$ \Psi_1 $ and $ \Psi_2 $ are two orthogonal states of a spin $ \frac{1}{2} $ system. It is given that $ \Psi_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \sqrt{\frac{2}{3}} \begin{pmatrix} 0 \\ 1 \end{pmatrix} $ where $ \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $ \begin{pmatrix} 0 \\ 1 \end{pmatrix} $ represent the spin-up and spin-down states, respectively. When the system is in the state $ \Psi_2 $, its probability to be in the spin-up state is ________.
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.
Important Questions from Spin Electron Spin Pauli Matrices
- Consider the Pauli matrices $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$, $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
The value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$ is - The Hamiltonian of two interacting spin-1/2 particles is $H = \frac{A}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$, where $\vec{S}_1$ and $\vec{S}_2$ are the spin angular momenta of particles 1 and 2, respectively. Here, $A = 10.56 \text{ eV}$. The energy in eV required to induce an excitation from the ground state to the excited state (rounded off to two decimal places) is _____
An electron with mass $m$ and charge $q$ is in the spin up state $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ at time $t = 0$. A constant magnetic field is applied along the y-axis, $\vec{B} = B_0 \hat{j}$, where $B_0$ is a constant. The Hamiltonian of the system is $H = -\hbar \omega \sigma_y$, where $\omega = \frac{q B_0}{2m} > 0$ and $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$. The minimum time after which the electron will be in the spin down state along the x-axis, i.e., $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$, is
Consider two non-identical spin $\frac{1}{2}$ particles labelled $1$ and $2$ in the spin product state $|\frac{1}{2}, \frac{1}{2}\rangle_1 |\frac{1}{2}, -\frac{1}{2}\rangle$. The Hamiltonian of the system is
$H = \frac{4\lambda}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$,
where $\vec{S}_1$ and $\vec{S}_2$ are the spin operators of particles $1$ and $2$, respectively, and $\lambda$ is a constant with appropriate dimensions. What is the expectation value of $H$ in the above state?- A spin $\frac{1}{2}$ particle is in a spin up state along the $x$-axis (with unit vector $\hat{x}$) and is denoted as $|\frac{1}{2}, \frac{1}{2}\rangle_x$. What is the probability of finding the particle to be in a spin up state along the direction $\hat{x}'$, which lies in the $xy$-plane and makes an angle $\theta$ with respect to the positive $x$-axis, if such a measurement is made?
