Hamiltonian
Nuclear Spin $I$ , orbital angular momentum $L$ , and $spin$ is
$$\begin{align*} I =& \frac{3}{2} \\ L =& 0 \\ S =& \frac{1}{2} \\ \end{align*}$$Hamiltonian is
$$\begin{align*} H_S =& A \vec{I} \cdot \vec{J} + C J_z +D I_z \\ =& A \left[ I_z J_z + \frac{1}{2}(I_{ +}J_- +I_-J_{ +}) \right] + C J_z + D I_z \end{align*}$$ $$\begin{align*} |1\rangle =& |\frac{3}{2},\frac{1}{2}\rangle \\ |2\rangle =& |\frac{1}{2},\frac{1}{2}\rangle \\ |3\rangle =& |-\frac{1}{2},\frac{1}{2}\rangle \\ |4\rangle =& |-\frac{3}{2},\frac{1}{2}\rangle \\ |5\rangle =& |\frac{3}{2},-\frac{1}{2}\rangle \\ |6\rangle =& |\frac{1}{2},-\frac{1}{2}\rangle \\ |7\rangle =& |-\frac{1}{2},-\frac{1}{2}\rangle \\ |8\rangle =& |-\frac{3}{2},-\frac{1}{2}\rangle \\ \end{align*}$$Hamiltonian in the Hilbert space spaned by above kets is
$\mid 1 \rangle$ | $\mid 2 \rangle$ | $\mid 3 \rangle$ | $\mid 4 \rangle$ | $\mid 5 \rangle$ | $\mid 6 \rangle$ | $\mid 7 \rangle$ | $\mid 8 \rangle$ | |||
$\langle 1 \mid$ | $\frac{3}{4}A+\frac{1}{2}C+\frac{3}{2}D$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | ||
$\langle 2 \mid$ | $0$ | $\frac{1}{4}A+\frac{1}{2}C+\frac{1}{2}D$ | $0$ | $0$ | $\frac{\sqrt{3}}{2}A$ | $0$ | $0$ | $0$ | ||
$\langle 3 \mid$ | $0$ | $0$ | $-\frac{1}{4}A+\frac{1}{2}C-\frac{1}{2}D$ | $0$ | $0$ | $A$ | $0$ | $0$ | ||
$\langle 4 \mid$ | $0$ | $0$ | $0$ | $-\frac{3}{4}A+\frac{1}{2}C-\frac{3}{2}D$ | $0$ | $0$ | $\frac{\sqrt{3}}{2}A$ | $0$ | ||
$\langle 5 \mid$ | $0$ | $\frac{\sqrt{3}}{2}A$ | $0$ | $0$ | $-\frac{3}{4}A-\frac{1}{2}C+\frac{3}{2}D$ | $0$ | $0$ | $0$ | ||
$\langle 6 \mid$ | $0$ | $0$ | $A$ | $0$ | $0$ | $-\frac{1}{4}A-\frac{1}{2}C+\frac{1}{2}D$ | $0$ | $0$ | ||
$\langle 7 \mid$ | $0$ | $0$ | $0$ | $\frac{\sqrt{3}}{2}A$ | $0$ | $0$ | $\frac{1}{4}A-\frac{1}{2}C-\frac{1}{2}D$ | $0$ | ||
$\langle 8 \mid$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $\frac{3}{4}A-\frac{1}{2}C-\frac{3}{2}D$ |
Numerical Results
Python code
import numpy as np
import sympy as sym
from matplotlib import pyplot as plt
# Hamiltonian is H = a*I*J + c*J_z + d*I_z
a = 4000000
def H(B):
c = 4000*B
d = 4*B
H = np.zeros([8,8]) #Hamiltonian in direct production Hilbert space
H[0,0] = 3/4*a + 1/2*c + 3/2*d
H[7,7] = 3/4*a - 1/2*c - 3/2*d
H[1,1] = 1/4*a + 1/2*c + 1/2*d
H[2,2] = -3/4*a - 1/2*c + 3/2*d
H[1,2] = np.sqrt(3)/2*a
H[2,1] = H[1,2]
H[3,3] = -1/4*a + 1/2*c - 1/2*d
H[4,4] = -1/4*a - 1/2*c + 1/2*d
H[3,4] = a
H[4,3] = H[3,4]
H[5,5] = -3/4*a + 1/2*c - 3/2*d
H[6,6] = 1/4*a - 1/2*c - 1/2*d
H[5,6] = np.sqrt(3)/2*a
H[6,5] = H[5,6]
# np.set_printoptions(precision=1)
[x,y] = np.linalg.eig(H) #Diagnolization the Hamiltonian
x = np.sort(x)
return x
N = 2000
bmax = 5000 #the upper limit of magnetic field
b = np.linspace(0,bmax,N)
E = np.zeros([8,N]) #the eight eigenvaluses of energy , as a function of magnetic field
for i in range(8):
for j in range(N):
E[i,j] = H(b[j])[i]
plt.plot(b,E[i])
plt.show()
#+RESULTS: : None
Results
Reference
C. J. Pethick, H. Smith, Bose-Einstein Codensation in Dilute Gases