theoriginal169 Posted June 11, 2010 Posted June 11, 2010 B=B[math]_{0}[/math]cos([math]\omega[/math]t)k an electron is at rest in an oscillating field bo and omega is constant and k is operator for z how to construct the Hamiltonian matrix ? Merged post follows: Consecutive posts mergedi tougth H=-b0[math]\gamma[/math]h/2[math]\sigma[/math][math]_{z}[/math]
timo Posted June 11, 2010 Posted June 11, 2010 The energy of a magnetic dipole [math]\vec m[/math](consider the spin as a magnetic dipole) in a magnetic field [math] \vec B [/math] is proportional to [math] - \vec m \cdot \vec B[/math] (such that energy is minimized for parallel orientation and maximized for anti-parallel one). From that, constructing the Hamiltonian should be straightforward (looking up the constants, replacing observables with operators, be happy). Depending on what you want to do with your Hamiltonian, a smart choice of coordinate system might be a good idea. Remark: Iirc might be a rather unexpected factor of 2 somewhere in the transition from the spin to the magnetic moment, but you'll probably see that yourself.
theoriginal169 Posted June 11, 2010 Author Posted June 11, 2010 The energy of a magnetic dipole [math]\vec m[/math](consider the spin as a magnetic dipole) in a magnetic field [math] \vec B [/math] is proportional to [math] - \vec m \cdot \vec B[/math] (such that energy is minimized for parallel orientation and maximized for anti-parallel one). From that, constructing the Hamiltonian should be straightforward (looking up the constants, replacing observables with operators, be happy). Depending on what you want to do with your Hamiltonian, a smart choice of coordinate system might be a good idea. Remark: Iirc might be a rather unexpected factor of 2 somewhere in the transition from the spin to the magnetic moment, but you'll probably see that yourself. you are talking about the quantum number for the composite systems there is no place for m at this situation . H=-gamma b0 Sz is the Hamiltonian matrix for the spin at the magnetic fields . and you can change the axis whit Schrodinger equation ih dx/x =H x where is x is eigenspinor .
timo Posted June 11, 2010 Posted June 11, 2010 you are talking about the quantum number for the composite systemsthere is no place for m at this situation . I have no idea what you are trying to tell me there. I did not talk about a composite system in any place. I do not necessarily use mainstream labels in case that irritated you. That's why I said what the respective variables stand for. It's the physics that matters, not the letters used to describe it. Are you asking a question in this thread or are you trying to teach something to us?
theoriginal169 Posted June 11, 2010 Author Posted June 11, 2010 you didnt get my question so i try to explain it ...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now