I was reading about teleportation I have some question about it. Can I keep normalized data (some customized data) on probability amplitudes (which are complex numbers) of a qubit (Alice's side)? Then after teleporting this qubit can I retrieve custom data on the receiving side (called as Bob's side)? Is this possible scenario or not? Is there any Qiskit code running on quantum computer available for this?


1 Answer 1


Yes, you can send normalized data across participants assuming you can convert them to probability amplitudes. To give a quick example, consider this single-qubit state: $$|\psi\rangle= \alpha|0\rangle + \beta|1\rangle$$

Where $\alpha$ and $\beta$ are normalized data ($\alpha^{2} + \beta^{2} = 1$) that Alice is trying to send to Bob.

Then the following steps are to be followed by them:

  1. Alice and Bob create an entangled pair of qubits and each holds one part of it. Usually, this is done via $|\phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ which is one of the four bell's state. The rest of the step assumes we use $|\phi^{+}\rangle$ to pass $|\psi\rangle$ from Alice to Bob.

  2. Alice applies Controlled-Not gate where the target is the qubit from the Bell's state that Alice holds and control is $|\psi\rangle$.

  3. Hadamard gate is applied on $|\psi\rangle$. Then the two qubits that Alice holds are measured. (Don't forget that $|\psi\rangle$ in our example is a single-qubit state)

  4. The two states that Alice sees is then classically communicated to Bob. (In our case it can either be 00, 01, 10 or 11)

  5. And depending on the results that Alice communicates with him, Bob does the following on his part of the entangled state(which essentially is a single qubit):

    (a) If 00, Apply Identity gate.

    (b) If 01, Apply X gate.

    (c) If 10, Apply Z gate.

    (d) If 11, Apply Z gate followed by X gate.

  6. The result of the above computations would be that Alice will lose her $|\psi\rangle$ and Bob's entangled part will now be equal to $|\psi\rangle$. Note that they both cannot keep $|\psi\rangle$ because of the No-Cloning Theorem.

While the above example is for a single-qubit case, the same logic applies for multiple qubit cases, you could pack your normalized data into a unitary matrix and then convert it to a gate.

However, you would need to repeat the experiment more times in order to retrieve the data from the amplitude states (In the above example, you would get $|0\rangle$ with $\alpha^2$ probability). Here's the qiskit implementation of it.

  • $\begingroup$ I am seeing unitary gates. It’s seems like we have to give angels theta and phi as input. If I want to give a input like 1 +4i ,2+3i as alpha and beta how can I get angels? And in qiskit I am seeing the circuit always initiate from state zero. Thank you responding. $\endgroup$
    – John Jones
    Commented Mar 3, 2021 at 5:02
  • $\begingroup$ A generalized Unitary gate takes three angles, $\theta$ $\phi$ and $\lambda$ and it is denoted by U3 gate in qiskit. You would need to convert complex numbers in their polar form to determines the angles. Following this way will be tedious since you would need to figure out $\theta$ $\phi$ and $\lambda$ via comparison, a better way could be directly feeding the matrix and let qiskit convert it into the gate. $\endgroup$ Commented Mar 3, 2021 at 5:50
  • $\begingroup$ Doing that requires you to make a unitary matrix on the normalized data(data in your example is not normalized). You can make any random matrix into a unitary matrix by doing Schmidt decomposition. And finally you would have to do some post-processing to get back the original (pre-Schmidt) values. $\endgroup$ Commented Mar 3, 2021 at 5:54
  • $\begingroup$ can anyone give an example for converting complex $\alpha$ and $\beta$ into a unitary matrix? for example, I have normalized $\alpha$ and$\beta$ as (0.24506+0.9633i,0.0046238+0.10943i) how can I convert this into a unitary gate and use it in qiskit? Thank You. $\endgroup$
    – User1086
    Commented Mar 3, 2021 at 20:57

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