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Recently suggested scheme [1] of quantum computing uses g-qubit states as circular polarizations from the solution of Maxwell equations in terms of geometric algebra, along with clear definition of a complex plane as bivector in three dimensions. Here all the details of receiving the solution, and its polarization transformations are analyzed. The results can particularly be applied to the problems of quantum computing and quantum cryptography. The suggested formalism replaces conventional quantum mechanics states as objects constructed in complex vector Hilbert space framework by geometrically feasible framework of multivectors.

The circular polarized electromagnetic waves are the only type of waves following from the solution of Maxwell equations in free space done in geometric algebra terms.

Let’s take the electromagnetic field in the form:

F = F 0 exp [ I S ( ω t − k ⋅ r ) ] (1)

requiring that it satisfies the Maxwell system of equations in free space, which in geometrical algebra terms is one equation:

( ∂ t + ∇ ) F = 0 (2)

where ∇ = ∂ ∂ x x ^ + ∂ ∂ y y ^ + ∂ ∂ z z ^ and multiplications are the geometrical algebra ones.

Element F 0 in (1) is a constant element of geometric algebra G 3 and I S is unit value bivector of a plane S in three dimensions, that is a generalization of the imaginary unit [

e I S φ = cos φ + I S sin φ , φ = ω t − k ⋅ r

Solution of (2) should be sum of a vector (electric field E) and bivector (magnetic field I 3 H ):

F = E + I 3 H

with some initial conditions:

E + I 3 H | t = 0 , r = 0 = F 0 = E | t = 0 , r = 0 + I 3 H | t = 0 , r = 0 = E 0 + I 3 H 0

In the magnetic field I 3 H the item I 3 is unit pseudoscalar in three dimensions assumed to be the right-hand screw oriented volume, relative to an ordered triple of orthonormal vectors.

Substitution of (1) into the Maxwell’s (2) will exactly show us what the solution looks like.

The derivative by time gives

∂ ∂ t F = F 0 e I S φ I S ∂ ∂ t ( ω t − k ⋅ r ) = F 0 e I S φ I S ω = F I S ω

The geometric algebra product ∇ F is:

∇ F = F 0 I S e I S φ ∇ ( ω t − k ⋅ r ) = − F 0 e I S φ I S k = − F I S k

or

∇ F = F 0 e I S φ ∇ ( ω t − k ⋅ r ) I S = − F 0 e I S φ k I S = − F k I S ,

depending on do we write I S ( ω t − k ⋅ r ) or ( ω t − k ⋅ r ) I S . The result should be the same since ω t − k ⋅ r is a scalar.

Commutativity I S k = k I S is true only if k × I 3 I S = 0 . The following agreement takes place between orientation of I 3 , orientation of I S and direction of vector k [^{1}.

^{1}For any vector we write a ^ = a / | a | .

Assuming that orientation is I 3 = k ^ I S , the Maxwell equation becomes:

F ( I S ω − I 3 | k | ) = F ( ω I S − | k | k ^ I S ) = 0

or

( E + I 3 H ) ω = ( E + I 3 H ) k

Left hand side is sum of vector and bivector, while right hand side is scalar E ⋅ k plus bivector E ∧ k , plus pseudoscalar I 3 ( H ⋅ k ) , plus vector I 3 ( H ∧ k ) . It follows that both E and H lie on the plane of I S and then:

ω E = I 3 H k , ω I 3 H = E k → ω 2 | k | 2 I 3 H k = ω E

Thus, ω = | k | and we get equation I 3 H k ^ = E from which particularly follows | E | 2 = | H | 2 and E ^ k ^ H ^ = I 3 .

The result for the case I 3 = k ^ I S is that the solution of (2) is

F = ( E 0 + I 3 H 0 ) exp [ I S ( ω t − k ⋅ r ) ]

where E 0 and H 0 are arbitrary mutually orthogonal vectors of equal length, lying on the plane S. Vector k should be normal to that plane, k ^ = − I 3 I S and | k | = ω .

In the above result the sense of the I S orientation and the direction of were assumed to agree with I 3 = k ^ I S . Opposite orientation, − I 3 = k ^ I S , that’s k and I S compose left hand screw and k ^ = I 3 I S , will give solution F = F 0 exp [ I S ( ω t − k ⋅ r ) ] with E ^ H ^ k ^ = I 3 .

Summary:

For a plane S in three dimensions Maxwell equation (2) has two solutions

・ F + = ( E 0 + I 3 H 0 ) exp [ I S ( ω t − k + ⋅ r ) ] , with k ^ + = I 3 I S , E ^ H ^ k ^ + = I 3 , and the triple { E ^ , H ^ , k ^ + } is right hand screw oriented, that’s rotation of E ^ to H ^ by π/2 gives movement of right hand screw in the direction of k + = | k | I 3 I S .

・ F − = ( E 0 + I 3 H 0 ) exp [ I S ( ω t − k − ⋅ r ) ] , with k ^ − = − I 3 I S , E ^ H ^ k ^ − = − I 3 , and the triple { E ^ , H ^ , k ^ − } is left hand screw oriented, that’s rotation of E ^ to H ^ by π/2 gives movement of left hand screw in the direction of k − = − | k | I 3 I S or, equivalently, movement of right hand screw in the opposite direction, − k − .

・ E_{0} and H_{0}, initial values of E and H, are arbitrary mutually orthogonal vectors of equal length, lying on the plane S. Vectors k ± = ± | k ± | I 3 I S are normal to that plane. The length of the wave vectors | k ± | is equal to angular frequency w.

Maxwell Equation (2) is a linear one. Then any linear combination of F + and F − saving the structure of (1) will also be a solution.

Let’s write:

{ F + = ( E 0 + I 3 H 0 ) exp [ I S ω t ] exp [ − I S [ ( I 3 I S ) ⋅ r ] ] F − = ( E 0 + I 3 H 0 ) exp [ I S ω t ] exp [ I S [ ( I 3 I S ) ⋅ r ] ] (3)

Then for arbitrary scalars λ and μ:

λ F + + μ F − = ( E 0 + I 3 H 0 ) e I S ω t ( λ e − I S [ ( I 3 I S ) ⋅ r ] + μ e I S [ ( I 3 I S ) ⋅ r ] ) (4)

is solution of (2). The item in second parenthesis is weighted linear combination of two states with the same phase in the same plane but opposite sense of orientation. The states are strictly coupled, entangled if you prefer, because bivector plane should be the same for both, does not matter what happens with it.

One another option is:

( λ 1 + I 3 μ 1 ) ( E 0 + I 3 H 0 ) exp [ I S ω ( t − ( I 3 I S ) ⋅ r ) ] + ( λ 2 + I 3 μ 2 ) ( E 0 + I 3 H 0 ) exp [ I S ω ( t + ( I 3 I S ) ⋅ r ) ] = [ λ 1 E 0 − μ 1 H 0 + I 3 ( μ 1 E 0 + λ 1 H 0 ) ] exp [ I S ω ( t − ( I 3 I S ) ⋅ r ) ] + [ λ 2 E 0 − μ 2 H 0 + I 3 ( μ 2 E 0 + λ 2 H 0 ) ] exp [ I S ω ( t + ( I 3 I S ) ⋅ r ) ]

which is just rotation, along with possible change of length, of electric and magnetic initial vectors in their plane.

Polarizations, in our approach, exponents in the solution of (3), have the form of states [

The basis bivectors satisfy multiplication rules (in the righth and screw orientation of I 3 ):

B 1 B 2 = − B 3 , B 1 B 3 = B 2 , B 2 B 3 = − B 1

One can identify basis bivectors with usual coordinate planes: B 1 = y ^ z ^ , B 2 = z ^ x ^ , B 3 = x ^ y ^ . Any one of these three bivectors can be taken as explicitly identifying imaginary unit, though any unit value bivector in three dimensions can take the role [

Thus:

α + I S β = α + β ( b 1 B 1 + b 2 B 2 + b 3 B 3 ) ≡ α + β 1 B 1 + β 2 B 2 + β 3 B 3

The difference between units of information in classical computational scheme, quantum mechanical conventional computations (qubits) and geometric algebra

scheme (g-qubits) with variable explicitly defined complex plane is seen from

Circular polarizations received as solutions of Maxwell Equation (2) is an excellent choice to have such g-qubits in a lab.

Commonly accepted idea to use systems of qubits to tremendously increase speed of computations is based on assumption of entanglement ? roughly speaking when touching one qubit all the other in the system react instantly, in no time. A bit strange, though you should not care about that because our paradigm is very different.

Assume we have some general state:

α + I S β = α + β ( b 1 B 1 + b 2 B 2 + b 3 B 3 ) ≡ α + β 1 B 1 + β 2 B 2 + β 3 B 3

The state can be identified as a point ( α 1 , β 1 , β 2 , β 3 ) on unit sphere S 3 . It can be subjected to a Clifford translation

α + I S β ⇒ e I C l Δ ψ ( α + I S β )

executing displacement Δ ψ at point ( α 1 , β 1 , β 2 , β 3 ) along intersection of S 3 with the unit bivector plane I C l .

Let’s make notations more like conventional quantum mechanical ones. I will write:

α + I S β ≡ | g 〉 ( α , β , I S ) , α + I S β ¯ = α − I S β ≡ 〈 g | ( α , β , I S )

and use Hamiltonian like form of the Clifford translation bivector.

Conventional Hamiltonian

( γ + γ 1 γ 2 − i γ 3 γ 2 + i γ 3 γ − γ 1 ) ,

with removed not important scalar γ, has the lift in G 3 + [

H = I 3 ( γ 1 B 1 + γ 2 B 2 + γ 3 B 3 )

Then the associated Clifford translation plane bivector is − I 3 H ( t ) . By normalizing the bivector to unit value we get generalization of imaginary unit

i ⇒ I 3 H ( t ) | H ( t ) | ,

that is critical for the whole approach. Therefore, for some Δt, Clifford translation for a given Hamiltonian is:

| g ( t + Δ t ) 〉 ( α ( t + Δ t ) , β ( t + Δ t ) , I S ( t + Δ t ) ) = e − ( I 3 H ( t ) | H ( t ) | ) | H ( t ) | Δ t | g ( t ) 〉 ( α ( t ) , β ( t ) , I S ( t ) ) (5)

For an arbitrary sequence of infinitesimal Clifford translations, the final state is integral^{2}

∫ e − I H ( l ) H ( l ) d l | g ( l ) 〉 ( α ( l ) , β ( l ) , I S ( l ) )

along the curve on unit sphere S 3 composed of infinitesimal displacements by

− ( I 3 H ( t ) | H ( t ) | ) | H ( l ) | d l

Let’s calculate the result of the right-hand side of (5) in general case when the plane of I 3 H ( t ) | H ( t ) | differs from S ( t ) .

To calculate the geometric algebra product of the two exponents in Clifford translation with not coinciding exponent planes, e I S 1 Δ φ 1 e I S 2 φ 2 , S 1 ≠ S 2 , let’s first expand I S 1 in original basis to get formulas for generators of Clifford translation. If I S 1 = γ 1 B 1 + γ 2 B 2 + γ 3 B 3 then a part of geometrical product e I S 1 Δ φ 1 e I S 2 φ 2 is:

I S 1 I S 2 = ( γ 1 B 1 + γ 2 B 2 + γ 3 B 3 ) ( β 1 B 1 + β 2 B 2 + β 3 B 3 ) = − ( γ 1 β 1 + γ 2 β 2 + γ 3 β 3 ) + ( γ 3 β 2 − γ 2 β 3 ) B 1 + ( γ 1 β 3 − γ 3 β 1 ) B 2 + ( γ 2 β 1 − γ 1 β 2 ) B 3 = − ( γ ⋅ β ) − I 3 ( γ × β ) = I S 1 ⋅ I S 2 + I S 1 ∧ I S 2

(see

^{2}In the case of constant plane of Hamiltonian, it easily follows the Schrodinger equation of conventional quantum mechanics with clearly defined imaginary unit.

where γ and β are vectors dual to bivectors I S 1 and I S 2 .

Thus, the full product is:

e I S 1 Δ φ 1 e I S 2 φ 2 = cos Δ φ 1 cos φ 2 + sin Δ φ 1 cos φ 2 I S 1 + cos Δ φ 1 sin φ 2 I S 2 + sin Δ φ 1 sin φ 2 I S 1 I S 2 = cos Δ φ 1 cos φ 2 − ( sin Δ φ 1 cos φ 2 I S 1 I S 2 ) I S 2 − I S 1 ( cos Δ φ 1 sin φ 2 I S 1 I S 2 ) + sin Δ φ 1 sin φ 2 I S 1 I S 2

= cos Δ φ 1 cos φ 2 + sin Δ φ 1 sin φ 2 ( I S 1 ⋅ I S 2 ) ︸ ( scalarpart ) + sin Δ φ 1 cos φ 2 I S 1 + cos Δ φ 1 sin φ 2 I S 2 + sin Δ φ 1 sin φ 2 I S 1 ∧ I S 2 ︸ ( bivectorpart , expansioninnon-orthonormalbasis I S 1 , I S 2 , I S 1 ∧ I S 2 ) ^{3}

Now we have everything to retrieve action of Clifford translation generated by a Hamiltonian on general solution (4):

e − ( I 3 H ( t ) | H ( t ) | ) | H ( t ) | Δ t ( E 0 + I 3 H 0 ) ( λ e I S ω ( t − ( I 3 I S ) ⋅ r ) + μ e I S ω ( t + ( I 3 I S ) ⋅ r ) )

To make expressions simpler I will use notations I 3 H ( t ) | H ( t ) | ≡ I H , ω ( t − ( I 3 I S ) ⋅ r ) ≡ φ + , and ω ( t + ( I 3 I S ) ⋅ r ) ≡ φ − . Then we get (see Sections 1.3 and 1.6 in [

e − I H | H ( t ) | Δ t ( E 0 + I 3 H 0 ) ( λ e I S φ + + μ e I S φ − ) = − ( E 0 + I 3 H 0 ) ( λ e − I H | H ( t ) | Δ t e I S φ + + μ e − I H | H ( t ) | Δ t e I S φ − )

= − ( E 0 + I 3 H 0 ) λ ( cos ( | H ( t ) | Δ t ) cos φ + − sin ( | H ( t ) | Δ t ) sin φ + ( I H ⋅ I S ) ) − ( E 0 + I 3 H 0 ) λ ( sin ( | H ( t ) | Δ t ) cos φ + I H + cos ( | H ( t ) | Δ t ) sin φ + I S + sin ( | H ( t ) | Δ t ) sin φ + ( I H ∧ I S ) ) − ( E 0 + I 3 H 0 ) μ ( sin ( | H ( t ) | Δ t ) cos φ − I H + cos ( | H ( t ) | Δ t ) sin φ − I S + sin ( | H ( t ) | Δ t ) sin φ − ( I H ∧ I S ) )

^{3}In the case I S 1 = I S 2 we trivially have rotation of e I S 2 φ 2 by angle Δ φ 1 .

Let’s take popular case of I S = B 3 = x ^ y ^ (plane orthogonal to z ^ axis) and I H = B 1 = y ^ z ^ (or I H = B 2 = z ^ x ^ , does not matter.) The above formula becomes:

− ( E 0 + I 3 H 0 ) [ λ ( cos ( | H ( t ) | Δ t ) cos φ + + sin ( | H ( t ) | Δ t ) cos φ + B 1 + sin ( | H ( t ) | Δ t ) sin φ + B 2 + cos ( | H ( t ) | Δ t ) sin φ + B 3 ) + μ ( cos ( | H ( t ) | Δ t ) cos φ − + sin ( | H ( t ) | Δ t ) cos φ − B 1 + sin ( | H ( t ) | Δ t ) sin φ − B 2 + cos ( | H ( t ) | Δ t ) sin φ − B 3 ) ]

It makes simpler if F + and F − are equally weighted, say both λ and μ are equal to one:

− ( E 0 + I 3 H 0 ) [ cos ( | H ( t ) | Δ t ) ( cos φ + + cos φ − ) + sin ( | H ( t ) | Δ t ) ( cos φ + + cos φ − ) B 1 + sin ( | H ( t ) | Δ t ) ( sin φ + + sin φ − ) B 2 + cos ( | H ( t ) | Δ t ) ( sin φ + + sin φ − ) B 3 ] = − 2 ( E 0 + I 3 H 0 ) cos ( z ^ ⋅ r ) [ cos ( | H ( t ) | Δ t ) cos ω t + sin ( | H ( t ) | Δ t ) cos ω t B 1 + sin ( | H ( t ) | Δ t ) sin ω t B 2 + cos ( | H ( t ) | Δ t ) sin ω t B 3 ) (6)

Since a state in the described formalism is operator that gives the result of measurement when acting on observable, which can be any element of geometric algebra G 3 , the following is detailed description of the case when the element in parenthesis of the (6) expression acts on some bivector. Such operation is generalization of the Hopf fibration and rotates the bivector in three dimensions.

Denoting4:

cos ( | H ( t ) | Δ t ) cos ω t + sin ( | H ( t ) | Δ t ) cos ω t B 1 + sin ( | H ( t ) | Δ t ) sin ω t B 2 + cos ( | H ( t ) | Δ t ) sin ω t B 3 ≡ α + β 1 B 1 + β 2 B 2 + β 3 B 3 ≡ e I H , ω ψ

^{4}Easy to see that the left-hand side is unit value element of G 3 + .

where

I H , ω = ( γ 1 B 1 + γ 2 B 2 + γ 3 B 3 )

γ 1 = sin ( | H ( t ) | Δ t ) cos ω t sin 2 ( | H ( t ) | Δ t ) + cos 2 ( | H ( t ) | Δ t ) sin 2 ω t

γ 2 = sin ( | H ( t ) | Δ t ) sin ω t sin 2 ( | H ( t ) | Δ t ) + cos 2 ( | H ( t ) | Δ t ) sin 2 ω t

γ 3 = cos ( | H ( t ) | Δ t ) sin ω t sin 2 ( | H ( t ) | Δ t ) + cos 2 ( | H ( t ) | Δ t ) sin 2 ω t

ψ = cos − 1 ( cos ( | H ( t ) | Δ t ) cos ω t )

and taking a bivector operand (observable) c 1 B 1 + c 2 B 2 + c 3 B 3 we get the result of measurement, action of the state on observable (see [

e − I H , ω ψ ( c 1 B 1 + c 2 B 2 + c 3 B 3 ) e I H , ω ψ = ( c 1 [ ( α 2 + β 1 2 ) − ( β 2 2 + β 3 2 ) ] + 2 c 2 ( β 1 β 2 − α β 3 ) + 2 c 3 ( β 1 β 3 + α β 2 ) ) B 1 + ( 2 c 1 ( α β 3 + β 1 β 2 ) + c 2 [ ( α 2 + β 2 2 ) − ( β 1 2 + β 3 2 ) ] + 2 c 3 ( β 2 β 3 − α β 1 ) ) B 2 + ( 2 c 1 ( β 1 β 3 − α β 2 ) + 2 c 2 ( β 2 β 3 + α β 1 ) + c 3 [ ( α 2 + β 3 2 ) − ( β 1 2 + β 2 2 ) ] ) B 3

= ( c 1 cos 2 ω t − sin 2 ω t ( c 2 cos ( 2 | H ( t ) | Δ t ) − c 3 sin ( 2 | H ( t ) | Δ t ) ) ) B 1 + ( c 1 sin 2 ω t + cos 2 ω t ( c 2 cos ( 2 | H ( t ) | Δ t ) − c 3 sin ( 2 | H ( t ) | Δ t ) ) ) B 2 + ( c 2 sin ( 2 | H ( t ) | Δ t ) + c 3 cos ( 2 | H ( t ) | Δ t ) ) B 3

One interesting remark. If the observable belongs only to the B 1 plane, that’s c 2 = c 3 = 0 , the result of measurement has only components in B 1 and B 2 , projections of the value c 1 due to rotation with angular velocity 2ω around the z ^ axis.

The core of quantum computing should not be in entanglement as it understood in conventional quantum mechanics, which only formally follows from representation of the many particle states as tensor products of individual particle states and not supported by really operating physical devices. The core of quantum computing scheme should be in manipulation and transferring of sets of states as operators decomposed in geometrical algebra variant of qubits (g-qubits), or four-dimensional unit sphere points, if you prefer. Such operators can act on observables, particularly through measurements. From the recent calculation we realize that the action of state, which depends on ( t , r ) , on an observable can be done only if observable is defined at the same point ( t , r ) where the state is defined [

Thus, if we have a state

α + I S β ≡ | g 〉 ( α , β , I S ) = | g 〉 ( α ( t , r ) , β ( t , r ) , I S ( t , r ) )

as in the case of polarization defined states, it becomes a state acting on a set of observables if the latter are defined at some given points:

| c n 〉 = | c 〉 ( c ( t n , r n ) , I C ( t n , r n ) ) = c 0 ( t n , r n ) + I C ( t n , r n ) 1 − c 0 2 ( t n , r n ) , n = 1 , ⋯ , N

Then the state transforms into multi-observable one:

| g ( α ( t 1 , r 1 ) , β ( t 1 , r 1 ) , I S ( t 1 , r 1 ) ) ⋯ g ( α ( t N , r N ) , β ( t N , r N ) , I S ( t N , r N ) ) 〉 ≡ | g 1 ⋯ g N 〉 = ∬ | g 〉 ( α ( t , r ) , β ( t , r ) , I S ( t , r ) ) δ ( r − r 1 ) δ ( t − t 1 ) d r d t + ⋯ + ∬ | g 〉 ( α ( t , r ) , β ( t , r ) , I S ( t , r ) ) δ ( r − r N ) δ ( t − t N ) d r d t = ∑ n = 1 n = N ∬ | g 〉 ( α ( t , r ) , β ( t , r ) , I S ( t , r ) ) δ ( r − r n ) δ ( t − t n ) d r d t

This formula for | g 1 ⋯ g N 〉 bears clear physical and geometrical sense, contrary to conventional quantum mechanics definition following formally from tensor product which does not have good physical interpretation but is the root of entanglement-based quantum computing.

The formula also prompts how quantum encryption decoding can be effectively implemented with the bivector value security key (see

The formula can also be applied to challenging area of anyons in three dimensions.

^{5}Good to remember that “state” and “wave function” are (at least should be) synonyms in conventional quantum mechanics.

Two seminal ideas―variable and explicitly defined complex plane in three dimensions, and the G 3 + states^{5} as operators acting on observables―allow to put forth comprehensive and much more detailed formalism appropriate for quantum mechanics in general and particularly for quantum computing schemes. The approach may be thought about, for example, as a far going geometric algebra generalization of some proposals for quantum computing formulated in terms of light beam time bins, see [

Soiguine, A. (2018) Polarizations as States and Their Evolution in Geometric Algebra Terms with Variable Complex Plane. Journal of Applied Mathematics and Physics, 6, 704-714. https://doi.org/10.4236/jamp.2018.64063