# How does superposition apply to quantum computing?

beginner here. I've always heard explanations on quantum computing, all about superposition, entanglement, etc.

But how does superposition actually apply to quantum computing? Yeah, its "in between" a 1 and a 0 but why does this matter? How do quantum computers actually use this feature, and why is it so important?

Now, a bit is the building block of classical computers, and it can be in the states 0 or 1, a classical two level system. And a qubit is the building block of quantum computers, it is a two-level quantum system, hence it is a unit vector in the space $$\mathbb{C}^2$$. Therefore it can be span by two orthogonal vectors, $$|e_1\rangle$$ and $$|e_2\rangle$$. When we talk about the state of a qubit, we usually pick the computational basis, a basis where the these two orthogonal vectors is specified as: $$|e_1\rangle = |0 \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \hspace{1 cm} |e_2 \rangle = |1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ So in the computational basis, the state of a qubit, $$|\psi \rangle$$ can be written as: $$|\psi \rangle = \alpha|0\rangle + \beta|1\rangle \hspace{1 cm} \alpha, \beta \in \mathbb{C} \hspace{1 cm} |\alpha|^2 + |\beta|^2 = 1$$ Note how the state of the qubit is different than that of a classical bit. It can be written in a linear combination or superposition between the state $$|0\rangle$$ and the state $$|1\rangle$$. This is fine because we must remember that a qubit, after all, is a quantum mechanical object, it inherently possess quantum mechanical properties like superposition. So if you have a quantum computer, it must be able to deal with superposition of states, because if it can't then it is not really a quantum computer.
How does entanglement play a role in quantum computing? Well, if there is no entanglement during the entirety of your computation, then this means you can describe the N qubit quantum state $$|\psi \rangle = \big(\alpha_1 |0 \rangle + \beta_1 |1\rangle \big) \otimes \big(\alpha_2 |0 \rangle + \beta_2 |1\rangle \big) \otimes \cdots \otimes \big(\alpha_N |0 \rangle + \beta_N |1\rangle \big)$$ Then you can operate on each qubit separately, so you can simulate this very efficiently classically... hence no point to perform quantum computation at all. Note that if all the qubits in the $$N$$ qubit system interact with each other then we can't deal with each qubit individually... so in a way, if you want to simulate such system, you have to simulate a $$2^N \times 2^N$$ unitary matrix which is very costly classically... whereas when they are not entangle, you just need to simulate $$N$$ different $$2 \times 2$$ matrix independently which you can do efficiently classically.