Understanding Spin Matrix: Pauli Matrices and 6x6 Matrices

[Rajini]

Hi everyone,
I now able to understand spin matrix (if i am correct in other words Pauli matrix).
For e.g.,
for S=5/2 systems the spin matrix (say for S_X) is given by:

S_x= 1/2[a 6X6 matrix]

I hope members will know what is this 6X6 matrix! Since i don't know how to type matrix in this forum]..
Actually i wanted to know how this 6X6 matrix is obtained??

thanks and wish u a very happy new yr '09.
Rajini

 

[malawi_glenn]

"just" use the definition of the rotation operator:

[tex] D^{(s)}_{M,m} = <s,M|e^{-i\vec{S}\cdot \vec{n} \phi / \hbar} |s,m> [/tex]

where [tex] \vec{S} = (S_x, S_y, S_z) [/tex]

See Sakurai, Modern Quantum mechanics, chapter 3 or similar textbooks.

 

[malawi_glenn]

or wait a minute, what i just wrote is crap, it is not what you are looking for.

You want matrix representation of operators S_x , S_y and S_z

Just rewrite S_x and S_y in terms of ladder operators and work out each entry in the matrix :-)

Easy but time-consuming!

 

[Rajini]

Hi thanks..
but how?
Can you please make one calculation..for one (S_x or S_y or S_z? for a S=5/2..
thanks

 

[malawi_glenn]

i mean, this is easy if you are a researcher, the ladder operators are defined as:

[tex]J_{\pm} \equiv J_x \pm i J_y[/tex]

And then use:

[tex] J_{\pm}|j,m> = \hbar \sqrt{j(j+1) - m(m\pm 1 )}|j,m \pm 1 > [/tex]


That's all you need man

(Here J is the standard symbol for angular momentum)

 

[Rajini]

i really don't understand of..how to operate for S=5/2
I am a expert..but started learning slowly...
or if possible can u explain each steps found in this page (they did for 1/2)
http://quantummechanics.ucsd.edu/ph130a/130_notes/node278.html
i hope if you explain me i can derive for S=5/2,MS> state
thanks (if u don't find time..i hope i can succeed..but takes at least 1 week)
rajini

 

[malawi_glenn]

1 week?

Why don't you just try the matrix element

< S = 5/2, M = 5/2 | S_y | S = 5/2, M = 3/2>

Write S_y = (S(+) - S(-))/(2i)

And use the formula in my latest post. It is a really straightforward calculation, good luck

 

[malawi_glenn]

a question, do you know how matrix representation of operators work at all? I am having a problem to understand where your lack of understanding is. You only say "explain each step" but it is really hard to know what you know and what you don't know.

Why don't you tell what part you don't understand?

 

[Rajini]

really i don't know how matrix rep. of operators work!
< S = 5/2, M = 5/2 | S_y | S = 5/2, M = 3/2>
Okay how to rep. the above relation as a matrix.also why m=3/2?
thanks

 

[malawi_glenn]

I just gave a particular matrix element to begin with.

Now look at the spin 1/2 matrix, let's look at S_z :

The first entry in the upper left corner (1) is this one: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = 1/2> = 1 (I pull out the factor hbar/2 ...) Make sure you can do this!

Then the entry in the upper right corner (0) is: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = -1/2> = 0

The entry in the lower left is < S = 1/2, M = -1/2 |S_z|S = 1/2, M = 1/2> = 0

and

The entry in the lower right corner is < S = 1/2, M = -1/2 |S_z|S = 1/2, M = -1/2> ...

So what one does is to put the states with decreasing value of M as rows and coloums, then work out each entry in the matrix (that's why one calls <A| H | B> matrix element, it is an element in the matrix representing the operator H)

Now make sure you can do the S_z for spin 1/2 and also the S_x and S_y for spin 1/2. Then try to do spin 5/2, which is same procedure, but many many more elements to calculate.

 

[CPL.Luke]

also remember that the s_z matrix is just a diagonal matrix with the eigenvalues going down the diagonal.

ie for spin 5/2 along the diagonal the values are 1,1,1,-1,-1,-1

 

[malawi_glenn]

yes, so in reality, only the S_x and S_y are "tricky" to work out.
but the diagonal values should be 5/2, 3/2, 1/2, -1/2, -3/2, -5/2 ...

 

[Rajini]

I just gave a particular matrix element to begin with.

So what one does is to put the states with decreasing value of M as rows and coloums, then work out each entry in the matrix (that's why one calls <A| H | B> matrix element, it is an element in the matrix representing the operator H)

Now make sure you can do the S_z for spin 1/2 and also the S_x and S_y for spin 1/2. Then try to do spin 5/2, which is same procedure, but many many more elements to calculate.

I think i have to read a bit!. Since u wrote 'M' decreases (by 1) in rows and columns...Then for S=5/2 (2S+1=6 i.e., a 6X6 matrix) one will not get a diagonal matrix..(but this is not true)
or i don't understand properly the <A|H|B> methods..

 

[malawi_glenn]

Yes you will obtain a diagonal matrix for S_z, it is trivial to see that.

 

[Rajini]

I just gave a particular matrix element to begin with.

Now look at the spin 1/2 matrix, let's look at S_z :

The first entry in the upper left corner (1) is this one: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = 1/2> = 1 (I pull out the factor hbar/2 ...) Make sure you can do this!

One question. How you say its 1. There are four 1/2 so which u take? (i understand u take 1/2 outside the matrix so its 1...but which 1/2 u take)
thanks

 

[malawi_glenn]

I said "upper left corner"...

 

[Rajini]

I just gave a particular matrix element to begin with.


Then the entry in the upper right corner (0) is: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = -1/2> = 0

Then here upper right corner means i assume u r talking upper right corner M=-1/2 (minus 1/2)...so decreasing means it should be -1/4..but u say zero!

 

[malawi_glenn]

eh? can you evaluate < S = 1/2, M = 1/2 |S_z|S = 1/2, M = -1/2>
for me and show how it could be zero...

 

[Rajini]

actually i don't know so i asked you how it could be zero! but you are asking me the same question..anyway i don't want to disturb you now..normally i like QM but only without lesser than, greater than and pipe symbols (i won't say it bra-ket)..since i don't understand all these properly...also i guess there is lots of books which deals with all these lesser than, greater than and pipe symbols but NOT WITH EXAMPLES (or only for 1/2=S)...Any way i think i can manage to search for some books (tomorrow) and hope to find the solution..
anyway thanks
rajini

 

[malawi_glenn]

But the bra-ket notation is as simple as wavefunctions. Look:

wave function:

[tex] S_z \chi _+ = +\frac{\hbar}{2}\chi _+ [/tex]

Bra-ket

[tex] S_z |+> = +\frac{\hbar}{2}|+>[/tex]

Wave function:

[tex] \int d\vec{x} \chi _+ ^*\chi _+ = 1 [/tex]
[tex] \int d\vec{x} \chi _- ^*\chi _+ = 0 [/tex]

Bra-Ket:
[tex] <+|+> = 1 [/tex]
[tex] <-|+> = 0 [/tex]

I also mean, why do you want matrix representation of operators if you don't even know what it is or how bra-ket works?

 

[Rajini]

[tex]
S_z \chi _+ = +\frac{\hbar}{2}\chi _+
[/tex]
[tex]
S_z = +\frac{\hbar}{2}\chi _+ /\chi _+ = ?
[/tex]
here if we take X_+ to otherside what will happen..i mean S_z=?
if the above relation is correct...how to find S_z
I think i have to read some basic of such notations..can you advice me some books which explains with a example of <A,M|H|B,M> with high order spins (not 1/2)

 

[malawi_glenn]

What?

 

[CPL.Luke]

woops that was obvious, I guess break has worn on me a bit.

 

[Rajini]

Now i am somewhat clear! [if something is wrong here...pls. correct me]
But may be for others (non experts like me) it will be helpful..
For S=5/2 system:
the matrix for S_z:
the symbolic notation is
<S,M'_s|S_z|S,M_s>
Now take S=5/2 (this is fixed)
and M_s=5/2,3/2,1/2,-1,2-3/2,-5/2 (i.e., 2S+1=2(5/2)+1=six M_s states)
|S,M_s> = 5/2,3/2,1/2,-1,2-3/2,-5/2 (other way |5/2 5/2>, |5/2 3/2>, ...|5/2 -5/2>) (this is a row)
and
<S,M'_s| = [5/2,3/2,1/2,-1,2-3/2,-5/2]^T (other way [|5/2 5/2>, |5/2 3/2>, ...|5/2 -5/2>]^T) note: this is a column
Now make table (matrix) see figure.
and then fill the matrix elements..
Here → represents |S=5/2,Ms>
And ↓ represents <S=5/2,Ms|
Whenever Ms=M’s => the matrix element is simply Ms or M’s (for Sx).
If they are not equal => just zero [not same states and they don't exist so zero]
And finally multiply with [tex]\ 6.62606876(52)/pi\times\ 10^{-34}\ J\ s[/tex]
Now i will do the same for S_y and S_x (i.e., Ms-1 and Ms+1)

 

[malawi_glenn]

Now we're talking! you got the S_z correct, congrat!

Now comes to tricky and hard part, to do all S_x and S_y

You might want to make sure you can reproduce those for the S = 1/2 case, just as a warm up

 

[Rajini]

Hi I solved S_x and S_y..(for S_x please see the jpg file)
Now i have another confusion...
I have seen many people just omit hbar (i am not sure whether they include some where for compensation for omitting.)!
I use (S_x)^2 in my Hamiltonian and so i get hbar^2.
Suppose my final matrix is of 6X6 with hbar^2 outside the matrix..Now if i multiply with some number (this number's unit is reciprocal centimeter)..then is there any chance of getting rid off one hbar..so that i will get my matrix with one hbar...
So that i can add another Hamiltonian (off course a 6x6 matric with hbar (not sqaure) outside)!
thanks
rajini

 

[malawi_glenn]

I don't know what your original hamiltonian look like, please write it down and I'll try to help you.

The h-bar issue is that subatomic physicsits usually work in units c = 1 and h-bar = 1, when going back to SI units or whatever, one then restores the c's and h-bar's

Now your S_x looks ok! congratulations

Now you can find S_y by using the commutator relation between angular momentum operators: [tex] [S_x, S_z] = -iS_y [/tex]

 

[Rajini]

Okay now i will give you brief idea of what i want to know..(Then my S_y should also be correct!)
I have my spin Hamiltonian
H=D[S_z²-(1/3)S(S+1)] + E(S_x²-S_y²) + H_i[tex]\vec{S}[/tex]
Now the 1st term
I get a 6x6 matrix (S=5/2) and S_z is the diagonal matrix and you know it. And 'D' is a number in some unit (usually cm^-1 or some similar units)
2nd term
Here also 6x6 matrix, (one can take E's unit same as D)
3rd term
This is also a 6x6 matrix. Here H_i=[Hx Hy Hz]_1x3[Sx Sy Sz]^T_3x1=[HxSx+HySy+HzSz]_1x1= a 6x6 matrix, which we will get by taking the Sx, Sy and Sz matrices. Since i am not taking any value for Hi...All the matrix elements will in terms of Hx, Hy and Hz.
So finally i need to add all the 3 terms to get the final Hamiltonian matrix (a 6x6 matrix)...
If i take hbar=1 then i can add and get the solution correctly..as mentioned in many science papers..the 1st and 2nd term has hbar squared and 3rd has hbar..how to add...but what i feel is..is there some ideas to play with the units of D and E ?..some science papers first they take hbar= 1 and then after finding the matrix they simply multiply a h (not h bar) to the whole 6x6 matrix...
Rajini

 

[malawi_glenn]

is H on the right side the same as H on left side so you have a iterative hamiltonian or what?

H=D[S_z^2-(1/3)S(S+1)] + E(Sx2-Sy2) + H_i*S_i

 

[Rajini]

No 'H' in left is the symbol for Hamiltonian..i.e., LHS H and RHS H (mag. field) are not same

 

[malawi_glenn]

what is unit of your H? Sorry but there are so many differnt conventions and also what kind of system does this apply to?

I usually call the magnetic field B, and B*S then is not equal energy..

Just go back to the place where you got the hamiltonian, if you can't solve it - start a new thread.