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| Let <math> f(x) \!</math> be a differentiable function, using
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| <math>[\hat{x},\hat{p}_{x}]=i\hbar</math>, prove:
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| (a) <math>[\hat{x},\hat{p}^{2}_{x}f(\hat{x}) ]=2i\hbar \hat{p}_{x}
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| f(\hat{x})</math>
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| (b)
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| <math>[\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})]</math>
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| (c) <math>[\hat{p}_{x},\hat{p}^{2}_{x}f(\hat{x})]=-i\hbar
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| \hat{p}^{2}_{x}\frac{df(\hat{x})}{dx}</math>
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| (d) <math>[\hat{p}_{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}]=-i\hbar
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| \hat{p}_{x}\frac{df(\hat{x})}{dx}\hat{p}_{x}</math>
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| sol:
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| (a) | | (a) |
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| </math> | | </math> |
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| | Back to [[Commutation Relations and Simultaneous Eigenvalues#Problems|Commutation Relations and Simultaneous Eigenvalues]] |
Latest revision as of 13:25, 18 January 2014
(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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