Parallelizable Calculation of Observables Values on Analog Quantum Computer

The superiority of hypothetical quantum computers is not due to faster calculations but due to different schemes of calculations running on special hardware. The core of quantum computing follows the way a state of a quantum system is defined when basic things interact with each other. In conventional approach it is implemented through tensor product of qubits. In the geometric algebra formalism simultaneous availability of all the results for non-measured observables is based on the definition of states as points on three-dimensional sphere.

• Geometric Algebra
• Wave Functions
• Entanglement
• Maxwell Equations
• Three-Dimensional Sphere
• States
• Observables
• Measurements
• GPU
• Multithreading
• OpenCL

1. Introduction: What Is Entanglement in Conventional Approach and in the Geometric Algebra Approach?

In conventional quantum mechanics entanglement means that multiple objects share a single quantum state. Entanglement is the amount by which multiple objects share a quantum state. They remain indeterminate until they are disentangled by a measurement.

The simplest quantum mechanical state, qubit, reads:

$C^2\ni \left(\begin{array}{c}{z}_1\\ z_2\end{array}\right)=z_1\left(\begin{array}{c}1\\ 0\end{array}\right)+z_2\left(\begin{array}{c}0\\ 1\end{array}\right)=z_1|0\rangle +z_2|1\rangle$

$z_1=x_1+i{y}_1$, $z_2=x_2+i{y}_2$, ${\Vert x_1+i{y}_1\Vert }^2+{\Vert x_2+i{y}_2\Vert }^2=1$

It has just two observable “things” after measurement, say “up” for $|0\rangle$ and “down” for $|1\rangle$, with probabilities $z_1^2$ and $z_2^2$. The qubit dimension two is the number of different observable things available after making a measurement on the particle.

In the case of two particles the vector space $C^2$ is generalized to density matrix defined on tensor product $C^2\otimes C^2$. The appropriateness of tensor products is that the tensor product itself captures all the ways that basic things can “interact” with each other [1] . Quantum entanglement means that aspects of one particle of an entangled pair depend on aspects of the other particle. Only when the measurement occurs does the quantum state “collapse” make instantaneously collapsing the other particle state.

That’s crucial words: “instantaneously collapsing”. Common wisdom considers that as physically real transformation of the other particle unknown state into the known one.

The scheme suggested in the geometric algebra approach is based on manipulation and transferring of quantum states as operators acting on observables. Wave functions act in that context on static $G_3^{+}$ elements through measurements, creating “particles”, see [2] .

When assuming that observables are identified by points on ${\mathbb{S}}^3$ we get that measurements of any number of observables by arbitrary set of wave functions are simultaneously available, see [1] [3] [4] .

This is a geometrically clear and unambiguous explanation of strict connectivity of the results of measurements instead of quite absurd “entanglement” in conventional quantum mechanics.

2. Maxwell Equation in Geometric Algebra

Without charges and currents the Maxwell equation is:

$\left({\partial }_t+\nabla \right)F=0$ (2.1)

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. Indeed, let’s take the electromagnetic field in the form:

$F=F_0\exp \left[I_S\left(\omega t-k\cdot r\right)\right]$(2.2)

requiring that it satisfies (2.1).

Element $F_0$ in (2.2) is a constant element of geometric algebra $G_3$ and $I_S$ is unit value bivector of a plane S in three dimensions, generalization of the imaginary unit [3] . The exponent in (2.2) is unit value element of $G_3^{+}$:

${\text{e}}^{I_S\phi }=\cos \phi +I_S\sin \phi$, $\phi =\omega t-k\cdot r$

Solution of (2.1) should be sum of a vector (electric field e) and bivector (magnetic field $I_{3}h$):

$F=e+I_{3}h$

For a plane S in three dimensions Maxwell Equation (2.1) has two solutions [5] :

Maxwell Equation (2.1) is a linear one. Then any linear combination of $F_{+}$ and $F_{-}$ saving the structure of (2.2) will also be a solution.

Let’s write:

$\{\begin{array}{l}{F}_{+}=\left(e_0+I_3{h}_0\right)\exp \left[I_S\omega \left(t-\left(I_3{I}_S\right)\cdot r\right)\right]=\left(e_0+I_3{h}_0\right)\exp \left[I_S\omega t\right]\exp \left[-I_S\left[\left(I_3{I}_S\right)\cdot r\right]\right]\\ F_{-}=\left(e_0+I_3{h}_0\right)\exp \left[I_S\omega \left(t+\left(I_3{I}_S\right)\cdot r\right)\right]=\left(e_0+I_3{h}_0\right)\exp \left[I_S\omega t\right]\exp \left[I_S\left[\left(I_3{I}_S\right)\cdot r\right]\right]\end{array}$ (2.3)

Then for arbitrary (real1) scalars $\lambda$ and $\mu$:

$\lambda F_{+}+\mu F_{-}=\left(e_0+I_3{h}_0\right){\text{e}}^{I_S\omega t}\left(\lambda {\text{e}}^{-I_S\left[\left(I_3{I}_S\right)\cdot r\right]}+\mu {\text{e}}^{I_S\left[\left(I_3{I}_S\right)\cdot r\right]}\right)$(2.4)

is solution of (2.1). The item in the 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 that plane.

Arbitrary linear combination (2.4) can be rewritten as:

$\lambda {\text{e}}^{I_{Plane}^{+}{\phi }^{+}}+\mu {\text{e}}^{I_{Plane}^{-}{\phi }^{-}}$(2.5)

where

${\phi }^{\pm }={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\cos \omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)\right)$,

$\begin{array}{c}{I}_{Plane}^{\pm }=I_S\frac{\sin \omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}{\sqrt{1+{\sin }^2\omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}}+I_{B_0}\frac{\cos \omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}{\sqrt{1+{\sin }^2\omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}}\\ +I_{E_0}\frac{\sin \omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}{\sqrt{1+{\sin }^2\omega \left(t\mp \left[\left(I_3{I}_S\right)\cdot r\right]\right)}}\end{array}$

The triple of unit value basis orthonormal bivectors $\left\{I_S,I_{B_0},I_{E_0}\right\}$ is comprised of the $I_S$ bivector, dual to the propagation direction vector; $I_{B_0}$ is dual to initial vector of magnetic field; $I_{E_0}$ is dual to initial vector of electric field. The expression (2.5) is linear combination of two geometric algebra states, g-qubits.

Linear combination of the two equally weighted basic solutions of the Maxwell equation $F_{+}$ and $F_{-}$, $\lambda F_{+}+\mu F_{-}$ with $\lambda =\mu =1$ reads:

$\begin{array}{l}{\lambda F_{+}+\mu F_{-}|}_{\lambda =\mu =1}\\ =2\cos \omega \left[\left(I_3{I}_S\right)\cdot r\right]\left(\frac{1}{\sqrt{2}}\cos \omega t+I_S\frac{1}{\sqrt{2}}\sin \omega t+I_{B_0}\frac{1}{\sqrt{2}}\cos \omega t+I_{E_0}\frac{1}{\sqrt{2}}\sin \omega t\right)\end{array}$ (2.6)

where $\cos \phi =\frac{1}{\sqrt{2}}\cos \omega t$ and $\sin \phi =\frac{1}{\sqrt{2}}\sqrt{1+{\left(\sin \omega t\right)}^2}$. It can be written in standard exponential form $\cos \phi +\sin \phi I_B={\text{e}}^{I_B\phi }$.2

I will call such g-qubits spreons because they spread over the whole three-dimensional space for all values of time and instantly change under Clifford translations over the whole three-dimensional space for all values of time, along with the results of measurement of any observable.3

3. Measurement of Observables by Spreons

Measurement of any observable $C_0+C_1{B}_1+C_2{B}_2+C_3{B}_3$ (actually Hopf fibration) by a state $\alpha +{\beta }_1{B}_1+{\beta }_2{B}_2+{\beta }_3{B}_3$ in the current formalism is, see, for example, [6] :

$\begin{array}{l}{C}_0+C_1{B}_1+C_2{B}_2+C_3{B}_3\stackrel{\alpha +{\beta }_1{B}_1+{\beta }_2{B}_2+{\beta }_3{B}_3}{\to }\\ C_0+\left(C_1\left[\left({\alpha }^2+{\beta }_1^2\right)-\left({\beta }_2^2+{\beta }_3^2\right)\right]+2C_2\left({\beta }_1{\beta }_2-\alpha {\beta }_3\right)+2C_3\left(\alpha {\beta }_2+{\beta }_1{\beta }_3\right)\right)B_1\\ +\left(2C_1\left(\alpha {\beta }_3+{\beta }_1{\beta }_2\right)+C_2\left[\left({\alpha }^2+{\beta }_2^2\right)-\left({\beta }_1^2+{\beta }_3^2\right)\right]+2C_3\left({\beta }_2{\beta }_3-\alpha {\beta }_1\right)\right)B_2\\ +\left(2C_1\left({\beta }_1{\beta }_3-\alpha {\beta }_2\right)+2C_2\left(\alpha {\beta }_1+{\beta }_2{\beta }_3\right)+C_3\left[\left({\alpha }^2+{\beta }_3^2\right)-\left({\beta }_1^2+{\beta }_2^2\right)\right]\right)B_3\end{array}$

²Good to remember that the two basic solutions $F_{+}$ and $F_{-}$ differ only by the sign of $I_3{I}_S$, which is caused by orientation of $I_S$ that in its turn defines if the triple $\left\{\hat{E},\hat{H},\pm I_3{I}_S\right\}$ is right-hand screw or left-hand screw oriented.

In the case of spreon (2.6):

$B_1=I_S$, $B_2=I_{B_0}$, $B_3=I_{E_0}$,

$\alpha =2\cos \omega \left[\left(I_3{I}_S\right)\cdot r\right]\frac{1}{\sqrt{2}}\cos \omega t$

${\beta }_1=2\cos \omega \left[\left(I_3{I}_S\right)\cdot r\right]\frac{1}{\sqrt{2}}\sin \omega t$

${\beta }_2=2\cos \omega \left[\left(I_3{I}_S\right)\cdot r\right]\frac{1}{\sqrt{2}}\cos \omega t$

${\beta }_3=2\cos \omega \left[\left(I_3{I}_S\right)\cdot r\right]\frac{1}{\sqrt{2}}\sin \omega t$,

and we get a $G_3^{+}$ element spreading through the three-dimensional space for all values of the time parameter t. It is $G_3^{+}$ element spreading through the three-dimensional space for all values of time parameter t:

$\begin{array}{l}4{\cos }^2\omega \left[\left(I_3{I}_S\right)\cdot r\right][C_0+C_3{I}_S+\left(C_1\sin 2\omega t+C_2\cos 2\omega t\right)I_{B_0}\\ +\left(C_2\sin 2\omega t-C_1\cos 2\omega t\right)I_{E_0}]\end{array}$ (3.1)

Geometrically, that means that the measured observable is rotated by $\frac{\pi }{2}$ in the $I_{B_0}$ plane, such that the $C_3{I}_S$ component becomes orthogonal to plane $I_S$ and remains unchanged. Two other components became orthogonal to $I_{B_0}$ and $I_{E_0}$ and continue rotating in $I_S$ with angular velocity $2\omega t$. The factor $4{\cos }^2\omega \left[\left(I_3{I}_S\right)\cdot r\right]$ defines the dependency of that transformed values through all points of the three-dimensional space.

In the current scheme any test observable can be placed anywhere into continuum of the $\left(t,r\right)$ dependent values of the spreon state. The above formula (3.1) gives the result of measurements simultaneously at all points $\left(t,r\right)$.

Assume we have two observables, $C_0^1+C_1^1{B}_1+C_2^1{B}_2+C_3^1{B}_3$ and $C_0^2+C_1^2{B}_1+C_2^2{B}_2+C_3^2{B}_3$. Write them in exponential form:

$\begin{array}{c}{C}^1\equiv C_0^1+C_1^1{B}_1+C_2^1{B}_2+C_3^1{B}_3\\ =\left|C^1\right|\left(\frac{C_0^1}{\left|C^1\right|}+\frac{\left|C_B^1\right|}{\left|C^1\right|}\frac{C_1^1{B}_1}{\left|C_B^1\right|}+\frac{\left|C_B^1\right|}{\left|C^1\right|}\frac{C_2^1{B}_2}{\left|C_B^1\right|}+\frac{\left|C_B^1\right|}{\left|C^1\right|}\frac{C_3^1{B}_3}{\left|C_B^1\right|}\right)\\ =\left|C^1\right|\left(\cos {\phi }_1+\sin {\phi }_1{S}_1\right)=\left|C^1\right|{\text{e}}^{S_1{\phi }_1}\end{array}$

where $\left|C^1\right|=\sqrt{{\left(C_0^1\right)}^2+{\left(C_1^1\right)}^2+{\left(C_2^1\right)}^2+{\left(C_3^1\right)}^2}$, $\left|C_B^1\right|=\sqrt{{\left(C_1^1\right)}^2+{\left(C_2^1\right)}^2+{\left(C_3^1\right)}^2}$,

$\cos {\phi }_1=\frac{C_0^1}{\left|C^1\right|}$, $\sin {\phi }_1=\frac{\left|C_B^1\right|}{\left|C^1\right|}$, $S_1=\frac{C_1^1{B}_1}{\left|C_B^1\right|}+\frac{C_2^1{B}_2}{\left|C_B^1\right|}+\frac{C_3^1{B}_3}{\left|C_B^1\right|}$

$\begin{array}{c}{C}^2\equiv C_0^2+C_1^2{B}_1+C_2^2{B}_2+C_3^2{B}_3\\ =\left|C^2\right|\left(\frac{C_0^2}{\left|C^2\right|}+\frac{\left|C_B^2\right|}{\left|C^2\right|}\frac{C_1^2{B}_1}{\left|C_B^2\right|}+\frac{\left|C_B^2\right|}{\left|C^2\right|}\frac{C_2^2{B}_2}{\left|C_B^2\right|}+\frac{\left|C_B^2\right|}{\left|C^2\right|}\frac{C_3^2{B}_3}{\left|C_B^2\right|}\right)\\ =\left|C^2\right|\left(\cos {\phi }_2+\sin {\phi }_2{S}_2\right)=\left|C^2\right|{\text{e}}^{S_2{\phi }_2}\end{array}$

where $\left|C^2\right|=\sqrt{{\left(C_0^2\right)}^2+{\left(C_1^2\right)}^2+{\left(C_2^2\right)}^2+{\left(C_3^2\right)}^2}$, $\left|C_B^2\right|=\sqrt{{\left(C_1^2\right)}^2+{\left(C_2^2\right)}^2+{\left(C_3^2\right)}^2}$, $\cos {\phi }_2=\frac{C_0^2}{\left|C^2\right|}$, $\sin {\phi }_2=\frac{\left|C_B^2\right|}{\left|C^2\right|}$, $S_2=\frac{C_1^2{B}_1}{\left|C_B^2\right|}+\frac{C_2^2{B}_2}{\left|C_B^2\right|}+\frac{C_3^2{B}_3}{\left|C_B^2\right|}$.

The observable $C^2$ can be written as:

$C^2=\frac{\left|C^2\right|}{\left|C^1\right|}{C}^1{\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}$,

and then its measurement by any g-qubit ${\text{e}}^{S\phi }$ reads:

${\text{e}}^{-S\phi }{C}^2{\text{e}}^{S\phi }={\text{e}}^{-S\phi }\frac{\left|C^2\right|}{\left|C^1\right|}{C}^1{\text{e}}^{S\phi }{\text{e}}^{-S\phi }{\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}{\text{e}}^{S\phi }$(3.2)

that is, up to the factor $\frac{\left|C^2\right|}{\left|C^1\right|}$, the result of measurement of $C^1$, multiplied by the result of measurement of ${\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}$.

4. Simultaneous Measurement of Observables by Spreons

Now let us use the result (3.2) of not made measurement in the case of state (2.6).

One remaining unspecified item is ${\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}$. Its $G_3^{+}$ expression follows, see e.g. [4] , from known formulas. If

${\text{e}}^{I_{S_1}{\phi }_1}={\alpha }_1+I_{S_1}{\beta }_1={\alpha }_1+{\beta }_1{b}_1^1{B}_1+{\beta }_1{b}_1^2{B}_2+{\beta }_1{b}_1^3{B}_3$

${\text{e}}^{I_{S_2}{\phi }_2}={\alpha }_2+I_{S_2}{\beta }_2={\alpha }_2+{\beta }_2{b}_2^1{B}_1+{\beta }_2{b}_2^2{B}_2+{\beta }_2{b}_2^3{B}_3$

where

${\left({\alpha }_1\right)}^2+{\left({\beta }_1\right)}^2\left({\left(b_1^1\right)}^2+{\left(b_1^2\right)}^2+{\left(b_1^3\right)}^2\right)={\left({\alpha }_1\right)}^2+{\left({\beta }_1\right)}^2=1$

${\left({\alpha }_2\right)}^2+{\left({\beta }_2\right)}^2\left({\left(b_2^1\right)}^2+{\left(b_2^2\right)}^2+{\left(b_2^3\right)}^2\right)={\left({\alpha }_2\right)}^2+{\left({\beta }_2\right)}^2=1$

in some bivector basis $B_1{B}_2{B}_3=1$, with multiplication rules $B_1{B}_2=-B_3$, $B_1{B}_3=B_2$, $B_2{B}_3=-B_1$, then the scalar part of ${\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}$ is:

${\alpha }_1{\alpha }_2-{\beta }_1{\beta }_2\left(b_1^1{b}_2^1+b_1^2{b}_2^2+b_1^3{b}_2^3\right)$,

and bivector part:

$\begin{array}{l}{B}_1\left({\beta }_1{\beta }_2\left(b_1^3{b}_2^2-b_1^2{b}_2^3\right)-{\alpha }_2{\beta }_1{b}_1^1-{\alpha }_1{\beta }_2{b}_2^1\right)\\ +B_2\left({\beta }_1{\beta }_2\left(b_1^1{b}_2^3-b_1^3{b}_2^1\right)-{\alpha }_2{\beta }_1{b}_1^2-{\alpha }_1{\beta }_2{b}_2^2\right)\\ +B_3\left({\beta }_1{\beta }_2\left(b_1^2{b}_2^1-b_1^1{b}_2^2\right)-{\alpha }_2{\beta }_1{b}_1^3-{\alpha }_1{\beta }_2{b}_2^3\right)\end{array}$

Now it remains to apply (3.1) with

$C_0^{Cl}={\alpha }_1{\alpha }_2-{\beta }_1{\beta }_2\left(b_1^1{b}_2^1+b_1^2{b}_2^2+b_1^3{b}_2^3\right)$

$C_1^{Cl}={\beta }_1{\beta }_2\left(b_1^3{b}_2^2-b_1^2{b}_2^3\right)-{\alpha }_2{\beta }_1{b}_1^1-{\alpha }_1{\beta }_2{b}_2^1$

$C_2^{Cl}={\beta }_1{\beta }_2\left(b_1^1{b}_2^3-b_1^3{b}_2^1\right)-{\alpha }_2{\beta }_1{b}_1^2-{\alpha }_1{\beta }_2{b}_2^2$

$C_3^{Cl}={\beta }_1{\beta }_2\left(b_1^2{b}_2^1-b_1^1{b}_2^2\right)-{\alpha }_2{\beta }_1{b}_1^3-{\alpha }_1{\beta }_2{b}_2^3$

The result of measuring of $C^1$ is, due to (3.1),

$\begin{array}{c}{C}_m^1=4{\cos }^2\omega \left[\left(I_3{I}_S\right)\cdot r\right][C_0^1+C_3^1{I}_S+\left(C_1^1\sin 2\omega t+C_2^1\cos 2\omega t\right)I_{B_0}\\ +\left(C_2^1\sin 2\omega t-C_1^1\cos 2\omega t\right)I_{E_0}]\end{array}$

And the result of measuring of ${\text{e}}^{-S_1{\phi }_1}{\text{e}}^{S_2{\phi }_2}$ is

$\begin{array}{c}{C}_m^{Cl}=4{\cos }^2\omega \left[\left(I_3{I}_S\right)\cdot r\right][C_0^{Cl}+C_3^1{I}_S+\left(C_1^{Cl}\sin 2\omega t+C_2^{Cl}\cos 2\omega t\right)I_{B_0}\\ +\left(C_2^{Cl}\sin 2\omega t-C_1^{Cl}\cos 2\omega t\right)I_{E_0}]\end{array}$

Finally, the result of not made measuring of $C^2$ is received from the result of measuring of $C^1$ as:

$\frac{\left|C^2\right|}{\left|C^1\right|}{C}_m^1{C}_m^{Cl}$,

where the product of two subalgebra $G_3^{+}$ elements is calculated by known formula (see, [4] ):

$g_1{g}_2=\left({\alpha }_1+I_{S_1}{\beta }_1\right)\left({\alpha }_2+I_{S_2}{\beta }_2\right)={\alpha }_1{\alpha }_2+I_{S_1}{\alpha }_2{\beta }_1+I_{S_2}{\alpha }_1{\beta }_2+I_{S_1}{I}_{S_2}{\beta }_1{\beta }_2$

5. Conclusion

The geometric algebra lift of conventional quantum mechanics qubits is the game-changing quantum leap forward. The approach brings into reality a kind of physical field spreading through the whole three-dimensional space and values of the time parameter. All measured observable values are simultaneously available all together, not through looking one by one. In this way the new type of quantum computer appeared to be a kind of analog computer keeping and instantly processing information by and on sets of objects possessing an infinite number of degrees of freedom.

NOTES

¹Remember, in the current theory scalars are real ones. “Complex” scalars have no sense.

3The two received solutions are similar in their form to the Majorana operators, though with a bivector instead of formal imaginary unit. Thus, not surprising that the following computational scheme represents more general software simulation which gets around the problem of physical implementation of Majorana fermions.

References

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3. Soiguine, A.M. (1996) Complex Conjugation-Relative to What? In: Abłamowicz, R., Parra, J.M. and Lounesto, P., Eds., Clifford Algebras with Numeric and Symbolic Computations, Birkhauser, Boston, 284-294. >https://doi.org/10.1007/978-1-4615-8157-4_19

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