# Slight issue with QFTing two qubits

Let's consider two qubits and the corresponding computational basis $$\{|0\rangle\, |1\rangle, |2\rangle, |3\rangle\}$$. In binary form, any of these vectors can also be written as a product $$|x_1\rangle\otimes |x_2\rangle$$ where $$x_1$$ and $$x_2$$ can be either $$0$$ or $$1$$. Let's take these two qubits and perform the following operation on them: $$|y\rangle=(H\otimes \mathbb I)(CR_2)(\mathbb I\otimes H)|x_1x_2\rangle=\frac{1}{2}(|0\rangle+e^{2\pi ix_2/2^2}e^{2\pi ix_1/2}|1\rangle)\otimes(|0\rangle+e^{2\pi ix_2/2}|1\rangle)$$where $$H$$ is the Hadamard gate and $$R_2$$ is the $$2\times 2$$ diagonal matrix $$[1, e^{2\pi i/2^k}]$$. Looking at that first exponential, if $$x=x_1x_2$$, then in decimal $$x=2x_1+x_2$$, which means that $$|y\rangle=\frac{1}{2}(|0\rangle+e^{2\pi i x_1/2^2}|1\rangle)\otimes (|0\rangle+e^{2\pi i x_2/2}|1\rangle)\tag 1$$ which is to be compared to the QFT of $$|x\rangle$$

$$|\tilde x\rangle=\frac{1}{2}(|0\rangle+e^{2\pi i x/2}|1\rangle)\otimes (|0\rangle+e^{2\pi ix/2^2}|1\rangle).\tag 2$$ These two are supposed to be the same thing, only with qubits in reversed order. However, something seems remiss to me: the second factor in $$|y\rangle$$ has $$x_2/2$$ in the exponential, while the first factor of $$|\tilde x\rangle$$ has $$x/2$$. Am I being colossally blind, or is there a mistake somewhere?

• Should it be $|y\rangle=\frac{1}{2}(|0\rangle+e^{2\pi i x_1/2^2}|1\rangle)\otimes (|0\rangle+e^{2\pi i x_2/2}|1\rangle)$? I think, as you hinted, you just had an endianness convention issue. The qubits are in the reverse order, aren't they? May 8 at 23:55
• @MarkS Well, yeah, I'm just comparing that $e^{2\pi i x_2/2}$ in $|y\rangle$ with the $e^{2\pi i x/2}$ in $|\tilde x\rangle$. Since $x\ne x_2$, doesn't that difference mean that $|y\rangle\ne |\tilde x\rangle$? May 9 at 18:21
• $x$ is two qubits (or one qudit, with $d=4$), while $x_2$ is a single qubit, right? May 9 at 18:25
• I just made a change to equation 1 - let me know if that's correct? May 9 at 18:27