To show that each vector is normalized, we need to verify that their norms (or magnitudes) are equal to 1. For a vector $|v⟩$ to be normalized, it must satisfy the condition:

$\langle v | v \rangle = 1$

Normalization of $|v(1)⟩$

The vector $|v(1)⟩$ is given by:

$|v(1)⟩ = \frac{1}{\sqrt{2}} |1(1)⟩ + \frac{i}{\sqrt{2}} |2(1)⟩$

Now, compute the inner product $\langle v(1) | v(1) \rangle$:

$\langle v(1) | v(1) \rangle = \left( \frac{1}{\sqrt{2}} \langle 1(1)| + \frac{-i}{\sqrt{2}} \langle 2(1)| \right) \left( \frac{1}{\sqrt{2}} |1(1)⟩ + \frac{i}{\sqrt{2}} |2(1)⟩ \right)$

Expanding this product:

$\langle v(1) | v(1) \rangle = \frac{1}{2} \langle 1(1) | 1(1) \rangle + \frac{i}{2} \langle 1(1) | 2(1) \rangle - \frac{i}{2} \langle 2(1) | 1(1) \rangle + \frac{1}{2} \langle 2(1) | 2(1) \rangle$

Since the basis vectors $|1(1)⟩$ and $|2(1)⟩$ are orthonormal, we have:

$\langle 1(1) | 1(1) \rangle = 1, \quad \langle 2(1) | 2(1) \rangle = 1, \quad \langle 1(1) | 2(1) \rangle = \langle 2(1) | 1(1) \rangle = 0$

Thus:

$\langle v(1) | v(1) \rangle = \frac{1}{2} \cdot 1 + 0 + 0 + \frac{1}{2} \cdot 1 = 1$

Hence, $|v(1)⟩$ is normalized.

Normalization of $|v(2)⟩$

The vector $|v(2)⟩$ is given by:

$|v(2)⟩ = \frac{1}{\sqrt{3}} |1(2)⟩ + \frac{\sqrt{2}}{\sqrt{3}} |2(2)⟩$

Now, compute the inner product $\langle v(2) | v(2) \rangle$:

$\langle v(2) | v(2) \rangle = \left( \frac{1}{\sqrt{3}} \langle 1(2)| + \frac{\sqrt{2}}{\sqrt{3}} \langle 2(2)| \right) \left( \frac{1}{\sqrt{3}} |1(2)⟩ + \frac{\sqrt{2}}{\sqrt{3}} |2(2)⟩ \right)$

Expanding this product:

$\langle v(2) | v(2) \rangle = \frac{1}{3} \langle 1(2) | 1(2) \rangle + \frac{\sqrt{2}}{3} \langle 1(2) | 2(2) \rangle + \frac{\sqrt{2}}{3} \langle 2(2) | 1(2) \rangle + \frac{2}{3} \langle 2(2) | 2(2) \rangle$

Again, since the basis vectors $|1(2)⟩$ and $|2(2)⟩$ are orthonormal, we have:

$\langle 1(2) | 1(2) \rangle = 1, \quad \langle 2(2) | 2(2) \rangle = 1, \quad \langle 1(2) | 2(2) \rangle = \langle 2(2) | 1(2) \rangle = 0$

Thus:

$\langle v(2) | v(2) \rangle = \frac{1}{3} \cdot 1 + 0 + 0 + \frac{2}{3} \cdot 1 = 1$

Hence, $|v(2)⟩$ is normalized.

Conclusion:

Both $|v(1)⟩$ and $|v(2)⟩$ are normalized, as their inner products with themselves yield 1.

---

Would you like further clarification or details?

Here are some related questions:

1. What are orthonormal vectors, and how are they used in quantum mechanics?
2. How do inner products work in the context of Hilbert spaces?
3. Why is normalization important for quantum states?
4. How can we generalize this process for vectors with more than two components?
5. What is the physical interpretation of these normalized vectors?

Tip: In quantum mechanics, normalization ensures that the total probability of finding a system in any state sums to 1.