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I can't seem to understand how to show that the Bell states for a basis. Should I explain that through the circuit and what gates are used or by the basic proof behind proving a set as a basis?

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    $\begingroup$ what's your understanding of what a "basis" is? See also quantumcomputing.stackexchange.com/q/15332/55 $\endgroup$ – glS Mar 30 at 9:56
  • $\begingroup$ I would definitely check out the link given by gIS in the above comment, but as to your precise question, the most straight forward way is to show it using the basic proof behind proving a set as a basis i.e. showing that the basis set form a linearly independent set and span the total set of vectors. $\endgroup$ – Rajiv Krishnakumar Mar 30 at 10:28
  • $\begingroup$ I did follow the link provided by gIS only. I found it just after posting my query and it solved my doubt. Thank you for both the solutions!! $\endgroup$ – Srijita Nandi Mar 30 at 17:06
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If you are taking the four Bell states

$|\Phi^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle \big) $
$|\Phi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle - |11\rangle \big) $
$|\Psi^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle + |10 \rangle \big) $
$|\Psi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle - |10 \rangle \big) $

and place them as a column of a matrix $U$, for instance:

\begin{align} U &= \bigg[ \hspace{0.2 cm} |\Phi^+ \rangle \hspace{0.2 cm} \bigg| \hspace{0.2 cm} |\Phi^- \rangle \hspace{0.2 cm} \bigg| \hspace{0.2 cm} |\Psi^+ \rangle \hspace{0.2 cm} \bigg| \hspace{0.2 cm} |\Psi^- \rangle \hspace{0.2 cm}\bigg] \\ &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0\\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix} \end{align}

Here you can see that $U$ is a unitary matrix. That is, $U\cdot U^*= U^* \cdot U = I$ where $U^*$ is the conjugate transpose of $U$. Since $U$ is unitary, its columns must form an orthonormal basis. These columns are the Bell states.

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  • $\begingroup$ How can you deduce this property for unitary matrices ? $\endgroup$ – BrockenDuck Mar 30 at 15:40
  • $\begingroup$ Let $U = [e_1 , e_2 , \cdots, e_n]$ then $U \cdot U^* = I$ implies $e_i \cdot e_j = 0 $ if $i \neq j$ and $1$ if $i = j$. $\endgroup$ – KAJ226 Mar 30 at 16:07
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    $\begingroup$ +1 In fact, you've shown more: that the Bell states form an orthonormal basis (i.e. an orthogonal basis made up of vectors of unit norm). $\endgroup$ – Adam Zalcman Mar 30 at 16:46
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The Bell states form an orthonormal basis of 2-qubit Hilbert space. The way to show it is to come back to the definition of what an orthonormal basis is:

  1. All vectors have length 1
  2. They are orthogonal to each other.
  3. The 2 qubit Hilbert space is 4 dimensional and you have 4 (orthonormal) vectors which implies linear independence. So the only thing you need to be able to do is compute <b|b'> where |b> and |b'> are Bell states.
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