# How to understand the idea of maximal entanglement intuitively?

I am trying to understand what it means for 2 qubits to be maximally entangled. When I look for information online or in books all I can find are rigorous mathematical definitions which I find a bit hard to understand and no explanation on the intuition behind them. I am hoping someone can explain it to me in more simpler terms and hopefully give me some examples of both maximally and non maximally entangled (but still entangled) qubits.

I am going to leave here my intuition of what I think it means in case it is of any help but note that it's probably wrong. Maybe 2 maximally entangled qubits means that the state of 1 of them is completely dependent on the other (such as $$|00\rangle+|11\rangle$$, normalization constant is neglected). While a state like $$|00\rangle+|01\rangle+|10\rangle$$ is maybe also considered somewhat entangled (but not maximally) since measuring a 1 in the first qubit tells you that the second one must be 0 but measuring a 0 in the first qubit tells you nothing about the second one.

Yes, your ideda is right. For example $$n$$-qubit GHZ state, i.e. $$\frac{1}{\sqrt{2}}(|0 \dots 0\rangle + |1 \dots 1\rangle)$$ is maximally entangled since knowing (measuring) state of only one qubit determines state of others.
On the other hand, so-called W-states (e.g. 3-qubit version $$\frac{1}{\sqrt{3}}(|100\rangle+|010\rangle+|001\rangle)$$) are also entangled but not maximally. If you measure one qubit and get $$|1\rangle$$, you know that others are in state $$|0\rangle$$. However, if the measurement returns $$|0\rangle$$, you have to measure other qubits until you get $$|1\rangle$$. Only then you can say that others are in state $$|0\rangle$$.
In other words, in maximally entangled state, information is saved with maximal possible redundancy. Decreasing amount of entanglement increases entropy. Once there is no entanglement you cannot infer with certainity states of other qubits by measuring just a few ones. An extreme case is uniformly distributed state $$\frac{1}{\sqrt{n}}\sum_{i=0}^{2^n-1}|i\rangle$$. In this case, all $$2^n$$ results are possible with same probability (like fair dice). This is a state generating maximally entropic distribution.
• What about the state $\frac{|+++>+|--->}{\sqrt(2)}=\frac{|000⟩+|011⟩+|110⟩+|101⟩}{2}? Looking at its representation in the Z basis it doesn't look maximally entangled while in the X basis it does. Apr 30 at 22:13 • That's the problem with intuition...see more here on relativity of entanglement: physics.stackexchange.com/questions/267293/…. And here how to calculate entanglement strength: en.m.wikipedia.org/wiki/Entropy_of_entanglement May 1 at 7:33 • Just one addition on intuition. If you measure one qubit say in state$|+\rangle\$, then others are also in that state. However, you still have a superposition of 0 and 1, not a classical state (i.e. 0 or 1) which should be result of the measurement. Again, this is intuition, not exact mathematical approach. May 1 at 7:42