# The physical meaning of Pauli Spin Matrices

1. Jun 7, 2008

### gn0m0n

OK, I understand the physical interpretation of spin and magnetic quantum number, so much as we can give one. That is, m_s = the component of s (or is it s^2? oh, dear....) in the arbitrarily chosen z direction.

Specifically, I am looking at Griffiths, p. 155-157. The eigenstates for spin 1/2 systems are easy enough to understand, although the derivations of them as actually presented in the context of magnetical orbital angular momentum through the lowering and raising operators was a bit obscure to me. I suppose I am even there struggling to recognize the physical significance of the operators (this is p. 147); if anyone can elucidate them at all I would much appreciate it.

Anyway, then we get to spin and Griffiths states that the "algebraic theory of spin is a carbon copy of the theory of orbital angular momentum," which I generally get. I also understand that we can then represent any spin 1/2 particle's state as a linear combination of the two vectors |1/2 1/2> and |1/2 -1/2>, ie, |s m> = ax(+) + bx(-) where x(+) is the spin-up eigenvector and x(-) is the spin-down eigenvector (right so far?). This can also be represented as the spinor (a b)^T (sorry, I'm making up how to convey the notation as I go along - this is the column vector with entries a, b) - now this essentially gives the coefficient of each basis vector? Ie, here, x(+) and x(-), right?

Fair enough, but then he says "the spin operators become 2x2 matrices"... can anyone expand on that? Is it that these operators are not conveying any action performed on the system but simply the observable, e.g. S_x=spin component in the x-direction? What do these operators act on? What about S(+) and S(-), the spin raising and lowering operators - these do not represent observables so what do they represent? I know they move us up and down the "ladder" of states but... how?

Again, it is mainly the Pauli spin matrices I want to understand better... For one thing, I don't understand how or why they are different, given that we choose the z-axis arbitrarily (although I do suspect that the indeterminacy principle has to do with it - perhaps once an axis is chosen we assume measurements are made first on that axis, which we call the z, then any subsequent spin measurements along either other axis will require these different matrices/operators?).

For x(+) we have simply (1 0)^T but for x(+)^(x) we have (1/sqrt(2) 1/sqrt(2)) ... why? What is physical meaning of this latter one? I know the physical meaning of first is that it is entirely "along the |1/2 1/2> vector", right? Now the latter one means we'd have equal probability of finding the spin in the x direction to be 1/2 or -1/2... but don't we also have equal probability (before we measure) to find spin in z direction to be 1/2 or -1/2?

As you can see, I'm going a little crazy here. Someone please save my sanity!

Sorry for the long post - I'm trying to be crystal clear.
Thanks.

2. Jun 7, 2008

### gn0m0n

We also first said that the generic spinor x= (a b)^T = ax(+) + bx(-) then also that x=((a+b)/sqrt(2))x(+)^(x) + ((a-b)/sqrt(2))x(-)^(x) So here we are representing the same generic spinor in terms of the same variable coefficients but with reference to different basis vectors, ie, eigenvectors for the different operator S_x? Actually, now that I think about it, were those first basis vectors eigenvectors for S_z?

Finally, Griffiths p. 157 says "if you measure S_x, the probability of getting h/2 is (1/2)|a+b|^2 and the probability of getting -h/2 is (1/2)|a-b|^2. You should check for yourself that these probabilities add up to 1." If anyone can elucidate this that would be great. I gather that the eigenvalues for the operators are possible values for the corresponding observables. But how do we use the quoted equations to obtain the probability for each?

I'm sorry if I posted in the wrong section - please help, and I'll get the hang of the forum - I'm new here! And any moderator can feel free to relabel the thread if desired.

Thanks...

3. Jun 7, 2008

### malawi_glenn

have you done matrix representation of operators yet?

My comment regarding your posts is that they are too long and have too many questions.

Myself can recomend Sakurai's book "Modern Quantum mechanics", the first chapter and the chapter about angular momentum are really really good.

4. Jun 7, 2008

### lbrits

It is important that you understand ordinary (orbital) angular momentum like the back of your hand. Then, I recommend that you read up on the Dirac equation as presented from a relativistic quantum mechanics point of view (rather than quantum field theory). If you take the non-relativistic limit of the Dirac equation, i.e., for particles moving slowly, then it reproduces the Schrodinger equation that contains the Pauli matrices. Hopefully someone can dig up a good reference (Wikipedia's looks like it was written by an infinite number of monkeys).

Furthermore, angular momentum is intricately related to the rotation group SO(3), and the angular momentum operators "generate" rotations. In the same way, a relativistic particle must respect the Lorentz group SO(1,3), and herein one gets various representations. One being the spinor representation, which is what the Dirac equation is all about.

Don't worry if you don't understand everything. Unfortunately getting the "why" behind spinors and the Pauli matrices takes quite a bit of math. You can think of it as a phenomenological thing for the time being.

5. Jun 7, 2008

### gn0m0n

Thanks for the replies. Apologies for the long post above.

I have only done matrix representation of operators as is presented in this Griffiths Quantum book.

I will order Sakurai's book today; I've been thinking about it for a while anyway :)

If I can follow up on Ibrits's comment: when you say the operators "generate" rotations, do you mean they represent physical rotations of a configuration in physical 3-space? Or are you referring to a more abstract "rotation" in some other space? I suppose asking whether these "rotations" are related to the various "representations" you refer to in the Lorentz group would be pushing my luck... :)

I appreciate the encouragement, Ibrits - I just always feel like I'd understand how to apply things better when I understand what they are doing or where they come from.

6. Jun 7, 2008

### lbrits

Generate in the Lie algebra sense. And rotations in the sense that
$$\psi'(\mathbf{r}') = \psi(\mathbf{O}\mathbf{r})$$, where O is some rotation matrix.

7. Jun 7, 2008

### Fredrik

Staff Emeritus
Another way of putting that is that if you describe the system as being in a particular state, there must be some operator $U(\mathbf O)$ that takes that state vector to the state vector you would have had to use if you had been turned some other way. The rotation matrix O can be defined as a function of three parameters, e.g. Euler angles. If you Taylor expand the operator U(O(parameters)) with respect to the parameters, the spin operators will appear in the first order terms. That's actually the proper way to define the spin operators.

8. Jun 8, 2008

### gn0m0n

That's great, thanks a lot! That's exactly what I wanted to know.

9. Jun 9, 2008

### Hans de Vries

There is a whole lot about Pauli spin matrices in the chapter on the Dirac equation
in my textbook (in progress). It uses the “modern” Weyl representation which is
now understood to be better expressing the underlying physics as the older
representation used by Griffits and Sakurai (it's also simpler !)
http://physics-quest.org/Book_Chapter_Dirac.pdf

The Pauli spin matrices can be used to:

1) Extract the spin of a particle. (from spinor notation to xyz notation)
2) Rotate the spin of the particle.
3) Boost a particle (!) "Rotation" in the Minkovski frame.

Point (3) needs two spinors combined into one 4-component Weyl spinor which
represent the left and right chiral components of the complete electron.

The Pauli spin matrices operate on spinors which represent spin.

A spinor represents:

4) A 3D pointer identifying the spin direction in space. (This is the SO(3) group)
5) It also differentiates between 0-360 degrees and 360-720 degrees with a sign inversion.
6) Lastly it contains a phase around the axis of the SO(3) spin pointer itself.

Point (6) is not so generally recognized. A spinor can be considered as a 'flag-pole'.
It has a direction in space but also a rotation around its own axis. This is something
what a 3d vector can not represent. You can use the Pauli spin matrices to rotate
the spinor around its own axis. This comes down simply to a multiplication of the spinor
with a complex value of $\exp(i\phi/2)$ because the eigenfunctions of rotations are the spinors
aligned along the direction of the rotation and $\exp(i\phi/2)$ is the eigenvalue of the rotation.
(See for instance figure 11.7)

Regards, Hans

10. Jun 11, 2008

### gn0m0n

Thank you all again, and this time especially Hans for the excerpt of the book! This kind of clear *physical* interpretation is what I was hoping for.

11. Jun 20, 2008

### Phrak

Here's something physical you can hang your hat on--at least partially so, where the elements of the Pauli matrices are in correspond with an external magnetic field.

$$i \hbar \frac{d}{dt} \left( \begin{array} {c} C_- (t) & C_+ (t) \end{array} \right) = -\mu \left( \begin{array} {cc} B_z (t) & B_x (t)-iB_y (t) & B_x (t)+iB_y (t) & -B_z (t) \end{array} \right) \left( \begin{array} {c} C_- (t) & C_+ (t) \end{array} \right)$$

$$C_-$$ and $$C_+$$ are the up and down z-components of spin. mu times the 2x2 matrix, is the Hamiltonian and can be rewritten as

$$\hat{H} = -\mu \left( \ \left( \begin{array} {cc} 0 & 1 & 1 & 0 \end{array} \right) B_x + \left( \begin{array} {cc} 0 & -i & i & 0 \end{array} \right) B_y + \left( \begin{array} {cc} 1 & 0 & 0 & -1 \end{array} \right) B_z \ \right)$$

The matrices should be identified as the x, y, and z Pauli spin matrices. The simplest thing you might want to do with this is zero Bx and By and set the Bz field constant. In solving the differential equations in C an interesting thing happens. The probabilites of C- and C+ remain constant in time but the amplitudes evolve with a complex phase having a frequency characteristic of the dipole energy mu Bz.

Last edited: Jun 20, 2008
12. Jun 21, 2008

### gn0m0n

Thanks again :) Every new way of hearing it and representing it helps.

13. Jun 23, 2008

### per.sundqvist

The Pauli matrices acts on two components of the wave function, arranged in a column vector like:

$$\vec{\phi}(\vec{r})=\left( \begin{array}{c} \psi_{\uparrow}(\vec{r}) \\ \psi_{\downarrow}(\vec{r}) \\ \end{array}\right)$$

What for example the $$\sigma_x$$ matrix do is then for example to "flip" the spin-up and the spin-down components:

$$\sigma_x\vec{\phi}(\vec{r})= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{c} \psi_{\uparrow}(\vec{r}) \\ \psi_{\downarrow}(\vec{r}) \\ \end{array}\right)= \left( \begin{array}{c} \psi_{\downarrow}(\vec{r}) \\ \psi_{\uparrow}(\vec{r}) \\ \end{array}\right)$$

14. Jun 23, 2008

### PhilDSP

The Pauli matrices essentially allow you to easily perform a parity/matching algorithm between the different subcomponents of a tetrad - most specifically tailored to the algebraic rules of the quaternion.

The quaternion tetrad values are folded into a 2 x 2 matrix in a very similar manner to mapping a bivector. Such a folded quaternion is in essence a spinor. Multiplication of the spinor by the Pauli matrices yields the quaternion algebra.

15. Jun 23, 2008

### PhilDSP

quaternion:

a, b, c, d

spinor:

a - id, b + ic
-b + ic, a + id

after some transformation, spinor:

x(a -id), y(b + ic)
z(b-ic), w(a + id)

after multiplication by the Pauli matrices

quaternion:

ra, sb, tc, ud

16. Jun 23, 2008

### reilly

I strongly suggest a little history;the idea of spin was created to explain the Stern-Gehrlach experiment, and to explain the fine structure of atomic spectra. There's little point in studying the formal properties of spin without a good sense of the physically based reasoning behind the concept. Do study the history. If you do, you'll be much less likely to be tied up in knots with the math.
Regards,
Reilly Atkinson

17. Dec 4, 2008

### turin

WARNING: My expressions may be "off" by factors of 2, pi, h, etc.. I don't think that these factors are important for the points that I wanted to make. If it bothers you very much, please let me know.

In order to start understanding spin, don't think of the Pauli matrices as labeled x,y,z. That makes them seem spooky, as if the particular axes have special meanings - THEY DON'T. For example, you could just as well swap the letters x,y,z around, and the physics would stay the same. (Of course, if you swap the letters by an odd permutation, then you change the handedness of the coordinate system, but that doesn't actually matter unless you want to talk about something really wierd - weak interactions.)

The starting point for understanding spin-1/2 is to realize what it is physically: it is a discrete property of (some) particles, that, when measured (i.e. in a magnetic field), the property will either be "alligned with" ("up") or "alligned against" ("down") the measurement. And, as reilly pointed out, you need to learn your history to appreciate this.

So, you know that in quantum mechanics, physically observable quantities are represented by operators. For example, position and momentum get operators that act continuously on a Hilbert space. Orbital angular momentum also gets an operator that acts on this Hilbert space, but discretely, due to the fact that the relevant coordinatization is cyclic. It turns out that spin gets an operator that does not act on this Hilbert space - it acts on a new finite-dimensional space (whereas the Hilbert space is infinite-dimensional), and the size of the space depends on the particle. That is not as wierd as it may sound - it basically just means that a particle has some structure not related to spatial configuration, and the number of possible observed values for this structure is determined by the particle whose spin is being measured. Well, OK, maybe that is wierd - a point particle can have structure. For the time being, forget that spin is a type of angular momentum - just think of it as some property that can be observed as either "with" ("up") or "against" ("down")

Since there are two independent states, the spin operator is a 2x2 matrix that acts on a 2-component object in this new space of 2-component objects. It turns out from experiment that the two possible observed values (say from magnetic splitting) look like quantitites of angular momentum. And the possible values are equal and opposite. So, if we call the magnitude of the possible value "|s|", then the matrix that measures the spin can be written in its own eigenbasis as

$$S=\left(\begin{array}{cc}|s|&0\\0&-|s|\end{array}\right)$$

where I have chosen to put the positive value in the upper left. Notice that if you factor out the "s", you will get one of the Pauli matrices. Let's do that, but lets just call it "S", for spin. And that's it, I'm done with spin.

Here's a quiz question: What is the canonical conjugate to spin?

...

"But wait!", you may say. "What about the other directions?" Well, as far as spin is concerned, THERE ARE NO OTHER DIRECTIONS! That is the main point I'm trying to make here, and the reason why I want to use the generic symbol "S". Of course, there are other linearly-independent 2x2 matrices, but this is quantum mechanics. Remember, all we need is a complete set of commuting observables. And, if we restrict ourselves to a single spin-1/2 particle, then S completes the set. However, this is where the angular momentum interpretation comes in. Remember from your treatment of angular momentum, you had the states |j,m>? Well, it turns out that this new space is interpretted as the space of objects of this form with j=1/2 and m=+/-1/2:

$$\alpha_{+\frac{1}{2}}\left|\frac{1}{2}\right.,\left.+\frac{1}{2}\right> +\alpha_{-\frac{1}{2}}\left|\frac{1}{2}\right.,\left.-\frac{1}{2}\right>$$
or, in our notation
$$\alpha_{up}\left(\begin{array}{c}1\\0\end{array}\right) +\alpha_{down}\left(\begin{array}{c}0\\1\end{array}\right)$$

Now, we identify |s| as j=1/2. (Again, you have to learn your history to decide if this identification makes sense from what I've said.)

Remember when you constructed angular momentum operators that, while not representing observables, allowed you to build the space of angular momentum states in a very similar way to building the states of the harmonic oscillator. These were the "raising" and "lowering" operators. They represent increasing or decreasing the number of quanta of the particular observable - in this case, angular momentum. For spin-1/2, the matrices for these operators are quite easy to guess. We need the raising operator, S+, to increase the spin by one unit, and the lowering operator, S-, to decrease the spin by one unit. If we imagine acting such an operator on an eigenstate, and then acting S on the result to measure the raised or lowered spin, then we easily induce that these operators must be (of the form)

$$S+=\left(\begin{array}{cc}0&1\\0&0\end{array}\right) \qquad\qquad S-=\left(\begin{array}{cc}0&0\\1&0\end{array}\right)$$

Also from your treatment of angular momentum, you should remember that

$$J_x=\frac{1}{2}\left(J_++J_-\right) \qquad\qquad J_y=-i\frac{1}{2}\left(J_+-J_-\right)$$

which is based on the abstract algebra of angular momentum, and did not depend on what the matrices actually were. This is the final clue needed to extract the Pauli matrices from our S,S+,S- based on the commutation relations of the angular momentum generators.

Last edited: Dec 4, 2008