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With 2 qubits, there are 4 ways to entangle the qubits - the four maximally entangled two-qubit Bell states, of which one is $1/\sqrt{2}(|00\rangle + |11\rangle)$.

Is it possible to entangle more than two qubits together? If so, is there any practical use to doing so?

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  • $\begingroup$ related from physics SE. $\endgroup$ – 吖奇说 Oct 21 '15 at 20:35
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Ok, entanglement just clicked for me and totally makes sense.

Yes it is completely possible to entangle more than two qubits!

Entanglement is a fancy word with a really simple meaning.

In quantum computing, you can modify the probabilities of values being read.

If you have two qubits, normally there is some probability for each of the possible states: reading a 0 for each qubit, reading a 1 for each qubit, reading a 0 for qubit a and a 1 for qubit b, and reading a 1 for qubit a and a 0 for qubit b.

When you modify probabilities such that any of these probabilities drop to zero, it limits the possibilities of what can happen when you measure the qubits.

Most specifically, in the two qubit case, you could eliminate the situations where they disagree (states $|01\rangle$ and $|10\rangle$), and make it a 50/50 chance between the states of them both being on, or both being off.

Now, they have NO CHANCE of disagreeing when you read their values.

Who knows how reality "implements" this situation, since distance doesn't matter, but there you go... it's pretty darn simple.

A very simple 3 qubit entanglement would be this: $|000\rangle$, or this: $1/\sqrt{2}(|000\rangle + |111\rangle)$

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  • $\begingroup$ The state $|000\rangle$ is not entangled, but the other one is. $\endgroup$ – user51117 May 26 '16 at 22:29
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Shor gave a pretty good and short introduction to quantum computation in his 1996 paper, Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer.

Consider a system with n components, each of which can have two states. Whereas in classical physics, a complete description of the state of this system requires only n bits, in quantum physics, a complete description of the state of this system requires $2^n − 1$ complex numbers. To be more precise, the state of the quantum system is a point in a $2^n$-dimensional vector space. For each of the $2^n$ possible classical positions of the components, there is a basis state of this vector space which we represent, for example, by $|011 · · · 0\rangle$ meaning that the first bit is 0, the second bit is 1, and so on. Here, the ket notation $|x\rangle$ means that x is a (pure) quantum state.

The idea of quantum computation is that it is done on qubits that are entangled (hence the state of the quantum system is a point in a $2^n$-dimensional vector space).

You'd probably also find Richard Jozsa's 1997 paper, Entanglement and Quantum Computation interesting:

Abstract: We argue that entanglement is the essential non-classical ingredient which provides the computational speed-up in quantum algorithms as compared to algorithms based on the processes of classical physics.

Here is an extract from it

All of the quantum algorithms utilise the process of computation by quantum parallelism [8]. This refers to a quantum computer’s capability to carry out many computations simultaneously in superposition if the input is set up in a suitable superposition of classically distinct inputs. Thus one might conclude that it is superposition that is at the root of the quantum computational speedup. However closer examination will show that entanglement is the essential feature rather than just superposition itself. Note that entanglement may be viewed as a special kind of superposition – superposition in the presence of a product structure on the state space – which arises from the system being made up of several subsystems. In our considerations these are the qubits comprising the computer. An entangled state is then a superposition of product states which cannot be expressed as a single product state.

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    $\begingroup$ Why does this answer the OP's question? $\endgroup$ – Ran G. Oct 21 '15 at 20:29
  • $\begingroup$ @RanG. I saw that OP has answered his own question so I am just writing here to provide more info regarding it. $\endgroup$ – 吖奇说 Oct 21 '15 at 20:37
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Yes, you can entangle any number of qubits. The classic example is the generalised GHZ (Greenberger–Horne–Zeilinger) state which is given by: $$\frac{|0^n \rangle + |1^n\rangle}{\sqrt{2}}$$

As for applications, an interesting application is in Quantum Secret Sharing.

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