> For the complete documentation index, see [llms.txt](https://lewisla.gitbook.io/learning-quantum/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://lewisla.gitbook.io/learning-quantum/qubits/quantum-bits.md).

# Quantum Bits

## Quantum Information

In the section about [*quantum mechanics*](/learning-quantum/physics/quantum-mechanics.md), we talked about the idea of [*superposition*](/learning-quantum/physics/quantum-mechanics.md#superposition), where more than one [*state*](/learning-quantum/physics/classical-mechanics.md#state) is overlapped over top of each other. The system is in [*multiple distinguishable states*](/learning-quantum/physics/quantum-mechanics.md#distinguishable-states) at the same time - the information is *superimposed.* We don't know what going on in there, we can only talk in [*probabilities* ](/learning-quantum/physics/quantum-mechanics.md#probability)- what state are we likely to see if we measure (or observe) the system at a particular moment?

### The Single Qubit System

A single quantum bit is the basic unit of quantum computation, but it's also the smallest (non-trivial) quantum system. Unlike a classical system, where a single bit can't do much at all, in the quantum system the individual bits can be used to do work alone [*(Nielsen, M. - p. 13)*](/learning-quantum/qubits/qubits-summary/qubit-references.md#the-idea-of-a-quantum-bit).

### Distinguishable States

Like a classical bit, the quantum bit has two observable states - 0 and 1. The difference here is very simple. In the quantum system, we have the potential for superposition. A quantum bit can be 1 *and* 0, at the *same time*. We measure this in between zone by talking about the [*probability* ](/learning-quantum/physics/quantum-mechanics.md#probability)of finding the qubit in a particular state *when we measure* it.

If you recall our discussion of [*vector spaces*](/learning-quantum/linear-algebra/space-dimension-and-span.md#space), the dimension of a system is defined by the number of [*basis*](/learning-quantum/linear-algebra/space-dimension-and-span.md#basis) vectors we have. So a single qubit system has *two basis vectors* - a 0 vector and a 1 vector. 

This means that a single qubit system defines a *plane.* The combinations of those two basis vectors all live together on this plane. The vectors that live on the plane represent all the possible answers we can get to using a single qubit.

## Notation

### General Vector

A single qubit is represented as a 2-dimensional [*quantum vector*](/learning-quantum/physics/quantum-mechanics.md#quantum-vectors). Recall that a quantum vector has as many dimensions as there are distinguishable states, so it makes sense that a single qubit vector has 2 elements. Like any quantum vector, we represent a qubit in an unknown state like this:

$$
|\psi\rangle
$$

### Specific States

The two basis vectors, which represent the pure "0" state and the pure "1" state respectively, are represented like this:

$$
|0\rangle = \begin{bmatrix}1\0\end{bmatrix}, |1\rangle=\begin{bmatrix}0\1\end{bmatrix}
$$

Note here that the elements of this column vector represent the particular state a qubit is in. The first element tells us if a qubit exists in the *first* state (which is 0), and the second element tells us if a qubit exists in the *second* state (which is 1). As we add more qubits to the system, we offer more qubits and more combinations of states, and more elements.

### Multiple Bits

As our system gets more complex and begins to contain more numbers, we can expand on this:

$$
|00\rangle=\begin{bmatrix}1\0\0\0\end{bmatrix},|01\rangle=\begin{bmatrix}0\1\0\0\end{bmatrix},|10\rangle=\begin{bmatrix}0\0\1\0\end{bmatrix},|11\rangle=\begin{bmatrix}0\0\0\1\end{bmatrix}
$$

{% hint style="success" %}
For multi-qubit systems like this, we order qubits from right to left. So the rightmost qubit $$|0\textcolor{red}1\rangle$$ is the *first*, and the leftmost qubit is the *last* $$|\textcolor{red}{0}1\rangle$$.
{% endhint %}

You can imagine that if we have a whole bunch of qubits, this could get very confusing very quickly. So we might use the decimal version of these binary numbers streamline things:

$$
|0\rangle=\begin{bmatrix}1\0\0\0\end{bmatrix},|1\rangle=\begin{bmatrix}0\1\0\0\end{bmatrix},|2\rangle=\begin{bmatrix}0\0\1\0\end{bmatrix},|3\rangle=\begin{bmatrix}0\0\0\1\end{bmatrix}
$$

{% hint style="info" %}
[Binary number review](https://www.khanacademy.org/math/algebra-home/alg-intro-to-algebra/algebra-alternate-number-bases/v/number-systems-introduction)
{% endhint %}

### Combinations of the Basis Vectors

Going back to our earlier discussion of $$|0\rangle$$ and $$|1\rangle$$ as our basis vectors, this means that all of the possible measurements for $$|\psi\rangle$$ are represented by a [*linear combination*](/learning-quantum/linear-algebra/vector-relationships.md#linear-combinations) of our basis vectors [*(Nielsen, M. - p. 13)*](/learning-quantum/qubits/qubits-summary/qubit-references.md#a-linear-combination-of-our-basis-vectors)*:*

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

Where $$a$$ and $$b$$ are [*complex numbers*](/learning-quantum/physics/quantum-mechanics.md#complex-numbers).

{% hint style="success" %}
Recall that the[ *inner product* ](/learning-quantum/linear-algebra/basics.md#inner-product)of our quantum vectors [*must equal 1*](/learning-quantum/physics/quantum-mechanics.md#notation), to allow for reversibility of operations: $$\langle \psi|\psi\rangle = 1$$
{% endhint %}

## Measuring Qubits

When we measure our quantum bit, the information [*collapses*](/learning-quantum/physics/quantum-mechanics.md#superposition) into one state or the other. That means that this linear combination of states...

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

...is a superposition of multiple possible states. We can solve this superposition two different ways - as a wave form, with the [*Schrodinger equation*](/learning-quantum/physics/quantum-mechanics.md#schrodingers-equation), or by literally looking at it.

{% hint style="success" %}
When we measure our qubit (and it collapses into one state or the other), we store the result in a classical bit. That means that measuring $$|0\rangle \rightarrow 0$$ and $$|1\rangle \rightarrow 1$$.
{% endhint %}

### The Probability of States

What will we get? It depends on what *kind* of superposition it is - that is to say it depends on the [*probabilities*](/learning-quantum/physics/quantum-mechanics.md#probability) of arriving at each state.

The probability of getting any *particular* state*,* added to the probability of getting any other *particular* state has to add up to 1, but as long as that true the *distribution* could be anything [*(Nielsen, M. - p. 13)*](/learning-quantum/qubits/qubits-summary/qubit-references.md#the-probability-of-measuring-a-qubit-in-a-particular-state)*:*

$$
\langle\psi|\psi\rangle=\text{prob}|0\rangle+\text{prob}|1\rangle=1
$$

That is to say that we may be 50% likely to see $$|0\rangle$$ and 50% likely to see $$|1\rangle$$. The probability *distribution* then looks like this:

![A graph showing an even probability distribution](/files/-M4Pwr_gyQUNB4b86Iwr)

The sum of the probabilities of finding the qubit in all possible states is therefore:

$$
50% + 50% = 100%
$$

Or:

$$
0.5+0.5=1
$$

### Quantum Results

Results from a classical system come in as a simple answer - for example, if we have two bits a and b:

$$
a = 0, b = 1
$$

And we *add* them together:

$$
a + b = 1
$$

This is the long and short of it, the answer is 1. I know the whole state of the system from top to bottom - I could measure it once, or ten thousand times, and I would always get the same answer.

Quantum systems are different - the state of the system is measured in probabilities. Thus, each time I measure it, I may get a different answer. The system may [*collapse* ](/learning-quantum/physics/quantum-mechanics.md#superposition)into a different state. Therefore, I can't measure just once. If I do, I won't see the whole picture. The result that I'm looking for isn't a simple number. It's a [*probability distribution*](/learning-quantum/qubits/quantum-bits.md#the-probability-of-states)*.*

This also means we need to run our calculations multiple times - as many times as we can get away with in fact. This way we'll get as close as possible to the real truth of the system.


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