Is the balanced superposition of all three-qubit computational basis states entangled?

Question

I am reading the "An introduction to quantum computing for non-physicists" by Rieffel and Polak. In page 308, they define an entangled state as a state that "cannot be described in terms of the state of each of its components." They used the following example, $$ \frac{1}{{\sqrt 2 }}\left| {\left. {00} \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. {11} \right\rangle } \right. = \left( {a_1 \left| {\left. 0 \right\rangle } \right. + b_1 \left| {\left. 1 \right\rangle } \right.} \right) \otimes \left( {a_2 \left| {\left. 0 \right\rangle } \right. + b_2 \left| {\left. 1 \right\rangle } \right.} \right) $$

In this example, there are no valid values for $a_1, a_2, b_1, b_2.$ As a result, this state is entangled.

My question is: with the above definition, is the following state entangled? $$ \frac{{\sqrt 2 }}{4}\left| {\left. {000} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {001} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {010} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {011} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {100} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {101} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {110} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {111} \right\rangle } \right. $$

Note that the above state can indeed be decomposed to $$ = \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) \otimes \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) \otimes \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) $$

Answer 1

$$ \frac{{\sqrt 2 }}{4}\left| {\left. {000} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {001} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {010} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {011} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {100} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {101} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {110} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {111} \right\rangle } \right. $$

Note that the above state can indeed be decomposed to $$ = \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) \otimes \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) \otimes \left( {\frac{1}{{\sqrt 2 }}\left| {\left. 0 \right\rangle } \right. + \frac{1}{{\sqrt 2 }}\left| {\left. 1 \right\rangle } \right.} \right) $$

My question is: with the above definition, is the [above] state entangled?

No. Because it can be decomposed into the direct/tensor product of individual qubits, it is not entangled.

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In general, an unentangled state of $n$ qubits can be written in term of just $2n$ complex parameters $a_i$ and $b_i$ as: $$ \left(a_1|0\rangle+b_1|1\rangle\right)\otimes \left(a_2|0\rangle+b_2|1\rangle\right)\otimes \ldots \left(a_{n}|0\rangle+b_{n}|1\rangle\right)\;, $$ whereas a general state of $n$ qubits requires $2^n$ complex parameters for its specification.

Answer 2

It may be worth noting that the states analogous to the Bell state $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ for 3 qubits quantum system is W state: $\frac{1}{\sqrt{3}}|001\rangle + \frac{1}{\sqrt{3}}|010\rangle + \frac{1}{\sqrt{3}}|100\rangle$ or GHZ state $\frac{1}{\sqrt{2}}|000\rangle + \frac{1}{\sqrt{2}}|111\rangle$. They are both non-separable and moreover non-biseparable, i.e. can't be decomposed into a product of three or two states.