Chapter 7 - Angular Momentum and Rotation



In [1]:

    
from numpy import sqrt, pi, exp, angle
from qutip import *

Our usual spin operators (spin-1/2)



In [2]:

    
pz = basis(2,0)
mz = basis(2,1)
px = 1/sqrt(2)*(pz + mz)
mx = 1/sqrt(2)*(pz - mz)
py = 1/sqrt(2)*(pz + 1j*mz)
my = 1/sqrt(2)*(pz - 1j*mz)
Sx = 1/2.0*sigmax()
Sy = 1/2.0*sigmay()
Sz = 1/2.0*sigmaz()
Splus = Sx + 1j*Sy
Sminus = Sx - 1j*Sy

Define the spin-1 operators. We use J just to keep them apart. Could be S instead.



In [8]:

    
Jx = 1/sqrt(2)*Qobj([[0,1,0],[1,0,1],[0,1,0]])
Jy = 1/sqrt(2)*Qobj([[0,-1j,0],[1j,0,-1j],[0,1j,0]])
Jz = Qobj([[1,0,0],[0,0,0],[0,0,-1]])
Jplus = Jx + 1j*Jy
Jminus = Jx - 1j*Jy



In [11]:

    
Jz









    Out[11]:





Quantum object: dims = [[3], [3]], shape = (3, 3), type = oper, isherm = True\begin{equation*}\left(\begin{array}{*{11}c}1.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0\\0.0 & 0.0 & -1.0\\\end{array}\right)\end{equation*}

Rotations of spin-1/2



In [12]:

    
Rz90 = (-1j*pi/2*Sz).expm()



In [13]:

    
Rz90









    Out[13]:





Quantum object: dims = [[2], [2]], shape = (2, 2), type = oper, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}(0.707-0.707j) & 0.0\\0.0 & (0.707+0.707j)\\\end{array}\right)\end{equation*}



In [14]:

    
Rz90*px









    Out[14]:





Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}(0.500-0.500j)\\(0.500+0.500j)\\\end{array}\right)\end{equation*}

Doesn't look like example 7.4 because it's not simplified. How to check:



In [15]:

    
exp(-1j*pi/4)*py









    Out[15]:





Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}(0.500-0.500j)\\(0.500+0.500j)\\\end{array}\right)\end{equation*}

Yes, this agrees.

Can also use inner product:



In [16]:

    
py.dag()*Rz90*px









    Out[16]:





Quantum object: dims = [[1], [1]], shape = (1, 1), type = bra\begin{equation*}\left(\begin{array}{*{11}c}(0.707-0.707j)\\\end{array}\right)\end{equation*}



In [18]:

    
(py.dag()*Rz90*px).norm()









    Out[18]:





0.9999999999999999



In [19]:

    
angle(0.707 - 0.707j) == -pi/4









    Out[19]:





True

Find the spin-1 states (the eigenstates of the corresponding matrix)

According to the postulates, the allowed values of an observable are the eigenvalues with corresponding eigenstates.

We know the matrix representation of Jx, Jy, Jz in the Z-basis so we can find all 9 states (in the Z basis). Why nine? There are three possible values $\hbar$, 0, $-\hbar$ for each of three directions.



In [19]:

    
yevals, (yd,y0,yu) = Jy.eigenstates()
zevals, (zd,z0,zu) = Jz.eigenstates()
xevals, (xd,x0,xu) = Jx.eigenstates()
xd = -xd  # fix the signs to match book
yu = -yu  # fix the signs to match book



In [23]:

    
y0









    Out[23]:





Quantum object: dims = [[3], [1]], shape = (3, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}0.707\\0.0\\0.707\\\end{array}\right)\end{equation*}



In [25]:

    
Rz90 = (-1j*pi/2*Jz).expm()



In [26]:

    
Rz90*x0









    Out[26]:





Quantum object: dims = [[3], [1]], shape = (3, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}-0.707j\\0.0\\-0.707j\\\end{array}\right)\end{equation*}



In [29]:

    
(y0.dag()*Rz90*x0).norm()









    Out[29]:





1.0

7.10 A spin-1 particle is measured to have $S_y=\hbar$. What is the probability that a subsequent measurement will yield $S_z=0$? $S_z=\hbar$? $S_z=-\hbar$?



In [56]:

    
(z0.dag()*yu).norm()**2









    Out[56]:





0.4999999999999999



In [30]:

    
(zu.dag()*yu).norm()**2









    Out[30]:





0.2500000000000002



In [31]:

    
(zd.dag()*yu).norm()**2









    Out[31]:





0.2500000000000001



In [32]:

    
yu









    Out[32]:





Quantum object: dims = [[3], [1]], shape = (3, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}0.500\\0.707j\\-0.500\\\end{array}\right)\end{equation*}



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