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Spin 1/2 particle in B field 
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#1
Nov407, 12:28 PM

P: 126

1. The problem statement, all variables and given/known data
Consider a spin 1/2 particle placed in a magnetic field [tex]\vec{B_0}[/tex] with components: [tex] B_x = \frac{1}{\sqrt{2}} B_0 [/tex] [tex] B_y = 0 [/tex] [tex] B_z = \frac{1}{\sqrt{2}} B_0 [/tex] 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 [tex] \omega_0 =  \gamma B_0 [/tex] [tex] H = \omega_0 S_z [/tex] [tex] S_z = \frac{\hbar}{2} \[ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\] [/tex] 3. The attempt at a solution I'm stuck on part a). My initial instinct is to do this: [tex] H = \omega_0 S_z [/tex] [tex] H =  \gamma \vec{B_0} S_z [/tex] [tex] H =  \gamma \vec{B_0} \frac{\hbar}{2} \[ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\] [/tex] But [tex]\vec{B_0}[/tex] 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. 


#2
Nov407, 01:08 PM

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The basical hamiltonian is on the form:
[tex] H = \vec{B} \cdot \vec{S} = B_x \cdot S_x + B_y \cdot S_y + B_z \cdot S_z [/tex] And the [tex] S_x = \frac{1}{2} \sigma _x [/tex] pauli matrix, etc (I use natural units, so dont bother)¨ I hope my hint helped you anyway to solve a) 


#3
Nov407, 02:11 PM

P: 126

So with this I get:
[tex] H = \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} \[ \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)\] [/tex] [tex]+ \frac{1}{\sqrt{2}} B_0 \frac{\hbar}{2} \[ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\] [/tex] [tex] H = B_0 \hbar \[ \left( \begin{array}{cc} \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \\ \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \\ \end{array} \right)\] [/tex] Right? 


#4
Nov407, 02:33 PM

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Spin 1/2 particle in B field
Yes, that looks correct. Now find eigenvalue and eigenvector of this matrix.



#5
Nov407, 02:46 PM

P: 126

For the eigenvalue:
[tex] (\frac{B_0 \hbar}{2\sqrt2})^2 (1\lambda)(1\lambda)  (\frac{B_0 \hbar}{2\sqrt2})^2 = 0 [/tex] [tex](1\lambda)(1\lambda) = 0[/tex] [tex] \lambda = 1 [/tex] And the eigenvector: [tex]\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 [/tex] This gives me [tex] c_2 = 0 [/tex] and [tex] c_1 = \frac{B_0 \hbar} {2\sqrt2} [/tex]. I think I did something wrong. 


#6
Nov407, 03:34 PM

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for matrix:
[tex] A = \[ \left( \begin{array}{cc} a & a\\ a & a \\ \end{array} \right)\] [/tex] The secular eq is [tex] (a \lambda )(a \lambda )  a^2 = 0 [/tex] if lambda is the eigenvalue. 


#7
Nov407, 03:38 PM

P: 126

Thanks, I found the problem, I should have had [tex] (1\lambda)(1\lambda) = 1 [/tex]



#8
Nov407, 03:50 PM

P: 126

Ok so I get the following system of equations for the case of the eigenvalue [tex]+\sqrt2[/tex]
[tex](1\sqrt2)c_1 + c_2 = 0[/tex] [tex]c_1 + (1\sqrt2)c_2 = 0 [/tex] which according to myself and my calculator has no solution.... 


#9
Nov407, 04:03 PM

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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) 


#10
Nov407, 04:14 PM

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.
[tex] \psi(t)> = 0.92388  + > + 0.382683   > [/tex] [tex] \psi(t)> = 0.382683  + >  0.92388   > [/tex] Is this correct? And for part c), which one do I use to find the probability? 


#11
Nov407, 04:21 PM

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But in order to get them analytically, just do substituion
[tex]c_1 = (1+\sqrt2)c_2[/tex] 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 [tex] a(t) \rangle = \exp (i E_a t/\hbar)a(0) \rangle [/tex] where E_a is the energy eigenvalue of that ket. So your egeinvectors are: [tex] \psi +> = 0.92388  + > + 0.382683   > [/tex] [tex] \psi > = 0.382683  + >  0.92388   > [/tex] Dont use time, as you did, it is not correct. The psi + has eigenvalue +sqrt2 etc. Now I have helped you very much. 


#12
Nov407, 04:23 PM

P: 126

Yes I've seen it. So the eigenvectors are just for [tex] \psi(0)>[/tex] ?



#13
Nov407, 04:34 PM

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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 > = aphi + > + bphi > 


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