I have been reading a lot about quantum entanglement when I came up to a part which says that before measuring one of the particles there is a 50:50 chance that a particle will be one of the pair but also before measuring both subatomic particles can be both variations at the same time up and down I believe is the correct terminology. Is this correct?

  • $\begingroup$ At the heart of quantum mechanics lies that it is meaningless to talk about observables having definite values before they are measured. What have you read about quantum mechanics before you read about entanglement? $\endgroup$ – ACuriousMind Jul 11 '14 at 21:54
  • $\begingroup$ Verry little I am still in high school so I get verry little education in the field from school.most of what I know is what I have read in books which is where I discovered quantum entanglement and this problem $\endgroup$ – William Cadman Jul 11 '14 at 21:57

I will try to present the case at hand as clear as possible, but I cannot avoid it using a bit of math. To clarify notation, I will also recap what entanglement and quantum states in their simplest cases are:

Let us only consider spin as the only degree of freedom our quantum mechanical 1-particle spin-$\frac{1}{2}$ state has. Then, a single particle has two possible definite states: $|\uparrow\rangle$ with spin up and $|\downarrow\rangle$ with spin down.

However, a basic fact in QM is that any state that is a sum of these is also allowed: $|\psi\rangle = a|\uparrow\rangle + b|\downarrow\rangle$ is, for $a,b \in \mathbb{C} \; \text{with} \; a^2+b^2 = 1 $, also a valid state of our system. Such a state has no definite spin. The first spin measurement you make on it has probability $a^2$ to yield spin up and probability $b^2$ to yield spin down.

Entanglement now occurs when you have two particles which are in a very peculiar state: First, the two particles have four definite states (the physicist would say that their space of states is the tensor product of the single-particle space of states):

$$ |\uparrow\rangle|\uparrow\rangle \; \text{and}\; |\uparrow\rangle|\downarrow\rangle \; \text{and} \; |\downarrow\rangle|\uparrow\rangle \; \text{and} \; |\downarrow\rangle|\downarrow\rangle$$

Here, the first ket (as these $|\rangle$ things are called) is the definite state of the first particle and the second ket is the state of the second particle, written as $|\uparrow\rangle|\downarrow\rangle$ we mean to say that the first particle has spin up and the second particle has spin down.

One popular entangled state is now

$$|\psi\rangle = \frac{1}{\sqrt{2}}|\uparrow\rangle|\downarrow\rangle + \frac{1}{\sqrt{2}}|\downarrow\rangle|\uparrow\rangle$$

which is a state that is with equal probability, i.e. 50%, $|\uparrow\rangle|\downarrow\rangle$ or $|\downarrow\rangle|\uparrow\rangle$ when measured. If you, by clever design, got yourself such a state in an experiment, you have produced a state that exhibits a very peculiar behaviour:

  1. Before the measurement, we know nothing about either particle - there is a 50:50 chance for $|\uparrow\rangle$ or $|\downarrow\rangle$ coming up for either of them if we measure only one of them, since they are oppositely aligned in both possible measurement results.

  2. Measure one of the particles, it doesn't matter which. Let's say you measure the first. You either get $|\uparrow\rangle$ or $|\downarrow\rangle$, with equal probability.

  3. Think: If you got $|\uparrow\rangle$, you know that the full state must be $|\uparrow\rangle|\downarrow\rangle$. If you got $|\downarrow\rangle$, you know that the full state must be $|\downarrow\rangle|\uparrow\rangle$

  4. You now know with 100% certainty which state the second particle is going to be measured as without having measured it.

So, in conclusion, before your first measurement, the particle spin can be anything, you have no information at all simply by knowing it is an entangled state. Only after the first measurement can you exploit your knowledge of the state to know without looking what state the other particle is in.

If I have not answered your question appropriately with this, feel free to ask for clarification.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.