Fall 2003 6 Qm

A pair of spin $1/2$ particles is prepared in the entangled state

\begin{align} | \psi \rangle = \frac{1}{\sqrt{2}} ( | \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle ) \end{align}

where the arrows refer to the $\hat{z}$ component of the particles' spins.

  1. The spin component of the first particle is measured along an axis $\hat{u}_\alpha$ making angles $\theta_\alpha$, $\phi_\alpha$ with the $\hat{z}$ axis. What is the probability of obtaining $+ \hbar / 2$?
  2. After the first measurement along $\hat{u}_\alpha$ yields spin up, the second particle's spin component is measured along $\hat{u}_\beta$ with angles $\theta_\beta$, $\phi_\beta$. What is the probability of obtaining $+ \hbar / 2$?




It does NOT matter what $\phi_a$ is for the first part of the problem. In spherical coordinates, $\phi_a$ measures the rotation about the z-axis (making a circle for $0<\phi<2\pi$ has the center of the circle as the z-axis for all $\phi$. Therefore all we need to do is calculate the probability of beuing spin up when rotating the y-axis by an angle $\theta_a$. This requires the rotation matrices. We calculate what each spin-1/2 particle rotates to and then multiply the 2 particles together. For 1 particle the new state will be

\begin{eqnarray} \chi \prime &=& \exp \left( i \theta_a \frac{S_y}{\hbar} \right) = \exp \left(i \frac{\theta_a}{2}\sigma_y \right) \chi \\ &=& \sum_{n=0}^{\infty} \frac{\left( i \frac{\theta_a}{2} \sigma_y \right)^n}{n!} \chi = \\ &=& \left( 1\cos\frac{\theta_a}{2} + i\sigma_y \sin\frac{\theta_a}{2} \right) \chi \\ &=& \left( \right) \chi \end{eqnarray}

Therefore $| \uparrow \rangle$ goes to $cos\frac{\theta_a}{2} | \uparrow \rangle - \sin\frac{\theta_a}{2} | \downarrow \rangle$ and $| \downarrow \rangle$ goes to $\sin\frac{\theta_a}{2} | \uparrow \rangle + cos\frac{\theta_a}{2} | \downarrow \rangle$. Therefore our new entangled state is

\begin{eqnarray} \chi \prime &=& \frac{1}{\sqrt{2}} \left[ \left( \cos\frac{\theta_a}{2} | \uparrow \rangle_1 - \sin\frac{\theta_a}{2} | \downarrow \rangle_1 \right) \left( \cos\frac{\theta_a}{2} | \uparrow \rangle_2 - \sin\frac{\theta_a}{2} | \downarrow \rangle_2 \right) \\ &+& \left( \sin\frac{\theta_a}{2} | \uparrow \rangle_1 + \cos\frac{\theta_a}{2} | \downarrow \rangle_1 \right) \left( \sin\frac{\theta_a}{2} | \uparrow \rangle_2 + \cos\frac{\theta_a}{2} | \downarrow \rangle_2 \right) \\ &=& \frac{1}{\sqrt{2}} \left( |\uparrow \rangle_1 | \uparrow \rangle_2 + | \downarrow \rangle_1 | \downarrow \rangle_2 \right) \end{eqnarray}

If we find the amplitude the first particle is spin up, we get $\frac{1}{\sqrt{2}} | \uparrow \rangle$. Therefore the probability of spin up for the first particle is $\frac{1}{2}$

Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-Share Alike 2.5 License.