Quantum-Computation-course-Basel/archives/exercises_2022/Exercise1.ipynb

Exercise 1

This exercises is based on the final part of Hello Qiskit. In it, you will construct a Bell test for classical and quantum variables.

Before the exercise

Before attempting the exercise, it is best to go through the full 5 levels of Hello Qiskit. The easiest way to do this is within the textbook itself, using this page.

You can also do it with the IBM Quantum Lab. This does which require a sign-in, but it's free. Then navigate to qiskit-textbook/content/ch-ex/hello-qiskit.ipynb. This will allow you to run the notebook and save your progress.

The exercise

You need to create a two-qubit state with the application of quantum gates that violates the Bell-test. Background on this can be found in the final part of Hello Qiskit, which is reproduced below. You can also read more here and here, here, and find inspiration for the right set of gates, here.

To complete this exercise, you'll need to run this notebook. Again you can do that using this page (though this won't let you save). But the best way is to upload it into the IBM Quantum Lab.

To submit, send the TA either:

• this notebook completed succesfully; or
• the output text of the cell containing "bell_test( P )" in the quantum case and the circuit you used to violate the test.

In [ ]:

print('Set up started...')
from qiskit_textbook.games import hello_quantum
print('Set up complete!')

Proving the Uniqueness of Quantum Variables

Bell test for classical variables

Here we'll investigate how quantum variables (based on qubits) differ from standard ones (based on bits).

We'll do this by creating a pair of variables, which we will call A and B. We aren't going to put any conditions on what these can be, or how they are initialized. So there are a lot of possibilities:

• They could be any kind of variable, such as
  • integer
  • list
  • dictionary
  • ...
• They could be initialized by any kind of process, such as
  • left empty
  • filled with a given set of values
  • generated by a given random process
    • indepedently applied to A and B
    • applied to A and B together, allowing for correlations between their randomness

If the variables are initialized by a random process, it means they'll have different values every time we run our program. This is perfectly fine. The only rule we need to obey is that the process of generating the randomness is the same for every run.

We'll use the function below to set up these variables. This currently has A and B defined as to be partially correlated random floating point numbers. But you can change it to whatever you want.

In [ ]:

import random
def setup_variables ():
    
    ### Replace this section with anything you want ###
    
    r = random.random()
    
    A = r*(2/3)
    B = r*(1/3)
    
    ### End of section ###
    
    return A, B

Our next job is to define a hashing function. This simply needs to take one of the variables as input, and then give a bit value as an output.

This function must also be capable of performing two different types of hash. So it needs to be able to be able to chew on a variable and spit out a bit in to different ways. We'll therefore also need to tell the function what kind of hash we want to use.

To be consistent with the rest of the program, the two possible hash types should be called 'H' and 'V'. Also, the output must be in the form of a single value bit string: either '0' or '1'.

In the (fairly arbitrary) example given, the bits were created by comparing A and B to a certain value. The output is '1' if they are under that value, and '0' otherwise. The type of hash determines the value used.

In [ ]:

def hash2bit ( variable, hash ):
    
    ### Replace this section with anything you want ###
    
    if hash=='V':
        bit = (variable<0.5)
    elif hash=='H':
        bit = (variable<0.25)
        
    bit = str(int(bit)) # Turn True or False into '1' and '0'
    
    ### End of section ###
        
    return bit

Once these are defined, there are four quantities we wish to calculate: P['HH'], P['HV'], P['VH'] and P['VV'].

Let's focus on P['HV'] as an example. This is the probability that the bit value derived from an 'H' type hash on A is different to that from a 'V' type has on B. We will estimate this probability by sampling many times and determining the fraction of samples for which the corresponding bit values disagree.

The other probabilities are defined similarly: P['HH'] compares a 'H' type hash on both A and B, P['VV'] compares a V type hash on both, and P['VH'] compares a V type hash on A with a H type has on B.

These probabilities are calculated in the following function, which returns all the values of P in a dictionary. The parameter shots is the number of samples we'll use.

In [ ]:

shots = 8192
def calculate_P ( ):
    
    P = {}
    for hashes in ['VV','VH','HV','HH']:
        
        # calculate each P[hashes] by sampling over `shots` samples
        P[hashes] = 0
        for shot in range(shots):

            A, B = setup_variables()

            a = hash2bit ( A, hashes[0] ) # hash type for variable `A` is the first character of `hashes`
            b = hash2bit ( B, hashes[1] ) # hash type for variable `B` is the second character of `hashes`

            P[hashes] += (a!=b) / shots
 
    return P

Now let's actually calculate these values for the method we have chosen to set up and hash the variables.

In [ ]:

P = calculate_P()
print(P)

These values will vary slightly from one run to the next due to the fact that we only use a finite number of shots. To change them significantly, we need to change the way the variables are initiated, and/or the way the hash functions are defined.

No matter how these functions are defined, there are certain restrictions that the values of P will always obey.

For example, consider the case that P['HV'], P['VH'] and P['VV'] are all 0.0. The only way that this can be possible is for P['HH'] to also be 0.0.

To see why, we start by noting that P['HV']=0.0 is telling us that hash2bit(A, H) and hash2bit(B, V) were never different in any of the runs. So this means we can always expect them to be equal.

hash2bit(A, H) = hash2bit(B, V)        (1)

From P['VV']=0.0 and P['VH']=0.0 we can similarly get

hash2bit(A, V) = hash2bit(B, V)        (2)

hash2bit(A, V) = hash2bit(B, H)        (3)

Putting (1) and (2) together implies that

hash2bit(A, H) = hash2bit(A, V)        (4)

Combining this with (3) gives

hash2bit(A, H) = hash2bit(B, H)        (5)

And if these values are always equal, we'll never see a run in which they are different. This is exactly what we set out to prove: P['HH']=0.0.

More generally, we can use the values of P['HV'], P['VH'] and P['VV'] to set an upper limit on what P['HH'] can be. By adapting the CHSH inequality we find that

\,\,\,\,\,\,\, P['HH'] \, \leq \, P['HV'] + P['VH'] + P['VV']

This is not just a special property of P['HH']. It's also true for all the others: each of these probabilities cannot be greater than the sum of the others.

To test whether this logic holds, we'll see how well the probabilities obey these inequalities. Note that we might get slight violations due to the fact that our the P values aren't exact, but are estimations made using a limited number of samples.

In [ ]:

def bell_test (P):
    
    sum_P = sum(P.values())
    for hashes in P:
        
        bound = sum_P - P[hashes]
        
        print("The upper bound for P['"+hashes+"'] is "+str(bound))
        print("The value of P['"+hashes+"'] is "+str(P[hashes]))
        if P[hashes]<=bound:
            print("The upper bound is obeyed :)\n")
        else:
            if P[hashes]-bound < 0.1:
                print("This seems to have gone over the upper bound, but only by a little bit :S\nProbably just rounding errors or statistical noise.\n")
            else:
                print("!!!!! This has gone well over the upper bound :O !!!!!\n")

In [ ]:

bell_test(P)

With the initialization and hash functions provided in this notebook, the value of P('HV') should be pretty much the same as the upper bound. Since the numbers are estimated statistically, and therefore are slightly approximate due to statistical noise, you might even see it go a tiny bit over. But you'll never see it significantly surpass the bound.

If you don't believe me, try it for yourself. Change the way the variables are initialized, and how the hashes are calculated, and try to get one of the bounds to be significantly broken.

Bell test for quantum variables

Now we are going to do the same thing all over again, except our variables A and B will be quantum variables. Specifically, they'll be the simplest kind of quantum variable: qubits.

When writing quantum programs, we have to set up our qubits and bits before we can use them. This is done by the function below. It defines a register of two bits, and assigns them as our variables A and B. It then sets up a register of two bits to receive the outputs, and assigns them as a and b.

Finally it uses these registers to set up an empty quantum program. This is called qc.

In [ ]:

from qiskit import QuantumRegister, ClassicalRegister, QuantumCircuit

def initialize_program ():
    
    qubit = QuantumRegister(2)
    A = qubit[0]
    B = qubit[1]
    
    bit = ClassicalRegister(2)
    a = bit[0]
    b = bit[1]
    
    qc = QuantumCircuit(qubit, bit)
    
    return A, B, a, b, qc

Before we start writing the quantum program to set up our variables, let's think about what needs to happen at the end of the program. This will be where we define the different hash functions, which turn our qubits into bits.

The simplest way to extract a bit from a qubit is through the measure gate. This corresponds to the Z output of a qubit in the visualization we've been using. Let's use this as our V type hash.

For the output that corresponds to the X output, there is no direct means of access. However, we can do it indirectly by first doing an h to swap the top and Z outputs, and then using the measure gate. This will be our H type hash.

Note that this function has more inputs that its classical counterpart. We have to tell it the bit on which to write the result, and the quantum program, qc, on which we write the gates.

In [ ]:

def hash2bit  ( variable, hash, bit, qc ):
    
    if hash=='H':
        qc.h( variable )
        
    qc.measure( variable, bit )

Now its time to set up the variables A and B. To write this program, you can use the grid below. You can either follow the suggested exercise, or do whatever you like. Once you are ready, just move on. The cell containing the setup_variables() function, will then use the program you wrote with the grid.

Note that our choice of means that the probabilities P['HH'], P['HV'], P['VH'] and P['VV'] will explicitly correspond to circles on our grid. For example, the circle at the very top tells us how likely the two X outputs would be to disagree. If this is white, then P['HH']=1 , if it is black then P['HH']=0.

Exercise

• Make it so that the X outputs of both qubits are more likely to disagree, whereas all other combinations of outputs are more likely to agree.

In [ ]:

initialize = []
success_condition = {'ZZ':+0.7071,'ZX':+0.7071,'XZ':+0.7071,'XX':-0.7071}
allowed_gates = {'0': {'bloch':0, 'x':0, 'z':0, 'h':0, 'cx':0, 'ry(pi/4)': 0, 'ry(-pi/4)': 0}, '1': {'bloch':0, 'x':0, 'z':0, 'h':0, 'cx':0, 'ry(pi/4)': 0, 'ry(-pi/4)': 0}, 'both': {'cz':0, 'unbloch':0}}
vi = [[], True, True]
qubit_names = {'0':'A', '1':'B'}
puzzle = hello_quantum.run_game(initialize, success_condition, allowed_gates, vi, qubit_names, mode='line')

Now the program as written above will be used to set up the quantum variables.

In [ ]:

import numpy as np
def setup_variables ( A, B, qc ):
    
    for line in puzzle.program:
        eval(line)

The values of P are calculated in the function below. In this, as in the puzzles in the rest of this notebook, this is done by running the job using Qiskit and getting results which tell us how many of the samples gave each possible output. The output is given as a bit string, string, which Qiskit numbers from right to left. This means that the value of a, which corresponds to bit[0] is the first from the right

a = string[-1]

and the value of b is right next to it at the second from the right

b = string[-2]

The number of samples for this bit string is provided by the dictionary of results, stats, as stats[string].

In [ ]:

shots = 8192
from qiskit import assemble, transpile

def calculate_P ( backend ):
    
    P = {}
    program = {}
    for hashes in ['VV','VH','HV','HH']:

        A, B, a, b, program[hashes] = initialize_program ()

        setup_variables( A, B, program[hashes] )

        hash2bit ( A, hashes[0], a, program[hashes])
        hash2bit ( B, hashes[1], b, program[hashes])
            
    # submit jobs
    t_qcs = transpile(list(program.values()), backend)
    qobj = assemble(t_qcs, shots=shots)
    job = backend.run(qobj)

    # get the results
    for hashes in ['VV','VH','HV','HH']:
        stats = job.result().get_counts(program[hashes])
        
        P[hashes] = 0
        for string in stats.keys():
            a = string[-1]
            b = string[-2]
            
            if a!=b:
                P[hashes] += stats[string] / shots

    return P

Now its time to choose and set up the actually device we are going to use. By default, we'll use a simulator. You could instead use a real cloud-based device by changing the backend accordingly.

In [ ]:

from qiskit import Aer
device = 'qasm_simulator'
backend = Aer.get_backend(device)

In [ ]:

P = calculate_P( backend )
print(P)

In [ ]:

bell_test( P )

If you prepared the state suggestion by the exercise, you will have found a significant violation of the upper bound for P['HH']. So what is going on here? The chain of logic we based the Bell test on obviously doesn't apply to quantum variables. But why?

The answer is that there is a hidden assumption in that logic. To see why, let's revisit point (4).

hash2bit ( A, H ) = hash2bit ( A, V )        (4)

Here we compare the value we'd get from an H type of hash of the variable A with that for a V type hash.

For classical variables, this is perfectly sensible. There is nothing stopping us from calculating both hashes and comparing the results. Even if calculating the hash of a variable changes the variable, that's not a problem. All we need to do is copy it beforehand and we can do both hashes without any issue.

The same is not true for quantum variables. The result of the hashes is not known until we actually do them. It's only then that the qubit actually decides what bit value to give. And once it decides the value for one type of hash, we can never determine what it would have decided if we had used another type of hash. We can't get around this by copying the quantum variables either, because quantum variables cannot be copied. This means there is no context in which the values hash2bit(A,H) and hash2bit(A,V) are well-defined at the same time, and so it is impossible to compare them.

Another hidden assumption is that hash2bit(A,hash) depends only on the type of hash chosen for variable A, and not the one chosen for variable B. This is also perfectly sensible, since this exactly the way we set up the hash2bit() function. However, the very fact that the upper bound was violated does seem to imply that each variable knows what hash is being done to the other, so they they can conspire to give very different behaviour when both have a H type hash.

Even so, we cannot say that our choice of hash on one qubit affects the outcome on the other. The effect is more subtle than that. For example, it is impossible to determine which variable is affecting which: You can change the order in which the hashes are done, or effectively do them at the same time, and you'll get the same results. What we can say is that the results are contextual: to fully understand results from one variable, it is sometimes required to look at what was done to another.

All this goes to show that quantum variables don't always follow the logic we are used to. They follow different rules, the rules of quantum mechanics, which will allow us to find ways of performing computation in new and different ways.

Visualize the circuit that achieves the Bell-test violation

In [ ]:

initialize = []
success_condition = {'ZZ':+0.7071,'ZX':+0.7071,'XZ':+0.7071,'XX':-0.7071}
allowed_gates = {'0': {'bloch':0, 'x':0, 'z':0, 'h':0, 'cx':0, 'ry(pi/4)': 0, 'ry(-pi/4)': 0}, '1': {'bloch':0, 'x':0, 'z':0, 'h':0, 'cx':0, 'ry(pi/4)': 0, 'ry(-pi/4)': 0}, 'both': {'cz':0, 'unbloch':0}}
vi = [[], True, True]
qubit_names = {'0':'q[0]', '1':'q[1]'}
puzzle = hello_quantum.run_game(initialize, success_condition, allowed_gates, vi, qubit_names, mode='line')

In [ ]:

puzzle.get_circuit().draw(output='mpl')

In [ ]:

import qiskit
qiskit.__qiskit_version__