johnsmi
Jul15-10, 06:11 AM
Hi, I am reading this article for homework about a ring in a megnetic field. It starts off by giving a hamiltonian (an adiabatic part -never mind)
H_{0}= \frac{1}{2M} [ \Pi -A]^{2} -\mu B( \phi) \cdot \sigma
A- is a known operator
where \Pi=\frac{1}{2a} \frac{d}{d \phi} -\frac{eB_{z} \pi a}{2c} is the generalized momentum operator
I know that the eigen states of \mu B( \phi) \cdot \sigma (Spinors) are:
| \uparrow ( \phi) > =(i \alpha e^{-i \phi} , -\beta)^{T}
| \downarrow ( \phi) > =(i\beta e^{-i \phi} , \alpha)^{T}
Now, in this article I have he sais that the eigen states of H0 can be written as
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n} and
| \downarrow ( \phi) > \otimes \psi ^{ \downarrow}_{n}
When, \psi \^{ \uparrow}_{n} for example is the eigenstate of a Hamiltonian:
H^{up}_{0}= \frac{1}{2M} [ \Pi -const]^{2} -\mu B
How did he get it (the last Hamiltonian)???
Someone told me to try and apply H0 on
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}
but I got something pretty awful
Can somone help me please???
H_{0}= \frac{1}{2M} [ \Pi -A]^{2} -\mu B( \phi) \cdot \sigma
A- is a known operator
where \Pi=\frac{1}{2a} \frac{d}{d \phi} -\frac{eB_{z} \pi a}{2c} is the generalized momentum operator
I know that the eigen states of \mu B( \phi) \cdot \sigma (Spinors) are:
| \uparrow ( \phi) > =(i \alpha e^{-i \phi} , -\beta)^{T}
| \downarrow ( \phi) > =(i\beta e^{-i \phi} , \alpha)^{T}
Now, in this article I have he sais that the eigen states of H0 can be written as
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n} and
| \downarrow ( \phi) > \otimes \psi ^{ \downarrow}_{n}
When, \psi \^{ \uparrow}_{n} for example is the eigenstate of a Hamiltonian:
H^{up}_{0}= \frac{1}{2M} [ \Pi -const]^{2} -\mu B
How did he get it (the last Hamiltonian)???
Someone told me to try and apply H0 on
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}
but I got something pretty awful
Can somone help me please???