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author | Prefetch | 2022-10-20 18:25:31 +0200 |
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committer | Prefetch | 2022-10-20 18:25:31 +0200 |
commit | 16555851b6514a736c5c9d8e73de7da7fc9b6288 (patch) | |
tree | 76b8bfd30f8941d0d85365990bcdbc5d0643cabc /source/know/concept/density-operator | |
parent | e5b9bce79b68a68ddd2e51daa16d2fea73b84fdb (diff) |
Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
Diffstat (limited to 'source/know/concept/density-operator')
-rw-r--r-- | source/know/concept/density-operator/index.md | 46 |
1 files changed, 23 insertions, 23 deletions
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} |