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\heading{E307: Magnetization of a ring with electrons}
\auname{Yaniv Vinish and Yotam Zmiri}


{\bf The problem:}

\Dn

A ring with radius ${R}$ is placed in a uniform magnetic field ${B}$.\\
There are 5 electrons with mass ${m}$, charge ${e}$ and spin ${\frac{1}{2}}$ .\\
Note that two electrons can't be at the same state.\\ \\
Let ${E}$ be the ground state energy \\
and ${M(B)=-dE/dB}$ be the magnetization \\
${B,R,m,e,g}$${\approx}$${2.0023}$ are given.\\ \\          						

If the electrons are spinless the magnetization is a periodic function with period ${B_0}$ \\
Later we will assume that ${B=lB_0+\Delta B}$\\
where l is an integer and 0${\textless}$${\Delta}$B${\ll}$${B_0}$.\\ \\

\Dn

(1) Write an expression for B${_0}$.

(2) Find ${M(B)}$ for ${B=0}$.

(3) Write the addition to ${M(B)}$ for small magnetic field.

(4) Find ${M(B)}$ for large ${l}$ when ${\Delta B=0}$.

(5) Write the addition to ${M(B)}$ for large ${l}$ when ${\Delta}$B${\neq}$0.

(6) Define large l. \\


\Dn\Dn


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{\bf The solution:}

\Dn

(1)
The Hamiltonian can be written as:\\ 
 
 ${H=\frac{1}{2m}(p-\frac{eRB}{2})^2}$\\
	
 The eigenstates of the H are the momentum states ${|k_n>}$ where: ${|k_n>= \frac{2\boldsymbol{\pi}}{2\boldsymbol{\pi}R}n}$\\ 
         
  The eigenvalues are: ${ En=\frac{1}{2MR^2}  (n-\frac{eR^2B}{2})^2}$\\	

${M(B)=-\frac{dE}{dB}=-\frac{e}{2M}(n-\frac{eR^2B}{2})}$\\

Now we can see that ${M(B)}$ is a periodic function and the period is:\\

 ${B_0=\frac{2}{eR^2}}$ .\\


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\Dn

(2)
Because of the spin of the electrons, the Hemiltonian include the zeeman term:\\

${H= \frac{1}{2m}(p-\frac{eRB}{2})^2 - geB \cdot \frac{S}{2m}}$\\

The Zeeman term split the energy levels so that the energies which has the same n is not degenerated anymore.\\

Let us assume that B is parallel to z axis.\\

When B ${\rightarrow}$ 0 every two electrons can have the same n, therefore the total energy is:\\

${E_{total}=\frac{5e^2R^2\Delta B^2}{8m} - (\frac{e}{2m}+\frac{eg}{4m})\Delta B+\frac{3}{2mR^2}}$ \\

and ${M(B)=\frac{e}{2m}(1 -\frac{5\Delta B}{B_0})+\frac{eg}{4m}}$\\

	
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\Dn

(3)
the addition to ${M(B)}$ for small magnetic field is the suseptability:\\

${\boldsymbol{\chi}=\frac{dM(B)}{dB}=-\frac{5e}{2mB_0}}$\\

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\Dn

(4)
When l is large enough, all the electrons would have spin ${|}$${\uparrow}$${>}$ because of the Zeeman splitting,\\

 Therefore: \\ 

${E_{total}=\frac{1}{2mR^2}(\frac{5e^2R^4\Delta B^2}{4}+10)-\frac{5eg\Delta B}{4m} }$\\ 

and ${M(B)=\frac{e}{2m}(0-\frac{5\Delta B}{B_0})+\frac{5eg}{4m}}$\\

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\Dn

(5)
the addition to ${M(B)}$ for small magnetic field is the suseptability:\\

${\boldsymbol{\chi}=\frac{dM(B)}{dB}=-\frac{5e}{2mB_0}}$\\

%%%%%%%%%%%%%
\Dn

(6)
l is large when the diffrence between the fifth energy level with spin ${|}$${\uparrow}$${>}$ to the lowest energy level\\
 
 with spin ${|}$${\downarrow}$${>}$ is smaller than ${\frac{egB}{2m}}$:\\

${\frac{2}{mR^2}}$ ${\textless}$ ${\frac{egB}{2m}}$ , where:  ${B=lB_0=\frac{2l}{eR^2}}$\\

${\frac{2}{g}}$ ${\textless}$ ${l}$


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