What is the difference between a Bell state and a Bell measurement? I am studying quantum computation, and Bell states have been introduced. I understand that Bell states can be prepared using the CNOT gate with various inputs, but it's not clear what a Bell measurement is. The reason I ask is that it came up in discussion without any explanation of what was meant by measurement!

For instance, if I prepare a bell state $ |\Psi_- \rangle$, what can I "measure?" What form do these measurements take? If possible, an answer in terms of quantum teleportation would be super helpful.

Speculation: It seems that perhaps all I can really measure are other, unknown, bell states, but I have general relativity diagrams from lecture that seem to indicate bell measurements can be performed on states of the form $| \psi \rangle = \alpha |0 \rangle + \beta |1 \rangle $. Any additional clarification in this regard is greatly appreciated.


1 Answer 1


The Wikipedia article on Bell measurement is pretty clear.

Consider $2$ qbits. While the most general situation is that these $2$-qbit are part of a more general $n$-qbit state ($n=3$ for teleportation), let us consider here only a simple $2$-qbit state .

A Bell measurement is a joint measurement performed on this $2$-qbit state, which result is always a projection onto a Bell state.

(The results of joint measurements just express correlations between the q-bits)

This is possible because the Bell states are a basis for the space of the $2$-qbit states, that is : any $2$-qbit state $|S\rangle$ could be written :

$|S\rangle = a_+|\Phi^+\rangle + a_-|\Phi-\rangle +b_+|\Psi^+\rangle + b_-|\Psi-\rangle$

If your prepare a $2$-qbit state $|S\rangle$, then, performing a Bell (joint) measurement on this state, will give you , applying the basic principles of Quantum Mechanics, for instance, the probability $|b_-|^2$ to find (or project to) the Bell state $|\Psi^-\rangle $

The case of teleportation is similar, while now, you have a $3$-qbit state, where the first and second qbits are local to a observer Alice, the third qbit being local to a (distant relatively to Alice) observer Bob, the second and third bits being entangled.

You way rewrite this $3$-qbit state as : $|S\rangle = a_+|\Phi^+\rangle |3_{1}\rangle + a_-|\Phi-\rangle|3_{2}\rangle +b_+|\Psi^+\rangle|3_{3}\rangle + b_-|\Psi-\rangle |3_{4}\rangle$,

where the $|3_i\rangle$ states represent the states of the third distant Bob's $q$-bit.

A Bell measurement on the first $2$-qbits of Alice, for instance $|\Psi-\rangle $, will project the third bit into the state $|3_{4}\rangle$


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