# Bypassing the Normalization

Get familiar with the concept of bypassing the normalization process.

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## Normalization

The angle $\theta$ controls the probabilities of measuring the qubit in either state `0`

or `1`

. Therefore, $\theta$ also determines $\alpha$ and $\beta$.

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

Any valid qubit state vector must be normalized:

$\alpha^2 + \beta^2 = 1$

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 $\angle ABC$ (the angle at point B) is right.

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

Now, the **Pythagorean theorem** states that the area of the square whose
side is opposite the right angle (hypotenuse, $c$) is equal to the sum of
the areas of the squares on the other two sides (legs $a$, $b$).

$c^2=a^2+b^2$

When looking at the figure 2-dimensional qubit system, again, we can see that $\alpha$ and $\beta$ 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:

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

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

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

Let’s look at the following figure.

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