Lab 8 - Simple Harmonic Oscillator states

Problems from Chapter 12



In [1]:

    
from numpy import sqrt
from qutip import *

Define the standard operators



In [2]:

    
N = 10  # pick a size for our state-space
a = destroy(N)
n = a.dag()*a

Problem 12.1:



In [3]:

    
a*a.dag() - a.dag()*a









    Out[3]:





Quantum object: dims = [[10], [10]], shape = (10, 10), 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 & 0.0 & 0.0\\0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.000 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.000 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.000 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & -9.0\\\end{array}\right)\end{equation*}

Problem 12.2:



In [4]:

    
n*a.dag() - a.dag()*n









    Out[4]:





Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 1.414 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.732 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 2.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 2.236 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.449 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.646 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.828 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 3.000 & 0.0\\\end{array}\right)\end{equation*}



In [5]:

    
n*a.dag() - a.dag()*n == a.dag()









    Out[5]:





True

Problem 12.3 (use n=2 as a test-case):

To define $|2\rangle$ use the basis(N,n) command where N is the dimension of the vector, and n is the quantum number.



In [6]:

    
psi = basis(N,2)



In [7]:

    
psi









    Out[7]:





Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}0.0\\0.0\\1.0\\0.0\\0.0\\0.0\\0.0\\0.0\\0.0\\0.0\\\end{array}\right)\end{equation*}



In [8]:

    
a.dag()*psi









    Out[8]:





Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}0.0\\0.0\\0.0\\1.732\\0.0\\0.0\\0.0\\0.0\\0.0\\0.0\\\end{array}\right)\end{equation*}



In [9]:

    
a.dag()*basis(N,2) == sqrt(3)*basis(N,3)









    Out[9]:





True

Problem 12.5 and 12.6:

These are simple, just view the matrix representation of the operators



In [10]:

    
a









    Out[10]:





Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.414 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 1.732 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 2.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.236 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.449 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.646 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.828 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 3.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\end{array}\right)\end{equation*}



In [11]:

    
a.dag()









    Out[11]:





Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = False\begin{equation*}\left(\begin{array}{*{11}c}0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 1.414 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.732 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 2.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 2.236 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.449 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.646 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.828 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 3.0 & 0.0\\\end{array}\right)\end{equation*}

Problem 12.7:

First, define $\hat{X}$ and $\hat{P}$ operators



In [12]:

    
X = 1/2 * (a + a.dag())
P = 1/2j * (a - a.dag())



In [13]:

    
psi = 1/sqrt(2)*(basis(N,1)+basis(N,2))
ex = psi.dag()*X*psi
exq = psi.dag()*X*X*psi
ep = psi.dag()*P*psi
epq = psi.dag()*P*P*psi



In [14]:

    
deltaX = sqrt(exq[0][0][0] - ex[0][0][0]**2)
deltaP = sqrt(epq[0][0][0] - ep[0][0][0]**2)



In [15]:

    
deltaX * deltaP * 2 # compare to uncertainty relation (ΔxΔp >= 1/2)
# the factor of two is to convert from the unitless version of the operator









    Out[15]:





(1.4142135623730951+0j)

Alternatively, we can find the indeterminacy bound for ΔX and ΔP (the unitless operators): $$\Delta X \Delta P \geq \frac{1}{2}\left|\left\langle \left[\hat{X},\hat{P}\right] \right\rangle\right|$$



In [31]:

    
1/2*(psi.dag()*commutator(X,P)*psi).norm()









    Out[31]:





0.24999999999999994

Which is also satisfied by the calculated value (1.41 > 0.25)

Problem 12.8:



In [16]:

    
psi = 1/sqrt(2)*(basis(N,2)+basis(N,4))
ex = psi.dag()*X*psi
exq = psi.dag()*X*X*psi
ep = psi.dag()*P*psi
epq = psi.dag()*P*P*psi



In [17]:

    
deltaX = sqrt(exq[0][0][0] - ex[0][0][0]**2)
deltaP = sqrt(epq[0][0][0] - ep[0][0][0]**2)



In [19]:

    
deltaX * deltaP * 2 # to compare to book solution which uses the full x and p operators with units









    Out[19]:





(3.0413812651491092+0j)



In [ ]: