Electron spin query

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I read this quote from an article and another like is is

http://scienceblogs.com/principles/2010/07/electron_spin_for_toddlers.php [Broken]

"and we see that the half-angle appears with a rather interesting consequence. To see this consequence let’s consider a rotation of 360 degrees – 2 radians. Substituting yields

|a>2pi = -|a>

In other words, upon a rotation of 2 pi the state ket does not return to it’s original state – there is an additional factor of a minus. In order to get back to the original state ket one must rotate through 720 degrees. This is the origin of the statement that an electron is unsual in the fact that when rotated through a full circle it does not “look the same”. "

Now I thought that |a> and -|a> represented the SAME state because -1 is just exp i(pi) which has square modulus of 1. According to the postulates of QM this should represent the same state?

Can anyone put me into the right way of thinking here?
 
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  • #2
Vanadium 50
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Think about difference between |a> and -|a> when interfering with something else.
 
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Not sure about this. I'm trying to understand the spin up and down states for electrons and deriving their forms. I want to understand David Alberts book "QM and experience" I can see that if you added two states like |a> + |b> with -|a> + |b> then it would affect the final state but according Albert ther is no difference between -|a> and |a> . I mean there is or there is not if it's in between then is the superposition with words? - I'm confused!
 
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For example in one book I have they show that one of the spin vector bases in the x dirn is

|-x> = - (1/sqrt2)(|+> - |->) where |+> is the spin base in the z direction and |-x> is one of the bases in the x dirn.

In Alberts book he gets |-x> = (1/sqrt2)(|+> - |->). The former case is derived using the rotation operator and Albert does it by using eigenvalue equation I think.

Can I ignore the sign because they both agree on the |+x> as = (1/sqrt2)(|+> + |->) ?

Many thanks for any help.
 
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I don't know what Albert says, but there is a difference between |a> and -|a>. However, there is no difference between their squares.
 
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Thanks Vanadium 50.

Can you give me an example of where and how it would make a difference? In all the books I have they seem to include in the postulates that not only is -|a> the same as |a> but so also is any c|a> where c is a complex number and |C|^2 =1.

Thanks for any suggestions.
 
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Yes, when interfering with something else. <a|2 and (-<a|)2 are equal, and <b|2 and (-<b|)2 are equal, but <a+b|2 and <a-b|2 are not equal.
 
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Yes, when interfering with something else. <a|2 and (-<a|)2 are equal, and <b|2 and (-<b|)2 are equal, but <a+b|2 and <a-b|2 are not equal.
I'm not sure what you mean by <a+b| and <a-b| how are they constructed and what could they represent in terms of an electron spin example? Sorry but I'm a bit slow on this. I understand that the state of a system could be a linear combination (ie superposition ) of basis vectors like |a> and |b>, say but what do you mean when you write them together as|a+b>?
 
  • #9
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For example if you mean

|a+b> = a|a> + b|b> and |a-b> = a|a> - b|b>

Then |<a+b|a+b>|^2 = |<a-b|a-b>|^2 just the same!

Now |a+b> + |a-b> = 2 a|a> and so expressing |a> in terms of the coposite vector will give

|a+b> + |a-b> =|a>
----------------
2 a

Is this what you meant?
 

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