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Say I have some qubit $|q\rangle = \alpha |0\rangle + \beta |1\rangle$. If I apply some unitary $U$, I get $U |q\rangle$; great. I can also think of $U$ as a change-of-basis matrix that maps the $|0\rangle, |1\rangle$ basis to the $U|0\rangle, U|1\rangle$ basis. If I just decide to measure the original $|q\rangle$ in that new basis, then all of my very limited knowledge about all of this would lead me to believe I get $U|q\rangle$. So is there any actual difference between the two situations? This seems like a well-known or obvious thing but I really don't know.

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    $\begingroup$ The question in its current form does not make sense to me: there is little or no in common between unitary transformations and measurement. Maybe OP wanted to ask about the difference between unitary operators and unitary basis transformation, but this is just my guess. $\endgroup$ – kludg Nov 18 at 6:57
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Application of a unitary transformation $U$ on a state $|q\rangle$ really leads to a new state $U|q\rangle$. What you get after a measurement is a one particular outcome of $U|q\rangle$ state because the measurement leads to colapse of the state wave function.

If you repeat measurement many times you will get probability distribution of possible outcomes of the quantum state.

Another difference is a reversability. While transformation of the quantum state is always reversible because the unitary matrix can be always inverted, the measurement is irreversible because of the wave function colapse. To reconstruct the quantum state you need to do many measurements of same state in different basis and employ so-called quantum tomography.

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  • $\begingroup$ So would you say in the situation I described the difference is essentially that if I measure in a different basis I will cause the superposition to collapse, whereas if I apply some unitary transformation I will not cause it to collapse? $\endgroup$ – Pedro Nov 18 at 17:36
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    $\begingroup$ Yes, if you apply an unitary transformation, the collapse will not occur. On the other hand, if you measure a quantum state, it will collapse. However, you can measure only some q-bits, in this case the collapse will be only partial (i.e. measured q-bits collapse). $\endgroup$ – Martin Vesely Nov 18 at 19:05
  • $\begingroup$ @Pedro: Hi, if my answer satisfied you, could you please accept it? $\endgroup$ – Martin Vesely Nov 19 at 5:18

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