Question

(a) Prove the following commutator identity:

$\mathbf{[}\mathbf{A}\mathbf{B}\mathbf{.}\mathbf{C}\mathbf{]}\mathbf{=}\mathit{A}\mathbf{[}\mathbf{B}\mathbf{.}\mathbf{C}\mathbf{]}\mathbf{+}\mathbf{[}\mathbf{A}\mathbf{.}\mathbf{C}\mathbf{]}\mathit{B}$

b) Show that

$\mathbf{[}{\mathbf{x}}^{\mathbf{n}}\mathbf{,}\mathbf{p}\mathbf{]}\mathbf{=}\mathit{i}\mathit{h}\mathit{n}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$

(c) Show more generally that

$\mathbf{[}\mathbf{f}\left(x\right)\mathbf{,}\mathbf{p}\mathbf{]}\mathbf{=}\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{d}\mathbf{f}}{\mathbf{d}\mathbf{x}}$

for any function$\mathit{f}\left(x\right)$.

Short answer

a) $[\mathrm{AB}.\mathrm{C}]=A[\mathrm{B}.\mathrm{C}]+[\mathrm{A}.\mathrm{C}]B$

b)$[{\mathrm{x}}^{\mathrm{n}},\mathrm{p}]=ihn{x}^{\mathrm{n}-1}$

c) $[\mathrm{f}\left(\mathrm{x}\right),\mathrm{p}]=i\sout{\mathrm{h}}\frac{\mathrm{df}}{\mathrm{dx}}$

Step-by-step solution

Concept used

Commutator of two quantities and is defined:

$\left[A,B\right]=AB-BA$

$\Rightarrow \left[AB,C\right]=ABC-CAB$ ……. (1)

 Prove Commutator quantity

We can add $\left(ACB-ACB\right)$to equation (1):

localid="1658124315043" $\begin{array}{rcl}\left[AB,C\right]& =& ABC-CAB+ACB-ACB\\ & =& A\left(BC-CB\right)+\left(AC-CA\right)B\\ & =& A\left[B,C\right]+\left[A,C\right]B\\ & =& A\left[B,C\right]+\left[A,C\right]B\end{array}$

Use mathematical induction

We use mathematical induction:

Mathematical induction:

$$\begin{gathered} n=1 \\ \left[x,p\right]=i\sout{h} \end{gathered}$$

We assume that following identity is valid for some $n$:

$\left[x^n,p\right]=i\sout{h}n{x}^{n-1}$

Induction step: $n+1$

localid="1658124506145" $\begin{array}{rcl}\left[x^{n+1},p\right]& =& \left[x·x^{n}p\right]\\ & =& \left[x^n,p\right]+\left[x,p\right]x^n\\ & =& x·i\sout{h}n{x}^{n-1}+i\sout{h}{x}^n\\ & =& i\sout{h}n{x}^n+i\sout{h}{x}^n\\ & =& i\sout{h}\left(n+1\right)x^n\end{array}$

To prove the given relation, after commutator have a test function

c)

After commutator have a test function:

$\begin{array}{rcl}\left[f\left(x\right),p\right]\psi \left(x\right)& =& \left[f\left(x\right),-i\sout{h}\frac{d}{dx}\right]\psi \left(x\right)\\ & =& -i\sout{h}\left[f\left(x\right),\frac{d}{dx}\right]\psi \left(x\right)\\ & =& -i\sout{h}\left(f\frac{d\psi }{dx}-\frac{d}{dx}\left(f\left(x\right)\psi \left(x\right)\right)\right)\\ & =& -i\sout{h}\left(f\frac{d\psi }{dx}-f\frac{d\psi }{dx}-\frac{df}{dx}\psi \right)\\ & =& i\sout{h}\frac{df}{dx}\psi .\end{array}$

Since this relation must be valid for any test function $\psi \left(x\right)$, it follows:

$\left[f\left(x\right),p\right]=i\sout{h}\frac{df}{dx}$