# Why can no pair of single qubits look like $\frac{1}{\sqrt2}(|00\rangle+|11\rangle$)?

I've just started to learn Quantum Computing and, to do it, I'm reading the course "Introduction from Quantum Computing" by IBM.

Now, I'm reading the chapter "Entangled states", section "Entangled states" where I have found the formula:

$$|\Phi\rangle=\frac{1}{\sqrt 2}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}=\frac{1}{\sqrt 2}(|00\rangle+|11\rangle).$$

They say about this state vector:

There are no pairs of single qubit states $$|a\rangle$$ and $$|b\rangle$$ whose product state would look like this.

But I don't understand it. Why there are no pairs of single qubits? It is because both qubits are on state 0 and 1 at the same time, isn't it?

## 3 Answers

Suppose there are two single qubit states $$|a\rangle = \alpha|0\rangle + \beta|1\rangle$$ and $$|b\rangle=\delta|0\rangle + \gamma|1\rangle$$ such that their product state is that entangled state. Thus, the product state would look like:

$$|a\rangle \otimes |b\rangle=\alpha\delta|00\rangle+\alpha\gamma|01\rangle+\beta\delta|10\rangle+\beta\gamma|11\rangle$$

Since the state we want is $$\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$, we have that $$\alpha\delta=\beta\gamma=\frac{1}{\sqrt{2}}$$ and $$\alpha\gamma=\beta\delta=0$$. It is easy to see a contradiction here. Since $$\alpha\gamma=0$$, either one of the terms in this product must be equal to $$0$$. However, if either is zero, $$\alpha\delta$$ or $$\beta\gamma$$ will also be equal to $$0$$, but we want them to be $$\frac{1}{\sqrt{2}}$$. Therefore, no such states $$|a\rangle$$ and $$|b\rangle$$ exist.

The tensor product of two vectors $$\begin{pmatrix} a \\ b \end{pmatrix}$$ and $$\begin{pmatrix}c \\ d \end{pmatrix}$$ is $$\begin{pmatrix} ac \\ ad \\ bc \\ bd \end{pmatrix}$$. In this case, determining the inability to represent the state as a product of two qubit vectors is simpler than some others: $$a$$ and/or $$d$$ as well as $$b$$ and/or $$c$$ must be 0 for $$ad$$ and $$bc$$ to be 0, but any variable being 0 will make either $$ac$$ or $$bd$$ 0 as well, so there's no way the state vector of the $$\left|\Phi\right>$$ state can ever be written in the form $$\begin{pmatrix} a \\ b \end{pmatrix} \otimes \begin{pmatrix}c \\ d \end{pmatrix} = \begin{pmatrix} ac \\ ad \\ bc \\ bd \end{pmatrix}$$.

$$\left|00\right>$$ by itself is $$\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix}$$, so $$a$$ and $$c$$ being 1 and $$b$$ and $$d$$ being 0 allows the tensor product, meaning it is not entangled.

To put in simple words, there are basically two types of quantum states - separable and entangled. Separable states can be writen as a tensor product of other states while entangled states cannot.

The state you are asking about, so-called Bell state $$|\beta_{00}\rangle$$, is an example of entangled state, hence it cannot be writen as tensor product of two others states (mathematically it is explained in other answers).

Disclaimer: My answer may seem oversimplifying. I am of course talking only about pure states and may be I neglected some details. However, for explanation to a beginner, I think the answer is sufficient and explain the issue asked about.

• Thanks for your answer. I'm going to ask you this because I want to do more courses about this: which course are you studying? I want to do it. Every course is good to learn more about this amazing subject. Thanks! Oct 7, 2021 at 18:51
• @VansFannel: More or less I study the subject on my own. Firstly, I went thought IBM Q manual, then I enrolled to on-line course on Coursera, followed by reading of countless articles. You can find a list of courses (including the one on Coursera I took) here: quantumcomputing.stackexchange.com/questions/2667/…. Concerning a good introductory book, try this: arxiv.org/abs/1907.09415, something on Qiskit: arxiv.org/abs/1903.04359 and list of algorithms: arxiv.org/abs/1804.03719 Oct 7, 2021 at 19:49