Commutation Problem

Let $f(x)\!$ be a differentiable function, using $[{\hat {x}},{\hat {p}}_{x}]=i\hbar$, prove:

(a) $[{\hat {x}},{\hat {p}}_{x}^{2}f({\hat {x}})]=2i\hbar {\hat {p}}_{x}f({\hat {x}})$

(b) $[{\hat {x}},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]=i\hbar [f({\hat {x}}){\hat {p}}_{x}+{\hat {p}}_{x}f({\hat {x}})]$

(c) $[{\hat {p}}_{x},{\hat {p}}_{x}^{2}f({\hat {x}})]=-i\hbar {\hat {p}}_{x}^{2}{\frac {df({\hat {x}})}{dx}}$

(d) $[{\hat {p}}_{x},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]=-i\hbar {\hat {p}}_{x}{\frac {df({\hat {x}})}{dx}}{\hat {p}}_{x}$

sol:

(a)

${\begin{aligned}&[{\hat {x}},{\hat {p}}_{x}^{2}f({\hat {x}})]\\&=[{\hat {x}},{\hat {p}}_{x}]{\hat {p}}_{x}f({\hat {x}})+{\hat {p}}_{x}[{\hat {x}},{\hat {p}}_{x}f({\hat {x}})]\\&=i\hbar {\hat {p}}_{x}f({\hat {x}})+{\hat {p}}_{x}^{2}[{\hat {x}},f({\hat {x}})]+{\hat {p}}_{x}[{\hat {x}},{\hat {p}}_{x}]f({\hat {x}})\\&=i\hbar {\hat {p}}_{x}f({\hat {x}})+i\hbar {\hat {p}}_{x}f({\hat {x}})\\&=2i\hbar {\hat {p}}_{x}f({\hat {x}})\end{aligned}}$

(b)

${\begin{aligned}&[{\hat {x}},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]\\&=[{\hat {x}},{\hat {p}}_{x}]f({\hat {x}}){\hat {p}}_{x}+{\hat {p}}_{x}[{\hat {x}},f({\hat {x}}){\hat {p}}_{x}]\\&=i\hbar f({\hat {x}}){\hat {p}}_{x}+{\hat {p}}_{x}[{\hat {x}},{\hat {p}}_{x}]f({\hat {x}})+{\hat {p}}_{x}[{\hat {x}},f({\hat {x}})]{\hat {p}}_{x}\\&=i\hbar [f({\hat {x}}){\hat {p}}_{x}+{\hat {p}}_{x}f({\hat {x}})]\end{aligned}}$

(c)

${\begin{aligned}&[{\hat {p}}_{x},{\hat {p}}_{x}^{2}f({\hat {x}})]\\&=[{\hat {p}}_{x},{\hat {p}}_{x}^{2}]f({\hat {x}})+{\hat {p}}_{x}^{2}[{\hat {p}}_{x},f({\hat {x}})]\\&={\hat {p}}_{x}^{2}[{\hat {p}}_{x},f({\hat {x}})]\end{aligned}}$

Now, consider

${\begin{aligned}&[{\hat {p}}_{x},f({\hat {x}})]\psi (x)\\&=-i\hbar {\frac {d}{dx}}(f(x)\psi (x))+i\hbar f(x){\frac {d\psi (x)}{dx}}\\&=-i\hbar {\frac {df}{dx}}\psi (x)-i\hbar f(x){\frac {d\psi (x)}{dx}}+i\hbar f(x){\frac {d\psi (x)}{dx}}\\&=-i\hbar {\frac {df}{dx}}\psi (x)\end{aligned}}$

So

$[{\hat {p}}_{x},f({\hat {x}})]=-i\hbar {\frac {df({\hat {x}})}{dx}}$

and so

$[{\hat {p}}_{x},{\hat {p}}_{x}^{2}f({\hat {x}})]=-i\hbar {\hat {p}}_{x}^{2}{\frac {df({\hat {x}})}{dx}}$

(d)

${\begin{aligned}&[{\hat {p}}_{x},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]\\&={\hat {p}}_{x}f({\hat {x}})[{\hat {p}}_{x},{\hat {p}}_{x}]+[{\hat {p}}_{x},{\hat {p}}_{x}f({\hat {x}})]{\hat {p}}_{x}\\&={\hat {p}}_{x}[{\hat {p}}_{x},f({\hat {x}})]{\hat {p}}_{x}+[{\hat {p}}_{x},{\hat {p}}_{x}]f({\hat {x}}){\hat {p}}_{x}\\&=-i\hbar {\hat {p}}_{x}{\frac {df({\hat {x}})}{dx}}{\hat {p}}_{x}\end{aligned}}$