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Complex phases in Stern-Gerlach filters.

  1. Sep 23, 2005 #1

    CarlB

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    I found an interesting way of calculating the complex phase changes induced by a series of Stern-Gerlach filters applied to a beam of spin-1/2 fermions. It's obvious enough that I'm sure someone else has seen it before, and I was hoping someone would give me a reference.

    A Stern-Gerlach filter allows passage of only a single spin state, for example, it may allow only spin+1/2 in the +z direction. Applied to spin-1/2 fermions, the appropriate operator for a Stern-Gerlach filter oriented in the u direction is:

    [tex]P_u = \frac{1+\vec{u}\cdot\vec{\sigma}}{2}[/tex]

    where [tex]\vec{\sigma}[/tex] is the usual vector of Pauli spin matrices.

    Consider a series of Stern-Gerlach filters that begin and end with filters with the same orientation. Such a sequence is equivalent to the projection operator for the end filter multiplied by a complex constant. For example, if the sequence consists of four filters oriented in the +z,+y,+x,+z directions (in that order), the operator for the product is:

    [tex]P_z P_x P_y P_z =
    \frac{1}{4}
    \left(\begin{array}{cc}1&0\\0&0\end{array}\right)
    \left(\begin{array}{cc}1&1\\1&1\end{array}\right)
    \left(\begin{array}{cc}1&-i\\i&1\end{array}\right)
    \left(\begin{array}{cc}1&0\\0&0\end{array}\right)[/tex]
    [tex]= \frac{1+i}{4}
    \left(\begin{array}{cc}1&0\\0&0\end{array}\right) = \frac{1+i}{4}P_z[/tex]

    In the above case, the complex constant is [tex](1+i)/4[/tex], which corresponds to a phase angle of [tex]\pi/4[/tex], and this is the subject of this note.

    The orientation vectors in the sequence of Stern-Gerlach filters define a path on a sphere. The vectors themselves define corners in the path. To get from one corner to another, the path travels on the shortest side of the great-circle (straight) paths. Note that the great circles are defined except when two consecutive corners are on exact opposite sides of the sphere, but the product of two projection operators oriented on opposite sides of the sphere is zero: [tex]P_uP_{-u}=0[/tex].

    A sequence of Stern-Gerlach filters that begins and ends with the same orientation corresponds to a path on the sphere that begins and ends with the same point. If the path is simple (that is, it doesn't cross itself), then the path splits the surface of the sphere into two portions, A and B. The total area of the surface of the sphere is [tex]4 \pi[/tex], so [tex]A+B=4\pi[/tex].

    Suppose a sequence of vectors on the sphere forms a simple path that defines a region of area A. The formula for the argument of the complex phase associated with such a sequence of Stern-Gerlach filters is:

    [tex]\arg(P_1P_2...P_nP_1) = A/2 = (4 \pi -A)/2.[/tex]

    The explicit calculation above used vectors in the directions +z,+y,+x,+z. The associated path is an octant of the sphere. The total surface area of the unit sphere is [tex]4 \pi[/tex], so an octant of it has area [tex]\pi/2[/tex]. According to the above formula, the complex phase associated with this series of Stern-Gerlach filters is [tex]\pi/4[/tex], which matches the explicit calculation.

    A simple proof of this is to consider two triangular regions that adjoin. One can show that the path that goes around the two triangular regions ends up with a phase angle equal to the sum of the phase angles for the two smaller triangles. In other words, phase angles are additive in areas, and there must be a linear relation between phase angle and area. When one considers infinitesimal paths, the constant of proportionality turns out to be 1/2.

    Carl
     
  2. jcsd
  3. Sep 23, 2005 #2
    Hello,

    Your piece reminds me Feynman, Quantum Mechanics, chap III.

    There, Feynman uses symmetry arguments to write down the probability amplitudes between two rotated states.
    It would be interresting to see if you can get these results from your starting point.

    Note, that your starting point is an "educated" starting point. Indeed, as you know very well the properties of the spin matrices, you can easily come with a projector. What has amazed me in the Feynman course is how Feynman is able to build the QM basics on so few fundamental prerequistites and coming up with so much physics.

    I will watch this threat!
     
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