- #1

maverick280857

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Hi,

I came across the following expression in Landau and Lifgarbagez's Quantum Mechanics (Non-relativistic Theory) book:

[tex]

\left(\cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)R_{nl}(r)Y_{l0}(\theta,\phi) = -\frac{i(l+1)}{\sqrt{4(l+1)^2-1}}\left[R_{nl}'-\frac{l}{r}R_{nl}\right]Y_{l-1,0} + \frac{i l}{\sqrt{4 l^2-1}}\left[R_{nl}' + \frac{l+1}{r}R_{nl}\right]Y_{l-1,0}

[/tex]

I don't see how this can be proved...is it using some cool recurrence relation or something, because I don't get it if I write

[tex]

\frac{\partial}{\partial \theta} = \frac{1}{2\hbar}(e^{-i\phi} \hat{L}_{+} - e^{i\phi}\hat{L}_{-})

[/tex]

Any [itex]\partial/\partial\theta[/itex] can only change the

How does the l value change, and the m value remain constant, as an outcome of the particular differential operator [itex]\partial/\partial z[/itex]?

Thanks in advance.

I came across the following expression in Landau and Lifgarbagez's Quantum Mechanics (Non-relativistic Theory) book:

[tex]

\left(\cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)R_{nl}(r)Y_{l0}(\theta,\phi) = -\frac{i(l+1)}{\sqrt{4(l+1)^2-1}}\left[R_{nl}'-\frac{l}{r}R_{nl}\right]Y_{l-1,0} + \frac{i l}{\sqrt{4 l^2-1}}\left[R_{nl}' + \frac{l+1}{r}R_{nl}\right]Y_{l-1,0}

[/tex]

I don't see how this can be proved...is it using some cool recurrence relation or something, because I don't get it if I write

[tex]

\frac{\partial}{\partial \theta} = \frac{1}{2\hbar}(e^{-i\phi} \hat{L}_{+} - e^{i\phi}\hat{L}_{-})

[/tex]

Any [itex]\partial/\partial\theta[/itex] can only change the

**m**value, but not the**l**value, because it involves [itex]\hat{L}_{\pm}[/itex].How does the l value change, and the m value remain constant, as an outcome of the particular differential operator [itex]\partial/\partial z[/itex]?

Thanks in advance.

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