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,...,s-1,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!
From where does [itex]S_z[/itex] come? Depending on how you found [itex]S_z[/itex], you might be able to use the same technique to find [itex]S_x[/itex] and [itex]S_y[/itex] first, or to find [itex]S_+[/itex] and [itex]S_-[/itex] first. Neither do I. What states? The eigenstates of [itex]S_z[/itex]? General states? Is this question from a course? From a book? If so which course and book?
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...
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,m-1 \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]
George Jones, you mean [tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex] ?
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_z|S,m> = m |S,m>[/tex] and is thus a diagonal (2S+1)x(2S+1) matrix with elements [tex]S, S-1,...,-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, S-1,..., -S[/tex] and [tex]\theta\in [0,\pi][/tex]): [tex]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')!}[/tex] and Uz is diagonal with ([tex]\phi\in [0,2 \pi][/tex]): [tex]U^z_{k,k}(\phi) = e^{i (k-1) \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.
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 ?
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.
If you know what the 3x3 ladder operators look like, start with the definitions [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]
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...
The equations I have posted solve your problem for any spin, integer or half-integer. 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]
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 ?
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.
I don't undersand why you are this square root ? It is in reality 1/2 and not [tex] \frac{1}{\sqrt{2}}[/tex]
[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.
Strange, you should get the square root and I don't see anything wrong in my eqs. Besides your results should check against: [tex] [S_i,S_j]=i \hbar\epsilon_{i,j,k} S_k [/tex]
[tex] S_+ \left| s m \right> = \sqrt{s \left( s + 1 \right) - m \left(m + 1 \right)} \left| s,m+1 \right>; [/tex] [tex] 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>; [/tex] That's where the square root comes from. When you multiply by 1/2, you get it in the denominator.