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Spin States

  1. Jan 8, 2015 #1
    I am reading a book The Theoretical Minimum I didnt understand spin and quantum states
    "All possible spin states can be represented in a two dimensional vector space."
    What it means ?
     
  2. jcsd
  3. Jan 8, 2015 #2
    Are you talking about a spin-1/2 system? Because if so, it means that all possible spin states for this system can be written in the form: ##|\chi_{1/2}\rangle=\alpha|+1/2\rangle+\beta|-1/2\rangle##, where ##\alpha## and ##\beta## are complex numbers. In other words, every possible spin configuration is given by a suitable superposition of a state with spin up and a state with spin down, i.e. it forms a two-dimensional vector space. If you want, you can identify the two basis vectors with the usual versors: ##|1/2\rangle=(1, 0)## and ##|-1/2\rangle=(0, 1)##, so a generic state will be of the kind ##\vec{\chi}_{1/2}=(\alpha, \beta)##.
     
  4. Jan 8, 2015 #3
    If you have The Theoritical Minimum book you can look it Lecture 2.2
     
  5. Jan 8, 2015 #4
    If you read a few line below the vector space sentece the author says exactly what I wrote before.
     
  6. Jan 8, 2015 #5
    I just dind understand how can we show left right spin states just use up and down states.
     
  7. Jan 9, 2015 #6
    According to the book, left and right states are the results of the measure of the spin when the system is aligned along the x-axis. In a vector form you can write them as: ##\vec{l}=\frac{1}{\sqrt{2}}(1,1)## and ##\vec{r}=\frac{1}{\sqrt{2}}(1,-1)## (you can show this, for example, diagonalizing the first Pauli matrix). But you can always write:

    \begin{align}
    |l\rangle=&\frac{1}{\sqrt{2}}\left(|1/2\rangle+|-1/2\rangle\right) \\
    |r\rangle=&\frac{1}{\sqrt{2}}\left(|1/2\rangle-|-1/2\rangle\right).
    \end{align}

    So, you can express the left/right states in terms of the up and down ones.
     
  8. Jan 13, 2015 #7
    In thinking of the Stern Gerlach experiment, I understand how the magnets can be aligned along the z-axis, and along the y axis (assuming y axis is perpendicular to propagation)..but how do you orient the magnets in the x-direction (along the propagation) to measure spin x? Wouldn't that block the beam?
     
  9. Jan 13, 2015 #8

    Nugatory

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    Staff: Mentor

    It works the same way that we can describe any compass direction using just two directions. Say we're allowed to use just north and east and we represent them as the vectors ##\vec{N}## and ##\vec{E}##.... Then south is ##-\vec{N}##, west is ##-\vec{E}##, northwest is ##\frac{\sqrt{2}}{2}(\vec{N}-\vec{E})## and so forth.

    The confusing thing about doing this with spin is that the associated magnetic moments point in opposite spatial directions, so you are tempted to think that spin-down, ##|D\rangle##, is equal to ##-|U\rangle##, the negative of spin-up. But it's not; spin-up and spin-down are orthogonal vectors in the abstract vector space. The easiest way to see this is to look at the representation of these vectors as 1x2 matrices. You'll see that not only is ##|U\rangle=|\psi_{z+}\rangle## not the negative of ##|D\rangle=|\psi_{z-}\rangle## but also that their product is zero and their sum is equal to ##|\psi_{x+}\rangle## which is a left-right spin state.
     
  10. Jan 13, 2015 #9

    Nugatory

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    Staff: Mentor

    A fairly common convention is that if you are standing at the source looking at the detector, the z-axis is up-down, the x-axis is left-right, and you need a slightly more ingenious setup if you're doing three-axis measurements.
     
  11. Jan 13, 2015 #10

    blue_leaf77

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    Probably using a magnet coil oriented along the beam propagation direction will help.
     
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