diff options
Diffstat (limited to 'content/know/concept/density-operator/index.pdc')
| -rw-r--r-- | content/know/concept/density-operator/index.pdc | 26 |
1 files changed, 14 insertions, 12 deletions
diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc index 39c2e85..5126f31 100644 --- a/content/know/concept/density-operator/index.pdc +++ b/content/know/concept/density-operator/index.pdc @@ -81,15 +81,17 @@ $$\begin{aligned} This can be used to find out whether a given $\hat{\rho}$ represents a pure or mixed ensemble. -Next, we define the ensemble average $\expval*{\expval*{\hat{L}}}$ -as the mean of the expectation values for states in the ensemble, -which can be calculated like so: +Next, we define the ensemble average $\expval*{\hat{O}}$ +as the mean of the expectation values of $\hat{O}$ for states in the ensemble. +We use the same notation as for the pure expectation value, +since this is only a small extension of the concept to mixed ensembles. +It is calculated like so: $$\begin{aligned} \boxed{ - \expval*{\expval*{\hat{L}}} - = \sum_{n} p_n \matrixel{\Psi_n}{\hat{L}}{\Psi_n} - = \mathrm{Tr}(\hat{L} \hat{\rho}) + \expval*{\hat{O}} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O}}{\Psi_n} + = \mathrm{Tr}(\hat{\rho} \hat{O}) } \end{aligned}$$ @@ -97,13 +99,13 @@ To prove the latter, we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so: $$\begin{aligned} - \mathrm{Tr}(\hat{L} \hat{\rho}) - &= \sum_{j} \matrixel{j}{\hat{L} \hat{\rho}}{j} - = \sum_{j} \sum_{n} p_n \matrixel{j}{\hat{L}}{\Psi_n} \braket{\Psi_n}{j} + \mathrm{Tr}(\hat{\rho} \hat{O}) + &= \sum_{j} \matrixel{j}{\hat{\rho} \hat{O}}{j} + = \sum_{j} \sum_{n} p_n \braket{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j} \\ - &= \sum_{n} \sum_{j} p_n \braket{\Psi_n}{j} \matrixel{j}{\hat{L}}{\Psi_n} - = \sum_{n} p_n \matrixel{\Psi_n}{\hat{I} \hat{L}}{\Psi_n} - = \expval*{\expval*{\hat{L}}} + &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \braket{j}{\Psi_n} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O} \hat{I}}{\Psi_n} + = \expval*{\hat{O}} \end{aligned}$$ In both the pure and mixed cases, |