2 + Quantum state evolution in C and G3
Alexander M. SOIGUINE
Abstract: It was shown, [1] - [3], that quantum mechanical qubit states as elements of C 2 can be generalized to elements of even subalgebra G3 of geometric algebra G3 over Euclidian space E3 . The construction critically depends on generalization of formal, unspecified, complex plane to arbitrary variable, but explicitly defined, planes in 3D, and of usual Hopf 3 2 B 2 fibration S S to maps g : g G3 , g 1S generated by arbitrary unit value bivectors B G3 . 2 2 Analysis of the structure of the map G3 C demonstrates that quantum state evolution in C gives only restricted information compared to that in .
1. Introduction
2 Qubits, unit value elements of the Hilbert space C of two dimensional complex vectors, can be 1 generalized to unit value elements I S of even subalgebra of geometric algebra over Euclidian space . I called such generalized qubits g -qubits [1].
Some minimal information about and is necessary. Algebraically, is linear space with canonical basis 1,e1,e2 ,e3,e2e3,e3e1,e1e2 ,e1e2e3, where 1 is unit value scalar,ei are orthonormal basis vectors in , eie j are oriented, mutually orthogonal unit value areas (bivectors) spanned by ei and e as edges, with orientation defined by rotation to by angle ; and e e e is unit value j 2 1 2 3 oriented volume spanned by ordered edges e1 , e2 and e3.
Subalgebra is spanned by scalar 1 and basis bivectors: 1,e e ,e e ,e e . Variables and in 2 3 3 1 1 2 IS G3 are scalars, I S is a unit size oriented area, bivector, lying in an arbitrary given plane
S E3 . Bivector is linear combination of basis bivectors ei e j , see Fig.1a.
1 This sum bears the sense “something and something”. It is not a sum of geometrically similar elements giving another element of the same type, see [3], [9].
1
Fig.1a Fig.1b
It was explained in [4], [5] that elements I S only differ from what is traditionally called “complex numbers” by the fact that S E3 is an arbitrary, though explicitly defined, plane in E3 . Putting it simply, are “complex numbers” depending on E2 embedded into . Traditional “imaginary unit” i is actually associated with some unspecified I S – everything is going on in one tacit fixed plane, not in 3D world. Usually i is considered just as a “number” with additional algebraic property i 2 1. That may be a source of ambiguities, as it happens in quantum mechanics.
2 I will denote unit value oriented volume e1e2e3 by I 3 . It has the property I3 1. Actually, there always are two options to create oriented unit volume, depending on the order of basis vectors in the product. They correspond to the two types of handedness – left and right screw handedness. In Fig.1a the variant of right screw handedness is shown. It is geometrically obvious that eie j e jei (bivector flips, changes orientation to the opposite, in swapping the edge vectors). Then changing of handedness, for example by e1e2e3 e2e1e3 , gives e2e1e3 e1e2e3 I3 .
Any basis vector is conjugate (conjugation means multiplication by ), or dual, to a basis bivector and inverse. For example I3e2e3 e2e3I3 e1 . By multiplying these equations from left or right by we get e2e3 I3e1 e1I3 . Such duality holds for arbitrary, not just basis, vectors and bivectors. Any vector b is conjugate to some bivector: b I3B BI 3 and for any bivector B there exists such that B I3b bI3 . Good to remember that in vector and bivector are in handedness opposite to while in they are in the same handedness as .
One can also think about as a right (left) single thread screw helix of the height one. In this way I3 is left (right) screw helix.
All above remains true (see Fig.1b) if we replace basis bivectors e2e3,e3e1,e1e2 by an arbitrary triple of unit bivectors B1, B2, B3 satisfying the same multiplication rules which are valid for (see Fig.2, right screw handedness assumed):
B1B2 B3 ; B1B3 B2 ; B2B3 B1 (1.1)
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Fig.2
2 3 2. Parameterization of unit value elements in G3 and C by points of S
Let’s take a g -qubit:
2 2 IS b1B1 b2B2 b3B3 1B1 2B2 3B3,i bi , 1, 2 2 2 b1 b2 b3 1
I will use notation so,,S for such elements. They can be considered as points on unit radius sphere in the sense that any point ,1,2 ,3 parameterizes a -qubit.
A pure qubit state in terms of conventional quantum mechanics is two dimensional unit value vector with complex components:
z 1 0 1 2 2 ~ ~ 1 2 z1 z2 , z1 z2 z1z1 z2 z2 1, zk zk iz k ,k 1,2 z2 0 1
We can also think about such pure qubit states as points on S 3 because:
3 1 2 1 2 2 2 ~ ~ 1 2 2 2 1 2 2 2 S z1 , z1 , z2 , z2 {z (z1, z2 )C ; z z1z1 z2 z2 z1 z1 z2 z2 1} (2.1)
Explicit relations, based on their parameterization by points, between elements and elements so,,S can be established through the following construction2.
Let basis bivector B1 is chosen as defining the complex plane S , then we have (see multiplication rules (1.1)):
2 Though both and can be parametrized by points of , it is not correct to say that
2 2 {z (z1, z2 )C ; z 1} is sphere or -qubit is sphere .
3
so(,,S) ( 1B1) 2B1B3 3B3 ( 1B1) (3 2B1)B3
Hence we get the map:
z1 i so,,S 1 1 , where i B 1 1 z2 3 i2
In the similar way, for two other selections of complex plane we get:
z2 i so, , S 1 2 ,i B , 2 2 z2 1 i3
z 3 i and so,,S 1 3 ,i B 3 3 z2 2 i1
Bi i i So we have three different maps so, , S(z1, z2 ) defined by explicitly declared complex planes Bi satisfying (1.1):
i 1 ,i B so(, , S) B B B 1 1 1 2 2 3 3 3 i2 1B1 (3 2B1)B3 i 2 ,i B2 (2.2) 2B2 (1 3B2 )B1 1 i3 B ( B )B i3 3 3 2 1 3 2 ,i B3 2 i1
There exists infinite number of options to select the triple Bi . It means that to recover a g -qubit z so,,S in 3D associated with 1 it is necessary, firstly, to define which bivector B in 3D i1 z2 should be taken as defining “complex” plane and then to choose another bivector B , orthogonal to i2 B . The third bivector B , orthogonal to both B and B , is then defined by the first two by i1 i3 i1 i2 orientation (handedness, right screw in the used case ): I B I B I B I . 3 i1 3 i2 3 i3 3
The conclusion is that to each single element (2.1) there corresponds infinite number of elements so,,S depending on choosing of a triple of orthonormal bivectors B1, B2, B3 in 3D satisfying (1.1), and associating one of them with complex plane. This allows constructing map, fibration 2 G3 C restricted to unit value elements.
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3. Fiber bundle
Take general definition of fiber bundle as a set E, M,,G, F where E is bundle (or total) space; M - the base space; F - standard fiber; G - Lie group which acts effectively on ; - bundle projection: 1 : E M , such that each space Fx (x) , fiber at xM , is homeomorphic to standard fiber .
Bi i i We can think about the map so, , S(z1, z2 ) as a fiber bundle. In the current case, the fiber 2 2 bundle will have so(,,S) g G3 : g 1 as total space and {z (z1, z2 )C ; z 1} as base
2 2 G | 3 C | 3 :G | 3 C | 3 space. I will denote them as 3 S and S respectively. The projection 3 S S depends on which particular Bi is taken from an arbitrary triple B1, B2, B3 satisfying (1.1) as associated complex plane of complex vectors of C 2 and explicitly given by (2.2), so we should write
Bi 2 :G | 3 C | 3 3 S S .
By some reasons that will be explained a bit later I will use complex plane associated with B3 , so by (2.2) the projection is:
i 3 : so(, , S) 1B1 2B2 3B3 (3.1) 2 i1
x1 iy1 2 Then for any z C | 3 the fiber in is comprised of all elements S x2 iy2
Fz x1 y2B1 x2B2 y1B3 with an arbitrary triple of orthonormal bivectors in 3D satisfying (1.1). That particularly means that standard fiber is equivalent to the group of rotations of the triple y2B1, x2B2 , y1B3 as a whole:
Fig 3a. Fig 3b.
Two example sections of a fiber. 3b received from 3a by 90 deg. counterclockwise rotation around vertical axis
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G | 3 All such rotations in G3 are also identified by elements of 3 S since for any bivector B the result of its rotation is 3 (see, for example [6], [7]):
~ ~ ~ so( ,, S) Bso( ,,S) , where so( ,, S) ( 1B1 2B2 3B3) 1B1 2B2 3B3
So, standard fiber is identified as and composition of rotations is:
I I I I I I ~ I I e S2 e S1 Be S1 e S2 e S1 e S2 Be S1 e S2
B3 2 G | 3 C | 3 All that means that the fibration 3 S S is principal fiber bundle with standard fiber and the group acting on it is the group of (right) multiplications by elements of .
Now the explanation why B3 was taken as complex plane. As was shown in [1], [2] the variant of classical Hopf fibration
2 2 ~ ~ ~ ~ 2 2 C | 3 S Pr z , z z z z z ,i z z z z , z z S : 1 2 1 2 1 2 1 2 1 2 1 2
can be received as one of the cases of generalized Hopf fibrations:
B 2 B ~ 2 G | 3 S : 3 S so(,,S) so(,,S) Bso(,,S) S .
The basis bivector B1 gives:
B1 so(, , S) ( IS )B1( IS ) ( 1B1 2B2 3B3)B1( 1B1 2B2 3B3) 2 2 2 2 ( 1 ) (2 3 )B1 2(3 12 )B2 2(13 2 )B3 2 2 2 2 ( 1 ) (2 3 ),23 12 ,213 2 (3.2) and for two other basis bivectors: so, , SB2 2 B 2 2 2 2 B 2 B 1 2 3 1 2 1 3 2 1 2 3 3 (3.3) 2 2 2 2 212 3 , 2 1 3 ,21 23 so, , SB3 2 B 2 B 2 2 2 2 B 2 1 3 1 2 3 1 2 3 1 2 3 (3.4) 2 2 2 2 22 13 ,223 1, 3 1 2
3 It is often convenient to write elements so,,S as exponents: so, , S eI S .
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In the literature mostly the third or the first variants are called Hopf fibrations. For my considerations it does not matter which variant to choose. I take the case (3.4) with B3 as complex plane:
i 3 so(,,S) 1B1 2B2 3B3 3B3 (2 1B3)B2 ,i B3 2 i1
4. Tangent spaces.
We need to temporarily get back to the case of G2 - geometric algebra on a plane [4].
Let an orthonormal basis e1,e2 is taken. It generates basis 1,e1,e2,e1e2 where e1e2 e1 e2 e1 e2 is usual geometric product of two vectors: first member is scalar product of the vectors and second member is oriented area (bivector) swept by rotating e1 to e2 by the angle which is less than .
The basis vectors satisfy particularly the properties:
2 2 I2 e1e2 1, I2e1 e2 (clockwise rotation), e1I2 e2 (counterclockwise rotation), I2e2 e1 , e2 I2 e1 . I am using notation I 2 for unit bivector e1e2 .
For further convenience, let’s construct a matrix basis isomorphic to . Commonly used 0 agreement will be that scalars are identified with scalar matrices, for example: . 0
For the case the second order matrices will suffice to get necessary isomorphism. Let’s take 1 0 0 1 4 1 and 2 as corresponding to geometric basis vectors e1 and e2 . They satisfy 0 1 1 0 2 1 0 0 1 1 0 2 2 2 the properties mentioned above, for example: i , I2 12 , 0 1 1 0 0 1 0 11 0 0 1 I 21 2 , etc. 1 00 1 1 0
The operation representing scalar product gives:
4 I begin with ordinary real component matrices. That is why the order of Pauli matrices in my following definition is different from usual one. An analogue of “imaginary” unit should logically be the last one, after pure real items.
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1 1 0 1 0 1 0 0 1 2 1 2 2 1 (4.1) 2 2 1 0 1 0 0 0
as it should be. If we take two arbitrary vectors expanded in basis 1,2:
0 0 1 2 1 2 a 11 22 0 1 2 0 2 1
0 0 1 2 1 2 b 11 22 0 1 2 0 2 1 then their product is:
0 0 1 0 1 1 2 2 2 1 1 2 ab (a b) 12 2 I2 0 11 22 12 21 0 0 1
The first member is usual scalar product and the second one is bivector of the value equal to the area of parallelogram with the sides a and b .
Important thing to keep in mind is that multiplication of any vector by unit bivector I 2 rotates the vector by : 2
0 1 1 2 2 1 aI2 21 12 (counterclockwise rotation) 2 1 1 0 1 2
0 1 1 2 2 1 I2a 21 12 (clockwise rotation) 1 02 1 1 2
This remains valid for any unit bivector of the same orientation as . We can conclude that such multiplications give basis vectors of the tangent spaces to the original vectors.
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This is identical to considering even elements 1 B2 corresponding to vectors and their multiplication by B : 1 B2 B 2 B1 . Elements and 2 B1 are orthogonal: B , B B B 0 (index 0 means scalar part). 1 2 2 1 0 1 2 2 1 0
General element of G2 in matrix basis 1,1,2,12 I2, isomorphic to geometrical basis
1,e1,e2,e1e2, is:
1 0 1 0 0 1 0 1 0 1 2 3 g2 0 1 2 3 0 1 0 1 1 0 1 0 2 3 0 1
This is arbitrary real valued matrix of the second order. Inversely, any matrix of second order can be uniquely mapped to the element of in the basis :
x y x t x t y z y z 1 2 I2 z t 2 2 2 2
To upgrade to G3 , we need basis matrices of a higher order because a second order non-zero matrix orthogonal in the sense of scalar product (4.1) to both 1 and 2 does not exist.
It is easy to verify that the three, playing the role of three-dimensional orthonormal basis, necessary matrices can be taken as 4th-order block matrices:
I 0 0 I 0 I 2 1 , 2 , 3 , 0 I I 0 I2 0
1 0 0 1 where I - scalar matrix corresponding to scalar 1 , I2 1 2 . Easy to see that 1 0 1 1 0 and 2 are received from 1 and 2 by replacing scalars by corresponding scalar matrices.
By multiplications in full 4 4 form one can easily proof that block-wise multiplications are correct, the basis 1, 2,3 is orthonormal, anticommutative, and 2 3 , 31 and 1 2 satisfy requirements (1.1).
I 0 I 0 I 0 2 2 Taking I3 1 2 3 , multiplications give: 23 I3 I31 , 0 I2 0 I2 0 I
0 I 0 I 0 I 0 I 2 2 2 31 I3 I3 2 , 1 2 I3 I3 3 . Then, since I3 1, I2 0 I 0 I 0 I2 0 we easily get:
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23 31 1 2 , 23 1 2 31 , 311 2 23
that is exactly (1.1) with basis bivectors B1 2 3 , B2 31 and B3 1 2 .
Following all that we can write general element g3 of algebra G3 expanded in formal matrix basis
1,1, 2,3,23,31,1 2,1 23 1,1,2,3, I31, I3 2, I33, I3 as:
g3 I 11 2 2 3 3 1 2 3 2 31 31 2 01 2 3
1 I2 0 1 2 3 I2 2 3 (4.2) 2 3 I2 2 3 1 I2 0 1
0 I This is arbitrary “complex” valued matrix where “imaginary unit” is matrix I 2 , I 0 corresponding to bivector expanded in the first two basis vectors 1 , 2 , that geometrically is bivector
B3 . Inversely, any “complex” valued matrix can be written as element in matrix basis
1,1, 2,3, I31, I3 2, I33, I3 isomorphic to geometrical basis 1,e1,e2 ,e3,e2e3,e3e1,e1e2 ,e1e2e3:
s I t u I v s z s z u x y v t w y v u x t w 2 2 1 2 3 2 3 31 1 2 I3 x I2 y z I2w 2 2 2 2 2 2 2 2 If belongs to even subalgebra G3 then, by identifying 2 3 B1 , 31 B2 , 1 2 B3, we have from (4.2) the correspondence:
I I 2 1 3 2 2 g3 1B1 2B2 3B3 (4.3) 3 I22 I21
In the same way as in the G2 case, multiplications of so(,1,2 ,3 ) 1B1 2 B2 3B3 by basis bivectors Bi give basis bivectors of tangent spaces to original bivectors:
T1 so(, 1, 2 , 3 )B1 1B1 2B2 3B3 B1 1 B1 3B2 2 B3
T2 so(,1,2,3)B2 1B1 2B2 3B3 B2 2 3B1 B2 1B3
T3 so(,1,2,3)B3 1B1 2B2 3B3 B3 3 2B1 1B2 B3
G | 3 These 3 S elements are orthogonal to 1B1 2 B2 3B3 and to each other, and are the tangent space basis elements at points .
2 T C | 3 Projections of i onto S are:
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i i i 1 2 2 1 3 (T1 ) , (T2 ) , (T3) 3 i i3 1 i2
2 2 C | 3 C These elements of S are orthogonal, in the sense of Euclidean scalar product in :
~ ~ i3 z, w Rez1w1 z2w2 , to each other and to the projection (so(, 1, 2 , 3 ) of 2 i1 G | 3 the original state in 3 S . They are the tangent space basis elements in at points
z1 i3 . z2 2 i1
5. Hamiltonians in and their lifts in G3
Take self-adjoint (Hamiltonian) linear operator in C 2 in its most general form:
a b c I 2d H aI b1 c 2 d 3 . Its corresponding G3 element, see (4.3), is c I 2d a b a I3bB1 cB2 dB3 and does not belong to , though the result of its action on an element from (qubit), has 1 preimage in . Let’s prove that. Take a qubit and a Hamiltonian in :
x iy r ei1 x y z 1 1 1 , r x2 y 2 , cos k , sin k , k 1,2 i k k k k k x iy r e 2 2 2 2 2 2 2 2 xk yk xk yk
a b Rei c d H , R c2 d 2 , cos , sin i Re a b c2 d 2 c2 d 2
Then:
a b Rei rei 1 Hz 1 i i 2 Re a br2e (5.1) (a b)r1 cos1 Rr2 cos( 2 ) i(a b)r1 sin1 Rr2 sin( 2 ) (a b)r2 cos 2 Rr1 cos( 1) i(a b)r2 sin 2 Rr1 sin( 1)
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x1 iy1 2 For each single qubit z C | 3 its fiber (full preimage) in G | 3 is S 3 S x2 iy2
Fz x1 y2B1 x2B2 y1B3 with an arbitrary triple of orthonormal bivectors B1, B2, B3 in 3D satisfying (1.1). In exponential form:
y x y F x y B x B y B x 1 x2 2 B 2 B 1 B z 1 2 1 2 2 1 3 1 1 2 1 2 2 2 3 1 x1 1 x1 1 x1
I S cos sinb1B1 b2B2 b3B3 e , (5.2) where
cos x ,sin 1 x 2 1 1 y x y b 2 ,b 2 ,b 1 (5.3) 1 2 2 2 3 2 1 x1 1 x1 1 x1 I S b1B1 b2 B2 b3B3
Using (5.2) we get from (5.3) the fiber at Hz :
IS ( H )(H ) FHz e , (5.4)
where
cos(H ) (a b)r1 cos1 Rr2 cos( 2 ); I b (H )B b (H )B b (H )B ; S (H ) 1 1 2 2 3 3 (a b)r sin Rr sin( ) (a b)r cos Rr cos( ) b (H ) 2 2 1 1 ,b (H ) 2 2 1 1 , 1 2 2 2 (5.5) 1 cos (H ) 1 cos (H ) (a b)r sin Rr sin( ) b (H ) 1 1 2 2 3 2 1 cos (H )
1 We can also explicitly write the element g3(H) which, acting (from the right) on (z) gives 1(Hz) :
1 1 (z)g3(H) (Hz)
From (5.1) and (5.4):
I S I S ( H )(H ) I S I S ( H )(H ) e g3(H) e g3(H) e e
12 where IS , , IS (H ) and (H) are defined by (5.3) and (5.5).
6. Clifford translations
2 i C | 3 Cl z e z Suppose, action on z in S is multiplication by an exponent: , often called Clifford translation5, that’s:
i1 i r1e r1 cos( 1 ) ir1 sin( 1 ) Cl (z) e i 2 r2e r2 cos( 2 ) ir2 sin( 2 )
G | 3 Then the fiber of Cl (z) in 3 S is:
F r cos( ) r sin( )B r cos( )B r sin( )B Cl ( z) 1 1 2 2 1 2 2 2 1 1 3
r1 cos( 1 ) r1 sin( 1 )B3 r2 cos( 2 ) r2 sin( 2 )B3 B2 (6.1)
B3 B31 B3 B3 2 B3 r1e e r2e e B2 e Fz
In other words, Clifford translations in are equivalent to multiplications of fibers in by standard fiber elements with bivector part B3 (which is associated with formal imaginary unit i ) and the same phase .
In commonly used quantum mechanics all states from the set Cl z are considered as identical to the state , they have the same values of observables. That is not commonly true for g -qubits. Only in the
2 6 case when all the observables common plane, say IS , (unspecified in the C model ) is the same as the
IS I S plane associated with complex plane in , the -qubit e Fz leaves any such observable e unchanged:
~ I S I S I S ~ I S Fze e e Fz Fze Fz
Suppose angle in is varying. Then the tangent to Clifford orbit is
Cl z ei z ie i z iCl z.
5 It is called “translation” because does not change distance between elements. 6 In the G3 model observables are also elements of G3 , see [1], [2]
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2 ~ ~ It is orthogonal to Cl z in the sense of usual scalar product in C : u,v Reu1v1 u2v2 , and
2 C | 3 remains on S : iCl (z) 1 (translational velocity of Clifford translation is one).
~ z1 2 z2 2 For any z C | 3 define w C | 3 . The elements iz , w and iw are orthogonal to S ~ S z2 z1 z and pairwise orthogonal. That means they span tangent space and the plane spanned by w,iw is orthogonal to the Clifford orbit plane spanned by z,iz.
The vector of translational velocity rotates while moving in the orbit plane. Both and also rotates with the same unit value angular velocity since Cl w ei w ei iw Cl iw and Cl iw ei iw ie i iw Cl w .
To make the planes of rotations clearly identified, that is difficult to do with formal imaginary unit, let’s G | 3 lift Clifford translation to 3 S using (6.1). Translational velocity, similar to the case, is F eB3 F B F and is orthogonal to F : Cl (z) z 3 Cl (z) Cl (z)
~ ~ ~ FCl (z) , B3FCl (z) FCl (z) FCl (z) B3 B3 0 0 0 0
Two other components of the tangent space, orthogonal to and B F at any point of the 3 Cl (z) orbit, are B F and B F 7. Their velocities while moving along Clifford orbit are: 1 Cl (z) 2 Cl (z)
B3 B1FCl (z) B1e Fz B1B3FCl (z) B2FCl (z) (derivative of is orthogonal to and looking in the direction )
B3 B2FCl (z) B2e Fz B2B3FCl (z) B1FCl (z) (derivative of B F is orthogonal to 2 Cl (z) and looking in the direction B F ) 1 Cl (z)
These two equations explicitly show that the two tangents, orthogonal to Clifford translation velocity, rotate in moving plane B1FCl (z) , B2FCl (z) with the same unit value rotational velocity. Interesting to notice that the triple of the translational velocity and two rotational velocities has orientation opposite
7 Clearly, the three B F are identical to earlier considered tangents T i Cl (z) i
14 to the triple of tangents: if the tangents B1FCl (z) , B2 FCl (z) , B3FCl (z) have right screw orientation, the speed triple B1FCl (z), B2FCl (z), B3FCl (z) is left screw.
If a fiber, g -qubit, makes full circle in Clifford translation: F F eB3 F , 0 2 , both z Cl (z) z B F and B F also make full rotation in their own plane by 2 . This is special case of the - 1 Cl (z) 2 Cl (z) qubit Berry phase incrementing in a closed curve quantum state path.
7. Conclusions
Evolution of a quantum state described in terms of G3 gives more detailed information about two state system compared to the C 2 Hilbert space model. It confirms the idea that distinctions between “quantum” and “classical” states become less deep if a more appropriate mathematical formalism is used. The paradigm spreads from trivial phenomena like tossed coin experiment [3] to recent results on entanglement and Bell theorem [8] where the former was demonstrated as not exclusively quantum property.
Works Cited
[1] A. Soiguine, "Geometric Algebra, Qubits, Geometric Evolution, and All That," January 2015. [Online]. Available: http://arxiv.org/abs/1502.02169.
[2] A. Soiguine, "What quantum "state" really is?," June 2014. [Online]. Available: http://arxiv.org/abs/1406.3751.
[3] A. Soiguine, "A tossed coin as quantum mechanical object," September 2013. [Online]. Available: http://arxiv.org/abs/1309.5002.
[4] A. Soiguine, Vector Algebra in Applied Problems, Leningrad: Naval Academy, 1990 (in Russian).
[5] A. Soiguine, "Complex Conjugation - Realtive to What?," in Clifford Algebras with Numeric and Symbolic Computations, Boston, Birkhauser, 1996, pp. 285-294.
[6] D. Hestenes, New Foundations of Classical Mechanics, Dordrecht/Boston/London: Kluwer Academic Publishers, 1999.
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[7] C. Doran and A. Lasenby, Geometric Algebra for Physicists, Cambridge: Cambridge University Press, 2010.
[8] X.-F. Qian, B. Little, J. C. Howell and J. H. Eberly, "Shifting the quantum-classical boundary: theory and experiment for statistically classical optical fields," Optica, pp. 611-615, 20 July 2015.
[9] J. W. Arthur, Understanding Geometric Algebra for Electromagnetic Theory, John Wiley & Sons, 2011.
[10] D. Chruscinski and A. Jamiolkowski, Geometric Phases in Classical and Quantum Mechanics, Boston: Birkhauser, 2004.
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