Entanglement also allows multiple states to be acted on simultaneously, unlike classical bits that can only have one value at a time. Entanglement is a necessary ingredient of any quantum computation that cannot be done efficiently on a classical computer.
–"Qubit", Wikipedia
I thought this was the definition of superposition. Why is quantum entanglement so important in quantum computing, to the point where advances are also measured in number of qubits entangled?
Best Answer
Both of these terms depend on what basis you're in, which can make them a little arbitrary. For example, I think of a single particle in the state $\left|\uparrow\right> + \left|\downarrow\right>$ as in a superposition, though someone else who prefers the $\hat x$ basis may disagree and call it their eigenstate (we're both right). Similarly, entanglement is something we only understand if we work in the single-particle basis. There, we define an entangled state as a state that cannot be written as a product of single-particle states. By a single-particle state, I mean something that can be written as $\left|\psi\right>_1 \otimes \left|\psi \right>_2$ for two particles.
For example, $\left|\uparrow\right>_1 \left|\uparrow\right>_2$ is not entangled, since it's just the product of $\left|\uparrow\right>_1$ and $\left|\uparrow\right>_2$. This is a very correlated state, but it's not entangled. Here, I'm using the subscript to denote which particle. An example of a state that isn't entangled is:
$$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\uparrow\right>_1 \left|\downarrow\right>_2 + \left|\downarrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$$
this is because I can write it as:
$$(\left|\uparrow\right>_1 + \left|\downarrow\right>_1) \otimes (\left|\uparrow\right>_2 + \left|\downarrow\right>_2)$$
On the other hand, the following states are entangled:
$$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\uparrow\right>_1 \left|\downarrow\right>_2 + \left|\downarrow\right>_1 \left|\uparrow\right>_2 - \left|\downarrow\right>_1 \left|\downarrow\right>_2$$ $$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$$
The latter is known as a Bell state. You cannot write either of these states as a generic $\left|\psi\right>_1 \otimes \left|\psi \right>_2$: go ahead and try!
A more practical metric I use is: does the result of measuring the state of one particle change my expectation of the state of another? If yes, the particles must be entangled. (Warning: if the answer is no, it doesn't rule out entanglement! You might need a more clever experiment). For example, if I start with $\left|\uparrow\right>_1 \left|\downarrow\right>_2$, and I know I will measure that the second particle is $\downarrow$, regardless of what basis I measure the first particle in. However, if I start with the state $\left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$, I have $50-50$ odds measuring that the second particle is $\uparrow$ or $\downarrow$. If I measure that the first particle is $\uparrow$, suddenly I know that I'll measure the second particle is $\uparrow$. I've learned something about the second particle only by measuring the first. A measurement on the first particle has changed the odds on the outcome of measuring the second, the hallmark of entanglement.
Side note: If a friend places two marbles in a bag, and promises that both are red or both are blue, you'd get a similar result as my second experiment. But this is not entanglement, that's just a classical lack of knowledge! My proposed experiment to measure entanglement of $\left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$, by measuring each particle in the $\hat z$ basis is convincing only if you know that you started in a pure, quantum mechanical state. In reality you need a more clever experiment to show that you have an entangled state. Examples include Bell's inequality and the CHSH inequality.