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Im not a scientist, so go easy on the explanation!

As I understand it we can create two entangled particles. The entangled particles have a spin property which is opposing. When we measure one of the particles for spin we get a result and we can deduce the spin direction of the unmeasured particle. All makes sense, but...

How do we know that the entangled particles didnt already have spin before being measured?

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Quantum mechanics states that the particles are in a superposition of states before observation, the particles are at every state at once. The wavefunction gives the probability of each state and when an observation is made, the wavefunction collapses to one single state. Moreover the question you put talks about something determining the state before observation and is called a local hidden variable, these are impossible, according to physicist Von Neumann, https://en.wikipedia.org/wiki/Local_hidden_variable_theory, this link will provide you with a bit more info.

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The answer is Bell inequalities. https://en.wikipedia.org/wiki/Bell%27s_theorem

Experimentally either Bell's theorem is true or there are non local (read: faster than light) communication of the "true value".

IIAOPSW
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    Answers which are just links are usually more suitable as comments... –  Jun 19 '18 at 05:12
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Suppose you have two electrons in a state of entangled spin. Suppose also that you can measure the spin for each electron along one of two axes the x axis or the z axis. Regardless of what axis you choose to measure you will get each of the possible results with probability 1/2. If you measure the spin of electron 1 along the x axis and the spin of electron 2 along the x axis, then you get opposite results: if electron 1 has a spin pointing up, electron 2 has a spin pointing down. If you measure the spin of electron 1 along the z axis and the spin of electron 2 along the z axis, then you get opposite results: if electron 1 has a spin pointing up, electron 2 has a spin pointing down. But if you measure electron 1 along the x axis and electron 2 along the z axis then the results will match with probability 1/2.

So the probability of getting a match when you compare the results depends on whether you do the same measurement on each electron. So if the electrons were set up to have the correlations in advance, the process for setting up the electron spins would somehow have to be able to work out what measurements you're going to do in advance, which is impossible. There is some maths for ruling out loopholes in this argument, but you don't need to know that to get the gist of the problem.

alanf
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  • This isn't actually answering the question the OP is asking. An answer that does is actually posted under the comments to the question. It's short and it's relevant. – Mozibur Ullah Jun 19 '18 at 08:16
  • One way we know they had no spin before the measurement is that this is ruled out by experiment, as explained in my answer. – alanf Jun 19 '18 at 11:30
  • but you begin your experiment with "suppose you have two electrons in a state of entangled spin"; so how can "they have no spin before" when you begin with them having spin? In fact, when a particle has spin it always has spin; it cannot not have spin; the only way this is possible is if you begin with a particle with no spin at all; that's why the comment above was interesting, because in their experiment they began with a pion which is not an elementary particle... – Mozibur Ullah Jun 19 '18 at 11:47
  • ... but a composite particle of two quarks - a meson; and as a composite particle it has total spin zero; this means when it decays into photons they will have opposite and equal spins. – Mozibur Ullah Jun 19 '18 at 11:48
  • The electrons are entangled with respect to their spin. Each individual electron has a state that does not correspond to it having a particular spin: there is no direction in which its reduced density matrix will be |up><up| or |down><down|. – alanf Jun 19 '18 at 13:13
  • Give it up Alan ... you might be able to fool other people with merely repeating half-digested jargon ad nauseam - but I'm afraid it doesn't fool me. – Mozibur Ullah Jun 19 '18 at 13:25
  • @MoziburUllah You seem to have some disagreement with me. If you would like to discuss that disagreement my email is alanmichaelforrester@googlemail.com. – alanf Jun 19 '18 at 14:44