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As part of a Quantum Theory project I have "constructed" an arbitrary 3-qubit basis:

$\left|B_0\right> =\left|000\right>$

$\left|B_1\right> = \frac{1}{\sqrt{2}}\cos(x)(\left|100\right> + \left|010\right>) + \sin(x)\left|001\right>$

and so on to $\left|B_6\right>$.

The states are orthogonal and form a proper basis set.

However, I would like to know if (and how) such a measurement can be performed.

I was looking at polarization of light and time-bins qubits experiments, but they only seem to look at creating those states, rather than performing a measurement in this basis.

Thank you in advance for your answers!

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    $\begingroup$ I do not see the "and so on" to the sixth basis state. Furthermore, measuring in a different basis is nothing more than doing some (inverted) operations on (a group of) qubit(s) and then measuring all in the $Z$-basis. $\endgroup$
    – nippon
    Mar 26 '20 at 14:28
  • $\begingroup$ If you can create those states from basis states, can you not invert the unitary to transform from those states back to the basis states, and then measure in the standard basis? $\endgroup$
    – DaftWullie
    Mar 26 '20 at 15:00
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    $\begingroup$ Why does your basis only consist of 7 states, not 8? $\endgroup$
    – DaftWullie
    Mar 26 '20 at 15:01
  • $\begingroup$ to measure in a given basis $\{|B_i\rangle\}_i$, you can implement an evolution sending the different elements of the basis to the different elements of the computational basis, e.g. $|B_i\rangle\to |i\rangle$, and then measure in the computational basis. What "measure in the computational basis" means in practice, depends on the implementation. How you would go in implementing this kind of evolution also depends on the architecture and the resources available. $\endgroup$
    – glS
    Mar 27 '20 at 10:26
  • $\begingroup$ DaftWullie, it is 8 states, I just wrote B6 to be the last one by mistake. Thanks for pointing that out though! As for the "so on", I am aware they do not follow too naturally, but since it's part of my project I am not sure how much I can give away, I'm sorry. Just be comfortable knowing that the rest are similar in construction. $\endgroup$
    – user10479
    Mar 27 '20 at 15:20

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