Chapter 7 - Angular Momentum and Rotation¶

We’ll use $\hbar=1$ for numerical results, this is standard practice, but it is important to remember.

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

Our usual spin operators (spin-1/2)

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.

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 # raising operator
Jminus = Jx - 1j*Jy # lowering operator

Jz

Rotations of spin-1/2¶

To define an exponentiated operator, we apply the .expm() method to the argument. So to write the rotation operator:

we do the following. It looks a little odd to read since you might expect the exp to come first, but looks like this:

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

Rz90

Rz90*px

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

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

Yes, this agrees.

Can also use inner product:

py.dag()*Rz90*px

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

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

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.

yevals, (yd,y0,yu) = Jy.eigenstates()
zevals, (zd,z0,zu) = Jz.eigenstates()
xevals, (xd,x0,xu) = Jx.eigenstates()

# Fix the signs to match book:
# (if first value of eigenstate is negative, flip all signs)
states = [xd, x0, xu, yd, y0, yu, zd, z0, zu]
states = [-s if s.full()[0][0]<0 else s for s in states]
# Python doesn't change the values in the list so we have 
# to replace the old values with the new ones:
[xd, x0, xu, yd, y0, yu, zd, z0, zu] = states

xd

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

Rz90*xu == -1j*yu

Rz90*xu

yd

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

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

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

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

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

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

abs(y0.dag()*xd)**2

abs(xd.dag()*y0)**2

7.13 $|\psi\rangle = \frac{1}{3}(2|1,1\rangle - i|1,0\rangle + 2|1,-1\rangle)$ Calculate expectation values of $S^2$, $S_y$, and $S_z$.¶

psi = (1/3)*(2*zu -1j*z0 +2*zd)
psi

Jsquared = Jx*Jx + Jy*Jy + Jz*Jz
Jsquared

psi.dag()*Jsquared*psi

psi.dag()*Jy*psi

psi.dag()*Jz*psi