Register to reply 
Spin matrices for particle of spin 1 
Share this thread: 
#1
Aug1407, 11:57 PM

P: 76

1. The problem statement, all variables and given/known data
Construct the spin matrices ([tex]S_{x}[/tex], [tex]S_{y}[/tex], and [tex]S_{z}[/tex]) for a particle of spin 1. Determine the action of [tex]S_{z}[/tex], [tex]S_{+}[/tex], and [tex]S_{}[/tex] on each of these states. 2. Relevant equations [tex]S=\sqrt{1(1+1)}\hbar [/tex] m=s,s+1,...,s1,s 3. The attempt at a solution m=1,0,0,0,0,0,0,0,1 [tex]S=\sqrt{2}\hbar [/tex] [tex]S_{z}=\hbar \[ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)\] [/tex] So I understand where [tex]S_{z}[/tex] 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 [tex]S_{x}[/tex] and [tex]S_{y}[/tex] 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
Aug1507, 06:59 AM

Mentor
P: 6,248

The eigenstates of [itex]S_z[/itex]? General states? Is this question from a course? From a book? If so which course and book? 


#3
Aug1507, 07:33 AM

P: 76

Thanks very much for your reply....
But I'm very confused now... This exact question "Construct the spin matrices ([tex]S_{x}[/tex], [tex]S_{y}[/tex], and [tex]S_{z}[/tex]) for a particle of spin 1. Determine the action of [tex]S_{z}[/tex], [tex]S_{+}[/tex], and [tex]S_{}[/tex] 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 [tex]S_{z}[/tex]. How would I find eigenstates of [tex]S_{z}[/tex]? 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
Aug1507, 08:03 AM

Mentor
P: 6,248

Spin matrices for particle of spin 1
Have you seen the general theory of angular momentum in your quantum course?
Stuff like ([itex]\hbar = 1[/itex]): [tex]J_z \left jm \right> = m \left jm \right>;[/tex] [tex]J_+ \left jm \right> = \sqrt{j \left( j + 1 \right)  m \left(m + 1 \right)} \left j,m+1 \right>;[/tex] [tex]J_ \left jm \right> = \sqrt{j \left( j + 1 \right)  m \left(m  1 \right)} \left j,m1 \right>.[/tex] If so, you can find the matices by considering stuff like [tex]\left< jm \right J_\pm \left jm \right>[/tex] for [itex]j=1[/itex] and [itex]m = 1, 0, 1.[/itex] 


#5
Aug1507, 08:13 AM

Sci Advisor
HW Helper
P: 4,738

George Jones, you mean
[tex]J_ \left jm \right> = \sqrt{j \left( j + 1 \right)  m \left(m  1 \right)} \left j,m1 \right>.[/tex] ? 


#7
Oct608, 08:22 AM

P: 5

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 [tex]S[/tex], [tex]S_z[/tex] comes from demanding [tex]S_zS,m> = m S,m>[/tex] and is thus a diagonal (2S+1)x(2S+1) matrix with elements [tex]S, S1,...,S[/tex] (Note: I assume [tex]S,S> = (1,0,...,0)[/tex] and [tex]S,S> = (0,0,...,1)[/tex] ) The [tex]S_x[/tex] and [tex]S_y[/tex] matrices  and consequently [tex]S_+[/tex] and [tex]S_[/tex]  come from rotations of Sz about the y and z axes, [tex]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}[/tex], where the Uy matrix is comprised by the elements ([tex]m, m' = S, S1,..., S[/tex] and [tex]\theta\in [0,\pi][/tex]): [tex]U^{y}_{m,m'}(\theta) = \sqrt{(Sm)! (S+m)! (Sm')! (S+m')!}\sum_{x=Max(0,m'm)}^{Min(S+m',Sm)}\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')!}[/tex] and Uz is diagonal with ([tex]\phi\in [0,2 \pi][/tex]): [tex]U^z_{k,k}(\phi) = e^{i (k1) \phi}, \text{ } k = 1,2,...,2 S+1[/tex] There is some ambiguity here however, since another Uz is quoted: [tex]U^z_{m,m}(\phi) = e^{i m \phi}, \text{ } m = S,...,S[/tex] In the former case, [tex]k[/tex] refer to matrix index (e.g. [tex]U_{1,1}[/tex] is the upper left element of the matrix), while in the latter case [tex]m[/tex] refer to magnetic number indexing (e.g. [tex]U_{S,S}[/tex] is the upper left element of the matrix). I use the former formula and it gives the expected results. 


#8
Dec408, 01:19 PM

HW Helper
P: 2,327

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



#9
Sep109, 05:20 AM

P: 9

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
Sep109, 05:48 AM

P: 5

In essence you are using combinations of spin1/2 to represent the behaviour of arbitrarily large spins. This way you can generate operators and wavefunctions of large spins starting from the known spin1/2 matrices. This was shown originaly by Majorana in 1932. I have retrieved the info from W.Thompson's Angular Momentum book. 


#11
Sep109, 08:20 AM

HW Helper
PF Gold
P: 3,440

[tex]S_{+}=S_{x}+iS_{y}[/tex] [tex]S_{}=S_{x}iS_{y}[/tex] and write the cartesian matrices as linear combinations of the ladder operators. [tex]S_{x} = ...[/tex] [tex]S_{y} = ...[/tex] 


#12
Sep109, 08:49 AM

P: 9

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... 


#14
Sep109, 09:23 AM

P: 5

Anyway, for spin = 1: [tex] 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) [/tex] [tex] 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) [/tex] [tex] S_z = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex] 


#15
Sep109, 09:23 AM

P: 9

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
Sep109, 09:27 AM

P: 5

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
Sep109, 09:28 AM

P: 9




#18
Sep109, 09:34 AM

P: 9

[tex]
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) [/tex] [tex] 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) [/tex] this is what I find with your relations. 


Register to reply 
Related Discussions  
The physical meaning of Pauli Spin Matrices  Quantum Physics  16  
The physical meaning of Pauli Spin Matrices  Advanced Physics Homework  2  
Spin 3/2 Rotation Matrices  Quantum Physics  6  
Particle spin  High Energy, Nuclear, Particle Physics  14 