In part 3 we’ll introduce some of the math behind qubits.

*Note: this post is part of my series, Programming the Multiverse*

## More on the H gate

Last time, we introduced some basic gates, including the H gate. We saw that the H gate puts a qubit into a **Superposition** where it has a 50% chance of being a \(\ket{0}\) or a 50% chance of being a \(\ket{1}\) when measured.

What happens if we combine two H gates?

If we input a qubit set to \(\ket{0}\) through the two H gates, we get… a \(\ket{0}\)! Likewise, if we input a qubit set to \(\ket{1}\) we get a \(\ket{1}\).

What’s going on? Here’s where we have to finally delve into a little bit of math.

## It’s all probability

The most important thing to remember is that qbits are never really representing actual data values, but they represent *probabilities* of seeing a 0 or a 1. And the gates are just manipulating those probabilities.

When we say a qubit is \(\ket{0}\) or \(\ket{1}\) what we’re saying is there’s a 100% probability of seeing a 0 or a 1.

Every qubit can be described by a formula:

\[\ket{\psi} = a\ket{0} + b\ket{1}\]If you recall, we pronounce it “ket-zero” and “ket-one.” We’re using “bra-ket” notation here. No, this does not have anything to do with anesthesia and intimate apparel, it’s just signifying that we’re talking about vectors (more on that later).

The variables *a* and *b* are called the **Probability Amplitude**. The way this formula works is that *a* and *b* have to add up, such that \(\vert{a}\vert^{2} + \vert{b}\vert^{2} = 1\). So in this formula, the total probability is 1 (100%) and the squares of the probabilities, *a* and *b* add up to 1.

In the case of a qubit that is definitely 0, we have \(1\ket{0} + 0\ket{1}\). We can drop the \(\ket{1}\) since it’s multiplied by 0. Similarly, a definite 1 is \(0\ket{0} + 1\ket{1}\) which simplifies to \(\ket{1}\).

For a qubit that is in superposition, we have a 50% chance of a 0 or a 1. So if we solve backward from:

\[\vert{a}\vert^{2} + \vert{b}\vert^{2} = 1\]and get:

\[\vert{\sqrt{\frac{1}{2}}}\vert^{2} + \vert{\sqrt{\frac{1}{2}}}\vert^{2} = 1\]So that the squares cancel out the square roots:

\[\frac{1}{2} + \frac{1}{2} = 1\]So then *a* and *b* are both equal to \(\sqrt{\frac{1}{2}}\) which simplifies to \(\frac{1}{\sqrt{2}}\).

Our final state vector is: \(\ket{\psi} = \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}\)

If we had a 25% chance of 0 and a 75% chance of 1, it would be \(\ket{\psi} = \frac{1}{\sqrt{4}}\ket{0} + \frac{\sqrt{3}}{\sqrt{4}}\ket{1}\)

## Back to the H gate

So the H gate is doing something to manipulate those probability amplitudes, *a* and *b*. But How? And how does it move each one from 0% or 100% to 50%? Stay tuned for Part 4.

Next article: Part 4 - The Guts of Gates