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

[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?

 

[vela]

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

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

 

[vwishndaetr]

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

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,

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

 

[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.

 

[vwishndaetr]

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.

Try the homework help forum, for calculus.

 

[vwishndaetr]

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

 

[vela]

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

 

[vwishndaetr]

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

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

 

[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 ? :)

 

[nickjer]

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,

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

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.

 

[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?

 

[vela]

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

 

[nickjer]

So you have:

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

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

 

[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! :)

 

[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. . .

 

[vela]

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

 

[vwishndaetr]

Yes and no.

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

 

[vwishndaetr]

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

 

[vela]

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

 

[vwishndaetr]

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

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>.

 

[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!" ?

 

[vela]

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!

 

[vwishndaetr]

Fantastic.

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

Thanks again.