In Stern–Gerlach experiment, where the spin-1/2 particle is fired in $y$ direction, the SG device can be oriented to measure the spin in $z$, $x$ or any other angle in $xz$-plane. What is then the state of the particle in $y$-states before measuring anything? If it is in the form; $$\mid\psi\rangle=\frac{1}{\sqrt2}\mid y_+\rangle+\frac{1}{\sqrt2}\mid y_-\rangle$$ and we know that $$\mid y_+\rangle=\frac{1}{\sqrt2}\mid z_+\rangle+\frac{i}{\sqrt2}\mid z_-\rangle$$ and $$\mid y_-\rangle=\frac{1}{\sqrt2}\mid z_+\rangle-\frac{i}{\sqrt2}\mid z_-\rangle$$, then substituting the last 2 equations in the first one yields $$\mid\psi\rangle=\mid z_+\rangle$$ But this is inconsistent with the observation that the state of the particle is a superposition of both $\mid z_+\rangle$ and $\mid z_-\rangle$ so, the initial state can not be a superposition of $\mid y_+\rangle$ and $\mid y_-\rangle$ which means the particle must be in a definite state in $y$ direction knowing that it will be measured in $xz$ plane in the future. If this logic is true, will we consider it as a kind of delayed choice quantum experiment?
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$\begingroup$ Could you break the logical argument in the last chapter into elementary sentences? This would make it easier to follow your reasoning. $\endgroup$– Stéphane RollandinCommented Feb 10, 2018 at 10:42
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$\begingroup$ If the particle state can not be in a superposition of eigen states in the direction where it is flying ($y$ direction), then it must be always in one of the eigen state $\mid y_+\rangle$ or $\mid y_-\rangle$. So, how could be this true even if we have not yet measured the spin of the particle along any direction? Does it mean the particle moving in particular direction has a well definite spin in that direction before measuring it? Or, does the particle knows that it will be measured in $xz$ plan so it decides to have a well defined state in $y$ direction even before doing the measurement? $\endgroup$– isaacCommented Feb 10, 2018 at 11:09
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$\begingroup$ Before measuring the state of the particle, it is generally not in a pure state, but in a mixed state. Furthermore, I don't understand how we know that $|y+\rangle = \frac{1}{\sqrt{2}} \left( |z+\rangle + i |z-\rangle \right)$ $\endgroup$– NotMeCommented Feb 10, 2018 at 11:14
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$\begingroup$ $|y+⟩=\frac{1}{\sqrt2} (|z+⟩+i|z−⟩)$ this can derived from the definition of Pauli matrices. $\endgroup$– isaacCommented Feb 10, 2018 at 11:21
1 Answer
First of all, your calculations are right, but there is a thing that doesn't work in the logic you followed. When you write: $$ |\psi>=\frac{1}{\sqrt{2}}\big(|y_+>+|y_->\big), \qquad \qquad \qquad (1) $$ you are saying that the system is in this state. In this case it is written in the base of $\sigma_y$ eigenstates, but the state is independent of the base chosen. $|\psi>$ clearly isn't an eigenstate for $\sigma_y$, but changing to $\sigma_z$ base you found that $|\psi>$ is an eigenstate for $\sigma_z$. This means that if you send this particle into SG apparatus and measure spin along $z$, you will necessarily get +! If you measure along $x$, instead, you'll have 50% chance of getting +, and 50% of getting -. Summing up, knowing the state $|\psi>$ of the paricle before it is sent in the SG apparatus, allows you to predict the probabilities of getting + or - along every axis you prefer.
So it is not true that "this is inconsistent with observation that the state of the particle is a superposition of both ∣z+⟩ and ∣z−⟩". You sent a particle in ∣z+⟩ state, so you must get + as result.
What happens in a real experiment is that we don't know $|\psi>$: the atoms we shot come from a source that produces them in a random state: for example it could be in $|\psi>$ OR $|\phi>$ (I'll try to explain why that "or" is so important later). So we can not answer to your question:
"What is then the state of the particle in y-states before measuring anything?"
And therefore we can not write (1).
When we measure spin along $z$ of particles coming directly from the source, we don't know their state, but the possible results of the measurement are still only 2: + and -. That's why we see two spots on the output screen. If now we take the beam associated to the spin + result, we know that those particles are surely in the state $|z_+>$ and can feed them to another SG apparatus, on which we can make predictions because we know the state of the input particles.
If we don't know the state of the input particles, can we still make some sort of prediction on the possible results? Revisiting the previous example, the state could be $|\psi>$ with probability 1/2 OR $|\phi>$ with probability 1/2 (the manufactorer of the source says me that the source I bought generates only this two states). The foundamental thing to understand is that you can not write: $$ |\gamma>= \frac{1}{\sqrt{2}}|\psi>+\frac{1}{\sqrt{2}}|\phi>, $$ because in this way you are assigning again a single, known state to the system $|\gamma>$, and you could make the usual predictions using QM postulates. You want to take account of the fact that if the system is in $|\psi>$, you have some predictions; if the system is in $|\phi>$, instead, you have different predictions.
In other words you can not create a average state vector. In this case you must use concepts of mixed state and density operator. Using this instruments you can make predictions even in cases like this (Cohen Tannoudji, Complements $E_{III}$ and $E_{IV}$, easy to find. It was really clear for me when I studied this, and there is also this particular example of SG).
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$\begingroup$ 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. $\endgroup$– isaacCommented Feb 10, 2018 at 15:21
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$\begingroup$ 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. $\endgroup$ Commented Feb 10, 2018 at 15:31
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$\begingroup$ The difference is the first one is not prepared in state $\mid x_+\rangle$, the second is. $\endgroup$– isaacCommented Feb 10, 2018 at 16:32
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$\begingroup$ 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. $\endgroup$– isaacCommented Feb 10, 2018 at 16:36
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$\begingroup$ 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. $\endgroup$ Commented Feb 10, 2018 at 17:21