# Deriving the pauli matrices from general su(2) matrix

1. Aug 1, 2013

### copernicus1

Hi, I think I need a sanity check, because I've been working on this for a while and I can't see what I'm doing wrong!

According to several authors, including Sakurai (Modern QM eq 3.3.21), a general way to write an operator from SU(2) is

$$\left(\begin{array}{cc}e^{i\alpha}\cos\gamma&-e^{-i\beta}\sin\gamma\\e^{i\beta}\sin\gamma&e^{-i\alpha}\cos\gamma\end{array}\right)$$

This clearly gives a determinant equal to one and has three parameters, as it should. It's basically identical to what Sakurai gives in his book, I've just redefined some variables.

But when I try to find the generators from this matrix by taking derivatives with respect to the parameters and then setting the parameters to zero, I get the following three matrices:

$$\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$$

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

and

$$\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$$

I can see how the first two are similar to a couple of the Pauli matrices (divide by i for the first and multiply by i for the second, which seems inconsistent), but the third has me completely stumped. Can anyone see what I'm doing wrong?

Thanks!

2. Aug 2, 2013

### Lanza52

I'm not familiar with that derivation. What tells you that taking the derivative and setting to zero would give you the Pauli matrices? I'm not too far ahead of you, but I don't see any reason that this would work other than a neat coincidence.

The "by hand" method in chapter one is the only method I can think of to give you the explicit form before you get to understanding the usefulness of the underlying algebra. However, if you are willing to take my word for it. The anti-commutation and commutation laws for this algebra define the representation for the pauli matrices. {σi,σj}=2δij etc. In fact, this is how Dirac found the relativistic analog to the pauli matrices, the gamma/dirac matrices.

3. Aug 2, 2013

### king vitamin

Well, if you look at the left-hand side of 3.3.21, those are exactly the pauli matrices and factors you would expect from differentiating the right-hand side as you've written it. I suppose that what you've shown is that, while any finite element in SU(2) can be written in Euler angle form, you cannot get any rotation in the x-direction using Sakurai's convention and using "small" Euler angles where one ignores angles larger than linear order (I think, I notice that you've changed definitions of alpha and beta a little bit).

For a simple generalization, write your matrix as:

$$\left(\begin{array}{cc}e^{i\alpha}( \cos\gamma + \sin\gamma)&-e^{-i\beta}(\sin\gamma\ - \cos\gamma) \\ e^{i\beta}(\sin\gamma - \cos\gamma)&e^{-i\alpha}(\cos\gamma + \sin\gamma)\end{array}\right)$$

Now clearly, I can redefine my gamma so that my matrix becomes your matrix, so I haven't really done anything. I get the same first matrix as you, but then I get

$$\left(\begin{array}{cc}1&-1\\1&1\end{array}\right)$$

and

$$\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)$$.

Of course, these are not linearly independent, but it's pretty obvious how to form a linearly independent combination of them. All you need is a set of normalization conventions and you have the Pauli matrices. Note that you wouldn't expect to get the Pauli matrices from the generators of an arbitrarily written SU(2) matrix unless you guessed the correct SU(2) rotation matrix from the beginning.

4. Aug 2, 2013

### copernicus1

Thanks for your responses. After working on it a bit, I did actually find a representation that works and will give the Pauli matrices as generators:

$$\left(\begin{array}{cc} -ie^{i\gamma} \cos\alpha \cos\beta - e^{-i\gamma} \sin\alpha \sin\beta & -ie^{-i\gamma}\cos\beta\sin\alpha - ie^{i\gamma}\cos\alpha\sin\beta\\ -ie^{i\gamma}\cos\beta\sin\alpha+ie^{-i\gamma}\cos\alpha\sin\beta &-ie^{-i\gamma}\cos\alpha\cos\beta+e^{i\gamma}\sin\alpha\sin\beta\end{array} \right)$$

@Lanza52: writing down a general operator for a group and then differentiating is (supposedly) a standard way of finding the generators for a group. I've read this in several texts on group theory. Maybe the matrix I was using before wasn't general enough to yield three distinct operators...I'm not entirely sure. But the method worked easily for SO(2) and SO(3), which was why I was frustrated with this matrix for SU(2)!

Last edited by a moderator: Aug 2, 2013
5. Aug 2, 2013

### Chopin

I think the reason it doesn't work here is that setting all of the parameters to 0 puts you at a singular point in your parameterization. Just like if you want to get the tangent vectors to a point on the earth by differentiating the spherical coordinate parameterization, but then you take the derivative at the north/south pole. The longitude parameter won't give you a valid tangent vector because at the poles, varying that parameter doesn't actually get you anywhere.

I'm not solid enough on my group theory to know for sure, but I suspect that if you take the derivative at a different point than (0,0,0), you'll get three non-zero vectors that are linearly independent and span the group. They probably won't be the Pauli matrices directly, though, they'll instead be some other basis for the space, but you could disentangle them to recover the Pauli matrices.

6. Aug 2, 2013

### Avodyne

Yes, this is exactly right. The original SU(2) matrix is independent of β when γ is zero, so γ=0 is a singular point of this coordinate system.

7. Aug 3, 2013

### copernicus1

Ah that makes sense. Thanks!

8. Aug 3, 2013

### vanhees71

The reason is that only that parametrizations of the group elements (at least in the neighborhood of the group identity) which are written as an exponential of the form
$$D(\alpha)=\exp(-\mathrm{i} \alpha_i T^i),$$
where $T^i$ are a basis of the Lie algebra gives you these basis elements.

In the case of the fundamental representation $T^i=\sigma^i/2$, where $\sigma^i$ are the three Pauli-spin matrices. The then three-dimensional vector $\vec{\alpha}$ with $\vec{\alpha} \in B_{\pi}^{3}$ has the usual meaning of an axial vector to describe a rotation. In this case you have
$$D(\vec{\alpha})=\cos(\alpha/2) I-\mathrm{i} \vec{n} \cdot \vec{\sigma} \sin(\alpha/2)$$
with $\alpha=|\alpha|$ and $\vec{n}=\vec{\alpha}/\alpha$.