Expectation value of operator - python

Question

I am suposed to solve following problem:

Calculate the probability $P_n(x > X)$ that a particle in then n-th eigenstate is found at a position with an x-value larger than X.  Here it is convenient to formulate  it  as  an  expectation  value  of  an  operator.   We  make  an  operator $P(x > X)$ which is a matrix with unity on the diagonal if $x > X$.

To do this  I  used  the  command > which  returns True if  the  number  on  the  left  is larger than the one on the right and False otherwise. To turn the True or False into 0 or 1 we use the method astype(int) on the NumPy array, which we then transform to an operator.T ake  a  few  values  of X,  e.g. X=  3,  5,  and  7,  and  plot $P_n(x > X)$  as  a function of $n$ (make sure that the box size $L$ is somewhat larger than $X$).
The excercise continues as follow

Explain what you see:  Why is it low when $n$ is small?  At which value ofnwould you expect $P_n(x > X)$ to grow from a small to a large value? Hint :which energy would it require classically?

My question is: How do I find the expectation value of the operator?

KAJ226 · Answer

$rho = | psi rangle langle psi |$ is the density matrix formulation of $|psirangle$.

Now, if giving a $|psirangle$ and $A$, and you want to calculate $langle psi | A| psi rangle$ in Python, you can do it as follows:

import numpy as np 
norm_psi = [1., 1., 1., 1.]/np.linalg.norm(psi)
A  = np.matrix( '3,0,0,1; 0,-1,0,0; 0,0,-1,0; 1,0,0,1' )
print('Operator A=n', A)
expectation  = np.inner(np.conj(norm_psi).T, np.matmul(A,norm_psi) ) #Calculate <norm_psi|A|norm_psi> 
print('n expectation value calculated by <norm_psi|A|norm_psi> :n ' , expectation)
#-------
rho =  np.outer(norm_psi, np.conj(norm_psi) )
expectation = np.trace(rho*A)
print('n expectation value calculated by Tr(rho*A):n ' , expectation)

The output would be something like:

Operator A=
 [[ 3  0  0  1]
 [ 0 -1  0  0]
 [ 0  0 -1  0]
 [ 1  0  0  1]]

 expectation value calculated by <norm_psi|A|norm_psi> :
  [[1.]]

 expectation value calculated by Tr(rho*A):
  1.0

You can use Qiskit to do this as well. And it is quite simply if your operator $A$ is straight forward decomposition of Pauli strings, like $A = Zotimes Z$ or $A = X otimes Z$ etc. For instance, if $$A = X otimes Z = begin{pmatrix} 0 & 1 1 & 0 end{pmatrix} otimes begin{pmatrix} 1 & 0 0 & -1 end{pmatrix} = begin{pmatrix} 0 & 0 & 1 & 0 0 & 0& 0 & -1 1 & 0 & 0 & 0 0 & -1 & 0 & 0 end{pmatrix}$$

and $psi$ is the same as previously: $|psi rangle = dfrac{1}{2}begin{pmatrix} 1 1 1 1 end{pmatrix}$. Note that this state can be prepared on a quantum circuit as:

then you can calculate the expectation as:

from qiskit import QuantumCircuit
from qiskit.aqua.operators import X, Y, Z, I
from qiskit.aqua.operators import StateFn

operator = X ^ Z  #Note that ^ represents tensor product 
qc = QuantumCircuit(2)
qc.h(0)
qc.h(1)
psi = StateFn(qc) 
expectation = (~psi @ operator @ psi).eval()
print('expectation value is:', expectation)

output: expectation value is: 0j

Which is what you would expect by looking at the matrix form of $A$.

Expectation value of operator - python

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