Quantum Computing without Complex Numbers

Question

p.s. I am trying to get a handle on what actual computing operations a quantum computer program does. Any information on that would be appreciated [noting the issue that that might count as a separate question].

I have been watching the video “Quantum Computing for Computer Scientists” (https://www.youtube.com/watch?v=F_Riqjdh2oM), by Microsoft. This video makes things easier by leaving out complex numbers — that is, I take it, using complex numbers that always have 0 as the coefficient for i.

I have been driven to looking at the Bloch Sphere and trying to understand why a one-dimensional concept is best expressed in three dimensions. (I understand that we use only the surface of the sphere — radius=1.) I take it that the sphere has two ordinary dimensions, and one imaginary/complex one — one i dimension — and that the outcome is 1 for i = 1 and 0 for i = -1… and that the probability of a 1 ranges from 0 at the bottom, through to 1 at the top of the sphere.

The issue is that, in the above-mentioned video, there is a “unit state circle machine”, where everything happens… and there are no i’s — it is all in the x and y dimensions. That should mean — I take it — that the entire displayed region represents the xy plane in a Bloch Sphere… meaning in turn that it is all 50/50 superposition. Conversely, however, the video talks about it all as though we can do quantum computing without needing i at all, and we can have 0’s and 1’s and superposition here.

That was the main question.

I have also read that rotations around the sphere — i.e. that do not move towards i=1 nor i=-1 — are immaterial to quantum computing. Conversely, there are operations on IBM’s {actual quantum computer!} web site, for making transformations for each of the three dimensions. I have been trying to get a handle on what the other phase changes mean, so any mention of this would be helpful (but maybe that is a separate question as well). (I am guessing that a transformation here would move to the opposite point on the sphere, maybe, but it is only a guess.)

Finally… I have nothing against maths, and I appreciate that the quantum computing works with a physics phenomenon that is describe using maths… but I would really appreciate answers that express intuitively what is going on [EDIT: presumably including some equations!], as opposed to just writing the equations that one would use to represent it. [Something like that; I am not sure offhand how to say it; the point is that, although I understand what a vector is, and so on, I am by no means fluent in that language.]

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Clarification: (I do understand matrices, complex numbers, vectors [slightly weak here], etc..) I am not against people writing equations. What bothers me is an answer that presents just the equations, as a complete account of what they [the equations] mean or do. For an example, look up absolutely anything mathematical, that you do not already understand, on Wikipedia. [What is a **** algorithm? A **** algorithm is any algorithm of the form ####. Divide both sides by &&&& to get a Stevenson Paradigm in the Toslo form ^^^^. Note that the Budgery Gars are subsumed by the Fridge Function, base Gamma, as is shown here. @@**@@##@@ So that is how Black Holes are formed. Also, Bill worked out, in 1887, that the Budgery Gars are actually inverse Cranston Gars, cubed, as is seen here. ]]##&&##[[ That is why the Billian Black Hole is inside-out (in Armadillo space).]

Answer 1

I have read and understood your question. I think you should refer here https://qiskit.org/textbook/ch-states/representing-qubit-states.html.

What they basically did was change the $|1\rangle$ and $|0\rangle$ basis vectors to $$ |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

$$ |-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix} $$ These are the x axis basis vectors.

This could have been $$ |\circlearrowleft\rangle, \quad |\circlearrowright\rangle$$ i.e. the y axis basis vectors.

as well. What i mean is that, in a 3D sphere in cartesian plane, we initially considered that z axis is 1,0 basis but in their consideration they chose to change the basis. This does simplify the concept to beginners but is not quite useful because most of the QC systems around the world (like the IBM) use the z basis.

As a matter of fact, you can take any diameter on the sphere and call it a basis, you'll be correct. Also, Linear algebra is very important as you progress ahead or else you wont understand the algorithms.

Answer 2

The Bloch sphere is a good way to visualize the (classical) polarizations of light, and a correct but in my opinion not especially enlightening way to visualize a spin ½ system. For the purpose of understanding quantum computing I think it's useless, because it only applies to a computer with a single qubit. The closest analogue to the Bloch sphere for $n>1$ qubits is not $n$ Bloch spheres, but a higher-dimensional complex sphere. It's true that quantum gates all represent rotations of this higher dimensional sphere, but I don't think the sphere or the rotations can be visualized in any useful way even when $n=2$.

I take it that the sphere has two ordinary dimensions, and one imaginary/complex one [...] and that the outcome is 1 for i = 1 and 0 for i = -1

No – in the standard parametrization, latitude on the Bloch sphere maps to the relative magnitude of the amplitudes of $\left|0\right\rangle$ and $\left|1\right\rangle$, and longitude maps to the complex phase of the amplitude of $\left|1\right\rangle$, and $\left|0\right\rangle$ has a complex phase of 0 by convention. In that parametrization it's correct that "the probability of a 1 ranges from 0 at the bottom, through to 1 at the top of the sphere" (or vice versa).

But it's better to think of the sphere as rotationally symmetric with no preferred coordinate system. The choice of particular antipodal points to represent 0 and 1 is arbitrary.

There's no useful distinction between real and imaginary values in a state vector. You can always swap them by multiplying the whole thing by $i$ with no effect on the outcome of the computation.

The issue is that, in the above-mentioned video, there is a “unit state circle machine”, where everything happens… and there are no i’s — it is all in the x and y dimensions. That should mean — I take it — that the entire displayed region represents the xy plane in a Bloch Sphere… meaning in turn that it is all 50/50 superposition.

Can you link to a particular time in the video? It's 90 minutes long.

If you restrict the amplitudes to real numbers then the Bloch sphere does restrict to a circle, but the circle is not the equator, it's a great circle containing the poles. The poles are $\left|0\right\rangle$ and $\left|1\right\rangle$, and the two points on the equator are $\tfrac{1}{\sqrt2}(\left|0\right\rangle \pm \left|1\right\rangle)$. The two equatorial points are the only ones with a 50-50 probability of 0 and 1.

I have also read that rotations around the sphere — i.e. that do not move towards i=1 nor i=-1 — are immaterial to quantum computing.

Your i=1, i=-1 notation is incorrect, but I assume you mean rotations that fix the poles. Such rotations are rotations of the phase of $\left|1\right\rangle$ in the complex plane, which doesn't affect the classical probability of measuring a 1 (or a 0). But again, this is just a single-qubit system, and any operation on it is basically trivial.

You can do any quantum computation on any number of qubits using only real amplitudes, at the cost of a single additional qubit and a constant-factor slowdown. The simplest way is to replace each $a+bi$ in a state vector by $\begin{pmatrix}a\\b\end{pmatrix}$, and each $a+bi$ in a unitary matrix representing a gate by $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$. You can think of the extra qubit as selecting the real (if 0) or imaginary (if 1) parts of the amplitudes.

But I think it's easier to understand the standard formalism with complex numbers. Don't think of them as a real part and an imaginary part, think of them as a magnitude and a phase.

Answer 3

As quantum computing (and quantum mechanics) is not intuitive concept, i.e. we do not encounter it in day-to-day life, it seems impossible to descibe it in plain words (I tried to do so many times). We still need some simplification and as a result we do not grasp whole concept of quantum computing.

If you wanted to take quantum computing seriously, you would need to learn at least basics of linear algebra and complex numbers, as those are essential math tools used in quantum mechanics.

If you are not familiar with these concept, I would recommend to look for some basics text books on matrices and complex numbers (maybe some YouTube videos are sufficient).

Concerning you question on Bloch spehere:

A qubit is described as linear combination of states $|0\rangle$ and $|1\rangle$ with complex coefficients (i.e. four real numbers). Since there is a requirements for normalization (i.e. sum of probabilities of measuring states $|0\rangle$ and $|1\rangle$ is one) and also because we do not have to care about global phase, we are left with two degree of freedom. Hence, two angles on Bloch spehere. (As you can see, it is difficult to not turn to math even without equations writen down).

Answer 4

In dealing with an AC circuit, the amplitude and phase of the voltage on a circuit element can be represented by a vector which rotates around the origin of a coordinate system, either x and y, or real and imaginary. There are some mathematical advantages in using the complex number plane. Normally in either system, only the horizontal component of the vector represents the instantaneous voltage on the circuit element. From what little I've seen of quantum wave equations, similar vectors are used to represent the amplitude and phase of the waves arriving at a point. Again only the real components of the vectors represent the strength of the wave at a given instant, but a vector sum is used when adding the waves in order to take into account the phase differences. This can be done using either type of coordinate system.