Show that the Bell states form a basis

Question

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?

KAJ226 · Accepted Answer

If you are taking the four Bell states
$|Phi^+ rangle = dfrac{1}{sqrt{2}}big(|00rangle + |11rangle big) $
$|Phi^- rangle = dfrac{1}{sqrt{2}}big(|00rangle - |11rangle big) $ 
$|Psi^+ rangle = dfrac{1}{sqrt{2}}big(|01rangle + |10 rangle big) $ 
$|Psi^- rangle = dfrac{1}{sqrt{2}}big(|01rangle - |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, $Ucdot U^*= U^* cdot U = I$ 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.

holl · Answer

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:

All vectors have length 1
They are orthogonal to each other.
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.

Show that the Bell states form a basis

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