- #1

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Using

J

_{-}|j,m>=[(j+m)(j-m+1)]

^{1/2}|j,m-1>

I can get from |++> to 1/[itex]\sqrt{2}[/itex](|-+> + |+->) but from here I am stuck since j+m=0 for |+->.

What is wrong with my working? Thanks

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- Thread starter semc
- Start date

- #1

- 354

- 2

Using

J

I can get from |++> to 1/[itex]\sqrt{2}[/itex](|-+> + |+->) but from here I am stuck since j+m=0 for |+->.

What is wrong with my working? Thanks

- #2

- 196

- 22

Perhaps you are not remembering that J_{-} has two pieces?

[tex]J_- = J_{1-} + J_{2-}[/tex]

[tex]J_- = J_{1-} + J_{2-}[/tex]

- #3

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[tex]J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>[/tex]

so [tex]J_{1-}|+-> + |-+> [/tex] is where I am stuck since [tex]m_1=1/2 , m_2=-1/2 [/tex] so everything will be reduced to 0.

- #4

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[tex]\left( J_{1-} + J_{2-} \right) \left(\lvert +- \rangle + \lvert -+ \rangle \right)[/tex]

There are 4 terms here, 2 of which survive the lowering operators.

- #5

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[tex]J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>[/tex]

If its correct then I should be able to evaluate this [tex]J_{1-}|+-> + |-+> [/tex]

I am aware that when the lowering operator acts on the lowest eigenvalue it will disappear but I cant continue since the formula has (j+m) term which is 0 for different spins. By the way I am doing revision for this so it would be better if anyone can just tell me what I did wrong here. This is not homework. Thanks

- #6

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Two of the terms vanish but two survive because you're acting on a + with the lowering operator.

- #7

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Using the lowering operator [tex]J_-|jm>=((j+m)(j-m+1))^{1/2}|jm-1>[/tex]

on |-+> I got this [tex]J_-|-1/2, 1/2>=((-1/2+1/2)(-1/2-1/2+1))^{1/2}|-1/2,-1/2>[/tex]

[tex](-1/2+1/2)(-1/2-1/2+1)=(0)(0)=0[/tex] and for |+-> the operator is also acting on the second entry which is - so it goes to 0 as well.

Clearly I am making a stupid mistake but I really don't see it myself. Hopefully someone can tell me whats wrong

- #8

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eg.

[tex]S_{1-} \lvert +- \rangle = \lvert -- \rangle[/tex]

[tex]S_{1-} \lvert +- \rangle = \lvert -- \rangle[/tex]

- #9

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Sorry, we cross-posted...

[tex]\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1[/tex]

[tex]\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1[/tex]

- #10

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[tex]\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1[/tex]

Yes but I believe this is for the case of |++> since j=m=1/2?

- #11

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Maybe that's the point you're missing? J

- #12

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- #13

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but j=1/2 and m=1/2.

- #14

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Think this is the part that I got confuse. For |+-> , does it mean that j=1/2 , m=-1/2?

- #15

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Think this is the part that I got confuse. For |+-> , does it mean that j=1/2 , m=-1/2?

Aha! I think we've nailed the problem. What you have here is the tensor product of two spin-1/2 states. So j

For instance |+-> would have j

- #16

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Haha I get it now. Thank you so much for explaining this to me!

- #17

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No problem

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