How are (u,v) and alpha+ related in the Eigenstates of Spin 1/2?

[Shackleford]

How did they get (u,v) in the second line and then alpha₊?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-11-20113911.jpg?t=1290274889

 

[MathematicalPhysicist]

Well, for alpha+, if you multiply (scalar multiplcation) the equation for 1/sqrt(2) (1,1), with the vector 1/sqrt(2) (exp(-i phi), exp(i phi)), the part that is multiplied by alpha minus is vanished, and you get the equality.

For u and v, you get u=v so obviously 2u^2=1, and the sign is arbitrary.

 

[Shackleford]

Well, for alpha+, if you multiply (scalar multiplcation) the equation for 1/sqrt(2) (1,1), with the vector 1/sqrt(2) (exp(-i phi), exp(i phi)), the part that is multiplied by alpha minus is vanished, and you get the equality.

For u and v, you get u=v so obviously 2u^2=1, and the sign is arbitrary.

Unless I'm really rusty in linear algebra, the first line yields (v,u) = (u,v).

 

[MathematicalPhysicist]

So?
That means v=u.

 

[Shackleford]

So?
That means v=u.

Dammit. You're right. I don't know why I was thinking it was a typo or something. Haha.

And the next line they just arbitrarily chose u and v to each be one to easily satisfy the normalization equation.

 

[MathematicalPhysicist]

Well, they arbitrarily chose u and v to be real, and from u=v you plug back to get 2u^2=1.
As I said the sign is arbitrary, and because we only interested in probabilities (eventually), the sign is irrelevant.

 

[Shackleford]

Well, they arbitrarily chose u and v to be real, and from u=v you plug back to get 2u^2=1.
As I said the sign is arbitrary, and because we only interested in probabilities (eventually), the sign is irrelevant.

Okay. Then of course you bring out the scalar multiple of 1/root 2.