Wolfram Alpha:

```Spin 1/2 Eigenvectors
---------------------

Consider z axis component of spin. Concentrate on spin and
nothing else.  Therefore, |ψ> = α|up> + β|down>.  This can
also be represented as a column vector:

- -
| α |
| β |
- -

The eigenvalues are +1 and -1.

+ and - can be considered as basis vectors with α and β as
the coefficients.

z-axis component:

-  -  - -      - -
|1  0|| α | = ±| α |
|0 -1|| β |    | β |
-  -  - -      - -

Positive:

-  -      - -
|  α | = +| α |
| -β |    | β |
-  -      - -

∴ β = -β so β must equal 0 and

-  -      - -
|  α | = +| α |
| -β |    | 0 |
-  -      - -

Negative:

-  -      - -
|  α | = -| α |
| -β |    | β |
-  -      - -

∴ α = -α so α must equal 0 and

-  -      - -
|  α | = -| 0 |
| -β |    | β |
-  -      - -

Therefore, the eigenvectors are,

- -       - -
| 1 | and | 0 |
| 0 |     | 1 |
- -       - -

x-axis component:

-  -  - -      - -
|0  1|| α | = ±| α |
|1  0|| β |    | β |
-  -  - -      - -

Positive:

- -      - -
| β | = +| α |
| α |    | β |
- -      - -

∴ β = α so

- -      - -
| β | = +| α |
| α |    | α |
- -      - -

Negative:

- -      - -
| β | = -| β |
| α |    | α |
- -      - -

∴ β = -α so

- -      -  -
| β | = -|  α |
| α |    | -α |
- -      -  -

Now,

|α|2 + |β|2 = 1

∴ 2α2 = 1 so

α = β = 1/√2

Therefore, the eigenvectors are,

-  -       -   -
|1/√2| and | 1/√2|
|1/√2|     |-1/√2|
-  -       -   -

y-axis component:

-  -  - -      - -
|0 -i|| α | = ±| α |
|i  0|| β |    | β |
-  -  - -      - -

Positive:

-   -      - -
| -iβ | = +| α |
|  iα |    | β |
-   -      - -

∴ -iβ = α so β = iα and

-   -      -  -
| -iβ | = +|  α |
|  iα |    | iα |
-   -      -  -

Negative:

-   -      - -
| -iβ | = -| α |
|  iα |    | β |
-   -      - -

∴ -iβ = -α so β = -iα and

-   -      -   -
| -iβ | = -|   α |
|  iα |    | -iα |
-   -      -   -

Therefore, the eigenvectors are,

-  -       -   -
|1/√2| and | 1/√2|
|i/√2|     |-i/√2|
-  -       -   -

Eigenvector Probabilities
-------------------------

If an electron is prepared in the up +z direction:

- -
|b> = | 1 |
| 0 |
- -

and we measure the positive x component:

-    -
|a> = | 1/√2 |
| 1/√2 |
-    -

Then the probability, Px = <a|b><a|b>* = |<a|b>|2

- -  -    -
<a|b> = | 1 || 1/√2 | = 1/√2
| 0 || 1/√2 |
- -  -    -

So, Px = (1/√2)(1/√2) = 1/2

Similarly, if an electron is prepared in the up +x direction:

-    -
|b> = | 1/√2 |
| 1/√2 |
-    -

and we measure the y component:

-    -
|a> = | 1/√2 |
| i/√2 |
-    -

Then Py = <a|b><a|b>* = |<a|b>|2

-    -  -    -
<a|b> = | 1/√2 || 1/√2 | = 1/√2
| i/√2 || 1/√2 |
-    -  -    -

So, Py = (1/2 + i/2)(1/2 - i/2) = 1/2

Therefore, if you line up spin in any direction and measure an
orthogonal component, it will have an equal probability of being
up or down.

What about the components of spin when the spin is lined up in an
arbitrary direction, n? (n is a unit vector with components nx,
ny, nz).  Classically, σ.n represents the component of σ along n.

σ.n = σxnx + σyny + σznz

-    -     -      -     -    -
= | 0 nx | + | 0 -iny | + | nz 0  |
| nx 0 |   | iny 0  |   | 0 -nz |
-    -     -      -     -    -
-         -
= |nz    nx-iny|
|nx+iny  -nz |
-         -
-     -
= |nz    n-|   where n- = nx - iny and n+ = nx + iny
|n+   -nz|
-     -

Which is Hermitian with eigenvalues of +/-1.

Now,

(σ.n)2 = (σxnx + σyny + σznz)(σxnx + σyny + σznz)

= nx2 + ny2 + nz2 + nxny(σxσy + σyσx)

+ nxnz(σxσz + σzσx) + nynz(σyσz + σzσy)

= nx2 + ny2 + nz2

= 1 (since σiσj = -σjσi)

Now consider perparing the electron in the +x direction and
measuring the z component as before.  We get, nx = 1 and
ny = nz = 0 which leaves us with the σx matrix and a positive
eigenvector equal to,

-    -
| 1/√2 |
| 1/√2 |
-    -

The probably of finding the electron in the positive z state
is given by:

Pz = |<a|b>|2

- -  -    -
<a|b> = | 1 || 1/√2 |
| 0 || 1/√2 |
- -  -    -

Therefore,

Pz = |<a|b>|2 = 1/2 as before.

The General Case
------------------

Now, what about the situation where we prepare the spin in direction
n and measure it in direction m where the angle between m and n is
θ.

m         n
\       /
\     /
\ θ /
\ /
.

The eigenvectors, α and β, associated with n can be found from,

-      - -  -    -  -
|nz    n-|| α | = | α |
|n+   -nz|| β |   | β |
-      - -  -    -  -

From which we get

nzα + n-β = α

n+α - nzβ = β

Therefore,

β = α(1 - nz)/n-

To normalize,

-             -  -           -
| α α(1 - nz/n+)||      α       |
-             - | α(1 - nz)/n- |
-           -
= α2 + α2(1 - nz)(1 - nz)/n+n-

= α2((n+n- + 1 - 2nz + nz2)/n+n-)

Now,

n+n- = n12 + n22

Write as

n12 + n22 + n32 - n32

So

n+n- = 1 - n32

Substiuting into the above gives,

= α2((1 - nz2 + 1 - 2nz + nz2)/n+n-)

= α2(2 - 2nz)/n+n-)

= α22(1 - nz)/(1 - nz)2

= α22(1 - nz)/(1 - nz)(1 + nz)

= 2α2/(1 + nz)

So, to get the normalization factor we need to find the inverse √
of this.  Thus, we get,

√{(1 + nz)2α2}

= (1/α)√{(1 + nz)2}

So the eigenvectors of n are,
-             -
(1/α)√{(1 + nz)/2}|     α         |
| α(1 - nz)/n-   |
-             -

Which is equivalent to,

-                           -
|        √{(1 + nz)/2}        |
| √{(1 + nz)/2}{(1 - nz)/n-}   |
-                           -

Now we can apply the same reasoning to get the eigenvectors of m

-                           -
|        √{(1 + mz)/2}        |
| √{(1 + mz)/2}{(1 - mz)/m-}   |
-                           -

To get the probability of measuring the state m given that the electron has
been prepared in the state n is gievn by <σ.m|σ.n> (note complex conjugate
for m),

/                                                                      \2
| -                                       -  -                         - |
|| √{(1 + mz)/2} √{(1 + mz)/2}{(1 - mz)/m+} ||       √{(1 + nz)/2}        ||
| -                                       - | √{(1 + nz)/2}{(1 - nz)/n-} ||
|                                            -                         - |
\                                                                      /

Which, after a lot of tedious math leads to,

Pm = (1 + nxmx + nymy + nzmz)/2

= (1 + n.m)/2

= (1 + cosθmn)/2 since |n| = |m| = 1

If an electron is prepared in a certain direction and measured in another,
the result of the measurement will be either spin up or spin down.  The
interesting thing is that if the exact same experiment is repeated a second
time, the result may or may not be the same.  If the experiment is repeated
multiple times, a probability distribution of the measured spins will be
obtained.  The above equation shows that the smaller the angle between
the prepared state and the measured state, the greater the probability of
spin up occurrences.  For example if the prepared and measured state
are in the same direction the probability will be 1 as expected.  If the
prepared state and the measured state are in opposite directions then the
probability will be 0 (θ = 180).  If the prepared state and the  measured
state are orthogonal then the probability will be 1/2 (θ = 180) as
previously found.

up
^
|  /
| / 1/2<Pup<1
|/
|-------> Pup = 1/2
|\
| \
|  \0<Pup<1/2
v
down

Once a measurment has been made the spin will collapse into either the up
or down state.

Thus,

|/> = α|u> + β|d>

Collapses into,

|+> = 1 |u> + 0|d> or |-> = 0 |u> + 1|d>

Subsequent measurements on the collpased state will always yield the
same result.  Thus, if the measured state is spin up, then repeated
measurements will always give spin up.

σn and σm cannot be measured simultaneously because they do not share
the same eigenvectors.```