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Diffstat (limited to 'source/know/concept/bell-state/index.md')
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1 files changed, 11 insertions, 9 deletions
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md index fa289de..d4508e6 100644 --- a/source/know/concept/bell-state/index.md +++ b/source/know/concept/bell-state/index.md @@ -24,14 +24,14 @@ $$\begin{aligned} } \end{aligned}$$ -Where e.g. $$\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$$ +Where e.g. $$\ket{0}_A \ket{1}_B \equiv \ket{0}_A \otimes \ket{1}_B$$ 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 $$\ket{\Phi^{+}}$$. +is that all four Bell states are *maximally entangled*. +For brevity, we will only show this for $$\ket{\Phi^{+}}$$ here. Consider the following pure [density operator](/know/concept/density-operator/): $$\begin{aligned} @@ -40,7 +40,8 @@ $$\begin{aligned} &= \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$$ is then calculated as follows: +The *reduced* density operator $$\hat{\rho}_A$$ of qubit $$A$$ +is then calculated like so, using a partial trace: $$\begin{aligned} \hat{\rho}_A @@ -54,12 +55,13 @@ $$\begin{aligned} = \frac{1}{2} \hat{I} \end{aligned}$$ -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 $$B$$. - +The same holds for qubit $$B$$. This result is *maximally mixed*, +therefore $$\ket{\Phi^{+}}$$ is maximally entangled. This means that a measurement of qubit $$A$$ -has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$. +has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$, +or in other words, no useful information can be gathered +from measuring just one of the qubits. + However, due to the entanglement, measuring $$A$$ also has consequences for qubit $$B$$: |