The problem asks us to find the probability that the spin component will be "up" along the Y-direction for the given quantum state $|\psi_3\rangle$.

Step-by-Step Solution

1. Quantum State of $|\psi_3\rangle$

We are given the state $|\psi_3\rangle$ as:

$|\psi_3\rangle = |+\rangle - 2 e^{-i\frac{\pi}{4}} |-\rangle$

This is a superposition of the eigenstates $|+\rangle$ and $|-\rangle$, which are typically associated with the spin component along the Z-direction.

2. Transformation to Y-basis

To find the probability for the spin component to be "up" along the Y-direction, we need to express $|\psi_3\rangle$ in terms of the spin-up and spin-down states along the Y-axis, denoted $|+\rangle_y$ and $|-\rangle_y$.

The Y-basis states can be written in terms of the Z-basis states $|+\rangle$ and $|-\rangle$ as:

$|+\rangle_y = \frac{1}{\sqrt{2}} (|+\rangle + i |-\rangle)$ $|-\rangle_y = \frac{1}{\sqrt{2}} (|+\rangle - i |-\rangle)$

3. Expanding $|\psi_3\rangle$ in Terms of $|+\rangle_y$ and $|-\rangle_y$

Now, let's express $|\psi_3\rangle$ in the Y-basis.

$|\psi_3\rangle = |+\rangle - 2 e^{-i\frac{\pi}{4}} |-\rangle$

Substitute the Z-basis states in terms of the Y-basis states:

[ |\psi_3\rangle = \left[ \frac{1}{\sqrt{2}} (|+\rangle_y + |-\rangle_y) \right] - 2 e^{-i\frac{\pi}{4}} \left[ \frac{1}{\sqrt{2}} (|+\rangle_y - i |-\rangle_y) \right] ]

Simplify the expression by distributing and combining like terms:

$|\psi_3\rangle = \frac{1}{\sqrt{2}} |+\rangle_y + \frac{1}{\sqrt{2}} |-\rangle_y - \frac{2 e^{-i\frac{\pi}{4}}}{\sqrt{2}} |+\rangle_y + \frac{2 i e^{-i\frac{\pi}{4}}}{\sqrt{2}} |-\rangle_y$

Now combine the terms with $|+\rangle_y$ and $|-\rangle_y$:

$|\psi_3\rangle = \frac{1 - 2 e^{-i\frac{\pi}{4}}}{\sqrt{2}} |+\rangle_y + \frac{1 + 2 i e^{-i\frac{\pi}{4}}}{\sqrt{2}} |-\rangle_y$

4. Probability for Spin "Up" Along Y-axis

The probability for the spin to be "up" along the Y-axis is the square of the amplitude of the $|+\rangle_y$ term. The amplitude of $|+\rangle_y$ is:

$\frac{1 - 2 e^{-i\frac{\pi}{4}}}{\sqrt{2}}$

Now, compute the probability:

$P_{\text{up}, y} = \left| \frac{1 - 2 e^{-i\frac{\pi}{4}}}{\sqrt{2}} \right|^2$

First, calculate $1 - 2 e^{-i\frac{\pi}{4}}$:

$e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$

$1 - 2 \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = 1 - \sqrt{2} + i \sqrt{2}$

Now, calculate the magnitude squared of this expression:

$\left| 1 - \sqrt{2} + i \sqrt{2} \right|^2 = (1 - \sqrt{2})^2 + (\sqrt{2})^2$

$= (1 - \sqrt{2})^2 + 2 = (1 - 2\sqrt{2} + 2) + 2 = 5 - 2\sqrt{2}$

Finally, the probability is:

$P_{\text{up}, y} = \frac{5 - 2\sqrt{2}}{2}$

Thus, the probability for the spin component to be "up" along the Y-direction is:

$P_{\text{up}, y} = \frac{5 - 2\sqrt{2}}{2}$

Do you need further details or clarifications?

Relative Questions:

1. What is the significance of the Y-axis in quantum spin measurements?
2. How can one interpret complex coefficients in quantum superposition?
3. Why are basis transformations important in quantum mechanics?
4. What role do Z-basis states play when measuring along different axes?
5. How do probabilities arise from amplitudes in quantum mechanics?

Tip:

In quantum mechanics, always check that the total probability sums to 1 when calculating probabilities in a given basis.