Quantum Spin System Basics

This problem concerns a spin $ \frac{1}{2} $ quantum system with two orthogonal states, $ \Psi_1 $ and $ \Psi_2 $. We are given the state $ \Psi_1 $ and need to find the probability of the system being in the spin-up state when it's in state $ \Psi_2 $. The spin-up state is represented by $ |\uparrow \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and the spin-down state by $ |\downarrow \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $.

Deriving the Orthogonal State $ \Psi_2 $

The given state $ \Psi_1 $ can be expressed in vector notation:

$ \Psi_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \sqrt{\frac{2}{3}} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \sqrt{\frac{2}{3}} \end{pmatrix} $

Let the state $ \Psi_2 $ be represented by the vector $ \begin{pmatrix} a \\ b \end{pmatrix} $. Since $ \Psi_1 $ and $ \Psi_2 $ are orthogonal, their inner product must be zero:

$ \langle \Psi_1 | \Psi_2 \rangle = 0 $

Calculating the inner product:

$ \begin{pmatrix} \frac{1}{\sqrt{3}}^* & \sqrt{\frac{2}{3}}^* \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \sqrt{\frac{2}{3}} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \frac{1}{\sqrt{3}} a + \sqrt{\frac{2}{3}} b = 0 $

From this equation, we find the relationship between $ a $ and $ b $: $ a = -\sqrt{2} b $.

For $ \Psi_2 $ to be a valid quantum state, it must be normalized, meaning $ |a|^2 + |b|^2 = 1 $. Substituting $ a = -\sqrt{2} b $:

$ |-\sqrt{2} b|^2 + |b|^2 = 1 $

$ 2|b|^2 + |b|^2 = 1 $

$ 3|b|^2 = 1 \implies |b|^2 = \frac{1}{3} $

Choosing $ b = \frac{1}{\sqrt{3}} $ (the phase is arbitrary and doesn't affect probability), we get $ a = -\sqrt{2} \left( \frac{1}{\sqrt{3}} \right) = -\sqrt{\frac{2}{3}} $.

Thus, a possible representation for $ \Psi_2 $ is:

$ \Psi_2 = \begin{pmatrix} -\sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} $

Calculating Spin-Up Probability in $ \Psi_2 $

The probability of finding the system in the spin-up state ($ |\uparrow \rangle $) when it is in state $ \Psi_2 $ is given by the square of the absolute value of the inner product $ \langle \uparrow | \Psi_2 \rangle $. This is also known as the Born rule.

$ \langle \uparrow | \Psi_2 \rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} -\sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} $

$ \langle \uparrow | \Psi_2 \rangle = (1) \times \left(-\sqrt{\frac{2}{3}}\right) + (0) \times \left(\frac{1}{\sqrt{3}}\right) = -\sqrt{\frac{2}{3}} $

The required probability $ P(\text{spin-up}) $ is the square of this amplitude:

$ P(\text{spin-up}) = |\langle \uparrow | \Psi_2 \rangle|^2 = \left|-\sqrt{\frac{2}{3}}\right|^2 = \frac{2}{3} $

The calculated probability $ \frac{2}{3} \approx 0.6667 $ falls within the specified range of 0.66 to 0.68.