Joy Christian wrote:Sure. Perhaps it is better to use a generic vector n, but it does not really mater. The vector "a" is supposed to span all directions as usual (Bell, for example, never uses n to indicate a generic vector --- he just writes "a" or "b" as generic vectors spanning all space). But it is easy to update the paper, so I will do so.
PS: The Academia.edu website sucks. It does not let me edit the paper for some reason. I often have problems with that website. It does not match their ambition.
Well, it is edited now. Looks perfect. Now..., before Lockyer and Reed (or any others) come along and want to mangle eq. (11) and cause confusion, here is some further explanation about it that I posted on
Albert Jan's blog.
FrediFizzx wrote:For a pair of particles that are in a singlet configuration, Nature has a 50-50 chance that they will be created as either a left handed system or right handed system. And that is just plain common sense. For a left handed system the expression is;
(I.a)(I.b) = -a.b - (a^b) = -a.b - I.(a x b) (1) LH System
For the right handed system we have;
(I.a)(I.b) = -a.b - (a^b) = -a.b - I.(a x b) (2) RH System
Note that both equations are the same. However, eq. (1) is in a left handed basis and eq. (2) is in a right hand basis so we can't add the left hand expression to the right hand expression properly. Now, translate eq. (1) to the right hand basis so that we can do that.
But first I want to stress, "However, eq. (1) is in a left handed basis and eq. (2) is in a right hand basis so we can't add the left hand expression to the right hand expression properly." The translation of eq. (1) to the right hand basis is simply;
(I.b)(I.a) = (-I.a)(-I.b) = -a.b - (-I).(a x b) = -a.b + a^b
GAViewer cofirms that this is correct.
>> -a.b + a^b
ans = 0.77 + 0.39*e2^e3 + -0.44*e3^e1 + -0.25*e1^e2
>> (I.b)*(I.a)
ans = 0.77 + 0.39*e2^e3 + -0.44*e3^e1 + -0.25*e1^e2
So that further explains eq.(11) and eq. (12) easily follows from that combined with eq. (10).