# Can we use entanglement for cascaded measurements

My question is as follows: Suppose we have two entangled states, for example, a GHZ state where $$|000\rangle$$ and $$|111\rangle$$ are entangled Now, suppose the $$|111\rangle$$ state is entangled with another state, say $$|110\rangle$$ (correct me if I make wrong assumptions along the way, for example, concerning the states, however, I am asking for the general concept) Wouldn’t measuring one of the qubits tell us the state of the other two qubits, and by knowing the state of the other 2 qubits we now know the state of the 4th and 5th qubit?

How would we represent this mathematically if possible?

For context (e.g.: there are three entangled qubits and one of these qubits is entangled with another 2 $$\to$$ also, would that make them all entangled? )

• How many qubits do you have? The GHZ state is conventionally only 3. Your third sentence sounds like you are violating the monogamy of entanglement, while your fourth sentence sounds like your state is something like $\frac {1}{\sqrt 2}(|00000\rangle + |111110\rangle)$. Commented Oct 1, 2023 at 0:02
• I have 5 qubits three of them are in a ghz state and one of those three is in a ghz state with 2 other qubits, therefore 5 qubits Commented Oct 1, 2023 at 0:20
• Then the other two first qubits are also entangled with the last two qubits. You have a state such as $1/\sqrt 2(|00000\rangle + |11111\rangle)$. Commented Oct 1, 2023 at 0:28

As explained in the comments, your comment about the $$|111\rangle$$ state being entangled with another state seems to suggest that you wish to violate the monogamy of entanglement. If a qubit is maximally entangled with some other qubits then it cannot be entangled with any other qubit.
But as suggested by the title of your question and further clarified in the comments you wish to know what happens when, given a state such as $$\frac{1}{\sqrt 2}(|00\cdots 0\rangle+|11\cdots 1\rangle)$$, we measure only one of the provided qubits. In this case if one of the qubits measures $$|0\rangle$$ in the computational basis then all other qubits will be $$|0\rangle$$. Contrariwise if the qubit measures $$|1\rangle$$ then all other qubits measure $$|1\rangle$$ as well.