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

S2...|1 0> = |1 0> ?It is a little confusing. :/Sorry, I'm a little lost. . .The S_-^{(1)} and S_-^{(2)} operators act on the respective parts of the state, not on the same part. So for the first term, you'd have:S_-^{(1)}|3/2\ \ \ 3/2\ \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangleand for the second term, you'd have:S_-^{(2
  • #1
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,

[tex] \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 [/tex]

Had no trouble with that.

Part D asks to apply the operator:

[tex]S_- = S_-^{(1)} + S_-^{(2)}[/tex],

on the composite state constructed in part A.

Now I have,

[tex] S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle [/tex]

and

[tex]S_- = S_x - iS_y[/tex]

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?
 
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  • #2
The individual S- operators act on the respective parts of the state. If |s m>=|s1 m1>|s2 m2>, you'd have

[tex]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)[/tex]
 
  • #3
vela said:
The individual S- operators act on the respective parts of the state. If |s m>=|s1 m1>|s2 m2>, you'd have

So for,

[tex]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)[/tex]

it would be,

[tex]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)[/tex]
 
  • #4
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
angie_liamzon said:
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.
 
  • #6
And if my above attempt is true, do I keep the sqrt(2/5) and sqrt(3/5)?
 
  • #7
Yes, you have to keep the coefficients. Sorry, I should have included them in the line I wrote above.
 
  • #8
vela said:
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!
 
  • #9
So,

[tex]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)[/tex]

so,

[tex]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)[/tex]

where,

[tex](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[/tex]

and,

[tex](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[/tex]

Substituting in the previous two, we get:

[tex]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[/tex]

which equals:

[tex]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[/tex]

To conclude:

[tex]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[/tex]

Done correctly ? :)
 
  • #10
vwishndaetr said:
So for,

[tex]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)[/tex]

it would be,

[tex]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)[/tex]

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
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
They'd be [itex]\sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle[/itex] and [itex]\sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle[/itex].
 
  • #13
So you have:

[tex] 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) [/tex]

[tex]= \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+...[/tex]
 
  • #14
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
Continuing with this,

[tex] 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) [/tex]

[tex]= \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[/tex]

Now applying:

[tex]S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle[/tex]

First,

[tex]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[/tex]

second,

[tex]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[/tex]

Substituiting:[tex] 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[/tex]

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

[tex] 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[/tex]

Crossing my fingers. . .
 
  • #16
You don't have [itex]S_-^{(2)}[/itex] acting on the right part of the states.
 
  • #17
Yes and no.

Don't S_^1 |3/2 3/2> and S_^2 |3/2 3/2> essentially equal the same thing?
 
  • #18
Wait are you saying that I am not acting on |1 0> for say in the first term?
 
  • #19
Yeah. S1 only acts on the first ket in a product, while S2 only acts on the second ket.
 
  • #20
vela said:
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>.
 
  • #21
Going for a correct attempt:

[tex] 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) [/tex]

[tex]= \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)[/tex]

Now applying:

[tex]S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle[/tex]

First,

[tex]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[/tex]

second,

[tex]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[/tex]

third,

[tex]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[/tex]

fourth,

[tex]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[/tex]

Substituiting:

[tex]= \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)[/tex]

Now, simplifying,

[tex]= \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[/tex]

Combining first and fourth,

[tex]= 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[/tex]

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
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
Fantastic.

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

Thanks again.
 

1. What is the "S" operator in quantum mechanics?

The "S" operator, also known as the spin operator, is a mathematical operator used in quantum mechanics to describe the intrinsic angular momentum of a particle. It is denoted by the symbol S and its eigenvalues represent the possible values of a particle's spin.

2. How is the "S" operator applied on a composite state?

The "S" operator acts on a composite state by operating on each individual state within the composite state. This means that the "S" operator will affect the overall spin of the composite state by changing the spin of each individual state.

3. What is the composite state |5/2 3/2>?

The composite state |5/2 3/2> represents a system with a total angular momentum of 5/2 and a spin of 3/2. This could be a combination of two particles with different spins or a single particle with multiple spin states.

4. What are the possible results when applying the "S" operator on |5/2 3/2>?

The possible results when applying the "S" operator on |5/2 3/2> are the different eigenvalues of the spin operator, which in this case would be 3/2, 1/2, -1/2, and -3/2. These represent the possible spin states of the composite system after the "S" operator is applied.

5. How is the "S" operator used in quantum computing?

The "S" operator is one of the fundamental operators used in quantum computing algorithms to manipulate and measure the spin states of qubits. It plays a crucial role in quantum information processing and quantum error correction.

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