# Born rule
**Born rule** is a postulate in quantum mechanics that gives an interpretation of [[probability-amplitude|probability amplitudes]]. Namely, the [[probability-amplitude|probability amplitude]] $c\in\mathbb{C}$ associated with some state $\lvert i\rangle$ encodes the probability of finding the quantum system in the state $\lvert i\rangle$. The value of this probability — according to the Born rule — equals the modulus squared of the probability amplitude $c$.

 $$P_i = |c|^2$$

## Qubit
A single [[qubit]] $\lvert\psi\rangle$ has two probability amplitudes, $a\in\mathbb{C}$ which is associated with the state $\lvert 0\rangle$, and $b\in\mathbb{C}$ which is associated with the state $\lvert 1\rangle$:

$$\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$$

If we attempt to measure the qubit, it will collapse to either state $\lvert 0\rangle$ with probability $P_0$, or state $\lvert 1\rangle$ with probability $P_1$. What is the value of these probabilities? That's exactly the question that the Born rule answers! It's the modulus square of the probability amplitudes.

$$ P_0 = |a|^2,\quad P_1 = |b|^2 $$

These are real probabilities and must satisfy the second axiom of probability i.e. they must add up to 1 (meaning 100%):

$$|a|^2 + |b|^2 = 1$$

### Concrete example
Let's use a concrete example for probability amplitudes $a\in\mathbb{C}$ and $b\in\mathbb{C}$.

$$\lvert\psi\rangle = \left(\frac{1}{2\sqrt{2}} + i\frac{1}{2\sqrt{2}}\right)\lvert 0 \rangle + \left(-\frac{\sqrt{3}}{2\sqrt{2}} + \frac{i\sqrt{3}}{2\sqrt{2}}\right)\lvert 1 \rangle$$

Recall that a modulus of a complex number written in Cartesian form is just the Pythagorean theorem $|z| = |x +iy| = \sqrt{x^2 + y^2}$. But Born rule simplifies this for us even further, because the square eliminates the square root $|z|^2 = |x + iy|^2 = x^2 + y^2$!

$$P_0 = |a|^2 = \left|\frac{1}{2\sqrt{2}} + i\frac{1}{2\sqrt{2}}\right|^2 = \frac{1}{4} = 25\%$$
$$P_1 = |b|^2 = \left|-\frac{\sqrt{3}}{2\sqrt{2}} + \frac{i\sqrt{3}}{2\sqrt{2}}\right|^2 = \frac{3}{4} = 75\%$$

Thus if we measure this qubit it will collapse to state $\lvert 0\rangle$ with 25% probability, or to state $\lvert 1\rangle$ with 75% probability. 

### Measurement after a measurement
What happens if we measure the qubit twice? Let's say we measure the previous qubit and -- against the odds -- it collapses to state $\lvert 0\rangle$, even though it only had 25% chance of doing so. What happens if we measure it again? Would it flip to $\lvert 1\rangle$ with 75% probability? The answer is no. No matter how many times we measure it across the $\lvert 0\rangle, \lvert 1\rangle$ basis, after collapse it's always going to retain the same state.

To see why, start from the collapsed qubit:
$$\lvert\psi\rangle = \lvert 0\rangle$$

This is actually just a short hand way of writing the following:
$$\lvert\psi\rangle = 1\cdot\lvert 0\rangle + 0\cdot\lvert 1\rangle$$

Which is also a short hand way of writing the following because probability amplitudes are still complex numbers:
$$\lvert\psi\rangle = (1 + i0)\lvert 0\rangle + (0 + i0)\lvert 1\rangle$$

Applying the Born rule, we get:
$$P_0 = |a|^2 = |1 + i0|^2 = 1 = 100\%$$
$$P_1 = |b|^2 = |0 + i0|^2 = 0 = 0\%$$

Therefore, after measuring the qubit across the $\lvert 0\rangle, \lvert 1\rangle$ basis twice, the probability of remaining in the state $\lvert 0\rangle$ is 100%, while the probability of recollapsing to state $\lvert 1\rangle$ is 0%. Meaning, its state will almost surely not change due to double measurement.

But this does not mean the qubit is permanently destroyed. Applying a gate such as the [[h-gate|Hadamard gate]] puts it back into superposition, and a subsequent measurement in the computational basis will again produce probabilistic outcomes.