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--- functions_inverses_and_unitary_operators [2021/03/01 20:55] – created admin
+++ functions_inverses_and_unitary_operators [2022/10/06 01:05] (current) – [Interaction of Functions with Commutators] admin
@@ Line 5: / Line 5: @@
 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
 \[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n.\]
 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 =====
@@ Line 15: / Line 15: @@
 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 $f$.
-  * 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 =====
@@ Line 28: / Line 28: @@
 ===== The Exponential Function =====
+The //**exponential function**// is an operator function that is used a lot in quantum mechanics.  It is defined via its power series as
+\[e^{a\hat{A}} = \sum_{n=0}^{\infty} \frac{a^n}{n!} \hat{A}^n = \hat{I} + a\hat{A} + \frac{a^2}{2}\hat{A}^2 + \cdots.\]
+Just like its classical counterpart, you can prove via the binomial theorem that this is equivalent to
+\[e^{a\hat{A}} = \lim_{n\rightarrow \infty} \left ( \hat{I} + \frac{a}{n} \hat{A}\right )^n\]
+If $\hat{A}$ and $\hat{B}$ commute then the usual rules for multiplying exponentials apply
+\[e^{a\hat{A}}e^{b\hat{B}} = e^{a\hat{A} + b\hat{B}},\]
+but in general this is not true.  Instead
+\[e^{a\hat{A}}e^{b\hat{B}} = e^{a\hat{A} + b\hat{B}} e^{ab[\hat{A},\hat{B}]/2},\]
+which can be proved by a somewhat laborious process of multiplying together the power series and collecting terms.
+As a special case of the rules for commutators, if $[\hat{A},\hat{B}] = 0$ then we will have $[e^{a\hat{A}},\hat{B}] = [\hat{A},e^{b\hat{B}}] = [e^{a\hat{A}},e^{b\hat{B}}] = 0$, but these will not be true if $\hat{A}$ and $\hat{B}$ do not commute.
+The Hermitian adjoint of an exponential can be taken as follows.
+\[\left ( e^{a\hat{A}}\right )^{\dagger} = \sum_{n=0}^{\infty} \frac{\left (a^*\right )^n}{n!} \left ( \hat{A}^{\dagger}\right )^n = e^{a^*\hat{A}^{\dagger}},\]
+i.e. you just take the Hermitian adjoint of the term inside the exponential.
+A special case of this is that if $\hat{A}$ is Hermitian then $\left ( e^{\hat{A}} \right )^{\dagger} = e^{\hat{A}}$, so $e^{\hat{A}}$ is Hermitian as well.  However, for the operator $e^{i\hat{A}}$, which is the type of operator that describes time evolution in quantum mechanics, we have $\left ( e^{i\hat{A}} \right )^{\dagger} = e^{-i\hat{A}}$, so this is not a Hermitian operator.
+====== Inverse Operators ======
+If it exists, the //**inverse**// $\hat{A}^{-1}$ of an operator $\hat{A}$ is defined by
+\[\hat{A}^{-1}\hat{A} = \hat{A}\hat{A}^{-1} = \hat{I}.\]
+Note that we //do not// use the notation $\frac{\hat{A}}{\hat{B}}$ for $\hat{A}\hat{B}^{-1}$ because, in general $\hat{A}\hat{B}^{-1} \neq \hat{B}^{-1}\hat{A}$, so $\frac{\hat{A}}{\hat{B}}$ would have no unique meaning.
+Taking inverses reverses the order of products and commutes with taking powers, i.e.
+\begin{align*}
+ \left ( \hat{A}\hat{B}\hat{C}\cdots\right )^{-1} & = \cdots\hat{C}^{-1}\hat{B}^{-1}\hat{A}^{-1}, & \left ( \hat{A}^{-1} \right )^{n} & = \left ( \hat{A}^n \right )^{-1}.
+\end{align*}
+You will prove the first of these in an in class activity.
+====== Unitary Operators ======
+An operator is //**unitary**// if $\hat{U}^{\dagger} = \hat{U}^{-1}$ or, equivalently $\hat{U}\hat{U}^{\dagger} = \hat{U}^{\dagger}\hat{U} = \hat{I}$.
+If $\hat{U}$ and $\hat{V}$ are unitary then so is their product $\hat{U}\hat{V}$, since
+\[\left ( \hat{U}\hat{V} \right )^{\dagger} \left ( \hat{U}\hat{V} \right ) = \hat{V}^{\dagger} \hat{U}^{\dagger} \hat{U}\hat{V} = \hat{V}^{\dagger} \hat{I} \hat{V} = \hat{V}^{\dagger}\hat{V} = \hat{I}.\]
+More generally, if $\hat{A}, \hat{B}, \hat{C}, \cdots$ are unitary then so is the product $\hat{A}\hat{B}\hat{C}\cdots$.
+If $a$ is real and $\hat{A}$ is Hermitian then $e^{ia\hat{A}}$ is unitary because
+\[\left ( e^{ia\hat{A}}\right )^{\dagger} e^{ia\hat{A}} = e^{-ia\hat{A}}e^{ia\hat{A}} = e^{i(-a\hat{A} + a\hat{A})} = e^{i0} = \hat{I}.\]
+{{:question-mark.png?direct&50|}}
+====== In Class Activities ======
+  - Prove that $\left ( \hat{A} \hat{B} \right )^{-1} = \hat{B}^{-1}\hat{A}^{-1}$