基于QED计算发送一对纠缠电子的成功概率技术问询
Alright, let's break this down step by step using Quantum Electrodynamics (QED) principles—this is a great question that ties together entanglement, particle transport, and QED's perturbative framework. First, let's clarify what we mean by "success" here: we want the probability that both entangled electrons reach their respective receivers and still maintain their initial spin entanglement when they get there. That's the core metric we're calculating.
1. Core Setup & Key Parameters
Let's ground this with concrete definitions to avoid ambiguity:
- Tube length: ( L ), electron initial velocity: ( v ) (we can handle relativistic cases with ( \gamma = 1/\sqrt{1-v2/c2} ), but non-relativistic works for most practical setups)
- Initial entangled spin state: We'll use the standard Bell state for quantum teleportation, ( |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|{\uparrow}{\uparrow}\rangle + |{\downarrow}{\downarrow}\rangle) )
2. Electron Survival Probability (Making It to the Receiver)
First, we need to calculate the chance each electron actually reaches the end of the tube without being scattered or absorbed. In QED, this relies on scattering cross-sections and attenuation rates:
- Vacuum tube scenario: The main risk is scattering off tube walls (metal or dielectric). For metal walls, electrons interact with free electrons and lattice ions—we use Mott scattering (QED's treatment of spinful electron scattering off charged targets) to compute the cross-section ( \sigma ). If the wall has atomic number density ( n ), the probability per unit length of scattering is ( \lambda = n\sigma ). The survival probability for one electron over length ( L ) is ( e^{-\lambda L} ), so both electrons making it is ( (e^{-\lambda L})^2 = e^{-2\lambda L} ).
- Medium-filled tube scenario: If there's gas inside the tube, add in scattering off gas atoms (ionization and elastic scattering cross-sections, calculated via QED perturbation theory) to update ( \lambda ).
3. Entanglement Retention Probability (Staying Entangled After Transport)
Even if both electrons arrive, entanglement can be lost to decoherence—this is where QED's spin-environment interactions come in:
- Spin-orbit coupling: As electrons move through the tube, they experience electric/magnetic fields from walls or medium atoms. This coupling twists spins randomly, breaking the correlated entanglement.
- Vacuum fluctuations: Virtual photons in the QED vacuum interact with electron spins, causing tiny random phase shifts—but this is usually negligible compared to wall/medium scattering unless in ultra-high vacuum with perfectly smooth walls.
To quantify this, we use density matrices: the initial entangled state's density matrix is ( \rho_{\text{initial}} = |\Phi+\rangle\langle\Phi+| ). After transport, the density matrix becomes mixed, and we calculate the fidelity ( F = \langle\Phi^+| \rho_{\text{final}} |\Phi^+\rangle )—this is the probability the state remains the original entangled Bell state.
If ( p ) is the probability a single electron's spin flips during transport (calculated via QED spin-flip scattering cross-sections), the entanglement stays intact if both spins flip or neither flips (since ( |\Phi^+\rangle ) is symmetric under joint spin flips). That probability is:
(1-p)^2 + p^2 = 1 - 2p + 2p^2
If only one spin flips, entanglement is broken—those cases contribute ( 2p(1-p) ) to the mixed state.
4. Total Success Probability
Combining the two factors above, the total success probability is the product of both electrons surviving transport and retaining their entanglement:
[
P_{\text{total}} = e^{-2\lambda L} \times (1 - 2p + 2p^2)
]
Idealized Edge Case
For a perfect vacuum tube with zero wall scattering (a thought experiment, since no such tube exists), ( \lambda = 0 ) and ( p = 0 ), so ( P_{\text{total}} = 1 )—100% success. In real setups, you'd plug in actual values for tube length, electron velocity, wall material, and medium to compute ( \lambda ) and ( p ) via QED perturbation calculations.
内容的提问来源于stack exchange，提问作者Lior
