Spin 1/2 particle in B field


by ultimateguy
Tags: 1 or 2, field, particle, spin
ultimateguy
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#1
Nov4-07, 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.
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malawi_glenn
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#2
Nov4-07, 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)
ultimateguy
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#3
Nov4-07, 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?

malawi_glenn
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#4
Nov4-07, 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.
ultimateguy
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#5
Nov4-07, 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.
malawi_glenn
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#6
Nov4-07, 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.
ultimateguy
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#7
Nov4-07, 03:38 PM
P: 126
Thanks, I found the problem, I should have had [tex] (1-\lambda)(-1-\lambda) = 1 [/tex]
ultimateguy
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#8
Nov4-07, 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....
malawi_glenn
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#9
Nov4-07, 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)
ultimateguy
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#10
Nov4-07, 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?
malawi_glenn
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#11
Nov4-07, 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.
ultimateguy
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#12
Nov4-07, 04:23 PM
P: 126
Yes I've seen it. So the eigenvectors are just for [tex] |\psi(0)>[/tex] ?
malawi_glenn
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#13
Nov4-07, 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 |-> = a|phi + > + b|phi ->


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