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Given Bell states $\mathcal{B} = \{\left\vert \phi^{\pm} \right\rangle, \left\vert \psi^{\pm} \right\rangle\}$, show that Quantum Fourier Transform(QFT) maps $\mathcal{B}\rightarrow \mathcal{B}$ by keeping $\left\vert \psi^{-} \right\rangle $ invariant.

After applying the QFT on Bell states, I found

$$ QFT \left\vert \phi^{+} \right\rangle = \frac{1}{2\sqrt{2}}(2\left\vert0 \right\rangle + (1-i)\left\vert 1 \right\rangle + (1+i) \left\vert3 \right\rangle) \\ QFT \left\vert \phi^{-} \right\rangle = \frac{1}{2\sqrt{2}}((1+i) \left\vert1 \right\rangle +2\left\vert2 \right\rangle + (1-i)\left\vert 3 \right\rangle ) \\ QFT \left\vert \psi^{+} \right\rangle = \frac{1}{2\sqrt{2}}(2\left\vert0 \right\rangle + (i-1)\left\vert 1 \right\rangle - (1+i) \left\vert3 \right\rangle) \\ QFT \left\vert \psi^{-} \right\rangle = \frac{1}{2\sqrt{2}}((1+i) \left\vert1 \right\rangle -2\left\vert2 \right\rangle + (1-i)\left\vert 3 \right\rangle ). $$

Can anyone please help, how to proceed from here ?

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Your calculation is correct. It is not true that QFT maps $\mathcal{B}$ to $\mathcal{B}$. It is also not true that QFT keeps $|\psi^-\rangle$ fixed.

For a proof by contradiction assume that QFT maps $\mathcal{B}$ to $\mathcal{B}$. Then QFT2 = CNOT also maps $\mathcal{B}$ to $\mathcal{B}$. However, CNOT maps every Bell state to a product state. The contradiction proves that QFT does not map $\mathcal{B}$ to $\mathcal{B}$.

The argument that QFT does not fix $|\psi^-\rangle$ is analogous.


Equality of QFT2 and CNOT

In order to see the equality between QFT2 on two qubits and CNOT first write down two copies of the QFT circuit

$$ QFT = SWAP \circ (I \otimes H) \circ CS \circ (H \otimes I). $$

where $CS = \mathrm{diag}(1, 1, 1, i)$. Next, move the final SWAP gate from the second QFT circuit backward to cancel the SWAP gate of the first QFT circuit. Next, cancel the Hadamards and combine the CS gates into the CZ gate. Finally, combine CZ with the two remaining Hadamards to obtain CNOT.

Alternatively, we can also see that QFT2=CNOT using matrix multiplication

$$ QFT^2 = \left(\frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \end{pmatrix}\right)^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}. $$

In any case, this is a restatement of the well-known property of the Discrete Fourier Transform that applying it twice fixes the first input and reverses the order of the other inputs.


Direct calculation

Direct calculation yields the same result. The matrix of QFT on two qubits is

$$ QFT = \frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \end{pmatrix} $$

and the column vector representations of the Bell states are

$$ |\phi^+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1\end{pmatrix} \,\,\,\, |\phi^-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ -1\end{pmatrix} \,\,\,\, |\psi^+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0\end{pmatrix} \,\,\,\, |\psi^-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0\end{pmatrix}. $$

By matrix multiplication we find

$$ QFT |\phi^+\rangle = \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ \frac{1-i}{2} \\ 0 \\ \frac{1+i}{2}\end{pmatrix} $$

in agreement with your result and in disagreement with the claim that QFT maps $\mathcal{B}$ to $\mathcal{B}$. Similarly, for the other vectors in $\mathcal{B}$.

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