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I think this question is more suitable for physics, but I am using a quantum computing example to understand why this should not work. Note that I am not a physicist and I know only the basics of quantum mechanics and computing, nothing at all about relativity.

Assume we have two qubits in a uniform state: $$ \frac{1}{2} ( |00\rangle + |01\rangle + |10\rangle + |11\rangle ). $$

Now, assume we send the second qubit to another planet by keeping the superposition, and we have the first qubit in our hands on Earth. The objective is to modify the value of the second qubit. We can achieve this by introducing

$$ H = 1/\sqrt{2} * \begin{bmatrix} 1& 1 \\ 1 & -1 \end{bmatrix}, $$ that is the Hadamard gate, and

$$ H’ = 1/\sqrt{2} * \begin{bmatrix} 1& -1\\ 1 & 1 \end{bmatrix}. $$

If we apply $H$ to our qubit, the final state becomes $$ \frac{ 1}{\sqrt{2}} ( |00\rangle + |10\rangle ). $$ That is, the second qubit is 0. While if we apply $H'$ to the first qubit, the final state becomes $$ \frac{ 1}{\sqrt{2}} ( |01\rangle + |11\rangle ). $$ That is, the second qubit is 1.

So, what's wrong in this example?


Edit: this is the Qiskit code which is showing the result by using the Hadamard gate. Maybe qubits are wrongly enumerated:

from qiskit import QuantumCircuit, transpile, assemble, Aer
import numpy as np

# Create a quantum circuit with two qubits
qc = QuantumCircuit(2)

# Apply Hadamard gate to both qubits: uniform state
qc.h(0)
qc.h(1)

simulator = Aer.get_backend('statevector_simulator')
compiled_circuit = transpile(qc, simulator)
job = assemble(compiled_circuit)
result = simulator.run(job).result()
final_statevector = np.array(result.get_statevector())
print("Initial State:")
for i, amp in enumerate(final_statevector):
    binary_str = format(i, '02b')  
    print(f"{amp:.2f} |{binary_str[0]}{binary_str[1]}>")

# Send the second qubit on Proxima Centauri and apply H to the first one
qc.h(0)

# Simulate the circuit to get the final statevector
simulator = Aer.get_backend('statevector_simulator')
compiled_circuit = transpile(qc, simulator)
job = assemble(compiled_circuit)
result = simulator.run(job).result()
final_statevector = np.array(result.get_statevector())
print()
print("Final State:")
for i, amp in enumerate(final_statevector):
    binary_str = format(i, '02b')  
    print(f"{amp:.2f} |{binary_str[0]}{binary_str[1]}>")

Edit 2: these are my calculation in the case of the application of the Hadamard gate the uniform state. We start with the following state vector: $$ \psi = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{bmatrix} $$ If we apply the Hadamard gate to the first qubit, we have to apply the following unitary $$ U = I \otimes H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1& 1 & 0 & 0\\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix} $$ to the state $\psi$, so that

$$ U\psi = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}. $$

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    $\begingroup$ I think you applied the gate to the wrong qubit (or maybe made another mistake in the calculation since you didn't give details). Applying H on the first qubit will give $1/\sqrt{2} (|00\rangle + |01\rangle)$ state, i.e., our qubit is in $|0\rangle$ state, but the other qubit is still in uniform state. $\endgroup$
    – EvgeniyZh
    Sep 13, 2023 at 7:35
  • $\begingroup$ @EvgeniyZh please have a look to the Qiskit code and the explicit calculations. Maybe I am enumerating the states in a wrong way. $\endgroup$
    – francler
    Sep 13, 2023 at 9:55
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    $\begingroup$ qiskit uses the opposite convention for qubit ordering compared to most other text books/courses $\endgroup$
    – DaftWullie
    Sep 13, 2023 at 11:43
  • $\begingroup$ For some historical background, Nick Herbert proposed FLASH, a scheme based on entanglement to achieve superluminal communication, in 1981. This led to the introduction of the famous no-cloning theorem, which can be shown to rule such schemes out. $\endgroup$
    – Kiro
    Sep 15, 2023 at 10:08

2 Answers 2

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The state that you start with can be written as $(|0\rangle+|1\rangle)\otimes (|0\rangle+|1\rangle)$. This means it is separable. Nothing you do on the first qubit will ever change anything on the second qubit (you have a calculation error which is probably that you applied Hadamard on the wrong qubit).

You could choose your initial state to be entangled. Then, sometimes, what you do on the first qubit does affect the second qubit. However, it does this non-deterministically. The only way that the second party can measure their qubit and extract (useful) information about what the outcome was is to first receive a message from the first person that tells them something about what happened. That message has to travel no faster than the speed of light, so the second person never has any opportunity to see any consequences of superluminal signalling.

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  • $\begingroup$ Thank you for your response. In fact, I do not understand how the initial uniform state could show any form of entanglement. In fact, I tried to make the same computations starting with the Bell state, but there is no unitary operator which can provide any of the both states I got, as such operator would not even be invertible! So, at this point the question is: is it possible to start from an entngled state and get some other state with an "arbitrarily" high probability of getting $0$ or $1$ on the second qubit, by only acting on the first qubit? $\endgroup$
    – francler
    Sep 13, 2023 at 10:02
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    $\begingroup$ @francler Simply, no. A unitary on one qubit has no effect on the other qubit. The only thing that can change it is to use a measurement. But measurement outcomes are random (50:50 if you use a maximally entangled state such as the Bell state). $\endgroup$
    – DaftWullie
    Sep 13, 2023 at 11:42
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Following your edit, I'll just clarify some things (and to just back-up @DaftWullie's answer and the comments). The final state is (following edit2): $$ |\psi\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle $$ But as you can see, this is a seperable state, because it can be expressed as a tensor product of individual qubit states. This means that there is no correlation between the first qubit and the second qubit. Hence, when any measurement is taken, there is no correlation.

Furthermore, applying $H'$ to the least significant qubit (on the right in little-endian) gives: $$ H'_0|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle +|1\rangle)\otimes \frac{1}{\sqrt{2}}(|0\rangle +|1\rangle) $$ If instead you meant to apply $H'$ to the most significant qubit in $|\psi\rangle$ then: $$ H_1'|\psi\rangle = |1\rangle\otimes|0\rangle $$

So in the given scenario:

  • After applying the Hadamard gate $H$ or $H'$ the two qubits are in a superposition but not entangled. They are described by a separable state, there is no correlation between the two qubits.
  • As a result, manipulating the qubit on Earth will not have any immediate effect on the qubit on the other planet. If we measure the qubit on Earth, it will not provide any information about the qubit on the other planet since they are not correlated.
  • The flaw in the example is the assumption that modifying the first qubit will directly change the state of the second qubit, even though the two qubits are in a separable state and are not entangled. However, quoting @DaftWullie verbatim: 'you could choose your initial state to be entangled', which leads to a different discussion.

If you're interested in a deeper mathematical insight into the entanglement of quantum states, density matrices provide a framework to determine whether a state is mixed or pure (note that a pure state is not the same as a separable state) and can be especially useful for discerning entanglement.

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