Represent qubit in a superposition

0

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

Aaron

Posted 2019-12-20T13:54:53.570

Reputation: 13

Question was closed 2019-12-21T13:50:07.467

Answers

3

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.

M. Al Jumaily

Posted 2019-12-20T13:54:53.570

Reputation: 628

3

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).

Filip Wudarski

Posted 2019-12-20T13:54:53.570

Reputation: 111

2

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

Martin Vesely

Posted 2019-12-20T13:54:53.570

Reputation: 7 763