# Quantum Chemistry

Chapter05 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32

Exercise. This is the solution to exercise 5.1 in the book.

We have that:

$\left[Â,\stackrel{̂}{B}\right]=Â\stackrel{̂}{B}-\stackrel{̂}{B}Â=-\left(\stackrel{̂}{B}Â-Â\stackrel{̂}{B}\right)=-\left[\stackrel{̂}{B},Â\right]$

We have that:

$\left[Â,{Â}^{n}\right]=Â{Â}^{n}-{Â}^{n}Â={Â}^{n+1}-{Â}^{n+1}=0$

We have that:

$\left[kÂ,\stackrel{̂}{B}\right]=kÂ\stackrel{̂}{B}-\stackrel{̂}{B}kÂ=kÂ\stackrel{̂}{B}-k\stackrel{̂}{B}Â=k\left(Â\stackrel{̂}{B}-\stackrel{̂}{B}Â\right)=k\left[Â,\stackrel{̂}{B}\right]$

$\left[Â,k\stackrel{̂}{B}\right]=Âk\stackrel{̂}{B}-k\stackrel{̂}{B}Â=kÂ\stackrel{̂}{B}-k\stackrel{̂}{B}Â=k\left(Â\stackrel{̂}{B}-\stackrel{̂}{B}Â\right)=k\left[Â,\stackrel{̂}{B}\right]$

We have that:

$\left[Â,\stackrel{̂}{B}+Ĉ\right]=Â\left(\stackrel{̂}{B}+Ĉ\right)-\left(\stackrel{̂}{B}+Ĉ\right)Â=Â\stackrel{̂}{B}+ÂĈ-\stackrel{̂}{B}Â-ĈÂ=\left[Â,\stackrel{̂}{B}\right]+\left[Â,Ĉ\right]$

Similarly,

$\left[Â+\stackrel{̂}{B},Ĉ\right]=\left(Â+\stackrel{̂}{B}\right)Ĉ-Ĉ\left(Â+\stackrel{̂}{B}\right)=ÂĈ+\stackrel{̂}{B}Ĉ-ĈÂ-Ĉ\stackrel{̂}{B}=\left[Â,Ĉ\right]+\left[\stackrel{̂}{B},Ĉ\right]$

We have that:

$\left[Â,\stackrel{̂}{B}Ĉ\right]=Â\stackrel{̂}{B}Ĉ-\stackrel{̂}{B}ĈÂ=Â\stackrel{̂}{B}Ĉ-\stackrel{̂}{B}ÂĈ+\stackrel{̂}{B}ÂĈ-\stackrel{̂}{B}ĈÂ=\left[Â,\stackrel{̂}{B}\right]Ĉ+\stackrel{̂}{B}\left[Â,Ĉ\right]$

Similarly,

$\left[Â\stackrel{̂}{B},Ĉ\right]=Â\stackrel{̂}{B}Ĉ-ĈÂ\stackrel{̂}{B}=ÂĈ\stackrel{̂}{B}-ĈÂ\stackrel{̂}{B}+Â\stackrel{̂}{B}Ĉ-ÂĈ\stackrel{̂}{B}=\left[Â,Ĉ\right]\stackrel{̂}{B}+Â\left[\stackrel{̂}{B},Ĉ\right]$