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| Let <math> f(x) \!</math> be a differentiable function, using <math>[x,p_{x}]=i\hbar</math>, prove:
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| (a) <math>[x,p^{2}_{x}f(x) ]=2i\hbar p_{x} f(x)</math>
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| (b) <math>[x,p_{x}f(x)p_{x}]=i\hbar[f(x)p_{x}+p_{x}f(x)]</math>
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| (c) <math>[p_{x},p^{2}_{x}f(x)]=-i\hbar p^{2}_{x}\frac{df}{dx}</math>
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| (d) <math>[p_{x},p_{x}f(x)p_{x}]=-i\hbar p_{x}\frac{df}{dx}p_{x}</math>
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| sol:
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| (a) | | (a) |
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| \begin{align} | | \begin{align} |
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| &[x,p^{2}_{x}f(x)] \\ | | &[\hat{x},\hat{p}^{2}_{x}f(\hat{x})] \\ |
| &=[x,p_{x}]p_{x}f(x)+p_{x}[x,p_{x}f(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 p_{x}f(x) + p^{2}_{x}[x,f(x)] + p_{x}[x,p_{x}]f(x) \\ | | &=i\hbar \hat{p}_{x}f(\hat{x}) + \hat{p}^{2}_{x}[\hat{x},f(\hat{x})] + \hat{p}_{x}[\hat{x},\hat{p}_{x}]f(\hat{x}) \\ |
| &=i\hbar p_{x}f(x)+ i\hbar p_{x}f(x) \\ | | &=i\hbar \hat{p}_{x}f(\hat{x})+ i\hbar \hat{p}_{x}f(\hat{x}) \\ |
| &=2i\hbar p_{x}f(x) | | &=2i\hbar \hat{p}_{x}f(\hat{x}) |
| \end{align} | | \end{align} |
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| \begin{align} | | \begin{align} |
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| &[x,p_{x}f(x)p_{x}] \\ | | &[\hat{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}] \\ |
| &=[x,p_{x}]f(x)p_{x}+p_{x}[x,f(x)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(x)p_{x} + p_{x}[x,p_{x}]f(x) + p_{x}[x,f(x)]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(x)p_{x}+p_{x}f(x)] | | &=i\hbar [f(\hat{x})\hat{p}_{x}+\hat{p}_{x}f(\hat{x})] |
| \end{align} | | \end{align} |
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| \begin{align} | | \begin{align} |
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| &[p_{x},p^{2}_{x}f(x)] \\ | | &[\hat{p}_{x},\hat{p}^{2}_{x}f(\hat{x})] \\ |
| &=[p_{x},p^{2}_{x}]f(x)+p^{2}_{x}[p_{x},f(x)] \\ | | &=[\hat{p}_{x},\hat{p}^{2}_{x}]f(\hat{x})+\hat{p}^{2}_{x}[\hat{p}_{x},f(\hat{x})] \\ |
| &= p^{2}_{x} [p_{x},f(x)] | | &= \hat{p}^{2}_{x} [\hat{p}_{x},f(\hat{x})] |
| \end{align} | | \end{align} |
| </math> | | </math> |
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| \begin{align} | | \begin{align} |
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| &[p_{x},f(x)]\psi(x) \\ | | &[\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{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)-i\hbar f(x)\frac{d\psi(x)}{dx} +i\hbar f(x)\frac{d\psi(x)}{dx} \\ |
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| </math> | | </math> |
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| So | | So |
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| <math>[p_{x},f(x)] | | <math>[\hat{p}_{x},f(\hat{x})] =-i\hbar \frac{df(\hat{x})}{dx} |
| =-i\hbar \frac{df}{dx} | |
| </math> | | </math> |
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| and so | | and so |
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| <math>[p_{x},p^{2}_{x}f(x)] | | <math>[\hat{p}_{x},\hat{p}^{2}_{x}f(\hat{x})] =-i\hbar |
| =-i\hbar p^{2}_{x}\frac{df}{dx} | | \hat{p}^{2}_{x}\frac{df(\hat{x})}{dx} </math> |
| </math> | |
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| \begin{align} | | \begin{align} |
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| &[p_{x},p_{x}f(x)p_{x}] \\ | | &[\hat{p}_{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}] \\ |
| &=p_{x}f[p_{x},p_{x}]+[p_{x},p_{x}f]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} \\ |
| &=p_{x}[p_{x},f]p_{x}+[p_{x},p_{x}]fp_{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 p_{x}\frac{df}{dx}p_{x} | | &=-i\hbar \hat{p}_{x}\frac{df(\hat{x})}{dx}\hat{p}_{x} |
| \end{align} | | \end{align} |
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| </math> | | </math> |
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| | Back to [[Commutation Relations and Simultaneous Eigenvalues#Problems|Commutation Relations and Simultaneous Eigenvalues]] |
(a)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fada588696df28b5b1031602a0b1efff400deee)
(b)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d763f255cba9b2b7132c1741dc4b039621e6d13)
(c)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c1c777d9c0cc7f3c47e157389c30321089a9904)
Now, consider
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cf0871e4d01d7747449bc159de1050f560c75e8)
So
![{\displaystyle [{\hat {p}}_{x},f({\hat {x}})]=-i\hbar {\frac {df({\hat {x}})}{dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecd7cc0f156b421c1bd4ec06dbb1209de9dc8469)
and so
![{\displaystyle [{\hat {p}}_{x},{\hat {p}}_{x}^{2}f({\hat {x}})]=-i\hbar {\hat {p}}_{x}^{2}{\frac {df({\hat {x}})}{dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/822113e688f98ede92348246d7b048fed4028ad7)
(d)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03114b8d07594651519a8c3feb4a30fc6915e884)
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