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+---
+title: "Bell state"
+date: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+In quantum information, the **Bell states** are a set of four two-qubit states
+which are simple and useful examples of [quantum entanglement](/know/concept/quantum-entanglement/).
+They are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \ket{\Phi^{\pm}}
+            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big)
+            \\
+            \ket{\Psi^{\pm}}
+            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where e.g. $\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$
+is the tensor product of qubit $A$ in state $\Ket{0}$ and $B$ in $\Ket{1}$.
+These states form an orthonormal basis for the two-qubit
+[Hilbert space](/know/concept/hilbert-space/).
+
+More importantly, however,
+is that the Bell states are maximally entangled,
+which we prove here for $\ket{\Phi^{+}}$.
+Consider the following pure [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+    \hat{\rho}
+    = \ket{\Phi^{+}} \bra{\Phi^{+}}
+    &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big)
+\end{aligned}$$
+
+The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows:
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    &= \Tr_B(\hat{\rho})
+    = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B
+    \\
+    &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big)
+    \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big)
+    \\
+    &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big)
+    = \frac{1}{2} \hat{I}
+\end{aligned}$$
+
+This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled.
+The same holds for the other three Bell states,
+and is equally true for qubit $B$.
+
+This means that a measurement of qubit $A$
+has a 50-50 chance to yield $\Ket{0}$ or $\Ket{1}$.
+However, due to the entanglement,
+measuring $A$ also has consequences for qubit $B$:
+
+$$\begin{aligned}
+    \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2
+    = \frac{1}{2}
+    \\
+    \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2
+    = 0
+    \\
+    \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2
+    = 0
+    \\
+    \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2
+    = \frac{1}{2}
+\end{aligned}$$
+
+As an example, if $A$ collapses into $\Ket{0}$ due to a measurement,
+then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$,
+even if it was not measured.
+This was a specific example for $\ket{\Phi^{+}}$,
+but analogous results can be found for the other Bell states.
+
+
+## References
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.