Let's solve **part (a)** step by step as requested: --- ### **Step 1: Calculate \( Y_{00}(\theta, \phi) \)** From the formula: \[ Y_{lm}(\theta, \phi) = \sqrt{\frac{2l + 1}{4\pi} \frac{(l-m)!}{(l+m)!}} (-1)^m e^{im\phi} P_{lm}(\cos\theta) \] For \( l = 0 \), \( m = 0 \): - \( (l-m)! = (0-0)! = 1 \) - \( (l+m)! = (0+0)! = 1 \) - \( P_{00}(z) = \frac{1}{2^0 0!}(1-z^2)^{0/2}\frac{d^0}{dz^0}(z^2-1)^0 = 1 \) - \( e^{i0\phi} = 1 \) - \( (-1)^0 = 1 \) So, \[ Y_{00}(\theta, \phi) = \sqrt{\frac{1}{4\pi}} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{\sqrt{4\pi}} \] **Explanation:** We substituted \( l = 0 \) and \( m = 0 \) into the general spherical harmonics formula and simplified all terms. --- ### **Step 2: Calculate \( Y_{1m}(\theta, \phi) \) for \( m = -1, 0, 1 \)** #### **For \( m = 0 \):** - \( (1-0)! = 1! = 1 \) - \( (1+0)! = 1! = 1 \) - \( (-1)^0 = 1 \) - \( e^{i0\phi} = 1 \) - \( P_{10}(z) = z \) (since for \( m = 0 \), \( P_{l0}(z) \) is the Legendre polynomial) So, \[ Y_{10}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos\theta \] #### **For \( m = 1 \):** - \( (1-1)! = 0! = 1 \) - \( (1+1)! = 2! = 2 \) - \( (-1)^1 = -1 \) - \( e^{i\phi} \) - \( P_{11}(z) = -\sqrt{1-z^2} \) (from associated Legendre function properties) So, \[ Y_{11}(\theta, \phi) = \sqrt{\frac{3}{8\pi}} (-1) e^{i\phi} (-\sin\theta) = -\sqrt{\frac{3}{8\pi}} e^{i\phi} \sin\theta \] #### **For \( m = -1 \):** - \( (1-(-1))! = 2! = 2 \) - \( (1+(-1))! = 0! = 1 \) - \( (-1)^{-1} = -1 \) - \( e^{-i\phi} \) - \( P_{1,-1}(z) = (-1)^1 (1-z^2)^{1/2} \) (associated Legendre property) So, \[ Y_{1,-1}(\theta, \phi) = \sqrt{\frac{3}{8\pi}} (-1) e^{-i\phi} \sin\theta = \sqrt{\frac{3}{8\pi}} e^{-i\phi} \sin\theta \] **Explanation:** Each \( Y_{1m} \) is calculated by plugging in the appropriate \( m \) value and using the properties of the associated Legendre functions. --- ## **Final Answers:** \[ \boxed{ \begin{aligned} Y_{00}(\theta, \phi) &= \frac{1}{\sqrt{4\pi}} \\ Y_{10}(\theta, \phi) &= \sqrt{\frac{3}{4\pi}} \cos\theta \\ Y_{11}(\theta, \phi) &= -\sqrt{\frac{3}{8\pi}} e^{i\phi} \sin\theta \\ Y_{1,-1}(\theta, \phi) &= \sqrt{\frac{3}{8\pi}} e^{-i\phi} \sin\theta \end{aligned} } \]

Let's proceed step-by-step as per the problem parts. --- ### **Part (a): Determining \(Y_{l,m}\) for \(m = -1, 0, 1\)** **Step 1:** Find \(Y_{00}(\theta, \phi)\) The general spherical harmonic formula: \[ Y_{lm}(\theta, \phi) = \sqrt{\frac{2l+1}{4\pi} \frac{(l-m)!}{(l+m)!}} \, e^{i m \phi} P_{lm}(\cos\theta) \] For \(l=0, m=0\): - \( (0-0)! = 1 \) - \( (0+0)! = 1 \) - \( P_{00}(\cos\theta) = 1 \) Thus, \[ Y_{00}(\theta, \phi) = \sqrt{\frac{1}{4\pi}} \times 1 \times 1 \times 1 = \frac{1}{\sqrt{4\pi}} \] **Explanation:** Since \(l=0\), the spherical harmonic reduces to a constant, normalized over the sphere. --- **Step 2:** Find \(Y_{1,m}(\theta, \phi)\) for \(m=-1,0,1\) - For \(l=1\): \[ Y_{1m}(\theta, \phi) = \sqrt{\frac{3}{4\pi} \frac{(1 - m)!}{(1 + m)!}} \, e^{i m \phi} P_{1m}(\cos\theta) \] **For \(m=0\):** - \(\displaystyle P_{10}(\cos\theta) = \cos\theta\) - The prefactor: \[ \sqrt{\frac{3}{4\pi} \times \frac{1!}{1!}} = \sqrt{\frac{3}{4\pi}} \] **Result:** \[ Y_{10}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos\theta \] --- **For \(m=1\):** - \(\displaystyle P_{11}(\cos\theta) = -\sqrt{1 - \cos^2\theta} = -\sin\theta \) - Prefactor: \[ \sqrt{\frac{3}{4\pi} \times \frac{0!}{2!}} = \sqrt{\frac{3}{4\pi} \times \frac{1}{2}} = \sqrt{\frac{3}{8\pi}} \] - Include phase factor \(e^{i\phi}\): \[ Y_{11}(\theta, \phi) = - \sqrt{\frac{3}{8\pi}} e^{i\phi} \sin\theta \] **For \(m=-1\):** - \(\displaystyle P_{1,-1}(\cos\theta) = (-1)^{1} \sqrt{1 - \cos^2\theta} = -\sin\theta \) - Prefactor: \[ \sqrt{\frac{3}{4\pi} \times \frac{2!}{0!}} = \sqrt{\frac{3}{4\pi} \times 2} = \sqrt{\frac{6}{4\pi}} = \sqrt{\frac{3}{2\pi}} \] Wait—careful: The associated Legendre functions for negative \(m\) are related by: \[ P_{l,-m}(z) = (-1)^m \frac{(l - m)!}{(l + m)!} P_{l,m}(z) \] But in the standard normalization for spherical harmonics, the \(Y_{l,-m}\) is related via complex conjugation to \(Y_{l,m}\): \[ Y_{l,-m} = (-1)^m Y_{l,m}^* \] Alternatively, directly: \[ Y_{1,-1}(\theta, \phi) = \sqrt{\frac{3}{8\pi}} e^{-i \phi} \sin \theta \] **Final expressions:** \[ \boxed{ \begin{aligned} Y_{00} &= \frac{1}{\sqrt{4\pi}} \\ Y_{10} &= \sqrt{\frac{3}{4\pi}} \cos\theta \\ Y_{1,1} &= - \sqrt{\frac{3}{8\pi}} e^{i\phi} \sin\theta \\ Y_{1,-1} &= \sqrt{\frac{3}{8\pi}} e^{-i\phi} \sin\theta \end{aligned} } \] --- ### **Summary:** | \(l\) | \(m\) | \(Y_{l m}(\theta,\phi)\) | |---|---|---| | 0 | 0 | \(\displaystyle \frac{1}{\sqrt{4\pi}}\) | | 1 | 0 | \(\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta\) | | 1 | 1 | \(\displaystyle -\sqrt{\frac{3}{8\pi}} e^{i\phi} \sin\theta\) | | 1 | -1 | \(\displaystyle \sqrt{\frac{3}{8\pi}} e^{-i\phi} \sin\theta\) | --- **Next steps (not requested here but for completeness):** Part (b) involves orthonormality verification, part (c) involves action of angular momentum operators, and part (d) involves forming orbitals.