How Does the Stern-Gerlach Experiment Illustrate Quantum Spin Measurement?

[rabbit44]

Hi, I've been trying to understand the Stern-Gerlach experiment. I've been reading this book which explains how to find the amplitudes of measuring a particle to have a spin 1/2 lined up along some direction.

Here is the book:

http://www-thphys.physics.ox.ac.uk/people/JamesBinney/QBc6.pdf

It's page 107 that confused me: that first matrix equation. The matrix representation of the spin matrix along theta seems to be in the basis of eigenstates of the Sz matrix. But then the object (1, 0) would be the state of the particle having spin along the z axis, in which case the equation doesn't seem to make sense, as it would suggest that the state of having spin along the z axis is an eigenstate of the spin matrix along theta.

Can anyone help here?

 

[Minich]

Can You explain your trouble more explicitly?
Which equations are initial and where is misunderstanding?
(6.114)?

 

[Fredrik]

It's page 107 that confused me: that first matrix equation. The matrix representation of the spin matrix along theta seems to be in the basis of eigenstates of the Sz matrix. But then the object (1, 0) would be the state of the particle having spin along the z axis, in which case the equation doesn't seem to make sense, as it would suggest that the state of having spin along the z axis is an eigenstate of the spin matrix along theta.

The equation would imply that if it had been true for any values of a and b, or at least specifically for a=1,b=0, but it isn't. See (6.117).

 

[rabbit44]

Can You explain your trouble more explicitly?
Which equations are initial and where is misunderstanding?
(6.114)?

Thanks for replying

Yes 6.114, I don't really get where this comes from. I think that

|+,z> = a*|+, p> + b*|-,p>

(where p is theta)

But I think that the matrix representation of sigma.n in 6.114 is in the basis of eigenstates of Sz, i.e. of |+,z> and |-,z>. I think this because the paragraph before 6.114 references 6.111, which is in the basis of eigenkets of Sz.

So if this is the case, is (1, 0) (a column vector), the representation of |+,z>? If this is the case then the left hand side of the equation surely reads:

sigma.n [ a|+,z> + b|-,z>] which isn't the same as the right hand side of [ a|+,z> + b|-,z>] as |+,z> isn't an eigenket of sigma.n

??ARGH?

Thanks

 

[rabbit44]

The equation would imply that if it had been true for any values of a and b, or at least specifically for a=1,b=0, but it isn't. See (6.117).

Sorry would you mind elaborating, I don't fully understand this. Sorry if I'm being very slow!

 

[Fredrik]

What 6.114-6.117 says is that if (a,b) is an eigenvector, then a and b must be specifically those numbers specified by 6.117. So (1,0) is clearly not an eigenvector (of $\hat n\cdot\vec\sigma$). (You said in #1 that the equation suggests that it is).

What you said in #4 is correct until the last statement before the argh. The two sides of the equations are the same, if a and b are as in 6.117. You seem to be assuming that a linear combination of two eigenvectors with different eigenvalues can't be an eigenvector of some other operator. That assumption is wrong.

 

[rabbit44]

What 6.114-6.117 says is that if (a,b) is an eigenvector, then a and b must be specifically those numbers specified by 6.117. So (1,0) is clearly not an eigenvector (of $\hat n\cdot\vec\sigma$). (You said in #1 that the equation suggests that it is).

What you said in #4 is correct until the last statement before the argh. The two sides of the equations are the same, if a and b are as in 6.117. You seem to be assuming that a linear combination of two eigenvectors with different eigenvalues can't be an eigenvector of some other operator. That assumption is wrong.

Ah OK so what I said where the two sides is correct, but I was incorrect in saying that they can't equal?

What is the reason for setting (a,b) as an eigenvector of sigma.n? (last question I promise!)

Thanks

 

[Fredrik]

Ah OK so what I said where the two sides is correct, but I was incorrect in saying that they can't equal?

If the word "where" is supposed to be "were", then yes. That's what the equation says, and the equality holds.

What is the reason for setting (a,b) as an eigenvector of sigma.n? (last question I promise!)

Because he wants to find out which linear combinations of |z+> and |z-> are eigenvectors of $\hat n\cdot\vec S$. He's not "setting" (a,b) as an eigenvector. He's determining what the eigenvectors are.