Trouble finding Clebsch-Gordan coefficient for deutron triplet state?

[semc]

Hi, for the deutron triplet state, I am trying to find the coefficient using the lowering operator but I kept getting 0 hope someone can help me.

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

What is wrong with my working? Thanks

 

[Oxvillian]

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

$$J_- = J_{1-} + J_{2-}$$

 

[semc]

$$J_- |+-> + |-+> = J_{1-} + J_{2-}|+-> + |-+>$$ I am not sure about this part but i split the operators into
$$J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>$$
so $$J_{1-}|+-> + |-+>$$ is where I am stuck since $$m_1=1/2 , m_2=-1/2$$ so everything will be reduced to 0.

 

[Oxvillian]

Don't forget the brackets!

$$\left( J_{1-} + J_{2-} \right) \left(\lvert +- \rangle + \lvert -+ \rangle \right)$$

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

 

[semc]

Yeah I am aware there are 4 terms but is this correct?
$$J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>$$
If its correct then I should be able to evaluate this $$J_{1-}|+-> + |-+>$$
I am aware that when the lowering operator acts on the lowest eigenvalue it will disappear but I can't 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

 

[Oxvillian]

You're not using the distributive law correctly: (A+B)(x+y) = Ax + Ay + Bx + By. Every state in the sum should have a lowering operator acting on it.

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

 

[semc]

I really still don't see what I did wrong there. I simply use (A+B)(x+y)=A(x+y)+B(x+y). I am having problem here because I am using the lowering operator on + but I am getting 0 which is clearly wrong but I just don't know why I got it wrong. I will show my working

Using the lowering operator $$J_-|jm>=((j+m)(j-m+1))^{1/2}|jm-1>$$
on |-+> I got this $$J_-|-1/2, 1/2>=((-1/2+1/2)(-1/2-1/2+1))^{1/2}|-1/2,-1/2>$$
$$(-1/2+1/2)(-1/2-1/2+1)=(0)(0)=0$$ 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 what's wrong

 

[Oxvillian]

eg.

$$S_{1-} \lvert +- \rangle = \lvert -- \rangle$$

 

[Oxvillian]

Sorry, we cross-posted...

$$\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1$$

 

[semc]

$$\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1$$

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

 

[Oxvillian]

It's also the case for |+-> because J_1- only acts on the +. It leaves the - alone.

Maybe that's the point you're missing? J_1- lowers the first index and leaves the second alone; J_2- acts on the second index and leaves the first alone.

 

[semc]

Yes I am aware of that but in the working I wrote a few post ago, the coefficient is 0 since we have (j+m) which is 0.

 

[Oxvillian]

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

 

[semc]

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

 

[Oxvillian]

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 j₁ = j₂= 1/2. The + and - refer to m₁ and m₂ respectively.

For instance |+-> would have j₁ = j₂= 1/2, m₁ = 1/2 and m₂ = -1/2.

 

[semc]

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

 

[Oxvillian]

No problem