥ =

[0 ±1∓1 0

]× z (56)

This is nothing but a multiplication by i or −i and looks very similar to the y-Pauli-matrix but only with real entries. Lets calculate the scalar product of two orthogonalcomplex numbers

z · z⊥ =[a b

]·[±b∓a

]= 0 (57)

In other words we need not define any metric to calculate the dual of a complex number,but we need only to transpose it.

6.2 Comparison of Complex Numbers

Now comes a more interesting question: How would you compare complex numbers?For example:

a+ ib > c+ id (58)

How do we compare numbers on the real axis? What we usually do is just see theposition of two numbers with respect to the origin and the one which is more to theright is greater than the other one. Actually we can also compare two real numbers withrespect to a third number applying the same procedure. We can do a similar thing withcomplex numbers. The geometric procedure would be as follows:

1. Define a line Γ through the complex numbers z and w

2. Lay an orthogonal line Υ through Γ and the Complex Number v with respect towhich z and w should be compared to.

3. Calculate the intersection point ς of Υ with Γ then calculate the distance from theintersection ς to z and w.

4. The complex number with the larger distance is the greater complex number.

Why did we define it this way? Suppose now if we were to rotate Υ and along withthat its orthogonal Γ then we could rotate it by an angle θ such that only a real partor imaginary part is left after the rotation. This would allow simple comparison. This

11

means that we are not concerned whether a number is to the left or to the right, becauseof isotropy of space. Let us treat this mathematically: Say we have two complex numbers:

w =

[ab

]z =

[cd

](59)

Then we can define a line Γ as:

Γ = w + (z − w) · t (60)

for some parameter t. Now we need an orthogonal line to Γ. For this purpose we mustfind an orthogonal zw⊥ to z − w. Which we can find using the procedure above:

zw⊥ =

[0 ±1∓1 0

]× (z − w) =

[0 ±1∓1 0

]×[c− ad− b

]=

[±(d− b)∓(c− a)

](61)

Now say that our complex with respect to which we want to compare the other two is:

v =

[ef

](62)

Then the line Υ is given by a line:

Υ = v + s · zw⊥ (63)

To find the intersection we must equation equations 60 and 63:

Γ = Υ→ v + s · zw⊥ = w + (z − w) · tv − w = t · (z − w)− s · CO(z − w)

v − w =(t · 1− s · CO

)(z − w)

(64)

This is a linear system of two equations with two variables and therefore has one solution.The signs of the solution depend on what convention we use for CO. If we write out thefull system of equation then we get:[

e− af − b

]=

[t ±s∓s t

] [c− ad− b

](65)

The determinant of this matrix vanishes if:

t2 + s2 = 0→ t = ± i s (66)

This would mean that the LH complex number is an ”eigen-complex-number” of the RHcomplex number. Which reminds us of rotation matrices. If it is non-vanishing thenthe system of equations can be solved. And it should be since s is a real parameter. Aspecial case is if all the three complex numbers lie on one line. In this case it turns out

12

that the orthogonal line is the line itself and reminds of the light-cone in Minkowksi-Space. Moreover the linear system of equations can be solved if s = 0, which makessense since in this case there doesn’t really exist a normal complex-vector and you alsoget two solutions which must as expected correspond to the two complex numbers whichare to be compared. As an example consider: w = γ · v z = δ · v. This would yield thefollowing system of equation:

(1− γ) v =(t · 1− s · CO

)(δ − γ) v (67)

and we get two equations:[(1− γ)vRe(1− γ)vIm

]=

[t ±s∓s t

] [(δ − γ)vRe(δ − γ)vIm

](68)

Due to the antisymmetry of the matrix this can be solved if s = 0 in other words: thereis no need to define an orthogonal vector since they are all on one line. Using this we getthe solution t = 1−γ

δ−γ . This obviously has a singularity if δ = γ. Which makes sense sincethis would mean that the two complex numbers we are trying to compare are equal andtherefore there is no need to compare them regardless with respect to which complexnumber we compare them. Now two special cases are if δ = 1 or γ = 1. In the first casewe get that the point of intersection is z and in the second case w. This makes sense.Note that we could have also defined the number which is more towards the right/leftof the line through the two complex numbers to be compared the larger one. The choiceof left and right is purely a convention.

7 Separability - Entanglement - Negentropy - Non-Locality

Let us consider a system of two Quantum objects given by a density matrix ρ in theHilbert Space HAB = HA ⊗ HB with dim(HA) = dim(HB) = 2 → dim(HAB) = 2 × 2The scalar product is defined by 〈A|B〉 = Tr

[A†B

]All Quantum states can be classified

as either seperable or entangled.

Definition 7.1 Separable States [7][6]The set of separable states S are defined as follows

S =

{ρ =

∑i

piρA,i ⊗ ρB,i∣∣∣∣ 0 ≤ pi ≤ 1,

∑i

pi = 1

}(69)

A state is therefore entangled if it is not separable. An important distinction is to bemade here: Entangled states are not necessarily non-local, hower all non-local states areentangled. Locality is broken iff the CHSH-inequality is violated. For density matricesthe CHSH-inequality [6] can be written as:

〈ρ∣∣2 1−BCHSH〉 ≡ Tr [ρ (2 1−BCHSH)] ≥ 0 (70)

13

Where the Bell operator BCHSH is:

BCHSH = ~a · ~σA ⊗(~b− ~b′

)· ~σB + ~a′ · ~σA ⊗

(~b+ ~b′

)· ~σB (71)

It turns out that the CHSH Inequality is violated for all pure entangled states andthat there are mixed entangled local states??

Theorem 7.2 PPT-criterion [7]Given a density matrix ρ in a Hilbert Space H with dim(H) = 2× 3 or dim(H) = 2× 2and denoting TX as the partial transposition operator in the subspace HX Then

(1A ⊗ TB) ρ ≥ 0 or (TA ⊗ 1B) ρ ≥ 0↔ ρ ∈ S (72)

Lemma 7.3 Werner State DecompositionThe Werner States for a Hilbert Space H = HA ⊗ HB with dim[H] = 2 × 2 can bedecomposed in the following way.

ρWerner =1

4(1⊗ 1− ασi ⊗ σi)

∣∣α ∈ [0, 1] (73)

Or in matrix notation using the z Basis:

ρWerner =1

4

1− α 0 0 0

0 1 + α −2α 00 −2α 1 + α 00 0 0 1− α

(74)

Using Theorem 7.2 the Werner states turn out to be separable for α ≤ 1/3 and thereforeentangled for α > 1/3 [6].

Theorem 7.4 Maximal violation of Bell inequality [6]Given a general density matrix H = HA ⊗HB with dim[H] = 2 × 2 in a Hilbert Spacein Bloch form

ρ =1

4(1⊗ 1 + riσi ⊗ 1 + ui1⊗ σi ⊗+tijσi ⊗ σj) (75)

and the Bell operator BCHSH = ~a · ~σA ⊗(~b− ~b′

)· ~σB + ~a′ · ~σA ⊗

(~b+ ~b′

)· ~σB Then the

maximal violation of the CHSH inequality is Bmax =√t21 + t22 > 1 where t1 and t2 are

the two largest eigenvalues of tij

For the Werner state all eigenvalues t of the matrix tij are t = α.

So√

2α2 > 1→ α > 1/√

2. Therefore the Werner state is non-local for α > 1√2

[6].

Hypothesis Do all non local states have Negentropy?Given a density matrix ρ ∈ (H = HA ⊗HB) with dim(H) = 2× 2 which can be decom-posed in Bloch form:

ρ =1

4

(1⊗ 1 +

∑i

αi σi ⊗ σi

)(76)

14

LetNL =

{ρ∣∣Tr [(BCHSH − 21) ρ] < 0

}(77)

i.e the set of all Non Local States and Let

NE ={ρ∣∣ (S(A|B) < 0) ∪ (S(B|A) < 0)

}(78)

i.e the set of all States with negative conditional Entropy.Then arises the question: Do all Non-Local states have Negentropy?

ρ ∈ NL→ ρ ∈ NE (79)

NL ⊂ NE (80)

In addition to finding this answer we would also like to geometrically locate the Separable,Mixed, Non-Local and Negentropic states. So let us first calculate the set of Non-Localstates. For this we can make use of Theorem 7.4. So we just need to find the two largesteigenvalues t1 and t2 of the density matrix ρ and apply the condition:

t2max,1 + t2max,2 > 1 (81)

Then we have to apply the condition that the Negentropy Entropy is negative.

S(AB)− S(A) < 0 (82)

For finding the Entropy of the subsystem let us take the partial trace of ρ. Since ρ canbe decomposed as in equation (76). We must take the partial trace of ~σ ⊗ ~σ and 1 onHB. Using the fact that the Pauli Spin Marices are traceless:

〈. ↑ |~σ ⊗ ~σ|. ↑〉+ 〈. ↓ |~σ ⊗ ~σ|. ↓〉 = 0

〈. ↑ |1⊗ 1|. ↑〉+ 〈. ↓ |1⊗ 1|. ↓〉 = 2 |↑〉 〈↑|+ 2 |↓〉 |↓〉(83)

Therefore:

TrB

[ρ] =1

2|↑〉 〈↑|+ 1

2|↓〉 |↓〉 (84)

With two equal eigenvalues:

λA =1

2(85)

Therefore the Entropy of the partial System A is given by:

S(A) = −(

1

2log2

(1

2

)+

1

2log2

(1

2

))= log2(2) = 1 (86)

So if the Negentropy Entropy has to be negative then:

S(AB) < 1 (87)

15

And for this matter we will now find its eigenvalues. The density matrix ρ can berepresented as a matrix using the z-Basis:

ρ =1

4

1 + α3 0 0 α1 − α2

0 1− α3 α1 + α2 00 α1 + α2 1− α3 0

α1 − α2 0 0 1 + α3

(88)

The eigenvalues of ρ turn out to be:

λ1 =1

4(1− α1 − α2 − α3)

λ2 =1

4(1− α1 + α2 + α3)

λ3 =1

4(1 + α1 − α2 + α3)

λ4 =1

4(1 + α1 + α2 − α3)

(89)

In other words if λk is an eigenvalues of ρ then:∑k

λk log2(λk) < 1 (90)

So the three conditions are:

αi ∈ [−1, 1]

λk ≥ 0 ∀k

NE →∑k

λk log2(λk) < 1

NL→ t2max,1 + t2max,2 > 1

Since the Matrix ti,j defining the decomposition of the density matrix of the qubit stateis diagonal. We just need to pick the two largest αi.To give a broader picture and to demonstrate the geometry of the Qubit state we wouldalso like to also find now a condition for separability and purity. Below Theorem 7.2we showed the conditions for which the Werner State is separable. We would like togeneralize this to all qubit states that can be decomposed as in equation (76). Soapplying the partial transposition we find:

ρTB = (1⊗ PT ) ρ ≥ 0 || ρTA = (PT ⊗ 1) ρ ≥ 0 (91)

This yields a matrix:

ρTB =1

4

1 + α3 0 0 α1 + α2

0 1− α3 α1 − α2 00 α1 − α2 1− α3 0

α1 + α2 0 0 1 + α3

(92)

16

Due to the symmetry of the density matrix positivity of the other partially transposeddensity matrix is ensured. In this case the sign of the components of α of all theeigenvalues change and invoking the condition for separability through positvity of itseigenvalues yields the following:

λ1,TP =1

4(1 + α1 + α2 + α3) ≥ 0

λ2,TP =1

4(1 + α1 − α2 − α3) ≥ 0

λ3,TP =1

4(1− α1 + α2 − α3) ≥ 0

λ4,TP =1

4(1− α1 − α2 + α3) ≥ 0

(93)

So the conditions for separability are:

λk ≥ 0 ∀k (94)

λi,TP ≥ 0 ∀i (95)

Now let us find a condition for purity. By definition of the density matrix it is mixedi.e. impure iff

Tr[ρ2] < 1 (96)

So let us use the this equation to find a condition on the set of αi:

ρ2 =

[1

4

(1⊗ 1 +

∑i

αiσi ⊗ σi

)]2

(97)

ρ2 =1

16

1⊗ 1 + 2∑i

αiσi ⊗ σi +∑ij

αiαj{σi, σj} ⊗ {σi, σj}

(98)

Using the anticommutator relation for the Pauli Matrices:

{σi, σj} = 2δij1 (99)

This simplifies to:

ρ2 =1

16

[1⊗ 1 + 2

∑i

αiσi ⊗ σi + 4∑i

α2i 1⊗ 1

](100)

And using the fact that the Pauli Matrices are Traceless so that the middle terms vanish:

Tr[ρ2] =1

16

[4 + 16

∑i

α2i

]< 1 (101)

→∑i

α2i <

12

16= 0.75 (102)

17

This implies that: ∑i

α2i < 0.75 (103)

Let us now summarize the conditions for the various classifications of qubit states:

Separability

αi ∈ [−1, 1]

λk ≥ 0 ∀kλi,TP ≥ 0 ∀i

Mixed

αi ∈ [−1, 1]

λk ≥ 0 ∀k∑iα2i < 0.75

Non-Local

αi ∈ [−1, 1]

λk ≥ 0 ∀kα2max,1 + α2

max,2 > 1

Negative Conditional Entropy

αi ∈ [−1, 1]

λk ≥ 0 ∀k∑k

λklog2 (λk) < 1

(104)

These conditions restrict values of αi to certain regions in the cube.

7.1 Geometry

We would now like to parametrize the states in such a way that it spans an equilateraltriangular plane between the state on the top-center (tc) state ρtc, the Bell-state ρ− →|Ψ−〉 and the other Bell-state ρ+ → |Ψ+〉. However thanks to symmetry we need notconsider the whole plane. We will infact split this equilateral triangular plane into tworectangular triangular planes by considering the following parametrization:

ρedge = (1− β) ρ− + β ρ+

ρplane = (1− γ) ρedge + γ ρtc

β ∈ [0, 1/2]

γ ∈ [0, 1]

γ ≤ 2β

(105)

18

Using the Bloch-decomposition of qubit density matrices.

ρ− =1

4

(1⊗ 1 +

∑i

α−i σi ⊗ σi

)

ρ+ =1

4

(1⊗ 1 +

∑i

α+i σi ⊗ σi

)

ρtc =1

4

(1⊗ 1 +

∑i

αtci σi ⊗ σi

)α−i = [−1,−1,−1] α+

i = [1, 1,−1] αtci = [0, 0, 1]

(106)

If we now plug in ρedge in ρplane we get the following:

ρplane = (1− γ)((1− β) ρ− + βρ+

)+ γρtc (107)

Simplifying this:

ρplane =1

41⊗ 1 +

1

4

∑i

[(βγ − β − γ)α−i + (1− γ)βα+

i + γ αtci]σi ⊗ σi (108)

Thus we can define a new αtci

αplanei =

−2βγ + 2β + γ−2βγ + 2β + γ

2γ

(109)

Such that we can describe the states lying on that plane by:

ρplane =1

4

(1⊗ 1 +

∑i

αplanei σi ⊗ σi

)(110)

Writing this density matrix in matrix form:

ρplane =1

4

2γ + 1 0 0 0

0 −2γ + 1 −4βγ + 4β + 2γ 00 −4βγ + 4β + 2γ −2γ + 1 00 0 0 2γ + 1

(111)

Thus this density matrix has eigenvalues:

λρplane=

γ/2 + 1/4γ/2 + 1/4

1/4 + β − βγ1/4− β − γ − βγ

(112)

Now we can plot the Geometry in Figure (1) of the states in the plane defined. It is clearfrom this figure that the negentropic states in red are a subset of the nonlocal states.

19

Therefore Negentropy is a not necessary condition for Non-Locality, but all Negentropystates are are also non-local. We have included in Figure (2) the Conditional Entropyas it varies along the lower edge.

Figure 1: Geometry of states in Bell-Tetrahaedon: The yellow states describe the sep-arable state, blue non-separable, orange non-local and red negentropic

20

[0.05, -0.71]

[0.15, -0.39]

[0.25, -0.19]

[0.5, 0]

[0.75, -0.19]

[0.85, -0.39]

[0.95, -0.71]

ρ-ρ+

0.2 0.4 0.6 0.8 1.0λ

-1.0

-0.8

-0.6

-0.4

-0.2

SAB -SA B

Figure 2: Conditional Entropy as it varies across the lower edge of the triangular planedefined above.

8 The Problem of Time

8.1 Principle of Least Action in Classical Mechanics

In Classical Mechanics we know that we can desribe any system in terms of its LagrangianL(q, q(t)). The principle of Least Action states that evolution of the system betweentwo configurations qa(t1) and qa(t2) is such that the action:

S :=

∫ t2

t1

L(q1, ..., qN , q1, ...qN , t)dt (113)

is minimzed over all possible trajectories between qa(t1) and qa(t2): Infact this conditionleads to the famous Euler-Lagrange equations:

∂L

∂qa− d

dt

(∂L

∂qa

)= 0

∣∣a ∈ {1, ..., N} (114)

8.2 Relativistic Generalization

The generalization of the action principle to a relativistic system faces serious challenges,because absolute time doesn’t exist. [8] Consider a system reduced to a single particle PWe also replace the absolute time by a parameter λ that increases along P’s worldline.

21

The Lagrangian is thus defined as a a differentiable function L : R8 → R such thatbetween any two events A1 and A2 of P ′s worldline the integral:

S :=

∫ λ2

λ1

L (xα(λ), xα(λ)) dλ (115)

has two properties:

• S has the dimension of Js

• S is independent of the parametrization λ

Requiring that S is independent of the parametrization induces a constaint on the Lag-rangian L. Considering a second parametrization λ of L:

xα(λ) = xα(λ) (116)

Therefore:˙xα :=

dxα

dλ=dxα

dλ

dλ

dλ= xα

dλ

dλ(117)

This implies that the invariance of the action is equivalent to

L (xα(λ), xα(λ)) dλ = L(xα(λ), ˙xα(λ)

)dλ (118)

Using the last two previous equations:

L (xα(λ), xα(λ))dλ

dλ= L

(xα, xα

dλ

dλ

)(119)

This equation must be fulfilled for any dλdλ

thus we can use the Euler Homogeneitytheorem:

xα∂L

∂xα= L (120)

Now we would like the use the definition for the canonical conjugate impuls:

pα =∂L

∂qα(121)

For us qα = xα Therefore:

xα∂L

∂xα= L (122)

And now using the definition of the Hamiltonian:

H = xαpα − L = 0 (123)

Therefore the Hamiltonian vanishes. In Quantum Mechanics however we know that theTime Evolution is governed by the Unitary Operator:

U = e−iHt (124)

This poses a serious problem in unifying Quantum Theory and Relativity and is oftenreferred to as the Problem of Time.

22

9 Clock Time and Entropy

9.1 Introduction

In this section we discuss the theoretical framework and recalculate the examples ofthe paper ”Clock Time and Entropy” [11] forward by Don N. Page. He makes a cleardistinction between measuring the probability of objects at different times and measuringdifferent records at the same time. The past cannot be known except thought its record inthe present, so we can only test the present. So we cannot directly test the conditionalprobability that the electron has spin up at t = tf given that it had spin down att = ti < tf , but only given that there are records at t = tf from which we can infer thatthe electron had spin up at t = ti. Wheeler states this even more strongly: ”The past hasno existence except as it is recorded in the present”. This principle should be extendedto say that we cannot directly compare things at different locations either. So thatall measurements are localized within the spatial extent of an observers senses and thetemporal extent of a single conscious moment. In the formalism of Quantum Mechanicson a spatial hypersurface is given by a density matrix ρ, the conditional probability ofthe result A, given a testable condition B, is:

p(A|B) =Tr [PAPBρPB]

Tr [PBρPB]=Tr [PBPAPBρ]

Tr [PBρ](125)

Because of the cyclicity of the trace and the hermitian and idempotent properties of theProjectors.

9.2 Inaccessibility of coordinate time

These testable conditional probability as described in 125 seem to be confined to a single”time”. However they cannot depend on the value of the coordinate time labelling thehypersurface, which is unobservable. The main arguments for this are:

• For a closed Universe the Wheeler-DeWitt equation, H |Ψ〉 = 0, implies that thewavefunction doesn’t depend on time.

• For an asymptotically-flat open universe it turns out that the phases between statesof different energy are unmeasureble, so that the coordinate tie dependence of thedensity matrix ρ is not detectble.

• One doesn’t have access to the coordinate time, so one should average over theinaccessible coordinate time dependence of the density matrix ρ.

ρ = 〈ρ(t)〉 = limT→∞

1

2T

∫ T

−Tρ(t)dt (126)

So without access to the coordinate time we can only test conditionaly probabilityof the form:

p(A|B) =〈Tr [PBPAPBρ(t)]〉〈Tr [PBρ(t)]〉

=Tr [PBPAPB ρ]

Tr [PB ρ](127)

23

Now consider the case where the condition is entirely the reading of a clock subsystem(C) with states |Ψ(T )〉. Let

PB → PT = |ΨC(T )〉 〈ΨC(T )| ⊗ IR (128)

Then

P (A|T ) =Tr [PTPAPT ρ]

Tr [PT ρ](129)

can vary with clock time.

9.3 Examples

9.3.1 2-Qubit System

Let us assume we have two coupled spin 12 systems in a vertical magnetic field. One

represents the clock and the second represents the rest. Its Hamiltonian is given by:

H =1

4~σC ⊗ ~σR +

1

2σCz ⊗ 1R +

1

21⊗ σRz +

3

41CR (130)

Using the z Basis we can write the Pauli Spin Matrices as:

σx = |↑〉 〈↓|+ |↓〉 〈↑|σy = i (− |↑〉 〈↓|+ |↓〉 〈↑|)σz = |↑〉 〈↑| − |↓〉 〈↓|1 = |↑〉 〈↑|+ |↓〉 〈↓|

(131)

Using these:

~σ ⊗ ~σ = + |↑↑〉 〈↓↓|+ |↑↓〉 〈↓↑|+ |↓↑〉 〈↑↓|+ |↓↓〉 〈↑↑|− |↑↑〉 〈↓↓|+ |↑↓〉 〈↓↑|+ |↓↑〉 〈↑↓| − |↓↓〉 〈↑↑|+ |↑↑〉 〈↑↑| − |↑↓〉 〈↑↓|− |↓↑〉 〈↓↑|+ |↓↓〉 〈↓↓|

= 2 |↑↓〉 〈↓↑|+ 2 |↓↑〉 〈↑↓|+ |↑↑〉 〈↑↑| − |↑↓〉 〈↑↓|− |↓↑〉 〈↓↑|+ |↓↓〉 〈↓↓|

(132)

~σ ⊗ 1 = |↑↑〉 〈↑↑|+ |↑↓〉 〈↑↓| − |↓↑〉 〈↓↑| − |↓↓〉 〈↓↓|1⊗ ~σ = |↑↑〉 〈↑↑|+ |↓↑〉 〈↓↑| − |↑↓〉 〈↑↓| − |↓↓〉 〈↓↓|1⊗ 1 = |↑↑〉 〈↑↑|+ |↑↓〉 〈↑↓|+ |↓↑〉 〈↓↑|+ |↓↓〉 〈↓↓|

(133)

Now with the help of these subcalculationas we can calculate the Hamiltonian as perequation (130):

H = 2 |↑↑〉 〈↑↑|+ 1

2(|↑↓〉+ |↓↑〉) (〈↑↓|+ 〈↓↑|) (134)

24

Let us now examine the time-dependant pure state:

|Ψ〉 =

√5

10

[(e−it + 1

)|↑↓〉+

(e−it − 1

)|↓↑〉

](135)

So we can calculate its time averaged density matrix and since limT→∞

12T

∫ T−T e

±itdt→ 0

ρ = limT→∞

1

2T

∫ T

−Tρdt = lim

T→∞

1

2T

∫ T

−T|Ψ〉 〈Ψ|

=1

10(|↑↓〉 〈↑↓|+ 2 |↑↓〉 〈↓↓|+ |↓↑〉 〈↓↑| − 2 |↓↑〉 〈↓↓|+ 2 |↓↓〉 〈↑↓| − 2 |↓↓〉 〈↓↑|+ 8 |↓↓〉 〈↓↓|)

(136)

Using the standard basis for Spin up and Spin down:

|↑〉 =

[10

]|↓〉 =

[01

] (137)

We can write the time averaged density matrix in this matrix form:

ρ =1

10

0 0 0 00 1 0 20 0 1 −20 2 −2 8

(138)

The eigenvalues of this Matrix are:

λ = {0, 0.1, 0.9} (139)

And since we defined 0 · log(0) ≡ 0 its Entropy is given by:

SCR = −0.1log(0.1)− 0.9log(0.9) = 0.3251 (140)

Now let

PT =1

2

(|↑〉+ e−iT |↓〉

) (〈↑|+ eiT 〈↓|

)⊗ 1R (141)

In Matrix form this is:

PT =1

2

1 0 eiT 00 1 0 eiT

e−iT 0 1 00 e−iT 0 1

(142)

So:

PTρPT =1

40

1 0 eiT 00 1 0 eiT

e−iT 0 1 00 e−iT 0 1

0 0 0 00 1 0 20 0 1 −20 2 −2 8

1 0 eiT 00 1 0 eiT

e−iT 0 1 00 e−iT 0 1

(143)

25

which turns out to be:

PTρPT =1

40

1 −2 eiT −2eiT

−2 9 + 2e−iT + 2eiT −2eiT 2 + 9eiT + 2ei2T

e−iT −2e−iT 1 −2−2e−iT 2 + 9e−iT + 2e−i2T −2 9 + 2e−iT + 2eiT

(144)

Therefore:

Tr [PT ρPT ] =1

40[20 + 8cos(T )] (145)

So:

ρT =PT ρPT

Tr [PT ρPT ]=

1

20 + 8cos(T )

1 −2 eiT −2eiT

−2 9 + 2e−iT + 2eiT −2eiT 2 + 9eiT + 2ei2T

e−iT −2e−iT 1 −2−2e−iT 2 + 9e−iT + 2e−i2T −2 9 + 2e−iT + 2eiT

(146)

This matrix can infact be split into a tensor product:

ρT = ρTC ⊗ ρTR (147)

where

ρTC =1

2

[1 eiT

e−iT 1

](148)

ρTR =1

4cosT + 10

[1 −2−2 9 + 4cos(T )

](149)

Therefore to find the eigenvalues of ρT we have to find the eigenvalues of ρTC and ρTRand multiply them in all possible combinations [12].The eigenvalues of ρTC and ρTR are:

λTC = {0, 1}

λTR = {5 + 2cosT ±√

20 + 16cosT + 4cos2T}(150)

So the eigenavlues of ρT are:

λT =

{0, 0,

1

2+

2√

(cosT + 2)2 + 1

10 + 4cosT,1

2−

2√

(cosT + 2)2 + 1

10 + 4cosT

}= {0, 0, a+b, a−b} (151)

We can now calculate the Entropy as follows:

S = − ((a+ b)log(a+ b) + (a− b)log(a− b)) = −[alog(a2 − b2) + blog

(a+ b

a− b

)]

S = log

(10 + 4cosT√

5 + 4cosT

)− 2

√(cosT + 2)2 + 1

5 + 2cosTlog

(5 + 2cosT + 2

√(cosT + 1)2 + 1√

4cosT + 5

)(152)

So this shows that although ρ is pure and its Entropy vanishes the Entropy of the timeaveraged density matrix ρ is finite however the measured Entropy changes with time.

26

9.3.2 3-Qubit System

Now let us turn our attention to a 3-qubit system. We would like to find the CourseGrained Entropy of the following system:Let the Hamiltonian be given by:

H =1

2σCz⊗1R1⊗1R2+

1

21C⊗σR1z⊗1R2+

1

21C⊗1R1⊗σR2z+

1

41C⊗~σR1⊗~σR2+

1

41⊗1⊗1

(153)Now we consider the pure time independent state:

|Ψ〉 =1

2(|↑↑↓〉 − |↑↓↑〉+ |↓↑↑〉+ |↓↓↑〉) (154)

Its density matrix can be written in Matrix form as:

|Ψ〉 〈Ψ| = 1

4

0 0 0 0 0 0 0 00 1 −1 0 0 1 1 00 −1 1 0 0 −1 −10 0 0 0 0 0 0 00 0 0 0 0 0 0 00 1 −1 0 0 1 1 00 1 −1 0 0 1 1 00 0 0 0 0 0 0 0

≡ 1

4

[ρ11 ρ12

ρ21 ρ22

](155)

Let us now apply the same Projector as in the first example:

PT =1

2

(|↑〉+ e−iT |↓〉

) (〈↑|+ eiT 〈↓|

)⊗ 1R1 ⊗ 1R2 =

1

2

[14×4 eiT 14×4

e−iT 14×4 14×4

](156)

PTρPT =

1

16

0 0 0 0 0 0 0 00 2 + 2cosT −2isinT 0 0 1 + 2eiT + ei2T 1− ei2T 00 2isinT 2− 2cosT 0 0 −1 + ei2T −1 + 2eiT − ei2T 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 1 + 2e−iT + e−i2T −1 + e−i2T 0 0 2 + 2cosT −2isinT 00 1 + e−i2T −1 + 2eiT − ei2T 0 0 2isinT 2− 2cosT 00 0 0 0 0 0 0 0

(157)

Its Trace:

Tr[PTρPT ] =1

2(158)

So:

27

PTρPTTr[PTρPT ]

=

1

8

0 0 0 0 0 0 0 00 2 + 2cosT −2isinT 0 0 1 + 2eiT + ei2T 1− ei2T 00 2isinT 2− 2cosT 0 0 −1 + ei2T −1 + 2eiT − ei2T 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 1 + 2e−iT + e−i2T −1 + e−i2T 0 0 2 + 2cosT −2isinT 00 1 + e−i2T −1 + 2eiT − ei2T 0 0 2isinT 2− 2cosT 00 0 0 0 0 0 0 0

(159)

This density matrix can be decomposed in the following way:

ρ = |ΨT 〉 〈ΨT | =

1

8

[1 eiT

e−iT 1

]⊗

0 0 0 00 2 + 2cosT −2isinT 00 2isinT 2− 2cosT 00 0 0 0

(160)

To find the coarse grained Entropy we have to compute the partial trace of ρT withrespect to all its subparts. Obviously if we Trace over HR1 ⊗ HR2 then we get ρTCbecause we can form the density matrix as a tensor product as indicated above. Tracingover HC we get the following density matrix:

ρT,R1,R2 =1

2[(1 + cosT ) |↑↓〉 〈↑↓|+ (1− cosT ) |↓↑〉 〈↓↑| − isinT |↑↓〉 〈↓↑|+ isinT |↓↑〉 〈↑↓|]

(161)Forming the Trace of this density matrix once over HR1 and once over HR2 we get thefollowing two density matrices

ρT,R1 =1

2

[1 + cosT 0

0 1− cosT

]ρT,R2 =

1

2

[1− cosT 0

0 1 + cosT

](162)

Therefore the eigenvalues of these matrices are:

λC = {0, 1}

λR1,2 =

{1

2(1 + cosT ) ,

1

2(1− cosT )

}(163)

And this the coarse grained Entropy is:

ST,coarse = −2 cos2(T/2) log(cos2(T/2))− 2 sin2(T/2) log(sin2(T/2)) (164)

This clearly varies periodically with time.

28

A Proposing an Extension of Einstein’s Energy-Mass equi-valence

Currently we are not able to unify General Relativity with Thermodynamics. It is how-ever very simple. Imagine that ”our” universe is part of a hypothetical larger universe.Then we can treat our universe according to the Grand-Canonical-Ensemble:

dE = TdS + pdV + µdV (165)

Dividing by dt where x we would get:

dE

dt= T (x)

dS(x)

dt+ p(x)

dV (x)

dt+ µ(x)

dN(x)

dt(166)

Now assume that our universe is actually isolated. This means that dE/dt = 0 by lawof conservation of energy.

T (x)dS(x)

dt+ p(x)

dV (x)

dt+ µ(x)

dN(x)

dt= 0 (167)

This equation describes some sort of energy flow in time and suggests that Entropy,Space and Mass are interconvertable forms of energy. Note that it might be possible,that the energy can not only flow in time but also in space. Now let us think about athought experiment and assume that T, p, c2 > 0: We could now interpret the pressureas the Gravitational Cosmological Constant and the chemical potential as µ = mc2.Relativistic corrections might be necessary. Further research and exact mathematicalformulation possibly through tensor fields is probably necessary.

29

B Using the Definition of Euler Mascheroni Constant tofind

∫∞0 1/xdx

The Euler Mascheroni Constant is known as [15]:

γ ≡ limk→∞

k∑i=1

1

i− ln(k) =

∫ ∞1

1

bxc− 1

xdx (168)

We want to find: ∫ ∞0

1/xdx (169)

For this matter we will take the first equation and change the summation index:

γ = limk→∞

k−1∑j=0

1

j + 1− ln(k) =

∫ ∞0

1

bx+ 1c−(

1

x− iπδ(x)

)dx

=

∫ ∞1

1

bxcdx−

∫ ∞0

(1

x− iπδ(x)

)dx

=

∫ ∞1

1

bxc− 1

xdx︸ ︷︷ ︸

γ

−∫ 1

0

1

xdx+

∫ ∞0

iπδ(x)dx

→∫ 1

0

1

xdx =

∫ ∞0

iπδ(x)dx

(170)

Since using the substitution u = 1/x it can be shown that∫ 1

01/xdx =

∫ ∞1

1/xdx (171)

We conclude that: ∫ ∞0

1

xdx =

∫ ∞0

i2πδ(x)dx (172)

Now using a sequence of normal distribution to approximate the Dirac-Delta we find,since we are integrating over half of its domain.∫ ∞

0

1

xdx = iπ (173)

This result can also be shown using complex integration.

30

References

[1] Claude E. Shannon, A Mathematical Theory of Communication, The Bell SystemTechnical Journal, Vol. 27, pp. 379–423, 623–656 (1948)

[2] J. von Neumann, Mathematical Foundations of Quantum Mechanics. Princeton Uni-versity Press. (1996)

[3] Reinhold A. Bertlmann and Nicolai Friis, Lecture Course Scripts, Theoretical PhysicsT2, Quantum Mechanics, Chapter 9.http://homepage.univie.ac.at/reinhold.bertlmann/lectures/

[4] Nicolas J. Cerf and Chris Adami, Negative Entropy in Quantum Information Theory,arXiv:quant-ph/9610005v1. (1996)

[5] Huzhiro Araki, Elliot H. Lieb, Entropy Inequalities, Commun. math. Phys. 18,160—170. (1970)

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[8] Eric Gourgolhon, Special Relativity in General Frames, Springer-Verlag BerlinHeidelberg. (2013)

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[12] Shaun M. Fallat, Algebraic Integers and Tensor Product of Matriceshttps://cms.math.ca/crux/v22/n8/page341-343.pdf.

[13] Paul A.M. Dirac, Principles of Quantum Mechanics, Oxford University Press, 4thEdition, Indian Edition. (1958)

[14] Euler’s Exponentials, Professor Raymond Floodavailable at https://www.youtube.com/watch?v=VIADThf8gjE

[15] Arfken, Weber and Harris, Mathematical Methods for Physicists, 7th Edition, El-sevier Inc. (2013)

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