**Qubit** (**Quantum Bit**) is the **fundamental unit of information** in **quantum computing** as like bit which is the basic unit of information in classical computing. However, unlike bit in the classical computing which can exist in one of two states such as 0 and 1 at any given point of time, the qubit can be found to exist in states 0, 1 or the **superposition state** which, simply speaking, can be said to be both 0 and 1.

In this post, you will learn some of the following:

- Qubit and Superposition State
- Superposition state explained with examples

Table of Contents

### Qubit and Superposition State

The following are some of the important points about the state of Qubits:

- The qubit can be found to exist in the state of 0, 1 or the superposition state which represents the state of both 0 and 1.
- However,
**when measured, Qubit would found to be in only one of the two states 0 and 1 and not the superposition state**. This is very important to understand.

The state in Quantum theory is represented using **|** and **>**. However, for the sake of understanding, we will use vector notation to represent the 0 and 1 state. Thus, the following holds true:

- The state of 0 can be represented using \(\vec{0}\)
- The state of 1 can be represented using \(\vec{1}\)
- The superposition state can be represented as the following:

\(\vec{\gamma}\) = \(\alpha\vec{0} + \beta\vec{1}\)where \(\alpha\) and \(\beta\) are complex numbers and \(\vec{\gamma}\) is the superposition state of state \(\alpha\vec{0}\) and \(\alpha\vec{1}\). Recall that the complex numbers are of the form,*c = a + ib*, where i = \(\sqrt{-1}\).

As per the quantum mechanics, the **modulus squared** of \(\alpha\) and \(\beta\) represents the probability of finding the qubit in state \(\vec{0}\) and \(\vec{1}\) respectively.

The fact that probability must sum upto 1 leads to following equation which could be used for determining value of \(\alpha\) and \(\beta\).

\(|{\alpha}|^2 + |{\beta}|^2 = 1\)One needs to note that if \({\alpha}, {\beta}\) are real complex numbers with imaginary coefficients such as (x + iy), the way to find the modulus squared is to multiple the complex number with its conjugate. This esssentially means that modulus squared for complex number \(x + iy\) is following:

\((x + iy)\times(x – iy)\)### Superposition states explained with examples

**Example 1:** **Let’s determine if the below can be said to represent a valid superposition state of a qubit:**

The coefficient of \(\vec{0}\) is the complex number, \(\frac{1}{2}\). The modulus squared of the coefficient, \(\frac{1}{4}\), is the probability that state 0 will happen.

The coefficient of \(\vec{1}\) is the complex number, \(\frac{\sqrt{3}}{2}\). The modulus squared of the coefficient, \(\frac{3}{4}\) is the probability that state 0 will happen.

Going by above, the following gets evaluated:

\((\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2\) \(= \frac{1}{4} + \frac{3}{4} = 1\)Thus, it could be said that \(\frac{1}{\sqrt{2}}\vec{0} + \frac{1}{\sqrt{2}}\vec{1} \) is a valid superposition state.

**Example 2**: **Let’s consider another example and check whether it is a valid superposition state:**

The sum of modulus squared of coefficients (probability) of states should be equal to 1. Let’s check this out.

\((\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2\) \(= \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \ne 1\)Thus, the superposition state represented by above is not a valid superposition state.

### Further Reading / References

- Quantum Superposition
- Book – Quantum Computing Explained (David McMahon). It is a great book for learning quantum computing. If you are planning to get deeper understanding of the book, you may not want to miss this one.

### Summary

In this post, you learned about some of the following:

- Introduction to Superposition states of Qubit
- Superposition state explained with examples

Did you find this article useful? Do you have any questions about this article or **understanding superposition states of Qubit in Quantum computing**? Leave a comment and ask your questions and I shall do my best to address your queries.

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