Bell State vs. Bell Measurement

Question

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.

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 $|3_{4}\rangle$

Answer 2

Bell state

A Bell state is a two-qubit state with maximal entanglement, sort of like two flipping coins that are 100% correlated. For example, two photons produced by parametric down-conversion are a Bell pair, and together the pair are in a Bell state.

Bell basis

The Concept of Basis Set

When you measure two qubits, you can get four possible measurements, so any 2-qubit state is described by a 4-vector. Any complete basis set for this space will have four basis vectors.

The Calculation Basis Set

In the calculation basis, these four basis vectors can each be written as integer kets, binary kets, or vectors:

$|0\rangle=|00\rangle= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, $|1\rangle=|01\rangle= \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $|2\rangle=|10\rangle= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, and $|3\rangle=|11\rangle= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$.

Although that particular basis set has four basis vectors that are unentangled states, there are infinitely many basis sets.

The Bell Basis Set

In the Bell basis, the basis vectors (the four Bell states) are each maximally entangled:

|$\Psi^\pm\rangle=\frac{1}{\sqrt 2}(|00\rangle\pm|11\rangle)=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 0 \\ 0 \\ \pm1 \end{bmatrix}$ and |$\Phi^\pm\rangle=\frac{1}{\sqrt 2}(|01\rangle\pm|10\rangle)=\frac{1}{\sqrt{2}} \begin{bmatrix} 0\\1 \\ \pm1 \\0\end{bmatrix}$.

Bell measurement

Quantum Measurement

Any two-qubit state can be expressed in any basis. A measurement in a particular basis has eigenvectors in that set, and a measurement will observe the eigenvalue of one of the basis vectors and leave the system in that eigenvector. The squared amplitudes of the basis vectors give the probabilities of measuring that basis vector's eigenvalue.

Measurement in the z Basis

For example if the state

$|\psi\rangle = 0.6 |0\rangle + 0.8 |3\rangle = \begin{bmatrix}0.6\\0 \\0 \\0.8 \end{bmatrix}$

is measured in the calculation basis, there is a $0.6^2$=36% probability of measuring 0 and 64% probability of measuring 3.

Measurement in the Bell Basis

That same state can be expressed in the Bell basis:

$|\phi\rangle = 0.6\cdot\frac{1}{\sqrt 2}(\Psi^++\Psi^-) + 0.8\cdot\frac{1}{\sqrt 2}(\Psi^+-\Psi^-) = 0.99\Psi^+ + 0.14\Psi^-$ with probabilities of 98% and 2% to measure $\Psi^+$ and $\Psi^-$ respectively

In practice

The usual way to measure in the Bell basis is to transform the basis set to calculation basis. Because Hadamard and CNOT are their own inverses, the four Bell states map to the four calculation basis states if H is applied to the conditional qubit after a CNOT is applied.

If the two qubit to be projected onto a Bell state are not in the same location, CNOT is not practical, but an ebit (the entanglement resource of a Bell pair, distributed to the two locations) can be consumed to make the measurement following https://arxiv.org/abs/1612.08578.