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Let's say I start with the following arbitrary qubits: $$ \color{red}{\vert Q_1 \rangle = \alpha_1 \vert 0 \rangle + \beta_1 \vert 1 \rangle}\\ \color{green}{\vert Q_2 \rangle = \alpha_2 \vert 0 \rangle + \beta_2 \vert 1 \rangle}\\ \color{blue}{\vert Q_3 \rangle = \alpha_3 \vert 0 \rangle + \beta_3 \vert 1 \rangle} $$ where $\forall i : \,\,|\alpha_i|^2 + |\beta_i|^2 = 1$.

I then decide to entangle all these qubits to obtain: $$ \vert \psi \rangle = \alpha \vert \color{red}{0}\color{green}{0}\color{blue}{0} \rangle + \beta \vert \color{red}{1}\color{green}{1}\color{blue}{1} \rangle $$ where $|\alpha|^2 + |\beta|^2 = 1$.

Now, I realize that I don't need the second qubit to be entangled with the others, so I unentangle it to obtain the following: $$ \begin{align} \vert \psi \rangle &= \alpha \vert \color{red}{0}\color{blue}{0} \rangle + \beta \vert \color{red}{1}\color{blue}{1} \rangle \end{align} $$ I don't really care about qubit two anymore. It can be measured to destroy its superposition or even be set to anything else. The point is I unentangle it from the system. Is such thing possible? If yes, what is the mathematics and the physical realization behind this?

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Yes, it is possible. To obtain the state $$\vert \phi \rangle = \alpha \vert 00 \rangle + \beta \vert 11 \rangle$$ from $$\vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 111 \rangle,$$ note that the latter can be written as $$\begin{aligned}\vert \psi\rangle=&\frac{1}{\sqrt{2}}\alpha \vert 00+ \rangle + \frac{1}{\sqrt{2}}\alpha \vert 00 -\rangle + \frac{1}{\sqrt{2}}\beta \vert 11 +\rangle-\frac{1}{\sqrt{2}}\beta \vert 11 -\rangle\\ =& \frac{1}{\sqrt{2}}(\alpha\vert 00\rangle + \beta\vert 11\rangle)\vert +\rangle +\frac{1}{\sqrt{2}}(\alpha\vert00\rangle -\beta \vert 11\rangle)\vert - \rangle\\ =& \frac{1}{\sqrt{2}} \vert \phi \rangle\vert+\rangle + (Z\otimes \text{Id}) \frac{1}{\sqrt{2}} \vert \phi \rangle\vert - \rangle.\end{aligned}$$ This shows how you can remove a qubit from this entangled state. If you measure this qubit in $X$-basis, then you get one of the following two outcomes:

  • $+1$ (i.e. $\vert +\rangle$), in which case you have successfully projected onto $\vert \phi \rangle$.
  • $-1$ (i.e. $\vert -\rangle$), in which case you have projected onto $(Z\otimes Id)\vert \phi\rangle$, which, however, can be transformed into $\vert \phi \rangle$ by applying a $Z$-gate on one of the remaining qubits.

In summary: measure the qubit that you want to remove in the $X$-Basis and perform a phase correction if neccessary.

This procedure is standard in entanglement-based quantum computation.

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  • $\begingroup$ It seems to me that $I$ and $Z$ gates in the last steps should be swapped. Application $Z$ on $|-\rangle$ changes it to $|+\rangle$ and measuring third qubit in Hadamard basis retrurns state $|0\rangle$. But I also feel that your approach does not unentangle qubits because you have to apply inverse operation to one making a entanglement, i.e. CNOT in this case. Or do I miss anything? $\endgroup$ – Martin Vesely Jan 26 at 13:13
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    $\begingroup$ @MartinVesely Z is applied to one of the remaining qubits, not the one that you measure. Also note that a measurement in X-Basis can only yield |+> or |->. This answer shows how you don't need to apply two-qubit gates to remove a qubit from the state. $\endgroup$ – M. Stern Jan 26 at 16:29
  • $\begingroup$ It is clear now, thanks. $\endgroup$ – Martin Vesely Jan 26 at 19:41
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Measuring the second qubit in the $X$ basis you get one of two results, associated with the two eigenstates of $X$, which are: $\newcommand{\ket}[1]{|#1\rangle}\sqrt2\ket\pm\equiv\ket0\pm\ket1$.

The post-measurement states corresponding to these two outcomes are given (up to normalisation) by the overlaps $\langle\pm_2|\psi\rangle$, where here $\ket{\pm_2}\equiv I\otimes\ket\pm\otimes I$. This gives $$ \ket{\psi_+}\equiv\frac{\langle+_2|\psi\rangle}{\sqrt{p_+}} = \alpha\ket{00} + \beta\ket{11}, \\ \ket{\psi_-}\equiv\frac{\langle-_2|\psi\rangle}{\sqrt{p_-}} = \alpha\ket{00} - \beta\ket{11}. $$ where $p_\pm\equiv\|\langle \pm_2|\psi\rangle\|^2=1/2$ are the outcome probabilities.

This tells you that, if you want to get the state $\alpha\ket{00}+\beta\ket{11}$ on the remaining qubits, you can do it by post-selecting on the outcome $|+\rangle$ when measuring one of the qubits (in your case the second one). By post-selection I mean that you will have the correct state on the two qubits only whenever you measure $\ket+$ on the second one. This can be used in computation by post-selecting the experimental runs in which the second qubit was found to be $\ket+$ (which will be roughly half of them), and discarding the rest.

If you also allow for a local operation be performed on the remaining qubits conditionally on the results of measuring the second one, then you even make this into a deterministic algorithm, by applying a $Z$ gate conditionally to finding $\ket{2_+}$ which as you can check converts $\ket{\psi_-}$ into $\ket{\psi_+}$.

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First of all, you cannot describe entanglement with tensor product. Resulting state of $q_0 \otimes q_1 \otimes q_2$ will be

$$ |\varphi\rangle = \sum_{i=000}^{111}a_{i}|i\rangle, $$ i.e. $i$ is $|000\rangle$,$|001\rangle$...$|111\rangle$; and $\sum_{i=0}^{n}|a_{i}|^2 = 1$.

To obtain state $\alpha|000\rangle + \beta|111\rangle$ you have to employ circuit similar to this:

Entanglement preparation

In this case $\alpha = \beta = \frac{1}{\sqrt{2}}$. To change $\alpha$ and $\beta$ you should replace Hadamard gate with for example $U3$ gate to set these constants according to your requirements.

To unentangle second qubit, you should appply inverse operation to the one which created the entanglement. In this case, it was CNOT gate. As inverse to CNOT is CNOT itself, the circuit where qubits $q_0$, $q_1$ and $q_2$ were firstly entangled and then qubit $q_1$ was unentangled is this:

Unentangling q1

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