1
$\begingroup$

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 $\lvert\psi\rangle_{B_H}$.

$\endgroup$
  • 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 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 at 8:32
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.