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Question

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

Solution

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

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

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

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

From Equation 6.1

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

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

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

Finally we may read off the operator

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

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