The operators associated with the radial component of the momentum pr and the radial coordinate r are denoted by Pr and R, respectively. Their actions on a radial wave function ψ(r) are given by Pr ψ(r) = -ih (1/r)(∂/∂r)(rψ(r)) and R ψ(r) = rψ(r). (a) Find the commutator [Pr , R] and ΔPr Δr , where

Question

The operators associated with the radial component of the momentum $P_{r}$ and the radial coordinate r are denoted by $\hat{P}_{r}$ and $\hat{R}$ , respectively. Their actions on a radial wave function $\psi (r)$ are given by $\hat{P}_{r}\psi (\vec{r})=-i\hbar (1/r)(∂/∂r)(r\psi (\vec{r}))$ and $\hat{R} \psi (\vec{r})=r\psi (\vec{r})$ .

(a) Find the commutator $[\hat{P}_{r} ,\hat{R}]$ and $\Delta P_{r}\Delta r$ , where $\Delta r=\sqrt{〈\hat{R}^{2} 〉-〈\hat{R}〉^{2} }$ and $\Delta P_{r}\sqrt{〈\hat{P}^{2}_{r}〉-〈\hat{P}_{r} 〉^{2} }$ .

(b) Show that $\hat{P}^{2}_{r}=-(\hbar ^{2} /r)(∂^{2}/∂r^{2})r$ .

(a) Find the commutator [latex][\hat{P}_{r} ,\hat{R}][/latex] and [latex] \Delta P_{r}\Delta r[/latex] , where [latex] \Delta r=\sqrt {〈\hat {R}^{2} 〉-〈\hat{R}〉^{2} }[/latex] and [latex]\Delta P_{r}\sqrt {〈\hat {P}^{2}_{r} 〉-〈\hat{P}_{r} 〉^{2} }[/latex].

(b) Show that [latex]\hat{P}^{2}_{r}=-(\hbar ^{2} /r)(∂^{2}/∂r^{2})r [/latex].

Accepted Answer

(a) Since [latex]\hat{R} \psi (\vec{r})=r\psi (\vec{r})[/latex] and

[latex]\hat{P}_{r}\psi (\vec{r})=-i\hbar \frac{1}{r} \frac{∂}{∂r} (r\psi (\vec{r}))=-i\hbar \frac{1}{r} \psi (\vec{r})-i\hbar \frac{∂\psi (\vec{r})}{∂r}[/latex], (6.269)

and since

[latex]\hat{P}_{r}(\hat{R} \psi (\vec{r}))=-i\hbar \frac{1}{r} \frac{∂}{∂r}\left(r^{2}\psi (\vec{r})\right) =-2i\hbar \psi (\vec{r})-i\hbar r\frac{∂\psi (\vec{r})}{∂r}[/latex], (6.270)

the action of the commutator [latex][\hat{P}_{r} ,\hat{R}][/latex] on a function [latex]\psi (\vec{r})[/latex] is given by

[latex][\hat{P}_{r} ,\hat{R}]\psi (\vec{r})=-i\hbar \left[\frac{1}{r} \frac{∂}{∂r}r,\hat{R}\right] \psi (\vec{r})=-i\hbar \frac{1}{r} \frac{∂}{∂r}\left(r^{2}\psi (\vec{r})\right)+i\hbar \frac{∂}{∂r}(r\psi (\vec{r}))[/latex]

[latex]=-2i\hbar \psi (\vec{r})-i\hbar r\frac{∂\psi (\vec{r})}{∂r}+i \hbar \psi (\vec{r})+i\hbar r\frac{∂\psi (\vec{r})}{∂r}[/latex]

[latex]=-i\hbar \psi (\vec{r})[/latex]. (6.271)

Thus, we have

[latex][\hat{P}_{r} ,\hat{R}]=-i\hbar [/latex]. (6.272)

Using the uncertainty relation for a pair of operators [latex]\hat{A} [/latex] and [latex]\hat{B}[/latex], [latex]\Delta A\Delta B\geq \frac{1}{2} |〈[\hat{A},\hat{B}]〉|[/latex], we can write

[latex]\Delta P_{r}\Delta r \geq \frac{1}{2}\left|\left\langle [\hat {P}_{r} ,\hat{R}]\right\rangle \right| [/latex], (6.273)

or

[latex]\Delta P_{r}\Delta r \geq \frac{\hbar }{2} [/latex]. (6.274)

(b) The action of [latex]\hat{P}^{2}_{r} [/latex] on [latex]\psi (\vec{r})[/latex] gives

[latex]\hat{P}^{2}_{r}\psi (\vec{r})=-\hbar ^{2} \frac{1}{r} \frac{∂}{∂r}\left[r\frac{1}{r} \frac{∂}{∂r}(r\psi )\right] =-\hbar ^{2} \frac{1}{r} \frac{∂^{2} }{∂r^{2} }(r\psi (\vec{r}))[/latex]; (6.275)

hence

[latex]\hat{P}^{2}_{r} =-\hbar ^{2} \frac{1}{r}\frac{∂^{2} }{∂r^{2} }(r)[/latex]. (6.276)

Question 6.P.8: The operators associated with the radial component of the mo...

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