# Represent qubit in a superposition [closed]

How do we represent a qubit $$\vert 1 \rangle$$ and put in a superposition (in dirac)?

## 3 Answers

How about I approach your question from computer science perspective. A bit can be only $$0$$ or only $$1$$. A qubit can be only $$0$$, or only $$1$$, or a combination (superposition) of $$0$$ and $$1$$.

We denote a zero bit as $$0$$ and zero qubit as $$\vert 0 \rangle$$. We also denote a bit of value one as $$1$$ and a qubit of value one as $$\vert 1 \rangle$$. Keep in mind that $$\vert 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{and } \vert 1 \rangle = \begin{bmatrix}0\\1\end{bmatrix}.$$ The question now is how to represent superposition? It is simple: it will be a combination of $$\vert 0 \rangle$$ and $$\vert 1 \rangle$$. Formally, a single-qubit $$\psi$$ is given as $$\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$$ where $$\alpha, \beta \in \mathbb{C}$$ and $$\vert \alpha \vert^2 + \vert \beta \vert^2 = 1$$.

Note that $$\alpha$$ denotes the probability of getting $$\vert 0 \rangle$$ and $$\beta$$ denotes the probability of getting $$\vert 1 \rangle$$.

For example, if $$\alpha = 0$$, then $$\beta = 1$$, hence, $$\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$$ $$\vert \psi \rangle = 0 \times \vert 0 \rangle + 1 \times \vert 1 \rangle$$ $$\vert \psi \rangle = \vert 1 \rangle$$

which means our qubit will always be "$$1$$" or $$\vert 1 \rangle$$ (i.e., our single-qubit collapses to $$\vert 0 \rangle$$ 0% of the time and $$\vert 1 \rangle$$ 100% of the time).

How about we make $$\alpha = \dfrac{1}{\sqrt{2}}$$, then, $$\beta = \dfrac{1}{\sqrt{2}}$$, hence, we have: $$\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$$ $$\vert \psi \rangle = \dfrac{1}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$$ So, when we measure our single-qubit, it collapses to $$\vert 0 \rangle$$ 50% of the time and $$\vert 1 \rangle$$ 50% as well.

In general, performing a measurement on a qubit destroys its superposition (i.e., the qubit will behave as a bit after measurement, it could be only 0 or only 1).

Furthermore, you should have a look at the Hadamard Gate which takes a qubit and turns it into superposition.

$$|1\rangle$$ is not in superposition, it is simply state 1. After measurement you will get 1 with 100 % probability.

However, generaly, qubit can be represented as $$|q\rangle = a|0\rangle + b|1\rangle$$, where $$a,b \in \mathbb{C}$$. So, you can think of $$|1\rangle$$ as a superposition with $$a=0$$ and $$b=1$$.

Concerning second question, $$|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$$

In your question, there is missing a key ingredient, which is: a superposition in which basis?

All pure (quantum) states are representable with only one non-zero coefficient in its native basis, and all pure states can be represented as a superposition in a different basis. Answer of @Martin Vesely gives you intuition how to represent $$|1\rangle$$ in a computational basis (which is its native basis). However, if you select a different basis set $$\{|\psi_1\rangle, |\psi_2\rangle\}$$, you can describe your state as:

$$|1\rangle = \alpha_1 |\psi_1\rangle + \alpha_2|\psi_2\rangle$$, where $$\alpha_k = \langle\psi_k|1\rangle$$, i.e. the overlap of your state with the basis that you're expanding in.

As an example, select $$|\pm\rangle$$ basis ($$|\pm\rangle = \frac{1}{\sqrt{2}}(|0\rangle\pm|1\rangle)$$, then $$|1\rangle = \frac{1}{\sqrt{2}}(|+\rangle -|-\rangle)$$, since $$\langle \pm|1\rangle = \frac{1}{\sqrt{2}}$$.

The same thinking you can apply to higher dimensional quantum states, with the difference, that you will have more basis elements (equal to dimensionality of the Hilbert space).