Sometimes we find that the result we want from a quantum algorithm is expressed in terms of a basis that is different from the usual computational basis, which I will call

$$ B_C = \left\{ \lvert 0 \rangle, \lvert 1 \rangle \right\} = \left\{ \left(\begin{array}{c}1\\0\end{array}\right),\left(\begin{array}{c}0\\1\end{array}\right) \right\}. $$

For example, at the end of the Deutsch's algorithm, the first qubits is in the state

$$ \lvert\psi\rangle = (-1)^{f(0)} \dfrac{\lvert 0 \rangle + (-1)^{f(0) \oplus f(1)} \lvert 1 \rangle}{\sqrt{2}}, $$

which can be expressed in terms of the Hadamard's basis

$$ B_H = \{ \lvert + \rangle, \lvert - \rangle \} = \left\{ \dfrac{1}{\sqrt{2}} \left( \begin{array}{c}1\\1\end{array} \right), \dfrac{1}{\sqrt{2}} \left( \begin{array}{c}1\\-1\end{array} \right) \right\} $$

$$ \begin{cases} f(0) \oplus f(1) = 0 \quad\Rightarrow\quad \lvert\psi\rangle = (-1)^{f(0)} \lvert + \rangle \\ f(0) \oplus f(1) = 1 \quad\Rightarrow\quad \lvert\psi\rangle = (-1)^{f(0)} \lvert - \rangle \end{cases} $$

Therefore, we can get the value of $f(0) \oplus f(1)$ just by measuring $\lvert\psi\rangle$ on $B_H$. Since we can only perform physical measurements on $B_C$, this can be achieved performing a change of basis.

Given two basis $A$ and $B$, if the matrix that transforms the elements of $A$ to the elements of $B$ is the matrix $M$, this is also the matrix that maps the coordinates of vectors with respect to $B$ to their coordinates with respect to $A$. Hence, which of these propositions is the correct to express the change of basis we must do to perform the measurement on $B_H$ knowing that physical measurements are actually performed in $B_C$?

  1. We must apply the Hadamard's gate, because the matrix $H$ maps $\left\{ \lvert 0 \rangle,\lvert 1 \rangle \right\}$ to $\left\{ \lvert + \rangle, \lvert - \rangle \right\}$, and therefore it transforms the components $\lvert\psi\rangle_{B_H}$ to $\lvert\psi\rangle_{B_C}$.

  2. We must apply the Hadamard's gate, because the matrix $H$ maps $\left\{ \lvert + \rangle,\lvert - \rangle \right\}$ to $\left\{ \lvert 0 \rangle, \lvert 1 \rangle \right\}$, and therefore it transforms the components $\lvert\psi\rangle_{B_C}$ to $\lvert\psi\rangle_{B_H}$.

  • 2
    $\begingroup$ This is boiling down to a confusion about when to use $U$ vs $U^{-1}$. You should give an example where $U \neq U^{-1}$ in order to get a case where you can actually tell the difference. $\endgroup$ – AHusain Apr 17 '19 at 21:01
  • $\begingroup$ That example would actually be the answer to this question, so if I had it, I wouldn't have come here $\endgroup$ – Jaime_mc2 Apr 18 '19 at 8:32

If you want to measure $|\phi\rangle$ in some basis $U|b_1\rangle,...,U|b_n\rangle$ instead of $|b_1\rangle,...,|b_n\rangle$, then you need to rotate the state "backward", i.e. measure $U^{-1}|\phi\rangle$ in $|b_1\rangle,...,|b_n\rangle$.
The simple rule to find the direction of rotation is to consider the state and the required measurement basis together $\{|\phi\rangle, U|b_1\rangle,...,U|b_n\rangle\}$. Then If you what to perform the same measurement in a different basis you need to rotate the whole system altogether.

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