Lab 1 - Vectors and Matrices¶
This notebook demonstrates the use of vectors and matrices in IPython. Note that the basis is not explicit in any of these operations. You must keep track of the basis yourself (using variable names, or notes etc).
In [1]:
from numpy import array, dot, outer, sqrt, matrix
from numpy.linalg import eig, eigvals
from matplotlib.pyplot import hist
In [2]:
%matplotlib inline
In [3]:
rv = array([1,2]) # a row vector
rv
Out[3]:
array([1, 2])
In [4]:
cv = array([[3],[4]]) # a column vector
cv
Out[4]:
array([[3],
[4]])
Two kinds of vector products we'll see: inner product (dot product) and outer product
1) Use the function dot(vector1, vector2) to find the dot product of rv and cv. Does the order of the arguments matter?¶
In [5]:
dot(rv,cv)
Out[5]:
array([11])
In [6]:
dot(cv,rv)
--------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-6-aa042602721b> in <module>() ----> 1 dot(cv,rv) ValueError: shapes (2,1) and (2,) not aligned: 1 (dim 1) != 2 (dim 0)
2) Use the function outer(vector1, vector2) to find the outer product of rv and cv. Does the order of the arguments matter?¶
In [7]:
outer(rv,cv)
Out[7]:
array([[3, 4],
[6, 8]])
In [8]:
outer(cv,rv)
Out[8]:
array([[3, 6],
[4, 8]])
II. Complex vectors¶
In [9]:
# Complex numbers in python have a j term:
a = 1+2j
In [10]:
v1 = array([1+2j, 3+2j, 5+1j, 4+0j])
The complex conjugate changes the sign of the imaginary part:
In [11]:
v1.conjugate()
Out[11]:
array([ 1.-2.j, 3.-2.j, 5.-1.j, 4.-0.j])
3) Use dot() and .conjugate() to find the dot product of v1 and it's own conjugate:¶
In [12]:
dot(v1.conjugate(),v1)
Out[12]:
(60+0j)
III. Matrices¶
In [13]:
# a two-dimensional array
m1 = array([[2,1],[2,1]])
m1
Out[13]:
array([[2, 1],
[2, 1]])
In [14]:
# can find transpose with the T method:
m1.T
Out[14]:
array([[2, 2],
[1, 1]])
In [15]:
# find the eigenvalues and eigenvectors of a matrix:
eig(m1)
Out[15]:
(array([ 3., 0.]), array([[ 0.70710678, -0.4472136 ],
[ 0.70710678, 0.89442719]]))
Can also use the matrix type which is like array but restricts to 2D. Also, matrix adds .H and .I methods for hermitian and inverse, respectively. For more information, see Stack Overflow question #4151128
In [16]:
m2 = matrix( [[2,1],[2,1]])
In [17]:
m2.H
Out[17]:
matrix([[2, 2],
[1, 1]])
In [18]:
eig(m2)
Out[18]:
(array([ 3., 0.]), matrix([[ 0.70710678, -0.4472136 ],
[ 0.70710678, 0.89442719]]))
In [19]:
# use a question mark to get help on a command
eig?
In [20]:
M14 = array([[0,1],[-2,3]])
In [21]:
eig(M14)
Out[21]:
(array([ 1., 2.]), array([[-0.70710678, -0.4472136 ],
[-0.70710678, -0.89442719]]))
Interpret this result: the two eigenvalues are 1 and 2 the eigenvectors are strange decimals, but we can check them against the stated solution:
In [22]:
1/sqrt(2) # this is the value for both entries in the first eigenvector
Out[22]:
0.70710678118654746
In [23]:
1/sqrt(5) # this is the first value in the second eigenvector
Out[23]:
0.44721359549995793
In [24]:
2/sqrt(5) # this is the second value in the second eigenvector
Out[24]:
0.89442719099991586
In [25]:
eigvals(M14)
Out[25]:
array([ 1., 2.])
Signs are opposite compared to the book, but it turns out that (-) doesn't matter in the interpretation of eigenvectors: only "direction" matters (the relative size of the entries).
Example: Problem 1.16 using Ipython functions¶
In [26]:
M16 = array([[0,-1j],[1j,0]])
In [27]:
evals, evecs = eig(M16)
In [28]:
evecs
Out[28]:
array([[-0.00000000-0.70710678j, 0.70710678+0.j ],
[ 0.70710678+0.j , 0.00000000-0.70710678j]])
In [29]:
evecs[:,0]
Out[29]:
array([-0.00000000-0.70710678j, 0.70710678+0.j ])
In [30]:
evecs[:,1]
Out[30]:
array([ 0.70710678+0.j , 0.00000000-0.70710678j])
In [31]:
dot(evecs[:,0].conjugate(),evecs[:,1])
Out[31]:
-1.6653345369377348e-16j
Part 2: Using QuTiP¶
Keeping track of row and column vectors in Ipython is somewhat artificial and tedious. The QuTiP library is designed to take care of many of these headaches
In [32]:
from qutip import *
In [33]:
# Create a row vector:
qv = Qobj([[1,2]])
qv
Out[33]:
Quantum object: dims = [[1], [2]], shape = (1, 2), type = bra\begin{equation*}\left(\begin{array}{*{11}c}1.0 & 2.0\\\end{array}\right)\end{equation*}
In [34]:
# Find the corresponding column vector
qv.dag()
Out[34]:
Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}1.0\\2.0\\\end{array}\right)\end{equation*}
In [35]:
qv2 = Qobj([[1+2j,4-1j]])
qv2
Out[35]:
Quantum object: dims = [[1], [2]], shape = (1, 2), type = bra\begin{equation*}\left(\begin{array}{*{11}c}(1.0+2.0j) & (4.0-1.0j)\\\end{array}\right)\end{equation*}
In [36]:
qv2.dag()
Out[36]:
Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}(1.0-2.0j)\\(4.0+1.0j)\\\end{array}\right)\end{equation*}
Vector products in QuTiP¶
Only need to know one operator: "*" The product will depend on the order, either inner or outer
In [37]:
qv2*qv2.dag() # inner product (dot product)
Out[37]:
Quantum object: dims = [[1], [1]], shape = (1, 1), type = bra\begin{equation*}\left(\begin{array}{*{11}c}22.0\\\end{array}\right)\end{equation*}
In [38]:
qv2.dag()*qv2 # outer product
Out[38]:
Quantum object: dims = [[2], [2]], shape = (2, 2), type = oper, isherm = True\begin{equation*}\left(\begin{array}{*{11}c}5.0 & (2.0-9.0j)\\(2.0+9.0j) & 17.0\\\end{array}\right)\end{equation*}
Matrix in QuTiP¶
In [39]:
qm = Qobj([[1,2],[2,1]])
qm
Out[39]:
Quantum object: dims = [[2], [2]], shape = (2, 2), type = oper, isherm = True\begin{equation*}\left(\begin{array}{*{11}c}1.0 & 2.0\\2.0 & 1.0\\\end{array}\right)\end{equation*}
In [40]:
qm.eigenenergies() # in quantum (as we will learn) eigenvalues often correspond to energy levels
Out[40]:
array([-1., 3.])
In [41]:
evals, evecs = qm.eigenstates()
In [42]:
evecs
Out[42]:
array([ Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket
Qobj data =
[[-0.70710678]
[ 0.70710678]],
Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket
Qobj data =
[[ 0.70710678]
[ 0.70710678]]], dtype=object)
In [43]:
evecs[0]
Out[43]:
Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket\begin{equation*}\left(\begin{array}{*{11}c}-0.707\\0.707\\\end{array}\right)\end{equation*}
Practice:¶
Problem 1.2 using the hist() function.¶
In [42]:
# Solution
n, bins, patches = hist([10,13,14,14,6,8,7,9,12,14,13,11,10,7,7],bins=5,range=(5,14))
In [43]:
# Solution
n
Out[43]:
array([ 1., 4., 3., 2., 5.])
In [44]:
# Solution
pvals = n/n.sum()
Problem 1.8¶
Hint: using sympy, we can calculate the relevant integral. The conds='none' asks the solver to ignore any strange conditions on the variables in the integral. This is fine for most of our integrals. Usually the variables are real and well-behaved numbers.
In [3]:
# Solution
from sympy import *
c,a,x = symbols("c a x")
Q.positive((c,a))
first = integrate(c*exp(-a*x),(x,0,oo),conds='none')
print("first = ",first)
second = integrate(a*exp(-a*x),(x,0,oo),conds='none')
print("second = ",second)
first = c/a second = 1