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

In summary, the conversation discusses how (u,v) was obtained in the second line and the significance of adding alpha+ in the equation. The speaker explains that multiplying the equation for 1/sqrt(2) with the vector 1/sqrt(2) results in the vanishing of the part multiplied by alpha minus, leading to the equality. They also mention that u and v are equal and the sign is arbitrary. The conversation concludes with a discussion on the normalization equation and the irrelevance of the sign in terms of probabilities.
  • #1
Shackleford
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2
How did they get (u,v) in the second line and then alpha+?

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  • #2
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.
 
  • #3
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).
 
  • #4
So?
That means v=u.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 

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

What are eigenstates of spin 1/2?

Eigenstates of spin 1/2 are quantum states that describe the spin of a particle with a spin quantum number of 1/2. They represent the possible orientations of the particle's spin in a given direction.

How are eigenstates of spin 1/2 related to the spin operator?

The spin operator is a mathematical tool used to describe the spin of a particle. Eigenstates of spin 1/2 are the eigenvectors of the spin operator, meaning they are the states that remain unchanged when acted upon by the spin operator.

What is the significance of the spin 1/2 eigenstates in quantum mechanics?

Eigenstates of spin 1/2 play a crucial role in quantum mechanics as they represent the fundamental states of a particle's spin. They are used to describe the spin of particles in various physical systems and have important implications in areas such as nuclear physics and quantum computing.

How are eigenstates of spin 1/2 different from other quantum states?

Eigenstates of spin 1/2 are unique in that they have a spin quantum number of 1/2, which is the minimum value for a particle's spin. They also have specific properties such as being orthogonal to each other and forming a complete basis for spin measurements.

Can eigenstates of spin 1/2 be observed in experiments?

No, eigenstates of spin 1/2 cannot be directly observed in experiments. However, their effects can be observed through measurements of spin properties such as spin projections and spin precession. These measurements confirm the existence of eigenstates and their importance in describing the spin of particles.

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