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Question

Show that the operator p^\hat{p}^{\prime} obtained by applying a translation to the operator p^\hat{p} is p^=T^p^T^=p^\hat{p}^{\prime}=\hat{T}^{\dagger} \hat{p} \hat{T}=\hat{p}.

Solution

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Using the definition of p^\hat{p}' and a test function f(x)f(x), we have

p^  f(x)=T^(a)  p^  T^(a)  f(x),\hat{p}'\;f(x) = \hat{T}^\dagger(a)\;\hat{p}\;\hat{T}(a)\;f(x),

and since T^(a)=T^(a)\hat{T}^\dagger(a) = \hat{T}(-a) (Equation 6.4),

p^  f(x)=T^(a)  p^  T^(a)  f(x).\hat{p}'\;f(x) = \hat{T}(-a)\;\hat{p}\;\hat{T}(a)\;f(x).

From Equation 6.1

p^  f(x)=T^(a)(iddx)f(xa).\hat{p}'\;f(x) = \hat{T}(-a)\left(-i\hbar\frac{d}{dx}\right)f(x - a).

Using Equation 6.1 again and the mathematical fact: d(x+a)=dxd(x+a) = dx, we find that

p^  f(x)=iddx  f(x)=p^  f(x).\hat{p}'\;f(x) = -i\hbar\frac{d}{dx}\;f(x) = \hat{p}\;f(x).

Finally we may read off the operator

p^=p^.\hat{p}' = \hat{p}.

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