# Quantum State in a Nutshell

Updated

There are thousands of articles trying to explain what exactly a quantum state is. Many of them boiled down to “the state of a qubit is 0, 1, or 0 and 1 at the same time”. This statement leads to both confusion and misinterpretation. The explanation I found on Quantum computing for the very curious is by far the most elegant and simplest:

The state of a qubit is a vector in a two-dimensional vector space. This vector space is known as state space.

I will use many of great content from Quantum computing for the very curious to explain things.

## Mapping qubits to classical bits

We’ve described what a qubit state is, but provided no link between a qubit state and a classical bit state. There are two possible states for a classical bit: 0 and 1. The corresponding states for a qubit is slightly fancier: $$|0\rangle$$ and $$|1\rangle$$.

The notation with $$|$$ and $$\rangle$$ is called a $$ket$$ notation. With a number wrapped between them, $$0$$ or $$1$$ are called $$kets$$. A $$ket$$ is a fancy term for a vector. In fact, $$|0\rangle$$ is really just $$\begin{bmatrix} 1 \newline 0 \end{bmatrix}$$; $$|1\rangle$$ is really just $$\begin{bmatrix} 0 \newline 1 \end{bmatrix}$$.

## States between 0 and 1

Both the states $$|0\rangle$$ and $$|1\rangle$$ are called computational basis states, which maps to classical 0 and 1 states. There are more states for a qubit. We’ve already learned that a quantum state is a two-dimensional vector. An example is given:

The state $$0.6|0\rangle + 0.8|1\rangle$$ is just a combination of the $$|0\rangle$$ vector and the $$|1\rangle$$ vector. A state like this is a superposition of $$|0\rangle$$ and $$|1\rangle$$, a fancy way of saying a linear combination of $$|0\rangle$$ and $$|1\rangle$$. $$0.6$$ is the amplitude for state $$|0\rangle$$, and $$0.8$$ is the amplitude for state $$|1\rangle$$.

Not all linear combination of vector $$|0\rangle$$ and $$|1\rangle$$ are qubit states. There is one constraint: the sums of the squares of the amplitudes must be 1. For example, we can compute $$0.6^2 + 0.8^2$$ and verify the result is 1.

For general quantum states, the amplitudes can be complex numbers as well. Denoting both amplitudes as $$\alpha$$ and $$\beta$$, a quantum state can be formally written as:

$\alpha |0\rangle + \beta |1\rangle \wedge \alpha^2 + \beta^2 = 1$

$$\alpha^2 + \beta^2 = 1$$ is called the normalization constraint.

If we think of $$|0\rangle$$ and $$|1\rangle$$ as orthonormal vectors, we can visualize the possible linear combination of these two vectors as a circle of radius 1:

Since amplitudes can be complex numbers, the state space really becomes a sphere. Summing all these up:

the quantum state of a qubit is a vector of unit length in a two-dimensional complex vector space known as state space.

Quantum computing for the very curious

## Measuring a qubit

Suppose we have qubit in a quantum state $$\alpha |0\rangle + \beta |1\rangle$$. We want to observe the state of this specific qubit. It turns out the law of physics prohibits us from figuring out the the amplitudes $$\alpha$$ and $$\beta$$ if they start out unknown. In short, the quantum state of any system is not directly observable.

To figure out the quantum state. We rely on a process called measurement in the computational basis. Suppose a qubit is in the state $$\alpha |0\rangle + \beta |1\rangle$$. Measuring the state of this qubit gives us the outcome $$0$$ with probability $$|\alpha|^2$$, or 1 with probability $$|\beta|^2$$. The state of the qubit after the measurement is thus either $$|0\rangle$$ or $$|1\rangle$$. After the measurement, $$\alpha$$ and $$\beta$$ are gone.