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Units, Constants and Useful Formulas

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Entangled States
----------------

Consider 2 electrons that are physically located
at different points in space.  Because they are
at different places, the Pauli exclusion principle
is not violated. For each electron we can write
their spin wavefunctions as:

|ψA> = α|uA> + β|dB>

|ψB> = γ|uA> + δ|dB>

Each of these states describes a qubit.

The states |u> and |d> are called PURE states.
The superposition of these states is also a pure
state.  Also, a superposition is said to be coherent
if there is an operator that, if applied to one
state, can turn it into another that is also
present in the superposition.  In the above case
σx(1/√2)|u> = (1/√2)|d> and vice versa so |ψ> is
a coherent superposition.  If a system is a
superposition of states it exists simultaneously
in each of these states until a measurement is

The joint state of the 2 qbits can be expressed
as the tensor product of the respective individual
qubits. Thus,

|ψAB> = (α|uA> + β|dA>) ⊗ (γ|uB> + δ|dB>)

= αγ|uAuB> + αδ|uAdB> + βγ|dAuB> + βδ|dAdB>
... (1)

Where α2γ2 + α2δ2 + β2γ2 + β2δ2 = 1

States belonging to a composite space that can
be factored into individual states belonging
to separate subspaces are also referred to as
SEPARABLE states.  Separable states correspond
to the situation in which each electron is prepared
independently, and can be measured independently.
This means that, for a separable state, there is
always a direction along which you can measure
the spin of the first electron to be ±1 with 100%
certainty, and there is always a direction along
which you will measure the spin of the second
electron to be ±1 with 100% certainty.

We can define a general state for 2 qubits as
follows:

|ψAB> = a|uAuB> + b|uAdB> + e|dAuB> + f|dAdB> ... (2)

As it stands, this state cannot be factored into
(1) unless a = αγ, b = αδ, e = βγ, f = βδ and the
condition af - be = 0  is enforced.  In general,
however, this is not the case.  A state that cannot
be factored (is not separable) is referred to as
an ENTANGLED state.  Entanglement requires coherence,
but coherence does not imply entanglement.  Note
that a qubit represents the basic state of a
single spin 1/2 particle and is therefore neither
separable nor entangled.

One other point to note is that (1) can be written
as αγ(|uu> + (δ/γ)|ud> + (β/α)|du> + (β/α)(δ/γ)|dd>
indicating 3 degrees of freedom (αγ, β/α, δ/γ) to
describe this state.  Whereas (2) has 4 degrees
of freedom.  A consequence of this is that there
are states where the 2 individual particles are
not independent of one another.

Now focus on the following state (the singlet):

|ψ> = 1/√2{|uAdB> - |dAuB>}

This state is not separable.  It is easy to see
this since it would imply that αγ = 0, αδ = 1/√2,
βγ = 1/√2 and βδ = 0, which is not possible.

For any direction we can show that the expectation
values(averages) of the associated spins are zero,
meaning that measurements along this direction are
equally likely to be +1 or -1.  First we need to
discuss basis states.

Basis States
------------

The eigenvectors of the σ matrices are:

σx:

-   -  - -       - -
| 0 1 || 1 | = +1| 1 |
| 1 0 || 1 |     | 1 |
-   -  - -       - -

-   -  -  -       -  -
| 0 1 ||  1 | = -1|  1 |
| 1 0 || -1 |     | -1 |
-   -  -  -       -  -

σy:

-    -  -  -       -  -
| 0 -i || -i | = +1| -i |
| i  0 ||  1 |     |  1 |
-    -  -  -       -  -

-    -  - -       - -
| 0 -i || i | = +1| i |
| i  0 || 1 |     | 1 |
-    -  - -       - -

σz:

-    -  - -       - -
| 1  0 || 1 | = +1| 1 |
| 0 -1 || 0 |     | 0 |
-    -  - -       - -

-    -  - -       - -
| 1  0 || 0 | = -1| 0 |
| 0 -1 || 1 |     | 1 |
-    -  - -       - -

The basis states |u> and |d> are chosen to be
the eigenvectors of σz.

- -             - -
|u> = | 1 | and |d> = | 0 |
| 0 |           | 1 |
- -             - -

- -     - -      - -     - -
σx| 1 | = | 0 |  σx| 0 | = | 1 |
| 0 |   | 1 |    | 1 |   | 0 |
- -     - -      - -     - -

- -     - -      - -     -  -
σy| 1 | = | 0 |  σy| 0 | = | -i |
| 0 |   | i |    | 1 |   |  0 |
- -     - -      - -     -  -

- -     - -      - -     -  -
σz| 1 | = | 1 |  σz| 0 | = |  0 |
| 0 |   | 0 |    | 1 |   | -1 |
- -     - -      - -     -  -

Consider 2 qubits and different spin matrices,
σ and τ.  σ acts on the first electron and does
nothing to the second electron.  τ does the
opposite.

For electron A we can write:

σx|uA> =   |dA>
σx|dA> =   |uA>
σy|uA> =  i|dA>
σy|dA> = -i|uA>
σz|uA> =   |uA>
σz|dA> =  -|dA>

- -             - -
Where, |u> = | 1 | and |d> = | 0 |
| 0 |           | 1 |
- -             - -

Similarly, for electron B we can write:

τx|uB> =   |dB>
τx|dB> =   |uB>
τy|uB> =  i|dB>
τy|dB> = -i|uB>
τz|uB> =   |uB>
τz|dB> =  -|dB>

Recall, that the expectation value, <>, of an operator
is given by <ψAB|σ|ψAB>.  For the singlet:

<σx> = <(<uAdB| - <dAuB|)(1/√2)|σx|(1/√2)(|uAdB - |dAuB>)>

= 0

<σy> = <(<uAdB| - <dAuB|)(1/√2)|σy|(1/√2)(|uAdb - |dAuB>)>

= 0

<σz> = <(<uAdB| - <dAuB|)(1/√2)|σz|(1/√2)(|uAdb - |dAuB>)>

= 0

Likewise, if we did the same analysis for τ we
would get the same result.

Therefore, in the entangled situation there is no
direction of definite spin for either A's or B's
electron (i.e. <σx> = <τx> = <σy> = <τy> = <σz> =
<τz> = 0).  For example, if the system is prepared
in the +z direction, A will never be able measure
the spin of her electron in the +z direction and
obtain +1 with 100% certainty.  Instead, she will
measure +1 50% of the time and -1 50% of the time.
B will encounter exactly the same situation with
his electron.  Because of this they both conclude
that their electrons must be in an entangled state.

A’s measurement will cause the wave function to
collapse.  If A finds her electron to be spin-up
then B will find his electron to be spin down i.e.

For the singlet state prior to measurement,
P(|uA>) = P(|uB>) = (1/√2)2 = 1/2

After measurement the system collapses into the
normalized eigenstate:

With P(|uA>) = 12 = 1 and P(|uB>) = 02 = 0.

We can compare this result to the one obtained for
the separable state (1/2(|uA> + |dA>) ⊗ (|uB> + |dB>).
We get:

<σx> = <(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

|σx|(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

= (1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

(1/2)(|dAuB> + |dAdB> + |uAuB> + |uAdB>)

= (1/4)(4) = 1

<σy> = <(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

|σy|(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

= (1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

(1/2)(i|dAuB> + i|dAdB> - i|uAuB> - i|uAdB>)

= 0

<σz> = <(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

|σz|(1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

= (1/2)(|uAuB> + |uAdB> + |dAuB> + |dAdB>)

(1/2)(|uAuB> + |uAdB> - |dAuB> - |dAdB>)

= 0

Likewise, if we did the same analysis for τ we
would get the same result.

Therefore, if a system of qubits is prepared in
the +z direction and A's electron is independent,
she would measure the spin in the +z direction to
be +1 (i.e. <σx> = 1, <σy> = <σz> = 0).  Likewise
for Bob's electron (i.e. <τx> = 1, <τy> = <τz> = 0).

We could also show that if the system of qubits
is prepared and measured in the ±x direction we get:

<σx> = (1/2)((<ψx|σx|ψx) = +1

<σx> = (1/2)((<ψ-x|σx|ψ-x) = -1

<σy> = <σz> = 0

and so on.

For the separable state:

|ψ> = (1/√2)(|uA> + |dA>) ⊗ (1/√2)(|uB> + |dB>)

= (1/2)|uAuB> + (1/2)|uAdB> + (1/2)|dAuB> + (1/2)|dAdB>

(Normalization:  (1/2)2 + (1/2)2 + (1/2)2 + (1/2)2 = 1)

For the separable state prior to measurement,
P(|uA>) = P(|uB>) = (1/2)2 + (1/2)2 = 1/2

After measurement by A the system collapses into
the normalized eigenstate:

(1/√2)|uAuB> + (1/√2)|uAdB> + 0|dAuB> + 0|dAdB>

∴ P(|uA>) = (1/√2)2 = 1/2

After measurement by B the system collapses into
the normalized eigenstate:

(1/√2)|uAuB> + 0|uAdB> + (1/√2)|dAuB> + 0|dAdB>

∴ P(|uB>) = (1/√2)2 = 1/2

We have focussed on the singlet state in our analysis
but there are 3 other entangled states of particular
importance.  These 3 states together with the singlet
are called the BELL STATES:

|Φ+> = 1/√2{|uAuB> + |dAdB>}

|Φ-> = 1/√2{|uAuB> - |dAdB>}

|Ψ+> = 1/√2{|uAdB> + |dAuB>}

|Ψ-> = 1/√2{|uAdB> - |dAuB>} - singlet

It should be pointed out that for the Φ states A
and B's measurements will result in identical spins
whereas in the Ψ states, measurements will result
in opposite spins.

Einstein - Podolsky - Rosen Correlation
---------------------------------------

The curious fact is that A and B's electrons will
remain entangled regardless of the separation between
them unless either is 'disturbed' by a measurement.
Once a measurement has been made, the state of the
original entangled system collapses and the entanglement
is destroyed.  The fact that the spin of B's electron
instantly "knows" what spin it is supposed to take
on (and vice versa), seemingly involves communication
between the two particles at speeds greater than the
speed of light, which is in conflict with Einstein's
special theory of relativity (violation of locality).
Einstein referred to the phenomenon as 'spooky; action
at a distance and supported an alternative idea
called 'hidden variables theory' which suggested
that quantum mechanics was incomplete.  However,
later on Bell's theorem would suggest that the
existance of local hidden variables is impossible.

Hidden Variables and Bell's Theorem
-----------------------------------

Bell's theorem showed that if local hidden variables
existed, certain experiments could be performed where
the result would satisfy a Bell inequality. If, on the
other hand, if Bell's inequality was violated local
hidden variables could not exist.

Bell's Theorem
--------------

Let N be the number of objects.

N(A not B) + N(B not C) ≥ N(A not C)

Therefore,

N(1 + 2) + N(4 + 7) ≥ N(1 + 4)

Therefore,

N(1 + 4) + N(2 + 7) ≥ N(1 + 4)

So that N(2 + 7) ≥ 0

This is generaly written as:

N(A,~B) + N(B,~C) ≥ N(A,~C)

Let ↑↗→ be possible measurement directions for A
and B's electrons.  Interpret P(↑,↗) as the
probability A finds her electron in the ↑ direction
and B finds his electron in the ↗ direction which,
for the singlet, is equivalent to the ~↗ = ↙
direction for A.  Returning to the above Venn diagram,
we can define A = ↑, B = ↗ and C = →.

Substituting into Bell's theorem gives:

N(↑,↗) + N(↗,→) ≥ N(↑,→)

We can equate N with a probability (i.e. P = n/NTotal).
To compute these probabilities it is helpful to
use the projection operators for spin.  These can
be written in terms of the σ matrices as follows:

-   -       -   -     -   -
ℙ↑ = (I + σz)/2 = | 1  0 | + | 1  0 | = | 1 0 |
| 0  1 |   | 0 -1 |   | 0 0 |
-   -       -   -     -   -

ℙ↗ = (I/2 + (σx + σ2)/2√2

-   -       -   -     -       -
ℙ→ = (I + σx)/2 = | 1  0 | + | 0  1 | = | 1/2 1/2 |
| 0  1 |   | 1  0 |   | 1/2 1/2 |
-   -       -   -     -       -

These act on electron 1.  There is an equivalent set
of operators that act on electron 2:

For ℙ↑:

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

Therefore, ℙ↑|u> = |u> and ℙ↑|d> = |0>

For ℙ→:

-       -  - -     -   -       -       -  - -     -   -
| 1/2 1/2 || 1 | = | 1/2 | and | 1/2 1/2 || 0 | = | 1/2 |
| 1/2 1/2 || 0 |   | 1/2 |     | 1/2 1/2 || 1 |   | 1/2 |
-       -  - -     -   -       -       -  - -     -   -

Therefore, ℙ→|u> = |u> and ℙ↑|d> = |0>

For ℙ↗:

-         -  - -            -    -
(1/2√2)| √2+1   1  || 1 | = (1/2√2)| √2+1 |
|   1  √2-1 || 0 |          |   1  |
-         -  - -            -    -

and,

-         -  - -     -    -
(1/2√2)| √2+1   1  || 0 | = |   1  |
|   1  √2-1 || 1 |   | √2-1 |
-         -  - -     -    -

The probabilities can then be computed as:

P = <ψ|ℙ1ℙ2|ψ>

Return to the singlet state and compute the
probability P(↑,↗).

<ψ|(I/2 + σz/2)(I/2 + (τx + τz)/2√2)|ψ>

Breaking it down piece by piece:

ℙ↑|(1/√2)(|ud> - |du>) = (1/√2)|ud>

ℙ↗|(1/√2)|ud>)> = (1/√2){|u> ⊗ ℙ↗|d>}

= (1/√2){|u> ⊗ {1|u> + (√2-1)/2√2|d>}

= (1/√2){|uu> ⊗ (√2-1)/2√2|ud>}

P(↑,↗) = <(<ud| - <du|)(1/√2)|(1/√2){|uu> ⊗ (√2-1)/2√2|ud>}

= <ud|(1/√2)(1/√2)(√2-1)/2√2|ud>

= (1/2)(√2-1)/2√2

= 0.075

Likewise, compute the probability P(↗,→).

<ψ|(I/2 + (σx + σz)/2√2)(I/2 + τx/2)|ψ>

P(↗,→) = 0.075

Finally, for P(↑,→).

<ψ|(I/2 + σz/2)(I/2 + τx/2)|ψ>

<(<ud| - <du|)(1/√2)|ℙ↑ℙ↗|(1/√2)(|ud> - |du>)

P(↑,→) = 0.25

P(↑,↗) + P(↗,→) ≱ P(↑,→)

There is another way to look at this.

A      B
↑↗→    ↑↗→
===    ===

N1  +++    ---   + = aligned, - = counter aligned

N2  ++-    --+

N3  +-+    -+-

N4  +--    -++

N5  -++    +--

N6  -+-    +-+

N7  --+    ++-

N8  ---    +++

Experimentally, we can compute the probabilities
as follows:

P(↑,↗) = (N3 + N4)/NT

P(↗,→) = (N2 + N6)/NT

P(↑,→) = (N2 + N4)/NT

P(↑,↗) + P(↗,→) ≥ P(↑,→)

∴ (N3 + N4)/NT + (N2 + N6)/NT ≥ (N2 + N4)/NT

∴ (N3 + N4) + (N2 + N6) ≥ (N2 + N4)

∴ (N2 + N4) + (N3 + N6) ≥ (N2 + N4)

Components of Spin
------------------

The eigenvectors of the Pauli matrices represents
an observable describing the spin of a spin 1/2
particle in the three spatial directions.

-    -      - -    - -
σz:  | 1  0 | => | 1 |, | 0 |
| 0 -1 |    | 0 |  | 1 |
-    -      - -    - -

-   -            - -          -  -
σx:  | 0 1 | => (1/√2)| 1 |, (1/√2)|  1 |
| 1 0 |          | 1 |        | -1 |
-   -            - -          -  -

- -                              - -          -  -
| 1 | = (1/√2)(χ+x + χ-x) = (1/2)| 1 | + (1/2)|  1 |
| 0 |                            | 1 |        | -1 |
- -                              - -          -  -

-    -            - -          -  -
σy:  | 0 -i | => (1/√2)| 1 |, (1/√2)|  1 |
| i  0 |          | i |        | -i |
-    -            - -          -  -

- -                              - -          -  -
| 1 | = (1/√2)(χ+y + χ-y) = (1/2)| 1 | + (1/2)|  1 |
| 0 |                            | i |        | -i |
- -                              - -          -  -

Therefore, P(|χ+x>) = P(|χ+y>) = 1/2

Alternatively, we can look at this as the overlap
of χ+x with χ+z.  Then:

P(χ+x) = <u+z|u+x><u+z|u+x>*

= |<u+z|u+x>|2

-   -  -    -
<u+z|u+x> = | 1 0 || 1/√2 | = 1/√2
-   - | 1/√2 |
-    -

So, P(χ+x) = (1/√2)(1/√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.  We can
generalize this to find the components of spin in
any direction when the electron is line up along
the +z axis.

Let n be a unit vector with components nx, ny, nz
that represents the measurements axis.  Therefore,
σ.n represents the component of σ along n, i.e.

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|
-     -

This 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)

Check:  If the electron spin is lined up along
the +z axis and measured in the x direction we
get nz = 1 and nx = ny = 0 which leaves us with the
σz matrix and a positive eigenvector equal to:

- -
| 1 |
| 0 |
- -

Then the probability, Px = <u+z|u+x><u+z|u+x>*

= |<u+z|u+x>|2

-   -  -    -
<u+z|u+x> = | 1 0 || 1/√2 | = 1/√2
-   - | 1/√2 |
-    -

So, P+x = (1/√2)(1/√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 /
\ /
o

Let |σ.n=1> be an eigenvector of σ.n with eigenvalue
+1 that satisfies:

σ.n|σ.n=1> = +1|σ.n=1>

This is interpreted as preparing the electron spin
in the up state along n and then measuring the
component of the spin along the same axis.  Likewise,
for direction, m:

σ.m|σ.m=1> = 1|σ.m=1>

Note: For the spin down case we would write:

|σ.n=-1> and σ.n|σ.n=-1> = -1|σ.n=-1>

The eigenvectors of σ.n=1 can be found from:

-      - -  -      -  -
|nz    n-|| α | = +1| α |
|n+   -nz|| β |     | β |
-      - -  -      -  -

From which we get

nzα + n-β = α

n+α - nzβ = β

These equations are equivalent.  Solving for β
gives:

n+α/(1 + nz) = α(1 - nz)/n-

∴ n+/(1 + nz) = (1 - nz)/n-

∴ n+n- = 1 - nz2

∴ nx2 + ny2 + nz2 = 1

Choose β = α(1 - nz)/n-

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-)

= 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, N, we need
to find the inverse √ of this.  Thus, we get:

N = √{(1 + nz)/2α2}

= (1/α)√{(1 + nz)2}

So the eigenvector of σ.n=1 is:

-            -
|σ.n=1> = (1/α)√{(1 + nz)/2}|     α        |
| α(1 - nz)/n- |
-            -

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

Now we can apply the same reasoning to get the
eigenvector of σ.m=1:

-                          -
|σ.m=1> = |        √{(1 + mz)/2}       |
| √{(1 + mz)/2}{(1 - mz)/m-} |
-                          -

The probability of measuring the state σ.m=1
given that the electron has been prepared in the
state σ.n=1 is:

P(σ.n=1 σ.m=1) = |<σ.n=1|σ.m=1>|2

Which, after a lot of tedious math leads to,

P(σ.n=1 σ.m=1) = (1 + nxmx + nymy + nzmz)/2

= (1 + n.m)/2

= (1 + cosθmn)/2 since |n| = |m| = 1

= cos2(θmn/2) (using trig identity)

Alternative derivation using polar coordinates:

nx = cosφsinθ

ny = sinφsinθ

nz = cosθ

-                         -
σ.n = |   cosθ       exp(-iφ)sinθ |   λ = ±1
| exp(iφ)sinθ     -cosθ     |
-                         -

det(σ.n - λI)v = 0.  For λ = +1 we get:

-                          -  - -
|   cosθ - 1    exp(-iφ)sinθ || α | = 0
| exp(iφ)sinθ     -cosθ - 1  || β |
-                          -  - -

∴ β = exp(iφ)(1 - cosθ)α/sinθ

= αexp(iφ)sin(θ/2)/cos(θ/2)

and,

β* = αexp(-iφ)sin(θ/2)/cos(θ/2)

Normalization:

α2 + β2 = α2 + β*β = α2(1 + sin2(θ/2)/cos2(θ/2)) = 1

∴ α2 = 1/(1 + sin2(θ/2)/cos2(θ/2))

Using the identity sinθ = 2sin(θ/2)cos(θ/2) this
becomes:

α2 = cos2(θ/2)

Which gives:

α = cos(θ/2) and β = sin(θ/2)

|+> = cos(θ/2)|u> + exp(iφ)sin(θ/2)|d>

This represents the qubit in terms of the Bloch
sphere.  Likewise:

|-> = sin(θ/2)|u> - exp(iφ)cos(θ/2)|d>

Density Operator Formulation:

ℙn = (I + σ.n)/2

-   -          -          -
= (1/2)| 1 0 | + (1/2)| cosθ  sinθ |
| 0 1 |        | sinθ -cosθ |
-   -          -          -

-   -          -         -          -          -
= (1/2)| 1 0 | + (1/2)|  0   sinθ | + (1/2)| cosθ   0   |
| 0 1 |        | sinθ  0   |        |  0   -cosθ |
-   -          -         -          -          -

-                     -
= |  cos2(θ/2) (1/2)sinθ  |
| (1/2)sinθ   sin2(θ/2) |
-                     -

This is the probability to measure |σ.n=1> if the
initial state is |σ.m=1> if the state is prepared
in a direction θ ...

|ψ> = (1/√2)|u> + (1/√)|d>

-     -
ρ = (1/2)|u><u| = | 1/2 0 |
|  0  0 |
-     -

∴ P(ℙn) = Tr(ρℙn) = (1/2)cos2(θ/2)

-      -
ρ' = (1/2)|d><d| = | 0   0  |
| 0  1/2 |
-      -

∴ P(ℙn) = Tr(ρ'ℙn) = (1/2)sin2(θ'/2)

σ.n=1
|   /
|θ /
| /
|/
o
|\
| \
|θ'\
|   \
σ.n=-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 = 90) as previously found.

The above is the result for a single electron.  In the
case of the singlet we need to modify this to account
for the normalization factors of (1/√2).  We get:

P(σ.n=1 σ.m=1) = (1/2)|<σ.m=1|σ.n=1>|2

= (1/2)(1 + cosθ)/2

≡ (1/2)cos2(θ/2) (trig identity)

We can now find the probabilities as follows:

P(↑↗) case:

↑|   /↗
|  /       θ = π/4
|θ/
|/
o
/
/
/~↗

Counterclockwise angle between ↑ and ~↗ = 5π/4

P(↑↗) = (1/2)cos2((5π/4)/2) = 0.075

P(↑→) case:

↑|
|
|          θ = π/2
|θ
------o------
~→

Counterclockwise angle between ↑ and ~→ = 3π/2

P(↑→) = (1/2)cos2((3π/2))/2 = 0.25

P(↗→) case:

↑|   /↗
|  /       θ = π/4
|θ/
|/
------o
~→

Counterclockwise angle between  ~→ and ↗ = 5π/4

P(↗→) = (1/2)cos2((5π/4))/2 = 0.075

Clearly, the last inequality cannot be true and Bell's
theorem is violated.  The implication of this is that
the hidden variables that says each particle carries
with it all of the required information at the time of
separation, and nothing needs to be transmitted from
one particle to the other at the time of measurement,
cannot be true.  Understanding 'spooky' action at a
distance is one of the major unsolved mysteries of
Physics. The study of String Theory and the idea of
special extra spatial dimensions may offer some hope
in this regard.