How Does Quantum Entanglement Affect Electron Spin Measurements?

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This is a pretty basic question I believe.
When you describe the state of a single particle, let's say its spin state, it can be in a superposition of different states (like up and down). When you do a measurement of its spin you'll either get up or down, since its spin state will collapse into one of them.
My question is about what happens with two particles. Let's take two electrons as an example.

1. Can you measure the spin z-components of the two electrons simultaneously?

2. Say you measure one electron with up spin and the other with down spin. What is the total spin of the electrons?

3. This is related to the last question. I am really confused by the so-called singlet state. The singlet state is a superposition of up-down and down-up states. When you have this specific superposition, the total spin is 0. I don't understand how one can measure the two-electron system to be in the singlet state, because wouldn't that imply that, when measured, the electrons are in a superposition of states? How is that even possible? I thought that when you made a measurement, the electrons chose either one or another configuration. Can somebody shed some light on this singlet state?

Thank you very much!
 
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If you apply the spin operator to the 2-electron wavefunction, what will you get?

1. the electrons are indistinguishable - so you cannot target a particular electron to measure it's spin - but you can discover individual spin states.

2. Total spin is zero.
If you find a 2-electron system with spin 0 then one must be spin up and the other spin down - and there are two ways this can happen. Since you don't know which is which - the electrons must be described by a superposition.

3. You can measure the total spin of the system - i.e. the electron spins, individually, contribute to the atomic dipole moment. If the net contribution is zero then you have a superposition of opposite spin states.

The above is a bit glib... the following may help:
http://farside.ph.utexas.edu/teaching/qm/Quantumhtml/node96.html
http://uw.physics.wisc.edu/~himpsel/449exch.pdf
 
Simon Bridge said:
If you apply the spin operator to the 2-electron wavefunction, what will you get?

1. the electrons are indistinguishable - so you cannot target a particular electron to measure it's spin - but you can discover individual spin states.

2. Total spin is zero.
If you find a 2-electron system with spin 0 then one must be spin up and the other spin down - and there are two ways this can happen. Since you don't know which is which - the electrons must be described by a superposition.

3. You can measure the total spin of the system - i.e. the electron spins, individually, contribute to the atomic dipole moment. If the net contribution is zero then you have a superposition of opposite spin states.

The above is a bit glib... the following may help:
http://farside.ph.utexas.edu/teaching/qm/Quantumhtml/node96.html
http://uw.physics.wisc.edu/~himpsel/449exch.pdf

Thank you, that was very helpful. When you have an up electron and a down electron, the total spin can either be 1 or 0. I'm having trouble understanding this. If one is pointing up and the other pointing down, why isn't the total spin just 0 everytime?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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