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Qubit in polar form $$\left|\psi\right\rangle=\cos(\theta/2)\left|0\right\rangle+\sin(\theta/2)\left|1\right\rangle $$

Now lets say i want to keep $\cos(\theta/2) = 50.400 $ degrees angle and $\sin(\theta/2) = 35.80 $ degrees angle into a qubit using qiskit? or how can i convert these angles into a gate so that state zero $\left|0\right\rangle$ which is default input in qiskit to get to desired state $\left|\psi\right\rangle=50.400\left|0\right\rangle+35.80\left|1\right\rangle $ using the angles

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    $\begingroup$ $cos(\theta) \leq 1 $ for all $\theta$. Similar with $\sin(\theta)$. $\endgroup$
    – KAJ226
    Apr 22 at 22:41
  • $\begingroup$ if you want to specify your $\theta$ in term of degree, just make a conversion from degree to radian. $\endgroup$
    – KAJ226
    Apr 22 at 22:43
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The first thing I will mention is that the target state you specified,

$$|\psi\rangle = 50.40|0\rangle + 35.80|1\rangle$$

is not a valid quantum state, because the probabilities do not add up to one.

$$|50.40|^2 + |35.80|^2 = 3821.80 \neq 1 $$

However, we can normalize your equation,

$$|\psi\rangle = \frac{50.40}{\sqrt{3821.80}}|0\rangle + \frac{35.80}{\sqrt{3821.80}}|1\rangle$$

which is a valid quantum state. Now, starting over with the general state representation,

$$|\psi\rangle = \cos{(\theta/2)}|0\rangle + \sin{(\theta/2)}|1\rangle$$

where $0 \leq \theta \leq \pi$. To solve for $\theta$ we do,

$$\cos{(\theta/2)} = \frac{50.40}{\sqrt{3821.80}}$$ $$\theta = 2 * \cos^{-1}{\bigg(\frac{50.40}{\sqrt{3821.80}}\bigg)} = 1.23523 \text{ radians}$$

We know that action of a quantum gate is found by multiplying the matrix representing the gate with the vector which represents the quantum state.

$$|\psi'\rangle = U|\psi\rangle$$

A general unitary must be able to take the $|0\rangle$ to the above state. Such a unitary is

$$ U = \begin{pmatrix} \cos{(\theta/2)} & -e^{i \lambda}\sin{(\theta/2)} \\ e^{i \phi}\sin{(\theta/2)} & e^{i\lambda + i\phi}\cos{(\theta/2)} \end{pmatrix} $$

where $0 \leq \phi < 2\pi$, and $0 \leq \lambda < 2\pi$. In Qiskit you have access to this general unitary using the $u$ gate:

$$u(\theta, \phi, \lambda) = U(\theta, \phi, \lambda)$$

To get our desired state, we will plug in $\phi = \lambda = 0$, and $\theta = 1.23523$:

from qiskit import QuantumCircuit, QuantumRegister, assemble, Aer
from qiskit.visualization import plot_histogram

q = QuantumRegister(1)
qc = QuantumCircuit(q)
qc.u(1.23523,0,0,q)

svsim = Aer.get_backend('statevector_simulator'). # Tell Qiskit how to simulate our circuit
qobj = assemble(qc) # Create a Qobj from the circuit for the simulator to run
result = svsim.run(qobj).result() # Do the simulation and return the result
out_state = result.get_statevector()

>>> print(out_state) # Display the output state vector
[0.81526191+0.j 0.57909241+0.j]

$\cos{(\theta/2)} = \frac{50.40}{\sqrt{3821.80}} \simeq 0.815$ and $\sin{(\theta/2)} = \frac{35.80}{\sqrt{3821.80}} \simeq 0.579$, which agrees with the results of the Qiskit Aer simulator. To visualize the probability distribution of your constructed state, you can run the following:

counts = result.get_counts()
plot_histogram(counts)

Good luck!

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    $\begingroup$ Thank You for responding. Can you tell me how we will get $\theta = 1.2353$? can I go back to my original $\psi$ form the output state vector(I can right)? $\endgroup$
    – John Jones
    Apr 23 at 15:49
  • $\begingroup$ Just edited my answer to explain how you calculate $\theta$. Which original $\psi$ are you referring to? If you are trying to reach the state $|\psi\rangle = 50.40|0\rangle + 35.80|1\rangle$, the answer is that it is not possible, because it is unnormalized. All physically realizable quantum states must satisfy $|\psi|^2 = 1$. For information on probability and normalization, I recommend "An Introduction to Quantum Mechanics" by David Griffiths, Sections 1.1 - 1.4. $\endgroup$
    – rjh324
    Apr 23 at 19:54
  • $\begingroup$ Thank You for the explanation. I am looking to go back to normalized one, If I pass the output state to through the same gate again? one more question sorry but can i use normalized complex numbers in place of 50.40 and 35.80, follow the same procedure? $\endgroup$
    – John Jones
    Apr 23 at 20:04
  • $\begingroup$ The output state is the normalized state $|\psi\rangle = \frac{50.40}{\sqrt{3821.80}}|0\rangle + \frac{35.80}{\sqrt{3821.80}}|1\rangle$. If you were add a second qc.u(1.23523,0,0,q) gate to the circuit, you would reach a completely different state. However, if you were to add a qc.u(-1.23523,0,0,q) gate, you would arrive back at the state $|\psi\rangle = |0\rangle$. And yes, you can use any combination of real and/or complex numbers and follow the same procedure, so long as $|\psi|^2 = 1$. $\endgroup$
    – rjh324
    Apr 23 at 20:37
  • $\begingroup$ Can I teleport the normalized $|\psi\rangle$ and see the state of the vector after teleporting? I have have been posting these question before ? great full if you help me with this. $\endgroup$
    – John Jones
    Apr 23 at 21:00

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