- #1

semc

- 361

- 3

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter semc
- Start date

- #1

semc

- 361

- 3

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

Oxvillian

- 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

semc

- 361

- 3

[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

Oxvillian

- 196

- 22

[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

semc

- 361

- 3

[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

Oxvillian

- 196

- 22

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

- #7

semc

- 361

- 3

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

Oxvillian

- 196

- 22

eg.

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

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

- #9

Oxvillian

- 196

- 22

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

semc

- 361

- 3

[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

Oxvillian

- 196

- 22

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

- #12

semc

- 361

- 3

- #13

Oxvillian

- 196

- 22

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

- #14

semc

- 361

- 3

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

- #15

Oxvillian

- 196

- 22

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

semc

- 361

- 3

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

- #17

Oxvillian

- 196

- 22

No problem

Share:

- Replies
- 2

- Views
- 764

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 20

- Views
- 3K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 0

- Views
- 38

- Replies
- 2

- Views
- 196

- Replies
- 5

- Views
- 287

- Last Post

- Replies
- 6

- Views
- 573