# Spin 1/2 particle in B field

by ultimateguy
Tags: 1 or 2, field, particle, spin
 P: 126 1. The problem statement, all variables and given/known data Consider a spin 1/2 particle placed in a magnetic field $$\vec{B_0}$$ with components: $$B_x = \frac{1}{\sqrt{2}} B_0$$ $$B_y = 0$$ $$B_z = \frac{1}{\sqrt{2}} B_0$$ a) Calculate the matrix representing, in the {| + >, | - >} basis, the operator H, the Hamiltonian of the system. b) Calculate the eigenvalues and the eigenvectors of H. c) The system at time t = 0 is in the state | - >. What values can be found if the energy is measured, and with what probabilities? 2. Relevant equations $$\omega_0 = - \gamma B_0$$ $$H = \omega_0 S_z$$ $$S_z = \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$ 3. The attempt at a solution I'm stuck on part a). My initial instinct is to do this: $$H = \omega_0 S_z$$ $$H = - \gamma \vec{B_0} S_z$$ $$H = - \gamma \vec{B_0} \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$ But $$\vec{B_0}$$ is a 3D column vector, and I can't multiply that into a 2x2 matrix. And I have to somehow express that with | + > and | - >... I have a feeling I'm on the wrong track.
 Sci Advisor HW Helper P: 4,738 The basical hamiltonian is on the form: $$H = \vec{B} \cdot \vec{S} = B_x \cdot S_x + B_y \cdot S_y + B_z \cdot S_z$$ And the $$S_x = \frac{1}{2} \sigma _x$$ pauli matrix, etc (I use natural units, so dont bother)¨ I hope my hint helped you anyway to solve a)
 P: 126 So with this I get: $$H = \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$$$ $$+ \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$$$ $$H = B_0 \hbar $\left( \begin{array}{cc} \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \\ \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} \\ \end{array} \right)$$$ Right?
 Sci Advisor HW Helper P: 4,738 Spin 1/2 particle in B field Yes, that looks correct. Now find eigenvalue and eigenvector of this matrix.
 P: 126 For the eigenvalue: $$(\frac{B_0 \hbar}{2\sqrt2})^2 (1-\lambda)(1-\lambda) - (\frac{B_0 \hbar}{2\sqrt2})^2 = 0$$ $$(1-\lambda)(1-\lambda) = 0$$ $$\lambda = 1$$ And the eigenvector: $$\frac{B_0 \hbar} {2\sqrt2} $\left( \begin{array}{cc} 0 & 1 \\ 1 & -2 \\ \end{array} \right)$ \times $\left(\begin{array}{c} c_1 \\ c_2 \\ \end{array} \right)$ = 0$$ This gives me $$c_2 = 0$$ and $$c_1 = \frac{B_0 \hbar} {2\sqrt2}$$. I think I did something wrong.
 Sci Advisor HW Helper P: 4,738 for matrix: $$A = $\left( \begin{array}{cc} a & a\\ a & -a \\ \end{array} \right)$$$ The secular eq is $$(a- \lambda )(-a- \lambda ) - a^2 = 0$$ if lambda is the eigenvalue.
 P: 126 Thanks, I found the problem, I should have had $$(1-\lambda)(-1-\lambda) = 1$$
 P: 126 Ok so I get the following system of equations for the case of the eigenvalue $$+\sqrt2$$ $$(1-\sqrt2)c_1 + c_2 = 0$$ $$c_1 + (-1-\sqrt2)c_2 = 0$$ which according to myself and my calculator has no solution....
 Sci Advisor HW Helper P: 4,738 If you can get eigenvalues to a matrix, then there exists corresponding eigenvectors. Eigen vectors are, by using Matlab: for sqrt2 = (-0.92388,-0.38268) for -sqrt2 = (0.38268,-0.92388)
 P: 126 Ok I found the two eigenvectors using Matlab. I'm not sure how to write them when applied to the system so that it makes sense. $$|\psi(t)> = 0.92388 | + > + 0.382683 | - >$$ $$|\psi(t)> = 0.382683 | + > - 0.92388 | - >$$ Is this correct? And for part c), which one do I use to find the probability?
 Sci Advisor HW Helper P: 4,738 But in order to get them analytically, just do substituion $$c_1 = (1+\sqrt2)c_2$$ From your second equation and put in into the first one and solve for c_2 Now how does a state evolve with time? Ever heard of "Time evolution operator" or similar? Time evolution of a ket is $$|a(t) \rangle = \exp (-i E_a t/\hbar)|a(0) \rangle$$ where E_a is the energy eigenvalue of that ket. So your egeinvectors are: $$|\psi +> = 0.92388 | + > + 0.382683 | - >$$ $$|\psi -> = 0.382683 | + > - 0.92388 | - >$$ Dont use time, as you did, it is not correct. The psi + has eigenvalue +sqrt2 etc. Now I have helped you very much.
 P: 126 Yes I've seen it. So the eigenvectors are just for $$|\psi(0)>$$ ?
 Sci Advisor HW Helper P: 4,738 no, just for the hamiltonian. at time = 0; the state is in |-> Then you must find out what just |-> is in superposition of the eigenvectors to the hamiltonian, in order to get the time evolution. i.e you should first write |-> = a|phi + > + b|phi ->

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