0
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

| improve this answer | |
$\endgroup$
2
$\begingroup$

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

| improve this answer | |
$\endgroup$
2
$\begingroup$

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).

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.