How does a general $n$-qubit entangled state deal with monogamy of entanglement?

Question

Now since only two electrons (or any quantum object) can be maximally entangled due to the monogamy of entanglement how does the general $n$-qubit states which has more than two qubits make use of entangled states to increase the computational space to $2^n$? so for example take this three qubit general state

$$a1|000⟩+a2|001⟩+a3|010⟩+a4|100⟩+a5|110⟩+a6|101⟩+a7|111⟩+a8|011⟩$$

If only 2 qubits out of these 3 are entangled how can we represent it in such a form?

same question for any $n$-qubit general state

$$∣ψ⟩_{n-qubit}= a1∣000…000⟩+ a2∣000…001⟩+a3 ∣000…010⟩+⋯.$$

i always understood general states as that all possible bits e:g l000>, l001>, ... (for example these three bits) cannot be computed independently since they are entangled but now that i know that only two can be maximally entangled how can we add a third non entangled qubit in that relation?. what part am i missing?

Answer 1

If only 2 qubits out of these 3 are entangled how can we represent it in such a form?

If none of the qubits are entangled, the state can be written as:

$(a_1|0\rangle + a_2|1\rangle) (a_3|0\rangle + a_4|1\rangle) (a_5|0\rangle + a_6|1\rangle)$

Taking the appropriate products of the $a_x$ coefficients gives the amplitudes of each 3-bit states in your original formula.

If 2-qbits are entangled while the third is independent, the formula can be written as:

$(a_1|00\rangle + a_2|01\rangle + a_3|10\rangle + a_4|11\rangle) (b_1|0\rangle + b_2|1\rangle)$

Where the first term is the formula for the entangled 2-bit state.

So in summary, if 2-qbits are entangled while the third is not, this will show up in the amplitudes of the final formula and it will be separable into the above equation. Whereas if all 3 are entangled, in general, it cannot be written in the above form.