Exercise 3.13 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.13}$

a) Show that the four-qubit state $\ket\psi = \frac{1}{2}(\ket{00} + \ket{11} + \ket{22} + \ket{33})$ of Example 3.2.3 is entangled with respect to the decomposition into two two-qubit subsystems consisting of the first and second qubits and the third and fourth qubits.

b) For the four decompositions into two subsystems consisting of one and three qubits, say whether $\ket\psi$ is entangled or unentangled with respect to each of these decompositions.

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(1)a. Assume $\ket{\psi}$ is not entangled with respect to this decomposition. Then (still using the 0/1/2/3 representation for each two-qubit part), $\ket{\psi}$ can be written asWe can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_1 = \frac{1}{2}$, so $b_1$ is non-zero. But we can also see that $a_0 b_1 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

b. Let us write $\ket{\psi}$ as $\frac{1}{2}(\ket{0000} +\ket{0101} +\ket{1010} + \ket{1111})$.First consider the one qubit / three qubit decomposition where the one qubit is the leftmost in this representation. Suppose $\ket{\psi}$ is not entangled with respect to this decomposition. Then $\ket{\psi}$ can be written as

(2)We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_2 = \frac{1}{2}$, so $b_2$ is non-zero. But we can also see that $a_0 b_2 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

For each of the other three possible positions for the single qubit, a similar argument can be constructed. Hence $\ket{\psi}$ is entangled with respect to any of these four decompositions.

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