Let f ( x ) {\displaystyle f(x)\!} be a differentiable function, using [ x , p x ] = i ℏ {\displaystyle [x,p_{x}]=i\hbar } , prove:
(a) [ x , p x 2 f ( x ) ] = 2 i ℏ p x f ( x ) {\displaystyle [x,p_{x}^{2}f(x)]=2i\hbar p_{x}f(x)}
(b) [ x , p x f ( x ) p x ] = i ℏ [ f ( x ) p x + p x f ( x ) ] {\displaystyle [x,p_{x}f(x)p_{x}]=i\hbar [f(x)p_{x}+p_{x}f(x)]}
(c) [ p x , p x 2 f ( x ) ] = − i ℏ p x 2 d f d x {\displaystyle [p_{x},p_{x}^{2}f(x)]=-i\hbar p_{x}^{2}{\frac {df}{dx}}}
(d) [ p x , p x f ( x ) p x ] = − i ℏ p x d f d x p x {\displaystyle [p_{x},p_{x}f(x)p_{x}]=-i\hbar p_{x}{\frac {df}{dx}}p_{x}}
sol:
(a)
[ x , p x 2 f ( x ) ] = [ x , p x ] p x f ( x ) + p x [ x , p x f ( x ) ] = i ℏ p x f ( x ) + p x 2 [ x , f ( x ) ] + p x [ x , p x ] f ( x ) = i ℏ p x f ( x ) + i ℏ p x f ( x ) = 2 i ℏ p x f ( x ) {\displaystyle {\begin{aligned}&[x,p_{x}^{2}f(x)]\\&=[x,p_{x}]p_{x}f(x)+p_{x}[x,p_{x}f(x)]\\&=i\hbar p_{x}f(x)+p_{x}^{2}[x,f(x)]+p_{x}[x,p_{x}]f(x)\\&=i\hbar p_{x}f(x)+i\hbar p_{x}f(x)\\&=2i\hbar p_{x}f(x)\end{aligned}}}
(b)
[ x , p x f ( x ) p x ] = [ x , p x ] f ( x ) p x + p x [ x , f ( x ) p x ] = i ℏ f ( x ) p x + p x [ x , p x ] f ( x ) + p x [ x , f ( x ) ] p x = i ℏ [ f ( x ) p x + p x f ( x ) ] {\displaystyle {\begin{aligned}&[x,p_{x}f(x)p_{x}]\\&=[x,p_{x}]f(x)p_{x}+p_{x}[x,f(x)p_{x}]\\&=i\hbar f(x)p_{x}+p_{x}[x,p_{x}]f(x)+p_{x}[x,f(x)]p_{x}\\&=i\hbar [f(x)p_{x}+p_{x}f(x)]\end{aligned}}}