Is there an easier way to predict output of 3-qubit circuit in Qiskit without using matrices?

Question

I'd like to know if there's an easy way to predict what is the output of the following code taken from this resource at question 13:

qc= QuantumCircuit(3)
qc.h(0)
qc.cx([0,0],[1,2])
backend_sv = BasicAer.get_backend('statevector_simulator') 
job = execute(qc, backend_sv,shots=1024)
result = job.result()
sv_ev = result.get_statevector(qc)

This is intended to be a practice question for the IBM Qiskit Certification. This means that one should be able to solve it without having access to a computer, and without pen and paper.

I'm able to find the solution on pen and paper, but the matrices are large $8$ by $8$. The correct answer is the following:

[0.70710678+0.j 0.        +0.j 0.        +0.j 0.        +0.j
 0.        +0.j 0.        +0.j 0.        +0.j 0.70710678+0.j]

I know that the output after the first three gates is the first Bell state, tensored with the $|0>$ state:

$$\frac{|00\rangle + |11\rangle}{\sqrt{2}} |0\rangle$$

But applying the last CNOT gate is not something I can do in my head. Is there an easier way to find the solution without dealing with 8x8 matrices in one's head?

Answer 1

Not sure what "without pen and paper" is supposed to mean (kind of a silly demand for any type of test), but if you're looking for an relatively easy way to explain the circuit without relying on vectors a matrices, you could perhaps use ket notation:

1. Three qubits start in one he all-zeros state:

$$ |\psi\rangle = |000\rangle $$

2. Hadamard is applied to qubit $q_0$, placing it in an equal super position of $|0\rangle$ and $|1\rangle$:

$$ |\psi\rangle =\frac{1}{\sqrt{2}} ( |000\rangle + |001\rangle)$$

3. This is followed by a CX gate with control in $q_0$ and targets on $q_1$ and $q_2$, so the first term of this superposition remains unchanged, second term of the superposition flips $q_1$ and $q_2$:

$$ |\psi\rangle =\frac{1}{\sqrt{2}} ( |000\rangle + |111\rangle)$$

4. This corresponds to a statevector with a $\frac{1}{\sqrt{2}}$ factor in the first and last positions:

$$ |\psi\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$