# Homework Help: Spin matrices for particle of spin 1

1. Aug 14, 2007

### genloz

1. The problem statement, all variables and given/known data
Construct the spin matrices ($$S_{x}$$, $$S_{y}$$, and $$S_{z}$$) for a particle of spin 1. Determine the action of $$S_{z}$$, $$S_{+}$$, and $$S_{-}$$ on each of these
states.

2. Relevant equations
$$S=\sqrt{1(1+1)}\hbar$$
m=-s,-s+1,...,s-1,s

3. The attempt at a solution
m=-1,0,0,0,0,0,0,0,-1
$$S=\sqrt{2}\hbar$$
$$S_{z}=\hbar $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right)$$$

So I understand where $$S_{z}$$ comes from... I know what the answers for the matrices are from http://en.wikipedia.org/wiki/Pauli_matrices but I don't know how to go about finding $$S_{x}$$ and $$S_{y}$$ even with the commutation laws etc... and I don't really understand what it means by 'the action of' in the second part despite numerous searchs on google....

Any help is much appreciated, thankyou!

2. Aug 15, 2007

### George Jones

Staff Emeritus
From where does $S_z$ come?

Depending on how you found $S_z$, you might be able to use the same technique to find $S_x$ and $S_y$ first, or to find $S_+$ and $S_-$ first.

Neither do I. What states?

The eigenstates of $S_z$? General states?

Is this question from a course? From a book? If so which course and book?

3. Aug 15, 2007

### genloz

But I'm very confused now... This exact question "Construct the spin matrices ($$S_{x}$$, $$S_{y}$$, and $$S_{z}$$) for a particle of spin 1. Determine the action of $$S_{z}$$, $$S_{+}$$, and $$S_{-}$$ on each of these states." is from a quantum mechanics course. The relevant equations were equations I thought were relevant... and I thought that the equation listed for m gave the matrix $$S_{z}$$.

How would I find eigenstates of $$S_{z}$$? And what is the usual way for determining spin matrices? I can't find this info out anywhere on the internet and am getting more and more confused by the second...

4. Aug 15, 2007

### George Jones

Staff Emeritus
Have you seen the general theory of angular momentum in your quantum course?

Stuff like ($\hbar = 1$):

$$J_z \left| jm \right> = m \left| jm \right>;$$

$$J_+ \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right>;$$

$$J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.$$

If so, you can find the matices by considering stuff like

$$\left< jm \right| J_\pm \left| jm \right>$$

for $j=1$ and $m = -1, 0, 1.$

Last edited: Aug 15, 2007
5. Aug 15, 2007

### malawi_glenn

George Jones, you mean

$$J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.$$

?

6. Aug 15, 2007

### George Jones

Staff Emeritus
Yes, thanks. I have corrected my previous post.

7. Oct 6, 2008

### kaltsoplyn

This is an old thread, but people are bound to come back looking for these answers, so here's my 2 cents.
For a given $$S$$, $$S_z$$ comes from demanding $$S_z|S,m> = m |S,m>$$ and is thus a diagonal (2S+1)x(2S+1) matrix with elements $$S, S-1,...,-S$$
(Note: I assume $$|S,S> = (1,0,...,0)$$ and $$|S,-S> = (0,0,...,1)$$ )
The $$S_x$$ and $$S_y$$ matrices - and consequently $$S_+$$ and $$S_-$$ - come from rotations of Sz about the y and z axes, $$S_x = U_y(\pi/2).S_z.U^{\dagger}_y(\pi/2) \text{ and } S_y = U_z(\pi/2).U_y(\pi/2).S_z.\left(U_z(\pi/2).U_y(\pi/2)\right)^{\dagger}$$,
where the Uy matrix is comprised by the elements ($$m, m' = S, S-1,..., -S$$ and $$\theta\in [0,\pi]$$):

$$U^{y}_{m,m'}(\theta) = \sqrt{(S-m)! (S+m)! (S-m')! (S+m')!}\sum_{x=Max(0,m'-m)}^{Min(S+m',S-m)}\frac{(-1)^x cos(-\theta/2)^{2 S + m' - m - 2 x} sin(-\theta/2)^{2 x + m - m'}}{(S + m' - x)! (S - m - x)! x! (x + m - m')!}$$

and Uz is diagonal with ($$\phi\in [0,2 \pi]$$):

$$U^z_{k,k}(\phi) = e^{i (k-1) \phi}, \text{ } k = 1,2,...,2 S+1$$

There is some ambiguity here however, since another Uz is quoted:

$$U^z_{m,m}(\phi) = e^{-i m \phi}, \text{ } m = S,...,-S$$

In the former case, $$k$$ refer to matrix index (e.g. $$U_{1,1}$$ is the upper left element of the matrix), while in the latter case $$m$$ refer to magnetic number indexing (e.g. $$U_{S,S}$$ is the upper left element of the matrix). I use the former formula and it gives the expected results.

8. Dec 4, 2008

### turin

How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?

9. Sep 1, 2009

### lydilmyo

Hello,
I've to construct Ix and Iy for I=1.

so, I can construct lowering and raising operator but how do you construct cartesian operator from this equation ?

there are no definition for Ix and Iy by their action on eigenstate vector like Iz, I+ and I-....

How can I do that easily ?

10. Sep 1, 2009

### kaltsoplyn

By using the spinor representation.
In essence you are using combinations of spin-1/2 to represent the behaviour of arbitrarily large spins. This way you can generate operators and wavefunctions of large spins starting from the known spin-1/2 matrices.

This was shown originaly by Majorana in 1932.
I have retrieved the info from W.Thompson's Angular Momentum book.

11. Sep 1, 2009

### kuruman

$$S_{+}=S_{x}+iS_{y}$$

$$S_{-}=S_{x}-iS_{y}$$

and write the cartesian matrices as linear combinations of the ladder operators.

$$S_{x} = ...$$

$$S_{y} = ...$$

12. Sep 1, 2009

### lydilmyo

but this relation is available only for pauli matrix (2x2) and so for spin 1/2....

please help me !!!! or give the matrix representation of Ix and Iy for I=1 if you know it...

13. Sep 1, 2009

### kuruman

The relations are good for any dimensionality.

14. Sep 1, 2009

### kaltsoplyn

The equations I have posted solve your problem for any spin, integer or half-integer.
Anyway, for spin = 1:

$$S_x = \left( \begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & 0 \end{array} \right)$$

$$S_y = \left( \begin{array}{ccc} 0 & -\frac{i}{\sqrt{2}} & 0 \\ \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & \frac{i}{\sqrt{2}} & 0 \end{array} \right)$$

$$S_z = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right)$$

15. Sep 1, 2009

### lydilmyo

so, why kaltsoplyn talk about a really hard way to determine Ix and Iy if it is so easy to construct its by this relations ?

16. Sep 1, 2009

### kaltsoplyn

Synthesizing high order spins starting from spin-1/2's is hard, although the procedure is straightforward.

The formulas I give are general that's why they look so complex. However, it's not very hard to produce any such matrix if you plug these formulas in a symbolic mathematical software package like, say, Mathematica or Maple.

17. Sep 1, 2009

### lydilmyo

I don't undersand why you are this square root ? It is in reality 1/2 and not $$\frac{1}{\sqrt{2}}$$

18. Sep 1, 2009

### lydilmyo

$$S_x = \frac{1}{2} \left(\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) + \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)\right)$$

$$S_y = \frac{i}{2} \left(\left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) - \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)\right)$$

this is what I find with your relations.

19. Sep 1, 2009

### kaltsoplyn

Strange, you should get the square root and I don't see anything wrong in my eqs.
Besides your results should check against:

$$[S_i,S_j]=i \hbar\epsilon_{i,j,k} S_k$$

20. Sep 1, 2009

### kuruman

$$S_+ \left| s m \right> = \sqrt{s \left( s + 1 \right) - m \left(m + 1 \right)} \left| s,m+1 \right>;$$

$$S_+ \left| 10 \right> = \sqrt{1* \left( 1 + 1 \right) - 0* \left(0 + 1 \right)} \left| 1,1 \right> = \sqrt{2}\left| 1,1 \right>;$$

That's where the square root comes from. When you multiply by 1/2, you get it in the denominator.