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Uncertainty in spin on multiple axes

  1. Jan 10, 2015 #1
    In my quantum mechanical studies, I came across the information that if you know an electron's spin on one axis, then you can not know its spin on another axis. For example, if you know that an electron is spin up on the z-axis, then apparently due to the Uncertainty Principle, you can not know whether it is spin right or left on the x-axis.

    However, I also found out that apparently you can still apply spin operators to eigenstates that don't correspond to that particular spin operator.

    Example: For the spin operator σ on the x-axis:

    σ11 and σ22= 0
    σ12 and σ21 = 1

    Now if you apply this operator to the spin up eigenstate <1 , 0> (which corresponds to the z-axis), then the product of your matrix vector multiplication comes out to be:

    <0 , 1> (which is the spin down state on the z-axis)

    Now what exactly does this mean physically? I initially thought that the fact that you can apply the x-axis spin operator to a z-axis eigenstate and simply get a flipped z-axis eigenstate meant that spin up on the z-axis would turn into spin left on the x-axis (since spin down and spin left are both -1 eigenvalues) if you were to redo a measurement of spin on the x-axis this time. However, wouldn't that interpretation violate the uncertainty principle since I would know that spin up on the z-axis = spin left on the x-axis? If this interpretation is incorrect, then what exactly is the implication of the result of applying a spin operator to an eigenstate for another axis?
     
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  3. Jan 10, 2015 #2

    Nugatory

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    Staff: Mentor

    You applied ##\sigma_x## to ##|z_+\rangle## (that is, the eigenvector of ##\sigma_z## with eigenvalue 1) and got a different vector, ##|z_-\rangle##, back. All that tells you is that ##|z_+\rangle## is not an eigenvector of ##\sigma_x##, and we already knew that.

    If you want to calculate the results of a spin measurement along the x-axis of a particle in the state ##|z_+\rangle##, you have to rewrite ##|z_+\rangle## in terms of the eigenvectors of ##\sigma_x##: ##|z_+\rangle = \frac{\sqrt{2}}{2}(|x_+\rangle + |x_-\rangle)##. Now you can see that the measurement along the x-axis will yield +1 or -1 (the eigenvalues of ##\sigma_x##), each with 50% probability.
     
  4. Jan 10, 2015 #3
    I do not think this has anything to do with the uncertainty principle. There are three independent variables the spin in the x-direction, the spin in the y-direction, and the spin in the z-direction. Knowing one of these tell you nothing about the other two.
     
  5. Jan 10, 2015 #4

    Nugatory

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    Staff: Mentor

    On the contrary, it is directly related to the uncertainty principle and is in some ways a clearer expression of the modern understanding of that principle than the older "measuring the position disturbs the momentum, and vice versa".

    To have a definite value of the spin along the z-axis we must prepare the system in an eigenstate of ##\sigma_z##. Because ##\sigma_z## does not commute with ##\sigma_x## that state cannot also be an eigenstate of ##\sigma_x## and therefore cannot have a definite value for the spin along the x-axis. If we tried something that did commute with ##\sigma_z##, we wouldn't have this problem; for example, we could prepare the system in such a way that the spin along the z-axis and the magnitude of the total spin both had definite values.
     
  6. Jan 12, 2015 #5

    vanhees71

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    2016 Award

    The general Heisenberg-Robertson uncertainty relation reads
    ##\Delta \sigma_z \Delta \sigma_x \geq \frac{1}{2} |\langle [\hat{\sigma_z},\hat{\sigma}_x \rangle|.##
    If you take an eigenvector of ##\sigma_z##, it doesn't tell you anything, because then the right-hand side is vanishing and thus the uncertainty relation trivially fulfilled.
     
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