Commutation Problem

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

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

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

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

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

sol:

(a)

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

(b)

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

(c)

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

Now, consider

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

So

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

and so

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

(d)

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