The problem asks to find vectors orthogonal to the given quantum states $|\psi_1\rangle$, $|\psi_2\rangle$, and $|\psi_3\rangle$, which are represented as superpositions of the basis kets $|+\rangle$ and $|-\rangle$.

Here are the given states:

$|\psi_1\rangle = \frac{2}{\sqrt{5}} |+\rangle + \frac{1}{\sqrt{5}} |-\rangle$ $|\psi_2\rangle = \frac{1}{\sqrt{3}} |+\rangle + \frac{i}{\sqrt{3}} |-\rangle$ $|\psi_3\rangle = \frac{1}{\sqrt{2}} |+\rangle + \frac{e^{i\pi/4}}{\sqrt{2}} |-\rangle$

Finding Orthogonal Vectors

To find a vector orthogonal to a given state, we can use the condition that the inner product between the two vectors must be zero:

$\langle \psi | \phi \rangle = 0$

Let $|\phi\rangle = a|+\rangle + b|-\rangle$, where we will solve for $a$ and $b$ to satisfy this orthogonality condition.

Step-by-Step for Each State

1. Orthogonal to $|\psi_1\rangle$: $\langle \psi_1 | \phi \rangle = \frac{2}{\sqrt{5}}a + \frac{1}{\sqrt{5}}b = 0$ Simplifying: $2a + b = 0$ This gives $b = -2a$. Therefore, one orthogonal state can be written as: $|\phi_1\rangle = a|+\rangle - 2a|-\rangle$ For normalization, we can take $a = 1$: $|\phi_1\rangle = |+\rangle - 2|-\rangle$

2. Orthogonal to $|\psi_2\rangle$: $\langle \psi_2 | \phi \rangle = \frac{1}{\sqrt{3}}a + \frac{i}{\sqrt{3}}b = 0$ Simplifying: $a + ib = 0$ This gives $b = -ia$. Thus, one orthogonal state can be: $|\phi_2\rangle = a|+\rangle - ia|-\rangle$ Setting $a = 1$ for normalization: $|\phi_2\rangle = |+\rangle - i|-\rangle$

3. Orthogonal to $|\psi_3\rangle$: $\langle \psi_3 | \phi \rangle = \frac{1}{\sqrt{2}}a + \frac{e^{i\pi/4}}{\sqrt{2}}b = 0$ Simplifying: $a + e^{i\pi/4}b = 0$ Therefore, $b = -e^{-i\pi/4}a$. So, the orthogonal state is: $|\phi_3\rangle = a|+\rangle - e^{-i\pi/4}a|-\rangle$ For normalization, taking $a = 1$: $|\phi_3\rangle = |+\rangle - e^{-i\pi/4}|-\rangle$

Conclusion

The orthogonal vectors are:

• For $|\psi_1\rangle$: $|\phi_1\rangle = |+\rangle - 2|-\rangle$
• For $|\psi_2\rangle$: $|\phi_2\rangle = |+\rangle - i|-\rangle$
• For $|\psi_3\rangle$: $|\phi_3\rangle = |+\rangle - e^{-i\pi/4}|-\rangle$

Would you like more details on any of these steps or another related question?

Related Questions:

1. How do you normalize a quantum state?
2. What is the significance of orthogonality in quantum mechanics?
3. How are superposition states constructed in quantum mechanics?
4. Can you explain the inner product and its role in quantum mechanics?
5. What is the geometric interpretation of orthogonal quantum states?

Tip: Orthogonal quantum states correspond to mutually exclusive outcomes in quantum measurements, meaning they have no overlap in probability.