# Comutators Identities

3. Prove the following identity

a) the comutator of two Hermitian operators is anti-Hermitian.

b) product of comutators $[AB, CD]=-AC\{D,B\}+A\{C.B\}D+-C\{D,A\}B+\{C,A\}DB$

a)

$A$ is hermitic if $A=A^\dagger$  ;   $A$ is antihermitic if $A=-A^\dagger$
$-[QR]=[RQ]=RQ-QR=R^\dagger Q^\dagger-Q^\dagger R^\dagger=(QR-RQ)^\dagger=[QR]^\dagger$  so $[QR]$  is antihemitic

We use above and the fact that $R$ and $Q$ are hermitic so that $R=R^\dagger$   and $Q=Q^\dagger$

b)

by definition $[AB]=AB-BA$  and $\{AB\}=AB+BA$

Therefore

$[AB,CD]=ABCD-CDAB$

$-AC\{DB\}+A\{CB\}D-C\{DA\}B+\{CA\}DB=$

$=-ACDB-ACBD+ACBD+ABCD-CDAB-CADB+CADB+ACBD=$

$=ABCD-CDAB$

6. Prove the Schwarz inequality i.e. $(<\alpha||\alpha>*<\beta||\beta>) \geq |<\alpha||\beta>|^2$

Schwartz inequality

$<a|a><b|b> ≥|<a|b>|^2$

Take

$|c>=|a>+λ*|b>$

$0 ≤ <c|c>=(<a|+λ^* |b>)(|a>+λ|b>)=<a|a>+λ<a|b>+λ^*<b|a>+|λ|^2<b|b>$

The expression above is minimum with respect to λ when its derivative is zero. This means

$λ=-(<b|a>)/(<b|b>)$

Place this value of λ in the expression above and one obtains the equation

$<a|a><b|b>-|<a|b>|^2≥0$    so that  $<a|a><b|b> ≥ |<a,b>|^2$

7. Prove that
a) $[\Delta A,\Delta B]=[A,B]$

b) If $A$ and $B$ are Hermitian $\{A,B\}^\dagger=\{A,B\}$

a)

$[ΔA,ΔB]=[(A-<A>,B-<B>]=$

$=(A-<A>)(B-<B>)-(B-<B>)(A-<A>)=$

$=(AB-A<B>-<A>B+<A><B>)-$

$-(BA-B<A>-<B>A+<B><A>)=$

$=AB-A<B>-<A>B+<A><B>-$

$-BA+<A>B+A<B>-<A><B> =$

$=AB-BA=[A,B]$

b)

$\{AB\}^\dagger=(AB+BA)^\dagger=(AB)^\dagger+ (BA)^\dagger=B^\dagger A^\dagger+A^\dagger B^\dagger=BA+AB=AB+BA=\{AB\}$