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I have read the following statement in some lecture notes:

The inner product of two n-qubit registers is taken by mirrored qubit pairs. Example: $\lvert ABC\rangle$ and $\lvert abc \rangle$ leading to $\langle CBA \lvert abc \rangle$

Is this statement correct? Assuming I'd take the inner product of the state $\lvert 01 \rangle$ with itself, it should result in $1$, since it is a normalized vector. But $\langle 10 \lvert 01 \rangle$ in my understanding would result in $0$.

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2 Answers 2

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Depends on the notation, I guess. For product states, when

$$|ABC\rangle=|A\rangle|B\rangle|C\rangle,\quad |abc\rangle=|a\rangle|b\rangle|c\rangle$$

the inner product is $$\langle ABC|abc\rangle =\langle A|a\rangle \langle B|b\rangle \langle C|c\rangle$$

while for non-product states it extends by linearity. If this rule is satisfied, whether to invert order of qubits in the bra vector or not is a matter of choice.

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    $\begingroup$ In every notation I've ever seen, the dagger of $|ABC\rangle$ is $\langle ABC|$, i.e. without the order reversed. That way $\langle 01|01\rangle = 1$ and $\langle 10|01\rangle = 0$ as your intuition expected. $\endgroup$
    – Chris E
    Commented Jul 20, 2022 at 22:30
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I suppose you can pick any notation you want, however, in any standard notation, anything inside the ket/bra is treated as the name of the state (a "string" object if you're OK with thinking in terms of programming languages ;)). So, the bra corresponding to $\vert{\psi}\rangle$ is $\langle\psi\vert$. So, if we consider the ket $\vert ABC\rangle$, the name of the state is "ABC", thus, the corresponding bra would be simply $\langle{ABC}\vert$.

Now, how you evaluate $\langle ABC\vert abc\rangle$ depends on what's the meaning of your notation. Usually, $\vert{xyz}\rangle$ is a shorthand for $\vert x\rangle_1\otimes\vert y\rangle_2\otimes z\rangle_3$ -- where the subscript $i=1,2,3$ suggests which subsystem of the composite system we are talking about. Thus, \begin{align*} \langle ABC\vert abc\rangle &= \big(\langle A\vert_1 \otimes \langle B\vert_2\otimes\langle C\vert_3 \big)\big(\vert a\rangle_1\otimes \vert b\rangle_2\otimes \vert c\rangle_3\big)\\ &=\langle A\vert a\rangle\langle B\vert b\rangle\langle C\vert c\rangle \end{align*}

However, if I am using a quirky notation where $\vert xyz\rangle\equiv \frac{\vert{xy}\rangle+(-1)^z\vert{yx}\rangle}{\sqrt{2}}$ and $\vert xy\rangle\equiv\vert{x}\rangle_1\otimes\vert y\rangle_2$. Then

\begin{align*} \langle ABC\vert abc\rangle &= \frac{1}{2}\big(\langle AB\vert ab\rangle+(-1)^C\langle BA\vert ab\rangle + (-1)^c \langle AB\vert ba\rangle+(-1)^{C+c}\langle BA\vert ba\rangle\big) \\ &= \frac{1}{2}\big(\langle A\vert a\rangle\langle B\vert b\rangle + (-1)^{C+c}\langle B\vert b\rangle \langle A\vert a\rangle\big) \\ &= \frac{1}{2}(1+(-1)^{C+c})\langle A\vert a\rangle\langle B\vert b\rangle \end{align*}

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