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Stern Gerlach

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    In an experiment, we've isolated silver atoms that are in a purely [itex] |\uparrow \rangle [/itex] state with respect to the z-axis, and feed these into a Stern-Gerlach apparatus oriented at an angle of [itex] \frac{\pi}{4} [/itex] between the x and z axes, which measures an observable we will call [itex] J_\frac{\pi}{4} [/itex]. Let the corresponding outcomes be represented by [itex] |+ \rangle, |-\rangle [/itex]. If we were to recombine these beams into a single beam, what is this new state?

    3. The attempt at a solution

    I imagine that the new state will be a mixed state of [itex] |+ \rangle, |-\rangle [/itex], or also as a mixed state up [itex] |\uparrow \rangle, |\downarrow \rangle [/itex]. Thus we can write [itex] | \psi \rangle = p_1 |+ \rangle + p_2 |-\rangle [/itex]. I'm not sure what p1 and p2 should be though, are they just [itex] \frac{1}{\sqrt{2}} [/itex] since they will occur with equal probabilities?
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  3. Oct 11, 2008 #2


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    If the apparatus measures an observable [itex]J_{\frac{\pi}{4}}[/itex], then shouldn't the final state be

    [tex] \left| \psi \right> = \hat{J}_{\frac{\pi}{4}} \left| \uparrow \right> [/tex]

    ??? (I've used a hat to denote the fact that J is an operator)

    But what is [itex]\hat{J}_{\frac{\pi}{4}}[/itex] in terms of |+ [itex]\rangle, |-\rangle [/itex] and the corresponding bras? And what is [itex] \left| \uparrow \right>[/itex] in terms of those basis vectors?
  4. Oct 11, 2008 #3
    There is a big difference between a mixed state and a superposition. [itex] | \psi \rangle = p_1 |+ \rangle + p_2 |-\rangle [/itex] is a superposition, not a mixed state. If you write it as a superposition, the coefficients would be square roots of probabilities, not probabilities.

    The probabilities are not equal. They would be equal if the magnetic field would be in the x direction.

    In any case, you have to think about how you could recombine the two "beams." For instance, one way would be to just go through the inverse process....
  5. Oct 11, 2008 #4

    No that's not the final state: instead you have to decompose the initial state in terms of the two eigenstates of
    [itex] \hat{J}_{\frac{\pi}{4}} [/itex].
  6. Oct 11, 2008 #5


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    I fail to see how [itex]\hat{J}_{\frac{\pi}{4}} \left|\uparrow \right>[/itex] isn't the final state, the initial state [itex] \left| \uparrow \right>[/itex] is acted upon by J to produce two beams which are then combined into a single beam/state [itex] \left| \psi \right>[/itex]. To get anything useful out of that, you will have to decompose both J and the initial state into the eigenbasis of J, but I had already hinted at that with my next two (rhetorical) questions.
  7. Oct 11, 2008 #6
    A measurement of the observable [itex] A [/itex] certainly does not create the state [itex] A|\psi\rangle [/itex] . It projects onto eigenstates instead.
    Last edited: Oct 11, 2008
  8. Oct 12, 2008 #7
    Well, in any case I should be able to write it either as a superposition or a mixed state. I do realize though my mistake in that the mixed state density matrices allow us to write probabilities in a linear sense, whereas superpositions are quadratic in nature (though I did use the correct coefficient in saying that if they were equal, it would be [itex] \frac{1}{\sqrt{2}} [/itex].

    In any case, we have the new mixed state in terms of [itex] \rho = p_1 | + \rangle \langle + | + p_2 |-\rangle \langle - | [/itex]. However, I'm still not entirely sure how to get the coefficients. Since these would correspond to probabilities, we want p1 to be the probability of the + state. I calculated the + state as being

    [tex] | + \rangle = \frac{1}{\sqrt{2}} | \uparrow \rangle + \frac{1+i}{2} | \downarrow \rangle[/tex]

    So then wouldn't we expect [itex] Pr(+) = |\langle + | + \rangle|^2 [/itex]? My problem is that this yields 1
  9. Oct 12, 2008 #8


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    Shouldn't you have [itex]Pr(+) = |\langle + | \uparrow \rangle|^2 [/itex] since the initial state is [itex]| \uparrow \rangle[/itex]?
  10. Oct 12, 2008 #9
    Yes, that does make sense, but doesn't that just yield equal probabilities? And I thought borgwal said they weren't equal.
  11. Oct 12, 2008 #10
    Well, there is a part in between where we are asked to find the Jx operator, and so in that I let [itex] | \rightarrow \rangle = \frac{1}{\sqrt{2}} \left( | \uparrow \rangle + e^{i\theta} | \downarrow \rangle \right) [/itex]. Then defining the relative phase with respect to the positive x-axis, I let [itex] \theta = 0 [/itex] (which also mean that [itex] | \leftarrow \rangle [/itex] had an angle associated to [itex] \theta = \pi [/itex]).

    Thus, now that I've defined the phase relative to the positive x-axis, then since we're now at an angle of [itex] \frac{\pi}{4} [/itex] I've set [itex] | + \rangle = \frac{1}{\sqrt{2}} \left( | \uparrow \rangle + e^{i\theta}| \downarrow \rangle \right) [/itex] where now [itex] \theta = \frac{\pi}{4} [/itex]. That's how I got that representation.
  12. Oct 12, 2008 #11
    What you need is the operator for angular momentum in the [itex] (\hat{x}+\hat{z})/\sqrt(2) [/itex] direction, and its eigenstates.

    The states you wrote down [equal superpositions of spin up and down with some relative phase] are eigenstates of components of the angular momentum operator perpendicular to z (i.e., in the x-y plane). In other words, your phase [itex] \theta [/itex] has to do with the angle between the x- and y-axes, not between the x- and z-axes.
  13. Oct 13, 2008 #12
    Yes, I made a stupid mistake in the beginning that propagated throughout my work and was confusing me terribly. My methods were correct, but my initial state was not.

    A question though, am I allowed to interchange between the mixed state, density matrix representation and the superposition representation?

    I ask because now that I've recombined the beams of [itex] J_\frac{\pi}{4} [/itex] I'm told to feed this beam back into [itex] J_x [/itex], and measure the probabilities of getting [itex] \pm\frac{\hbar}{2} [/itex]. I want to do this by applying Jx onto the combined state, but since it was done in density matrix form, I would need to transfer it to the superposition state first.
  14. Oct 13, 2008 #13
    Technically, the output state of the spin has become entangled with its center-of-mass motion. So the output is no longer a pure state for the spin, it's a mixed state.

    There is really only one way to recombine the beams again, and that is by the inverse transformation (assuming you don't measure where the Ag atom ended up before recombining), which acts on both spin and center-of-mass degree of freedom. If your professor meant a different solution than applying the inverse transformation then I suppose he might mean that you should take the mixed state of the spin and put that on the original Stern Gerlach device. But you cannot recombine the two outputs into one and the same beam while keeping the spin in the same mixed state!

    Edit: sorry to make this more difficult than it should be. I think the question about recombining is a bit unfortunate.
    Last edited: Oct 13, 2008
  15. Oct 13, 2008 #14
    Okay, so given that I have the mixed state, and pass this into an apparatus measuring Jx, how would I go calculating the associated probabilities of the outcomes?
  16. Oct 13, 2008 #15
    Given a mixed state [itex] \rho [/itex] and the operator [itex] J_x [/itex], you'd first find its eigenstates |k>, and then calculate [itex] P_k= <k|\rho|k> [/itex]
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