From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/density-operator/index.md | 46 +++++++++++++--------------
 1 file changed, 23 insertions(+), 23 deletions(-)

(limited to 'source/know/concept/density-operator')

diff --git a/source/know/concept/density-operator/index.md b/source/know/concept/density-operator/index.md
index d2042ef..ffc5444 100644
--- a/source/know/concept/density-operator/index.md
+++ b/source/know/concept/density-operator/index.md
@@ -9,18 +9,18 @@ layout: "concept"
 ---
 
 In quantum mechanics, the expectation value of an observable
-$\expval{\hat{L}}$ represents the average result from measuring
-$\hat{L}$ on a large number of systems (an **ensemble**)
-prepared in the same state $\Ket{\Psi}$,
+$$\expval{\hat{L}}$$ represents the average result from measuring
+$$\hat{L}$$ on a large number of systems (an **ensemble**)
+prepared in the same state $$\Ket{\Psi}$$,
 known as a **pure ensemble** or (somewhat confusingly) **pure state**.
 
 But what if the systems of the ensemble are not all in the same state?
 To work with such a **mixed ensemble** or **mixed state**,
-the **density operator** $\hat{\rho}$ or **density matrix** (in a basis) is useful.
-It is defined as follows, where $p_n$ is the probability
-that the system is in state $\Ket{\Psi_n}$,
+the **density operator** $$\hat{\rho}$$ or **density matrix** (in a basis) is useful.
+It is defined as follows, where $$p_n$$ is the probability
+that the system is in state $$\Ket{\Psi_n}$$,
 i.e. the proportion of systems in the ensemble that are
-in state $\Ket{\Psi_n}$:
+in state $$\Ket{\Psi_n}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -29,18 +29,18 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Do not let is this form fool you into thinking that $\hat{\rho}$ is diagonal:
-$\Ket{\Psi_n}$ need not be basis vectors.
-Instead, the matrix elements of $\hat{\rho}$ are found as usual,
-where $\Ket{j}$ and $\Ket{k}$ are basis vectors:
+Do not let is this form fool you into thinking that $$\hat{\rho}$$ is diagonal:
+$$\Ket{\Psi_n}$$ need not be basis vectors.
+Instead, the matrix elements of $$\hat{\rho}$$ are found as usual,
+where $$\Ket{j}$$ and $$\Ket{k}$$ are basis vectors:
 
 $$\begin{aligned}
     \matrixel{j}{\hat{\rho}}{k}
     = \sum_{n} p_n \Inprod{j}{\Psi_n} \Inprod{\Psi_n}{k}
 \end{aligned}$$
 
-However, from the special case where $\Ket{\Psi_n}$ are indeed basis vectors,
-we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian,
+However, from the special case where $$\Ket{\Psi_n}$$ are indeed basis vectors,
+we can conclude that $$\hat{\rho}$$ is positive semidefinite and Hermitian,
 and that its trace (i.e. the total probability) is 100%:
 
 $$\begin{gathered}
@@ -58,26 +58,26 @@ $$\begin{gathered}
 \end{gathered}$$
 
 These properties are preserved by all changes of basis.
-If the ensemble is purely $\Ket{\Psi}$,
-then $\hat{\rho}$ is given by a single state vector:
+If the ensemble is purely $$\Ket{\Psi}$$,
+then $$\hat{\rho}$$ is given by a single state vector:
 
 $$\begin{aligned}
     \hat{\rho} = \Ket{\Psi} \Bra{\Psi}
 \end{aligned}$$
 
-From the special case where $\Ket{\Psi}$ is a basis vector,
+From the special case where $$\Ket{\Psi}$$ is a basis vector,
 we can conclude that for a pure ensemble,
-$\hat{\rho}$ is idempotent, which means that:
+$$\hat{\rho}$$ is idempotent, which means that:
 
 $$\begin{aligned}
     \hat{\rho}^2 = \hat{\rho}
 \end{aligned}$$
 
-This can be used to find out whether a given $\hat{\rho}$
+This can be used to find out whether a given $$\hat{\rho}$$
 represents a pure or mixed ensemble.
 
-Next, we define the ensemble average $\expval{\hat{O}}$
-as the mean of the expectation values of $\hat{O}$ for states in the ensemble.
+Next, we define the ensemble average $$\expval{\hat{O}}$$
+as the mean of the expectation values of $$\hat{O}$$ for states in the ensemble.
 We use the same notation as for the pure expectation value,
 since this is only a small extension of the concept to mixed ensembles.
 It is calculated like so:
@@ -91,7 +91,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 To prove the latter,
-we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
+we write out the trace $$\mathrm{Tr}$$ as the sum of the diagonal elements, so:
 
 $$\begin{aligned}
     \mathrm{Tr}(\hat{\rho} \hat{O})
@@ -104,7 +104,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 In both the pure and mixed cases,
-if the state probabilities $p_n$ are constant with respect to time,
+if the state probabilities $$p_n$$ are constant with respect to time,
 then the evolution of the ensemble obeys the **Von Neumann equation**:
 
 $$\begin{aligned}
@@ -115,7 +115,7 @@ $$\begin{aligned}
 
 This equivalent to the Schrödinger equation:
 one can be derived from the other.
-We differentiate $\hat{\rho}$ with the product rule,
+We differentiate $$\hat{\rho}$$ with the product rule,
 and then substitute the opposite side of the Schrödinger equation:
 
 $$\begin{aligned}
-- 
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