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?

  • $\begingroup$ Hello! As you must know the expectation value of an operator $A$ in the state $\psi$ is $\langle A \rangle_{\psi} = \langle \psi |A| \psi \rangle$. I believe from this you can use matrix multiplication in Python, did you try? Also, if you are familiar with Qiskit, there are some ways to get the expectation value, if you are interested :) $\endgroup$
    – Lena
    Commented Jan 7, 2021 at 14:02

1 Answer 1


The expectation of an observable $A$ with respect to the state $|\psi \rangle$ can be calculated as: $$\langle \psi |A| \psi \rangle = Tr( \langle \psi | A| \psi \rangle) = Tr( A|\psi \rangle \langle \psi|) = Tr(A \rho) = Tr(\rho A)$$

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

Now, if giving a $|\psi\rangle$ 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> :

 expectation value calculated by Tr(rho*A):

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 $A = Z\otimes Z$ 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:

enter image description here

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)
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 $A$.


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