Chapter 7: Q42E (page 281)

In Appendix G. the operator for the square of the angular momentum is shown to be
${\hat{L}}^{2} = - h^{2} [c s c θ \frac{\partial}{\partial θ} (s i n θ \frac{\partial}{\partial θ}) + c s c^{2} θ \frac{\partial^{2}}{\partial ϕ^{2}}]$
Use this to rewrite equation (7-19) as ${\hat{L}}^{2} ⚇ Φ = - C h^{2} ⚇ Φ$

Short Answer

Expert verified

${\hat{L}}^{2} θ Φ = - C h^{2} θ Φ$

Step by step solution

Given data

The angular momentum operator is given as:

${\hat{L}}^{2} = - h^{2} [c s c θ \frac{\partial}{\partial θ} (s i n θ \frac{\partial}{\partial θ}) + c s c^{2} θ \frac{\partial^{2}}{\partial ϕ^{2}}]$

Where, $h$ is the reduced Planck’s constant.

Calculation

The separation of the variable of the above equation can be written as:

$(c s c θ \frac{\partial}{\partial θ} (s i n θ \frac{\partial}{\partial θ}) + c s c^{2} θ \frac{\partial^{2}}{\partial ϕ^{2}}) Θ Φ = C Θ Φ (c s c θ \frac{\partial}{\partial θ} (s i n θ \frac{\partial Θ Φ}{\partial θ}) + c s c^{2} θ \frac{\partial^{2} Θ Φ}{\partial ϕ^{2}}) = C Θ Φ \frac{1}{Θ} c s c θ \frac{\partial}{\partial θ} (s i n θ \frac{\partial Θ}{\partial θ}) + c s c^{2} θ \frac{1}{Φ} \frac{\partial^{2} Φ}{\partial ϕ^{2}} = C$

Now we can write the operator as:

${\hat{L}}^{2} = - h^{2} [c s c θ \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) c s c^{2} θ \frac{\partial^{2}}{\partial ϕ^{2}}] {\hat{L}}^{2} ⊝ Φ = - h^{2} [c s c θ \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) c s c^{2} θ \frac{\partial^{2}}{\partial ϕ^{2}}] ⊝ Φ = - h^{2} Φ c s c θ \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) - h^{2} Φ c s c^{2} θ \frac{\partial^{2} Φ}{\partial ϕ^{2}} = - h^{2} ⊝ Φ \frac{1}{⊝} c s c θ \frac{\partial}{\partial θ} (\sin θ \frac{\partial Φ}{\partial θ}) c s c^{2} θ \frac{1}{Φ} \frac{\partial^{2} Φ}{\partial ϕ^{2}} {\hat{L}}^{2} ⊝ Φ = - C h^{2} ⊝ Φ$

Conclusion

Thus, it can be written as ${\hat{L}}^{2} ⊝ Φ = - C h^{2} ⊝ Φ$ .

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In Appendix G. the operator for the square of the angular momentum is shown to be L^2=-h2[cscθ∂∂θsinθ∂∂θ+csc2θ∂2∂ϕ2]Use this to rewrite equation (7-19) asL^2⚇Φ=-Ch2⚇Φ

Short Answer

Step by step solution

Given data

Calculation

Conclusion

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Most popular questions from this chapter

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Geometrical and Physical Optics

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Study anywhere. Anytime. Across all devices.