I Why ##S_z|+\rangle=\frac{\hbar}{2}|+\rangle##

[Kashmir]

I've just started Quantum mechanics from McIntyre, and I've got a doubt.
Here is the relevant passage

How does the author say using the preceding section that
$S_z|+\rangle=\frac{\hbar}{2} |+\rangle$?

 

[PeroK]

I've just started Quantum mechanics from McIntyre, and I've got a doubt.
Here is the relevant passage
View attachment 296047
View attachment 296046

How does the author say using the preceding section that
$S_z|+\rangle=\frac{\hbar}{2} |+\rangle$?

That holds by definition. $| + \rangle$ is the vector with precisely that property.

 

[Kashmir]

That holds by definition. $| + \rangle$ is the vector with precisely that property.

Thank you for your response.

The author says "In the eigenvalue equation, the observable is represented by an
operator, the eigenvalue is one of the possible measurement results of the observable, and the eigenvector is the ket corresponding to the chosen eigenvalue of the operator. The eigenvector appears on
both sides of the equation because it is unchanged by the operator"

So this is the definition?

 

[PeroK]

Thank you for your response.

The author says "In the eigenvalue equation, the observable is represented by an
operator, the eigenvalue is one of the possible measurement results of the observable, and the eigenvector is the ket corresponding to the chosen eigenvalue of the operator. The eigenvector appears on
both sides of the equation because it is unchanged by the operator"

So this is the definition?

That's the definition of eigenthings. The definition of $|+\rangle$ is that it is the vector that satisfies that equation.

 

[Kashmir]

That's the definition of eigenthings. The definition of $|+\rangle$ is that it is the vector that satisfies that equation.

I reread the passage and think that
$S_z|+\rangle=\frac{\hbar}{2} |+\rangle$ follows from the passage I've quoted above. I'll try to explain it.

Keeping in mind Stern Gerlach experiment measuring spin along z direction we know that we've two measurements $\pm \frac{\hbar}{2}$ corresponding respectively to
$|\pm\rangle$.

Then the authors quote
"In the eigenvalue equation, the observable is represented by an
operator, the eigenvalue is one of the possible measurement results of the observable, and the eigenvector is the ket corresponding to the chosen eigenvalue of the operator. The eigenvector appears on
both sides of the equation because it is unchanged by the operator" implies then
$S_z|+\rangle=\frac{\hbar}{2} |+\rangle$

 

[PeroK]

I can't argue with that!

 

[Kashmir]

I can't argue with that!

Am I wrong sir? Then I'll reread the chapter again.

 

[PeroK]

Am I wrong sir? Then I'll reread the chapter again.

https://idioms.thefreedictionary.com/can't+argue+with+that

 

[Kashmir]

https://idioms.thefreedictionary.com/can't+argue+with+that

Thank you :)
See you next time :)