Homework Help: Applying S_ operator on composite state |5/2 3/2>

1. Feb 28, 2010

vwishndaetr

Original question:

A system contains two particles: the spin of particle 1 is 3/2 and the spin of particle 2 is 1. Motion of particles can be ignored.

Part A asked to find the total spin state |5/2 3/2> using Clebsch-Gordon coefficients.

I did so, and came up with,

$$\mid 5/2\ \ \ 3/2\ \rangle = \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle$$

Had no trouble with that.

Part D asks to apply the operator:

$$S_- = S_-^{(1)} + S_-^{(2)}$$,

on the composite state constructed in part A.

Now I have,

$$S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle$$

and

$$S_- = S_x - iS_y$$

Kinda don't know where to go from here. I understand that since we are using minus operator, spin will be decreasing. But how do values of s and m change accordingly?

Last edited: Feb 28, 2010
2. Feb 28, 2010

vela

Staff Emeritus
The individual S- operators act on the respective parts of the state. If |s m>=|s1 m1>|s2 m2>, you'd have

$$S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)$$

3. Feb 28, 2010

vwishndaetr

So for,

$$S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)$$

it would be,

$$S_-|5/2\ \ 3/2\rangle = (S_-^{(1)}|3/2\ \ 3/2\rangle)|1\ \ 0\rangle+|3/2\ \ 1/2\rangle(S_-^{(2)}|1\ \ 1\rangle)$$

4. Feb 28, 2010

angie_liamzon

hi! i'm a new member. just got a problem of getting the integral of e to the negative x squared from a to b. not the typical error function from 0 to infinity. please help me out with this. thank you.

5. Feb 28, 2010

vwishndaetr

Try the homework help forum, for calculus.

6. Feb 28, 2010

vwishndaetr

And if my above attempt is true, do I keep the sqrt(2/5) and sqrt(3/5)?

7. Feb 28, 2010

vela

Staff Emeritus
Yes, you have to keep the coefficients. Sorry, I should have included them in the line I wrote above.

8. Feb 28, 2010

vwishndaetr

Awesome! I will post my result in a few min, if you can just check it that'd be great!

9. Feb 28, 2010

vwishndaetr

So,

$$S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)$$

so,

$$S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ (S_-^{(1)}|3/2\ \ 3/2\rangle)|1\ \ 0\rangle+\sqrt{3/5}\ |3/2\ \ 1/2\rangle(S_-^{(2)}|1\ \ 1\rangle)$$

where,

$$(S_-^{(1)}|3/2\ \ 3/2\rangle) = \sqrt{3/2(3/2+1) - 3/2(3/2-1)}\ \hbar\mid 3/2\ \ 3/2-1 \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle$$

and,

$$(S_-^{(2)}|1\ \ 1\rangle) = \sqrt{1(1+1) - 1(1-1)}\ \hbar\mid 1\ \ 1-1 \rangle = \sqrt{2}\ \hbar\mid 1\ \ 0 \rangle$$

Substituting in the previous two, we get:

$$S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle+\sqrt{3/5}\ |3/2\ \ 1/2\rangle\sqrt{2}\ \hbar\mid 1\ \ 0 \rangle$$

which equals:

$$S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle+\sqrt{3/5}\sqrt{2}\ \hbar\ |3/2\ \ 1/2\rangle\mid 1\ \ 0 \rangle$$

To conclude:

$$S_-|5/2\ \ 3/2\rangle = (\sqrt{2/5}\ \sqrt{3}+\sqrt{3/5}\sqrt{2}\ )\ \hbar\\ \mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle$$

Done correctly ? :)

10. Feb 28, 2010

nickjer

You are forgetting to expand the whole equation out. It is similar to:

(s1+s2)*(v1+v2) = s1*v1+s1*v2+s2*v1+s2*v2

So you should have 4 parts and not just 2.

11. Feb 28, 2010

vwishndaetr

OK. I get what you're saying, but little los at the same time.

In this case, what would my "v1" and "v2" be?

12. Feb 28, 2010

vela

Staff Emeritus
They'd be $\sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle$ and $\sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle$.

13. Feb 28, 2010

nickjer

So you have:

$$S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \left( S_{-}^{1}+S_{-}^{2}\right) \cdot \left( \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle \right)$$

$$= \sqrt{2/5}(S_{-}^{1}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{1}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle+....$$

14. Feb 28, 2010

vwishndaetr

Oooo. For some reason when I saw (s1+s2)*(v1+v2) = s1*v1+s1*v2+s2*v1+s2*v2, I was thinking s1*(v1+s1)*(v2+s2)*(v1+s2)*v2 and it made me think in circles. Silly mistake on my part.

Thanks to the both of you! I will be back in a couple to make sure I got it! :)

15. Feb 28, 2010

vwishndaetr

Continuing with this,

$$S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \left( S_{-}^{1}+S_{-}^{2}\right) \cdot \left( \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle \right)$$

$$= \sqrt{2/5}(S_{-}^{1}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{1}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}(S_{-}^{2}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{2}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle$$

Now applying:

$$S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle$$

First,

$$S_-\mid 3/2\ \ 3/2 \rangle = \sqrt{3/2(3/2+1) - 3/2(3/2-1)}\ \hbar\mid 3/2\ \ 3/2-1 \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle$$

second,

$$S_-\mid 3/2\ \ 1/2 \rangle = \sqrt{3/2(3/2+1) - 1/2(1/2-1)}\ \hbar\mid 3/2\ \ 1/2-1 \rangle = \sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle$$

Substituiting:

$$S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \sqrt{2/5}(\sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(\sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}\sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(\sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle)\mid 1\ \ \ 1\ \rangle$$

Now, simplifying, I want to combine to two terms:

$$S_{-}\mid 5/2\ \ \ 3/2\ \rangle = 2\sqrt{6/5}\ \hbar\mid 3/2\ \ 1/2 \rangle\mid 1\ \ \ 0\ \rangle+2\sqrt{3/5}\ \hbar\mid 3/2\ \ -1/2 \rangle\mid 1\ \ \ 1\ \rangle$$

Crossing my fingers. . .

16. Feb 28, 2010

vela

Staff Emeritus
You don't have $S_-^{(2)}$ acting on the right part of the states.

17. Feb 28, 2010

vwishndaetr

Yes and no.

Don't S_^1 |3/2 3/2> and S_^2 |3/2 3/2> essentially equal the same thing?

18. Feb 28, 2010

vwishndaetr

Wait are you saying that im not acting on |1 0> for say in the first term?

19. Feb 28, 2010

vela

Staff Emeritus
Yeah. S1 only acts on the first ket in a product, while S2 only acts on the second ket.

20. Feb 28, 2010

vwishndaetr

Ok. So let me get this. . .

S1 acts on the |3/2 3/2> and |3/2 1/2> and the S2 acts on the |1 0> and |1 1>.

21. Feb 28, 2010

vwishndaetr

Going for a correct attempt:

$$S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \left( S_{-}^{1}+S_{-}^{2}\right) \cdot \left( \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle \right)$$

$$= \sqrt{2/5}(S_{-}^{1}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{1}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}\mid 3/2\ \ \ 3/2\ \rangle(S_{-}^{2}\mid 1\ \ \ 0\ \rangle)+\sqrt{3/5}\mid 3/2\ \ \ 1/2\ \rangle(S_{-}^{2}\mid 1\ \ \ 1\ \rangle)$$

Now applying:

$$S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle$$

First,

$$S_-^{(1)}\mid 3/2\ \ 3/2 \rangle = \sqrt{3/2(3/2+1) - 3/2(3/2-1)}\ \hbar\mid 3/2\ \ 3/2-1 \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle$$

second,

$$S_-^{(1)}\mid 3/2\ \ 1/2 \rangle = \sqrt{3/2(3/2+1) - 1/2(1/2-1)}\ \hbar\mid 3/2\ \ 1/2-1 \rangle = \sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle$$

third,

$$S_-^{(2)}\mid 1\ \ 0 \rangle = \sqrt{1(1+1) - 0(0-1)}\ \hbar\mid 1\ \ 0-1 \rangle = \sqrt{2}\ \hbar\mid 1\ \ -1\rangle$$

fourth,

$$S_-^{(2)}\mid 1\ \ 1 \rangle = \sqrt{1(1+1) - 1(1-1)}\ \hbar\mid 1\ \ 1-1 \rangle = \sqrt{2}\ \hbar\mid 1\ \ 0\rangle$$

Substituiting:

$$= \sqrt{2/5}(\sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(\sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}\mid 3/2\ \ \ 3/2\ \rangle(\sqrt{2}\ \hbar\mid 1\ \ -1\rangle)+\sqrt{3/5}\mid 3/2\ \ \ 1/2\ \rangle(\sqrt{2}\ \hbar\mid 1\ \ 0\rangle)$$

Now, simplifying,

$$= \sqrt{6/5}\ \hbar\mid 3/2\ \ 1/2 \rangle\mid 1\ \ \ 0\ \rangle+\2sqrt{3/5}\ \hbar\mid 3/2\ \ -1/2 \rangle\mid 1\ \ \ 1\ \rangle+ \sqrt{4/5}\ \hbar\mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ -1\rangle+\sqrt{6/5}\ \hbar\mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ 0\rangle$$

Combining first and fourth,

$$= 2\sqrt{6/5}\ \hbar\mid 3/2\ \ 1/2 \rangle\mid 1\ \ \ 0\ \rangle+\2\sqrt{3/5}\ \hbar\mid 3/2\ \ -1/2 \rangle\mid 1\ \ \ 1\ \rangle+ \sqrt{4/5}\ \hbar\mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ -1\rangle$$

I am eager to say there is no room for mistakes, but I thought I had it 3 hrs ago. Can I get a "You got it!" ?

22. Feb 28, 2010

vela

Staff Emeritus
I didn't check all the details, but your set-up looks fine. As long as you didn't make any arithmetic mistakes, I'd say you got it!

23. Feb 28, 2010

vwishndaetr

Fantastic.

Thanks so much for the prompt responses! I can now go to sleep!

Thanks again.