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I Complex coefficients in QM

  1. Sep 5, 2016 #1
    Ok. I will try to do my best to explain you what is my doubt. I'm not a native English Speaker and the book I was reading is not in English Lang, but I've translated it to English.

    In a spin-1/2 Stern Gerlach-like experiment, we can express the ket representing the spin component of the electrons in one direction as a linear combination of the ket spin components in z direction (z choosen to be standard basis), by multiplying them by a coefficient and summing the terms. My book says each coefficient is complex and thus have an amplitude and phase. Ok. To express the spin component in x direction, my book says, choose the phase angle to be zero for both coefficients: "we are free to choose the value of the phase. This freedom comes from the fact that we have required only that the x-axis be perpendicular to the z-axis, which limits the x-axis only to a plane rather than to a unique direction.".

    I dont understand this reasoning. I thought the phase angle of complex numbers were just a form of locate them in a complex plane, but according to the book that angle determines direction (in this case, the direction along the spin component is alligned to).

    Having doing that, they derived the spin component in y direction as follows:

    "having arbitrarily choosen the phase to be zero for the x states, we are no longer free to make that same choice for the y states.

    After they derived the equations, they founded the angle to be ± π/2 and come to their conclusion:

    "The two choices for the phase correspond to the two possibilities for the direction of the y-axis relative to the already determined x- and z-axes."

    So what can I conclude about all those things? Phase angle of a complex number determine its direction in 3 dimensional space as well its location in the complex plane? Can we represent both complex plane and "cartesian plane" of x,y,z-direction together? How could the direction of a spin component be determined only by a coefficient?
     
    Last edited: Sep 5, 2016
  2. jcsd
  3. Sep 5, 2016 #2

    mfb

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    What do you mean by "location"? Spin has a direction only. The particle has a location but that is irrelevant in this context.

    The phase difference between the two z-components determines the orientation in the x-y-plane. You can freely choose how to call the direction of 0 phase difference as this 0 is arbitrary.
     
  4. Sep 5, 2016 #3
    Thanks for the answer. By location I mean the coordinates of the complex number written in polar form, for example the location of z = 1 (cos(θ) + i sin(θ)) is (1, θ).
     
  5. Sep 5, 2016 #4

    mfb

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    Well, the value of a complex number determines where it is in the complex plane, sure. That complex plane is not associated to spatial directions.
     
  6. Sep 5, 2016 #5
    So what do you think the autor were trying to say when he wrote

    "we are free to choose the value of the phase. This freedom comes from the fact that we have required only that the x-axis be perpendicular to the z-axis, which limits the x-axis only to a plane rather than to a unique direction.".

    "The two choices for the phase correspond to the two possibilities for the direction of the y-axis relative to the already determined x- and z-axes."
     
  7. Sep 6, 2016 #6

    mfb

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    The same thing I said in post #2, probably.
     
  8. Sep 6, 2016 #7

    stevendaryl

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    I'm not sure I understand how much of the following is purely conventional (that is, depending on somewhat arbitrary choices), but a pretty standard way to represent a spin-1/2 particle's spin state is this:

    For a particle that is spin-up in direction [itex]\vec{r}[/itex], write [itex]\vec{r}[/itex] in polar coordinates:
    [itex]r_x = r sin(\theta) cos(\phi)[/itex]
    [itex]r_y = r sin(\theta) sin(\phi)[/itex]
    [itex]r_z = r cos(\theta)[/itex]

    Then the spin state for spin-up in direction [itex]\vec{r}[/itex] is given by:

    [itex]|\psi\rangle = cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}} |U\rangle + sin(\frac{\theta}{2}) e^{+i \frac{\phi}{2}} |D\rangle[/itex]

    where [itex]|U\rangle[/itex] and [itex]|D\rangle[/itex] are the states corresponding to spin-up and spin-down in the z-direction. So the phase of the coefficients determines the angle [itex]\phi[/itex], while the modulus determines the angle [itex]\theta[/itex].
     
  9. Sep 6, 2016 #8
    Thanks. Can you show your work in relating phase angle of the coefficients with the angles of direction that describes [itex]\vec{r}[/itex]?
     
  10. Sep 6, 2016 #9

    stevendaryl

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    I'm not sure what's the shortest, simplest path to this answer. Define the following matrices:
    • [itex]|U_z\rangle = \left( \begin{array}\\ 1 \\ 0 \end{array} \right)[/itex] (meaning the state with spin-up in the z-direction)
    • [itex]|D_z\rangle = \left( \begin{array}\\ 0 \\ 1 \end{array} \right)[/itex] (meaning the state with spin-down in the z-direction)
    • [itex]\sigma_z = \left( \begin{array}\\ 1 & 0 \\ 0 & -1 \end{array} \right)[/itex]
    • [itex]\sigma_x = \left( \begin{array}\\ 0 & 1 \\ 1 & 0 \end{array} \right)[/itex]
    • [itex]\sigma_y = \left( \begin{array}\\ 0 & -i \\ i & 0 \end{array} \right)[/itex]
    Then a state that is spin-up in direction [itex]\vec{r}[/itex] (a unit vector, meaning that [itex](r_x)^2 + (r_y)^2 + (r_z)^2 = 1[/itex]) satisfies the eigenvalue equation:

    [itex]\vec{\sigma} \cdot \vec{r} \left( \begin{array}\\ \alpha \\ \beta \end{array} \right) = \left( \begin{array}\\ \alpha \\ \beta \end{array} \right)[/itex]

    where [itex]\vec{\sigma} \cdot \vec{r} = \sigma_x r_x + \sigma_y r_y + \sigma_z r_z = \left( \begin{array}\\ r_z & r_x - i r_y \\ r_x + i r_y & - r_z \end{array} \right)[/itex]

    In polar coordinates, [itex]\vec{\sigma} \cdot \vec{r} = \left( \begin{array}\\ cos(\theta) & e^{-i \phi} sin(\theta) \\ e^{i \phi} sin(\theta) & - cos(\theta) \end{array} \right)[/itex]. So that eigenvalue equation implies:

    [itex]\alpha cos(\theta) + \beta e^{-i\phi} sin(\theta) = \alpha[/itex]

    or

    [itex]\alpha (cos(\theta)-1) + \beta e^{-i\phi} sin(\theta) = 0[/itex]

    Using the half-angle formula, [itex]cos(\theta)-1 = -2 sin^2(\frac{\theta}{2})[/itex], and [itex]sin(\theta) = 2 sin(\frac{\theta}{2}) cos(\frac{\theta}{2})[/itex], so we can write the eigenvalue equation as:

    [itex]2 sin(\frac{\theta}{2}) e^{-i \frac{\phi}{2}} (-\alpha sin(\frac{\theta}{2}) e^{+i\frac{\phi}{2}}+ \beta e^{-i\frac{\phi}{2}} cos(\frac{\theta}{2})) = 0[/itex]

    where I factored out [itex]2 sin(\frac{\theta}{2}) e^{-i \frac{\phi}{2}}[/itex] to make the answer obvious. Clearly a solution to the above equation is: [itex]\alpha = cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}}[/itex] and [itex]\beta = sin(\frac{\theta}{2}) e^{+i \frac{\phi}{2}}[/itex]
     
  11. Sep 7, 2016 #10
    Thank you!
     
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