Any thought on I11.4

Codebook
1

dev = qml.device(‘default.qubit’, wires=2)

@qml.qnode(dev)
def circuit_1(theta):
“”"Implement the circuit and measure Z I and I Z.

Args:
    theta (float): a rotation angle.
    
Returns:
    float, float: The expectation values of the observables Z I, and I Z
"""
##################
# YOUR CODE HERE #
##################  
qml.RX(2*theta,wires=0)
qml.RY(4*theta,wires=1)
return qml.expval(qml.PauliZ(0)), qml.expval(qml.PauliZ(1))

@qml.qnode(dev)
def circuit_2(theta):
“”"Implement the circuit and measure Z Z.

Args:
    theta (float): a rotation angle.
    
Returns:
    float: The expectation value of the observable Z Z
""" 

##################
# YOUR CODE HERE #
##################  
qml.RX(2*theta,wires=0)
qml.RY(4*theta,wires=1)

return qml.expval(qml.PauliZ(0)@qml.PauliZ(1))

def zi_iz_combination(ZI_results, IZ_results):
“”"Implement a function that acts on the ZI and IZ results to
produce the ZZ results. How do you think they should combine?

Args:
    ZI_results (array[float]): Results from the expectation value of 
        ZI in circuit_1.
    IZ_results (array[float]): Results from the expectation value of 
        IZ in circuit_2.

Returns:
    array[float]: A combination of ZI_results and IZ_results that 
    produces results equivalent to measuring ZZ.
"""

combined_results=np.zeros(len(ZI_results))

##################
# YOUR CODE HERE #
##################

b=np.array(ZI_results)
c=np.array(IZ_results)

Z=np.kron(b,c)
combined_results.append(Z)

return combined_results

theta = np.linspace(0, 2 * np.pi, 100)

Run circuit 1, and process the results

circuit_1_results = np.array([circuit_1(t) for t in theta])

ZI_results = circuit_1_results[:, 0]
IZ_results = circuit_1_results[:, 1]
combined_results = zi_iz_combination(ZI_results, IZ_results)

Run circuit 2

ZZ_results = np.array([circuit_2(t) for t in theta])

Plot your results

plot = plotter(theta, ZI_results, IZ_results, ZZ_results, combined_results)

2

if I increases Len by 2 it doesn’t work to accommodate the tensor product combined_results=np.zeros(2*len(ZI_results))

3

Hi @Vedant_Dwivedi, thanks for your question! You’re on the right track, but let’s take a closer look at the data types in the zi_iz_combination function.

def zi_iz_combination(ZI_results, IZ_results):
    "”"Implement a function that acts on the ZI and IZ results to
    produce the ZZ results. How do you think they should combine?

    Args:
        ZI_results (array[float]): Results from the expectation value of 
            ZI in circuit_1.
        IZ_results (array[float]): Results from the expectation value of 
            IZ in circuit_2.

    Returns:
        array[float]: A combination of ZI_results and IZ_results that 
        produces results equivalent to measuring ZZ.
    """

[EDIT: While it’s true that the observables ZI and IZ need to be tensored together to produce the observable ZZ] The expectation values being input to this function are just lists (arrays) of floats, and this function must return an array of floats of the same shape of the two inputs arrays. In the code snippet you posted, the two arrays are being tensored together using np.kron:

b=np.array(ZI_results)
c=np.array(IZ_results)
Z=np.kron(b,c)

Is the shape of Z the same as ZI_results and IZ_results? (It might be helpful to try a small test example locally to see what happens here). What else might you do with the two arrays?

Hope that helps, please let us know if you have any further questions!

4

hoped I understood correctly but still no correct answer.

combined_results=np.zeros(2*len(ZI_results))

##################
# YOUR CODE HERE #
##################
b=np.array(ZI_results)
c=np.array(IZ_results)
combined_results=np.kron(b,c)
5

Hi @Vedant_Dwivedi,

The aim here is, given ZI_results and IZ_results, to combine them (mathematically) to create an array of the same length that represents the ZZ results. What the function returns must have the same dimension as the input;

combined_results=np.zeros(2*len(ZI_results))

will give a result with twice the length of the inputs (and using kron will similarly give something with twice the length).

[BTW I have made an error in my earlier post, which I have edited; ZI and IZ do not tensor together to create ZZ, but rather they simply multiply. I’ve amended the original post.]

6

Got it Thanks :+1::v:@glassnotes

7

@Vedant_Dwivedi awesome, glad to hear it!