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Timeline for What is the quantum state of spin 1/2 particle in $y$ direction?

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Feb 16, 2018 at 19:16 comment added isaac Agreed, so if SG can not measure the spin along the direction of the propagation, then we can not use the arbitrarily base along that axis because it won`t give any results. This seems to have a deeper physical implication than just an experimental limitation. Therefore the principle should read:"No particle would have a meaningful measured spin-component along the axis of motion". Correct! this seems a beautiful new phrase of a known phenomena.
Feb 16, 2018 at 18:42 comment added Chronicler Allows us to predict probabilities of getting a up or down in ANY direction. You are right when you say that SG can not measure spin along the direction of propagation, but that is only due to how the experiment is set up. It has nothing to do with the state of the particle
Feb 16, 2018 at 18:39 comment added Chronicler In quantum mechanics we don't talk about the state of a particle in a certain direction. This has no sense, because the spin state of the particle is one and only one. So we can not say that the state of the particle is a simultaneous superposition of 50% up and 50% down along $z$ ,$x$ and $y$, this is completely meaningless. Knowing that the spin state space of a 1/2 spin particle has only two dimensions, we arbitrarly choose a basis given by the possible results of a measure of angular momentum along an arbitrary axis. Than we use that basis to decompose the state of the particle and this
Feb 16, 2018 at 18:04 comment added isaac I thought about that again and found the following problem; how can the particle emitted in y-direction be in a spin-state that can not be measured in that direction? Because it is impossible to align the SG-device to give information about the up-down position in y-direction. In other words, it is impossible to align the magnetic field of the device in the same direction of the motion and still expecting any deflection, correct. If this reality is physically inaccessible, how meaningful be the state of the particle in the direction of motion?
Feb 11, 2018 at 19:46 comment added isaac I think the answer will be; no. Because even the (-) y-state will also give 50% up and 50% down in z-base, so there is no way to tell which state the particle is.
Feb 11, 2018 at 19:10 comment added isaac Many thanks for the explanation. I was missing that the probability is necessarily a modulus square. So, if the particle is moving in y-direction toward SG-device oriented to measure xz-state, will it have a + in y-state from the above equation? or still we do not know it for sure?
Feb 10, 2018 at 20:40 comment added Chronicler There is a possibility that I am not understanding what you meant, but the sum of the squared moduli of the two coefficients is 1. That state is perfectly permitted and tells that the particle is in a state in which, if you measure along $y$ you'll get + with certainty,. If you measure along $z$ you'll have 50-50. I am repeating myself, it seems I can't understand your problem and help you.
Feb 10, 2018 at 19:15 comment added isaac But what is the point of writing a random state describing a particle emitted from a source as long as that state can not be the general one? Also, this (despite helpful to me) is a departure from my point in this thread. I have a reference wrote the state equation $\mid y_+\rangle=\frac{1}{\sqrt2}\mid z_+\rangle+\frac{i}{\sqrt2}\mid z_-\rangle$ and my question; if the sum of probability amplitude in this state=0 (because of imaginary part), then the state can not be a superposition in those bases which makes either a contradiction or implies that it must be always in one of the base states.
Feb 10, 2018 at 17:21 comment added Chronicler The fact that we measure 50% up and 50% down doesn't necessarily imply that the particles were all in the state $∣\psi⟩=\frac{1}{\sqrt{2}}∣z_+⟩+\frac{1}{\sqrt{2}}∣z_−⟩$. The source can produce a particle in a completely random state $∣\psi⟩=a∣z_+⟩+\sqrt{(1-a^2)}∣z_−⟩$ (omit the phase for simplicity). $a\in[0,1]$ and every value of $a$ has an equal probability to show up. I won't calculate the total probability of getting spin up and down here, but you can immagine that if we shoot many particles in this states, since every $a$ has equal probability, we will get 50% up and 50% down again.
Feb 10, 2018 at 16:36 comment added isaac How come the particle isn`t necessarily in $∣\psi⟩=\frac12 ∣z_+⟩+\frac12 ∣z_−⟩$.! the coefficient in front of the base defines the probability amplitude to find the particle in that state. So if we repeat the experiment many times, we should get spin up 50% of the time and spin down 50% as well. The general state can not be in any other superposition.
Feb 10, 2018 at 16:32 comment added isaac The difference is the first one is not prepared in state $\mid x_+\rangle$, the second is.
Feb 10, 2018 at 15:31 comment added Chronicler When the particle enters the apparatus, it isn't necessarily in $∣\psi⟩=\frac12 ∣z_+⟩+\frac12 ∣z_−⟩$. It could be in $∣\psi⟩=∣z_+⟩$, as well as in $∣\psi⟩=\sqrt{\frac{2}{3}} ∣z_+⟩+\sqrt{\frac{1}{3}} ∣z_−⟩$. We don't know. Those two states you wrote ARE the same, they will give the same physical predictions on measurements. I can't get where is the difference in physical interpretation.
Feb 10, 2018 at 15:21 comment added isaac But the state along z-direction has the form: $\mid\psi\rangle=\frac{1}{\sqrt2}\mid z_+\rangle+\frac{1}{\sqrt2}\mid z_-\rangle$ which is exactly the same as if the particle has been prepared in a state $\mid x_+\rangle=\frac{1}{\sqrt2}\mid z_+\rangle+\frac{1}{\sqrt2}\mid z_-\rangle$ , does it mean the particle has a well defined state along x-direction before doing the measurement on z-direction? In other words, both states are mathematically equivalent but they may have different physical interpretation which does not make sense.
Feb 10, 2018 at 13:43 history answered Chronicler CC BY-SA 3.0