Bypassing the Normalization

Get familiar with the concept of bypassing the normalization process.

Normalization

The angle θ\theta controls the probabilities of measuring the qubit in either state 0 or 1. Therefore, θ\theta also determines α\alpha and β\beta.

Let’s take a look at the figure 2-dimensional qubit system.

Any valid qubit state vector must be normalized:

α2+β2=1\alpha^2 + \beta^2 = 1

It states that vectors have the same magnitude (length). Since they all originate in the center, they form a circle with a radius of their magnitude, that is,half of the circle diameter.

In such a situation, Thales’ theorem states that if two conditions, where the first condition states that A, B, and C are distinct points on a circle, and the second condition states that the line AC is a diameter, are met, then the angle ABC\angle ABC (the angle at point B) is right.

In our case, the heads of 0|0\rangle, 1|1\rangle, and ψ|\psi\rangle represent the points A, B, and C, respectively. This satisfies the first condition. The line between 0|0\rangle and 1|1\rangle is the diameter, which satisfies the second condition. Therefore, the angle at the head of ψ|\psi\rangle is a right angle.

Now, the Pythagorean theorem states that the area of the square whose side is opposite the right angle (hypotenuse, cc) is equal to the sum of the areas of the squares on the other two sides (legs aa, bb).

c2=a2+b2c^2=a^2+b^2

When looking at the figure 2-dimensional qubit system, again, we can see that α\alpha and β\beta are the two legs of the rectangular triangle and the diameter of the circle is the hypotenuse.

Therefore, we can insert the normalization as follows:

c=α2+β2=1=1c=\sqrt{\alpha^2+\beta^2}=\sqrt{1}=1

The diameter cc is two times the radius and is therefore two times the magnitude of any vector ψ|\psi\rangle. The length of ψ|\psi\rangle is thus c2=12\frac{c}{2}=\frac{1}{2}.

Since all qubit state vectors have the same length, including 0|0\rangle and 1|1\rangle, there are two isosceles triangles (M0ψ\triangle M|0\rangle|\psi\rangle and Mψ1\triangle M|\psi\rangle|1\rangle).

Let’s look at the following figure.

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