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Diffstat (limited to 'source/know/concept/bell-state')
| -rw-r--r-- | source/know/concept/bell-state/index.md | 24 |
1 files changed, 12 insertions, 12 deletions
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md index 5f333a2..f454264 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. $\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$ -is the tensor product of qubit $A$ in state $\Ket{0}$ and $B$ in $\Ket{1}$. +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 $\ket{\Phi^{+}}$. +which we prove here for $$\ket{\Phi^{+}}$$. Consider the following pure [density operator](/know/concept/density-operator/): $$\begin{aligned} @@ -40,7 +40,7 @@ $$\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$$ is then calculated as follows: $$\begin{aligned} \hat{\rho}_A @@ -54,14 +54,14 @@ $$\begin{aligned} = \frac{1}{2} \hat{I} \end{aligned}$$ -This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled. +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 $B$. +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 $\Ket{1}$. +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 $B$: +measuring $$A$$ also has consequences for qubit $$B$$: $$\begin{aligned} \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 @@ -81,10 +81,10 @@ $$\begin{aligned} = \frac{1}{2} \end{aligned}$$ -As an example, if $A$ collapses into $\Ket{0}$ due to a measurement, -then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$, +As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement, +then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$, even if it was not measured. -This was a specific example for $\ket{\Phi^{+}}$, +This was a specific example for $$\ket{\Phi^{+}}$$, but analogous results can be found for the other Bell states. |