How to get a qubit into superposition?

Question

I have a qubit
$$\left| \psi \right> = (\alpha_1 + i\alpha_2 ) \left|0\right> + (\beta_1 + i\beta_2 )\left|1\right>$$

so if i give values $\alpha_1 + i\alpha_2 = 1 + 4i$ and $\beta_1 + i\beta_2 = 3 + 4i$ i will nomralise these and give it to psi

now when I measure using qiskit and plot it will be like 0.6 in state zero and 0.4 in state is there any way that I can make it an equal probability?

I tried passing it through Hadamard gate but it does not give me equal probability.

Answer 1

The Hadamard gate put the states $|0\rangle$ and $|1\rangle$ into uniform superposition but not for any arbitrary state.

You can check that: $H|0\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 0\end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ and similarly, you can show that $H|1\rangle = \dfrac{|0\rangle - |1\rangle}{\sqrt{2}}$. In each case, the probability to see a $|0\rangle$ or $|1\rangle$ is $\big| 1/\sqrt{2}\big|^2 = 1/2. $

If you have an arbitrary single qubit state, $|\psi \rangle = \alpha|0\rangle + \beta|1\rangle $ then you can show that $H|\psi\rangle = (\alpha + \beta)|0\rangle + (\alpha - \beta)|1\rangle$, so the probability of seeing the state $|0\rangle$ is $\big| \alpha + \beta|^2$, and the probability of seeing the state $|1\rangle$ is $\big| \alpha - \beta \big|^2$, thus they are not the same. So applying the Hadamard gate to an arbitrary state does not bring it to a uniform superposition state.

Of course, you can find a Unitary matrix $U$ that maps the state $|\psi\rangle$ to the uniform superposition state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$. But from all your previous questions, it seems like you are using Qiskit for the work that you are doing. What you can do is to reset the particular qubit you are interesting in to the $|0\rangle$ state through the reset option, then apply the Hadamard gate. This is probably the easiest. Below is an example of how to do reset:

from qiskit import QuantumRegister, ClassicalRegister, QuantumCircuit
qreg_q = QuantumRegister(1, 'q')
creg_c = ClassicalRegister(1, 'c')
circuit = QuantumCircuit(qreg_q, creg_c)
circuit.ry(1.213, qreg_q[0])  #A quantum state that is not in uniform superposition
circuit.reset(qreg_q[0]) # reset the qubit to |0>
circuit.h(qreg_q[0]) # Apply the Hadamard  gate 
print(circuit)

     ┌───────────┐     ┌───┐
q_0: ┤ RY(1.213) ├─|0>─┤ H ├
     └───────────┘     └───┘
c: 1/═══════════════════════