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Question

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VIPSolutions · Accepted Answer

Let's break down the problem step by step:

---

**Step 1: Understand the Integral**

You are asked to compute:

$$
\int_{0}^{2\pi} \int_0^{\pi} |Y_{lm}(\theta, \phi)|^2 \sin\theta \, d\theta \, d\phi
$$

for $ l = 1 $ and all allowed values of $ m $.

---

**Step 2: Properties of Spherical Harmonics**

Spherical harmonics $ Y_{lm}(\theta, \phi) $ are normalized such that:

$$
\int_{0}^{2\pi} \int_0^{\pi} |Y_{lm}(\theta, \phi)|^2 \sin\theta \, d\theta \, d\phi = 1
$$

for any allowed $ l $ and $ m $.

---

**Step 3: Allowed Values of $ m $ for $ l = 1 $**

For $ l = 1 $, $ m = -1, 0, 1 $.

---

**Step 4: Calculation**

For each $ m $ (i.e., $ m = -1, 0, 1 $):

$$
\int_{0}^{2\pi} \int_0^{\pi} |Y_{1m}(\theta, \phi)|^2 \sin\theta \, d\theta \, d\phi = 1
$$

This is **true for all normalized spherical harmonics**.

---

**Step 5: Second Part of the Question**

> Show that we can use $ P_r(r) = r^2 |R_{nl}(r)|^2 $ to compute the radial probability density. Find it for $ n = 2 $, $ l = 1 $ and plot it.

- The total probability density is given by the modulus squared of the wavefunction:
  $$
  |\Psi_{nlm}(r, \theta, \phi)|^2 = |R_{nl}(r)|^2 |Y_{lm}(\theta, \phi)|^2
  $$
- Integrating over all angles (as above) leaves:
  $$
  P_r(r) = r^2 |R_{nl}(r)|^2
  $$

**For $ n = 2, l = 1 $:**
- Use the known hydrogen atom radial function for $ n = 2, l = 1 $:
  $$
  R_{21}(r) = \frac{1}{\sqrt{24 a_0^3}} \left(\frac{r}{a_0}\right) e^{-r/2a_0}
  $$
- The radial probability density is:
  $$
  P_r(r) = r^2 |R_{21}(r)|^2 = r^2 \left[ \frac{1}{\sqrt{24 a_0^3}} \left(\frac{r}{a_0}\right) e^{-r/2a_0} \right]^2
  $$
  $$
  = \frac{1}{24 a_0^3} \left(\frac{r}{a_0}\right)^2 r^2 e^{-r/a_0}
  $$
  $$
  = \frac{1}{24 a_0^5} r^4 e^{-r/a_0}
  $$

---

### **Summary / Final Answers**

1. **Integral value:**  
   $$
   \int_{0}^{2\pi} \int_0^{\pi} |Y_{1m}(\theta, \phi)|^2 \sin\theta \, d\theta \, d\phi = 1
   $$ 
   for each $ m = -1, 0, 1 $.

2. **Radial probability density for $ n=2 $, $ l=1 $:**
   $$
   P_r(r) = \frac{1}{24 a_0^5} r^4 e^{-r/a_0}
   $$

---

**Plot:**  
To plot $ P_r(r) $, use the above formula for various values of $ r $ (in units of $ a_0 $). The curve starts at 0, peaks at some $ r $, and then decays to 0 as $ r \to \infty $.

If you want the explicit plot, let me know!

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