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functions_inverses_and_unitary_operators [2021/03/01 22:02] – [The Exponential Function] adminfunctions_inverses_and_unitary_operators [2022/10/06 01:05] (current) – [Interaction of Functions with Commutators] admin
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 and radius of convergence $|z| \leq r$. and radius of convergence $|z| \leq r$.
  
-We define the function $f$ as a function of operators as+We can extend $f$ to be a function on linear operators by defining 
 \[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n.\] \[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n.\]
  
 It is possible to prove that this series converges if It is possible to prove that this series converges if
-\[\sup_{\{\ket{\psi}| \| \psi \| \}} \Abs{\sand{\psi}{\hat{A}}{\psi}} leq r.\]+\[\sup_{\{\ket{\psi}| \| \psi \| = 1 \}} \Abs{\sand{\psi}{\hat{A}}{\psi}} \leq r.\]
  
 ===== Interaction of Functions with Commutators ===== ===== Interaction of Functions with Commutators =====
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 Since $[\hat{A}+\hat{B},\hat{C}] = [\hat{A},\hat{C}] + [\hat{B},\hat{C}]$ and $[\hat{A}^{n},\hat{B}] = 0$ whenever $[\hat{A},\hat{B}] = 0$, the fact that functions are defined in terms of power series means that Since $[\hat{A}+\hat{B},\hat{C}] = [\hat{A},\hat{C}] + [\hat{B},\hat{C}]$ and $[\hat{A}^{n},\hat{B}] = 0$ whenever $[\hat{A},\hat{B}] = 0$, the fact that functions are defined in terms of power series means that
   * If $[\hat{A},\hat{B}] = 0$ then $[f(\hat{A}),\hat{B}] = 0$ for any function $f$.   * If $[\hat{A},\hat{B}] = 0$ then $[f(\hat{A}),\hat{B}] = 0$ for any function $f$.
-  * In particular, since $[\hat{A},\hat{A}] = 0$, we have $[f(\hat{A}),g(\hat{A})] = 0$ for any functions $f$ and $g.+  * In particular, since $[\hat{A},\hat{A}] = 0$, we have $[f(\hat{A}),g(\hat{A})] = 0$ for any functions $f$ and $g$.
  
 ===== Interaction of Functions with Hermitian adjoints ===== ===== Interaction of Functions with Hermitian adjoints =====
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 If $a$ is real and $\hat{A}$ is Hermitian then $e^{ia\hat{A}}$ is unitary because If $a$ is real and $\hat{A}$ is Hermitian then $e^{ia\hat{A}}$ is unitary because
 \[\left ( e^{ia\hat{A}}\right )^{\dagger} e^{ia\hat{A}} = e^{-ia\hat{A}}e^{ia\hat{A}} = e^{i(-a\hat{A} + a\hat{A})} = e^{i0} = \hat{I}.\] \[\left ( e^{ia\hat{A}}\right )^{\dagger} e^{ia\hat{A}} = e^{-ia\hat{A}}e^{ia\hat{A}} = e^{i(-a\hat{A} + a\hat{A})} = e^{i0} = \hat{I}.\]
 +
 +{{:question-mark.png?direct&50|}}
 +====== In Class Activities ======
 +
 +  - Prove that $\left ( \hat{A} \hat{B} \right )^{-1} = \hat{B}^{-1}\hat{A}^{-1}$