5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $|3_{4}\rangle$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.