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

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Homework Help Overview

The discussion revolves around the relationship between the components (u, v) and the state alpha+ in the context of spin-1/2 eigenstates. Participants are examining the mathematical expressions and transformations involved in this quantum mechanics topic.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of (u, v) and its connection to alpha+, questioning how scalar multiplication leads to certain equalities. There is also exploration of the implications of setting u equal to v and the normalization conditions that follow.

Discussion Status

The conversation is actively exploring the implications of the relationships between u and v, with some participants noting the arbitrary choices made in their values. There is recognition of the normalization condition and its relevance to the probabilities involved, but no consensus has been reached on the underlying assumptions.

Contextual Notes

Participants are considering the implications of choosing u and v to be real numbers and the significance of the arbitrary sign in the context of probability calculations. The discussion reflects a focus on the mathematical properties rather than definitive conclusions.

Shackleford
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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
 
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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.
 
MathematicalPhysicist said:
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).
 
So?
That means v=u.
 
MathematicalPhysicist said:
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.
 
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.
 
MathematicalPhysicist said:
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.
 

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