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B A question about Electron spin

  1. Jul 22, 2016 #1

    rede96

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    I was wondering just what physical properties of an electron influence it's probability of being detected in a certain spin state?

    For example, if I prepare an electron in the spin up state along say the way y axis (vertical axis) and pass it through a detector that is at 0 degrees, then using the formula Cos2θ/2, I know there is 100% probability of detecting spin up. If I rotate my detection device to say 90 degrees then I know there is 50% chance of my equipment detecting spin up and if I rotate my device to say 60 degrees then I know there is a 75% chance of detecting a spin up.

    But what is it exactly (if known) about the electron that leads to these probabilities? Why does an angle of 60 degrees lead to a 75% probability of detection?

    By the way I have no background in physics at all, sorry. I'm just a very interested layman and my math isn't great either!
     
  2. jcsd
  3. Jul 22, 2016 #2
    Spin is another kind of probabilistic property, like position and momentum, in the quantum world.
    Just like position and momentum have an uncertainty principle between them, spin directions do as well.
    Look up the Stern-Gerlach experiment.
     
  4. Jul 22, 2016 #3

    rede96

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    Thanks for the reply. I've read up and watched some video lectures on the Stern-Gerlach experiment. I don't fully understand the Math but I do have a basic understanding of how it all works.

    However it doesn't explain just how the probability works. So for example when an electron is 'spin up' and the equipment set at a 60 degree angle, why is the probability 75%? Is it something to do with only 75% of the wave intersecting with magnetic pole of the equipment that detects the up spin for example?

    So I am trying to visualize just what is going on physically (or mathematically) that correlates the angle of the equipment to the probability of detection. If that makes sense.
     
  5. Jul 22, 2016 #4
    An electron is represented by a wavefunction which has some (complex-valued) amplitude of spin up and some amplitude of spin down. The probability of measuring spin up is given by the absolute value of the spin up amplitude squared. This is a consequence of Born's rule https://en.wikipedia.org/wiki/Born_rule, which you should just accept as a fact of nature. When you rotate the electron, you change the amplitudes of spin up and spin down. There are mathematical rules for how to change the amplitudes. The abstract mathematical object which represents the up and down amplitudes is called a spinor. It's not that different from a typical vector, but less intuitive because you can visualize a vector, but you just have to accept the mathematics for a spinor.
     
  6. Jul 22, 2016 #5
    It can be visualised as a flag.
    https://users.physics.ox.ac.uk/~Steane/teaching/rel_C_spinors.pdf
    The introduction of the extra degree of freedom makes all the difference. Just as the angle between two vectors in three dimensional space is less when viewed in two dimensional space so the angle between four dimensions is less when viewed in three.
     
  7. Jul 24, 2016 #6

    rede96

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    Thanks for the reply. I've had a look at the paper but it is beyond my level and doesn't really help me understand my question about how angle of spin (Or the equipment) is related to the probability of detection. As I mentioned before I have no physics background and only basic Math. Sorry.
     
  8. Jul 25, 2016 #7

    blue_leaf77

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    Quantum physics is, since its discovery, less intuitive relative to human being. Many physical phenomenon that take place in quantum level exhibits nature which seems to contradict our experience related to physical occurrences which are directly sensible, which we call classical physics. This may be the reason why the concepts and principles in quantum physics can be hard to visualize for beginners. In "Schroedinger picture" of quantum mechanics, the quantum state of a system is described by a wavefunction, this object is introduced as a vector in linear algebra. Apart from wavefunction, we also use operators to translate physical quantities like position, momentum, angular momentum, and energies into quantum mechanical language. With these two mathematical objects being essential in learning quantum mechanics, it's indeed difficult to comprehend a topic in this subject of physics without having any background in physics and math. To those without math and physics, quantum mechanics sounds just like a fairy tale.

    As pointed above, the quantum state of a system is described by a wavefunction. When one considers system of one electron with spin and neglecting the spatial degree of freedom, the wavefunction of this system is a vector in 2D complex vector space, i.e. any arbitrary wavefunction ##|\psi\rangle## can be written as
    $$
    |\psi\rangle = c_1 |u_1\rangle + c_2 |u_2\rangle
    $$
    where ##|u_1\rangle## and ##|u_2\rangle## are basis vectors in this space and ##c_1## and ##c_2## are complex numbers. Born's rule states that the probability of finding this system in state ##|u_i\rangle## is given by ##|c_i|^2##. Now, considering an operator called y component of spin ##S_y##, one can find the eigenvectors of this operator and having found these vectors denote them by ##|y;+\rangle## (spin up in y direction) and ##|y;-\rangle## (spin down in y direction). These vectors are usable as basis vectors for the space being considered. Therefore, any spin state of one electron system can be written as
    $$
    |\psi\rangle = c_{1y} |y;+\rangle+ c_{2y} |y;-\rangle
    $$
    This is why when you prepare the system in spin up along y axis and measure the state using a device aligned at zero degree with respect to y axis, you got 100 % chance of measuring ##|y;+\rangle##. This is so because ##c_{1y} = 1## and ##c_{2y} = 0##.

    Now spin is a vector operator, beside ##S_y## there are also ##S_x## and ##S_z## operators. When the measuring device is rotated 90 degree from the initial axis, it can represent either ##S_x## or ##S_z##. Take it to be ##S_z##, then the measurement will yield either ##|z;+\rangle## (spin up in z direction) or ##|z;-\rangle## (spin down in z direction). In turn, you can also use these last two vectors as a new basis and hence ##|\psi\rangle = c_{1z} |z;+\rangle+ c_{2z} |z;-\rangle##. Your initial state is ##|y;+\rangle##, it can be proven (using math of course) that the following equation holds
    $$
    |y;+\rangle = \frac{1}{\sqrt{2}}|z;+\rangle + \frac{i}{\sqrt{2}}|z;-\rangle
    $$
    You see from above that the probability of your device, after a rotation of 90 degrees, to detect spin up in the new (z) direction is ##|\frac{1}{\sqrt{2}}|^2 = 0.5## (50%).

    If the device was rotated 60 degree from y axis, the calculation gets slightly more complicated because it involves applying another operator called rotation operator ##\exp (i\frac{S_z}{\hbar} \frac{\pi}{3})##. Has any of what I have made as examples made sense to you (assuming you still have no background in math and physics)?
     
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