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Let's say I have n qbits each in a superposition $\begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix}$ so each possible outcome has a probability of $\frac{1}{2^n}$. Is it possible for me to suppress the probability of one outcome?

E.g. I have 3qbits, I want to remove the possibility of reading 101 when I measure the qbits. Possible outcomes:$\begin{bmatrix} 000\\ 001\\010\\011\\100\\101\\110\\111 \end{bmatrix}$ probabilities of outcomes: $\begin{bmatrix} 0.125\\ 0.125\\0.125\\0.125\\0.125\\0.125\\0.125\\0.125\end{bmatrix}$ -> $\begin{bmatrix} 0.143\\ 0.143\\0.143\\0.143\\0.143\\0\\0.143\\0.143\end{bmatrix}$

Is this possible? How could I do that? Furthermore, is it possible for me to remove the possibility of reading 2 outcomes: say 101 and 110.

Possible outcomes:$\begin{bmatrix} 000\\ 001\\010\\011\\100\\101\\110\\111 \end{bmatrix}$ probabilities of outcomes: $\begin{bmatrix} 0.125\\ 0.125\\0.125\\0.125\\0.125\\0.125\\0.125\\0.125\end{bmatrix}$ -> $\begin{bmatrix} 0.167\\ 0.167\\0.167\\0.167\\0.167\\0\\0\\0.167\end{bmatrix}$

Note: I don't actually care about the remaining probabilities of the outcomes, as long as they are nonzero.

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  • $\begingroup$ Welcome to QCSE. If you know what Grover’s algorithm is, then this might achieve what you are looking for. Alternatively you can evaluate in a dummy ancilla register that is $\vert 1\rangle$ iff the first register is $\vert 101\rangle$ and $\vert 0\rangle$ otherwise. and post-select states wherein the ancilla is $\vert 0\rangle$. Your success probability would depend on how many basis states you want to exclude. $\endgroup$ – Mark S Apr 19 at 3:32
  • $\begingroup$ Thanks for the welcome :) Yeah I'll look into the dummy ancilla register thing. Still kind of a noob with quantum computing. $\endgroup$ – Benedict Bien Apr 19 at 3:59
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    $\begingroup$ You're getting close to understanding, and to having the right intuition. Just initially envision concatenating another "ancilla" register using your first register to control when the first register is $101$. You can prepare a state uniformly distributed over: $$\begin{bmatrix} 000|0\\ 001|0\\010|0\\011|0\\100|0\\101|1\\110|0\\111|0 \end{bmatrix},$$ then measure the second register. If the second register is $1$, then the first register collapses to your desired state. This happens with probability $7/8$. $\endgroup$ – Mark S Apr 19 at 14:54
  • $\begingroup$ *If the second register is $0$ I mean. $\endgroup$ – Mark S Apr 19 at 17:06
  • $\begingroup$ Thanks a lot! I think I get how to do it! $\endgroup$ – Benedict Bien Apr 20 at 0:02

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