Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Wick rotation makes sense for Dirac algebra - it has +--- and ++++ and ---- signatures. Just wondering if one can Wick rotate Pauli algebra from the standard +++ pauli matrices to +--.

share|improve this question
add comment

3 Answers 3

Wick rotation is a tool to simplify certain calculations, by defining certain functions by analytic continuation from Euclidean space where they can be evaluated more simply. Turns out that rotating the spinors (as well as the momenta) usually does not simplify calculations involving those spinors. The reason is that the structure of the spinor representations of the groups so(p,q) depends on the signature p-q in an interesting and intricate way. Since the Wick rotation changes the signature, it would also change the structure of spinors in a way that is not all that straightforward to keep track of. So, nothing is stopping you from rotating the spinors as well as all the positions and momenta, but usually it is not useful.

share|improve this answer
    
-1: It is not simple at all--- the reps are different in Euclidean space. You absolutely have to rotate the reps, because the Euclidean theory must be rotationally invariant. I am sorry for downvoting an answer which is better than the others, but it is just no good. –  Ron Maimon Feb 22 '12 at 0:56
add comment

Since the original question asks about the Dirac algebra and Pauli matrices, I think the problem might best be answered by considering just these objects. This ignores the usual use of a Wick rotation in terms of changing the time variable by $t\to it$, but in the sense that the Dirac algebra and Pauli matrices define the Minkowski geometry I think this is a useful simplification.

There are two common sorts of Clifford algebras, real and complex. In both cases one begins with a set of "basis vectors". For the Dirac algebra the basis vectors are usually written as $\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}$ and for the Pauli algebra the basis vectors are $\{\sigma_x,\sigma_y,\sigma_z\}$. These basis vectors all anti-commute, and square to the various choices of signature.

To get a real(complex) Clifford algebra one considers the real(complex) vector space defined with a basis given by all possible products of basis vectors. For the Dirac algebra, one ends up with sixteen basis elements. These are sometimes called the bilinear terms. There are $2^4 = 16$ and they come in five varieties, one scalar (S), one pseudoscalar (P), four vectors (V), four axial vectors (A), and six tensors (T).

The Dirac algebra is a complex Clifford algebra, so its elements are 16-dimensional complex objects. Most people work with representations using 4x4 complex matrices. In this setting, a Wick rotation amounts to replacing a basis vector with $i$ times the basis vector.

Since there are only three basis vectors in the Pauli algebra, there are only eight $= 2^3$ basis elements in the Clifford algebra. The eight basis elements are a single scalar 1, three vectors $\sigma_x,\sigma_y,\sigma_z$, three bivectors $\sigma_x\sigma_y$, $\sigma_x\sigma_z$, $\sigma_y\sigma_z$, and a single pseudoscalar $\sigma_x\sigma_y\sigma_z$.

Now the Pauli algebra pseudoscalar is somewhat strange. It squares to -1 and it commutes with all the other basis elements (and therefore with everything in the algebra). As such it acts just like an imaginary unit $i$. And so with the Pauli algebra, we use the real Clifford algebra instead of the complex one. Our usual representation has 8 real dimensions and these are kept in 2x2 complex matrices.

So the difference between the Pauli and Dirac algebras, when applying a Wick rotation, is that Pauli algebra is real while the Dirac algebra is complex. Hmmmm.


As a sort of aside on Wick rotations of Pauli spin matrices, the post here came to mind when I read Witten's latest paper (January 17, 2011). See equation (2.4), page 19:

$B_0 = \left(\begin{array}{cc}0&+1\\-1&0\end{array}\right)$, $B_1 = \left(\begin{array}{cc}0&+1\\+1&0\end{array}\right)$, $B_2 = \left(\begin{array}{cc}+1&0\\0&-1\end{array}\right)$

Fivebranes and Knots Edward Witten (2011)

http://arxiv.org/abs/1101.3216

This is simply $B_0 = i\sigma_y, B_1=\sigma_x, B_2=\sigma_z$. These sorts of things happen constantly when one messes around with Clifford algebra as will be clear from the context.

share|improve this answer
    
While it's true that Pauli algebra is taken real, Dirac algebra found in physics is a complexification of a real algebra that is also used. This is of course in line with usual physicist's obsession to introduce complex numbers everywhere. E.g. in Lie reps. But often this loses information. Complex Lie algebras can contain both compact and split real forms (and sometimes also other forms). (cont.) –  Marek Jan 19 '11 at 8:57
    
In a similar way, complex Clifford algebras depend only on the number $n$ of dimensions. So this carries no information about the space-time signature. One has to work with real $(p,q)$ algebras. The way this is done in physics is that one works with the (1,3) real algebra and complexifies it (so that (1,3) algebra sitting inside still has a prominent place) but I don't think this is at all necessary (or even needed). It's just a historical baggage. –  Marek Jan 19 '11 at 8:59
    
Comment on the answer(v2): It seems there is a typo $\sigma_x \leftrightarrow \sigma_y$ in the next-to-last sentence. –  Qmechanic Feb 21 '12 at 13:53
add comment

This is meant to segue further from the answer above.

The $SL(2,~{\bf C})$ and $SU(1,~1)$ groups are defined on four and three dimensional spacetime respectively. The generators of these groups exist on the principal bundles $P(\Sigma,~{\cal G})$, where $\Sigma$ is a Cauchy surface (spatial surface) of dimensions 3 and 2 for $SL(2,~{\bf C})$ and $SU(1,~1)$. The set of connections $A$ on the principle bundle define Wilson loops $\oint Adx$ are maps $\mu:[0,~1]~\rightarrow~\sigma$ for $\mu(0)~=~\mu(1)$. An element of $\cal G$ assigned to $\mu$ by the holonomy map $H(\mu,~A)$ defines a function $$ F_\mu(A)~=~{1\over 2}Tr~H(\mu,~A) $$ that is invariant with respect to the group action of $\cal G$ Thus $F_\mu(A)$ is an element of ${\cal M}~=~A/{\cal G}$ or the moduli space. The rotation $\sigma_3~\rightarrow~i\sigma_3$ will carry the separability condition to the noncompact case.

The group $SU(1,1)$ is related to $SU(2)$ by the signature change on the basis elements $\sigma_1,~\sigma_2,~\sigma_3$ of $SU(2)$. For $\sigma_\pm~=~\sigma_1~\pm~\sigma_2$, the basis for $SU(1,~1)$ are then $\sigma_+,~\sigma_-,~\tau_3~=~i\sigma_3$. Now consider a connection one-form $$ A~=~A^+\sigma_+~+~A^3\sigma_3 $$ and a gauge transformation determined by the group action of $g~\in~{\cal G}$, $g~=~e^{i\lambda\tau_3}$. The gauge transformed connection is then $$ A^\prime~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2\lambda}A^+\sigma_+~+~A^3\sigma_3, $$ where $d\lambda~=~A^3$. Thus $\lambda$ is a parameterization of the gauge orbit for this connection. This leads to the observation $$ \lim_{\lambda\rightarrow\infty}A(\lambda)~\rightarrow~A^3\sigma_3, $$ where $ A^+\sigma_+~+~A^3\sigma_3$ and $A^3\sigma_3$ have distinct holonomy groups and thus represent distinct points in the moduli space $\cal M$. However by the last equation this gives $$ F_\mu(A^+\sigma_+~+~A^3\sigma_3)~=~ F_\mu(A^3\sigma_3), $$ which obtains similarly for any gauge invariant function. Hence there exist two distinct points in the moduli space that define the same set of gauge invariant functions. Hence there does not exist a measure over these two points that separates them and $\cal M$ is then nonHausdorff with a Zariski topology.

share|improve this answer
    
@Lawrence you seem to stray off the target a little at the end but otherwise you're correct. The simplest way to understand a wick rotation is as multiplication by $i=\sqrt{-1}$. –  user346 Jan 18 '11 at 4:14
    
How is that string of jargon related in any way to the question asked is beyond me. –  user566 Jan 18 '11 at 4:16
3  
I know the jargon well enough to recognize that the string of letters and symbols above is non-responsive, why you think it is a good idea to encourage non-responsive answers is also beyond me. –  user566 Jan 18 '11 at 4:56
1  
I completely agree with @Moshe so I am afraid I have to down-vote this. Most of this answer deals with technicalities completely unrelated to the question (like connections and algebraic geometry) and the place where it accidentally mentions Wick rotation are completely obscure and miss the question as well in my opinion. –  Marek Jan 18 '11 at 10:03
2  
I deleted a string of argumentative comments. Let's not have any more comments unless somebody has something constructive to add. –  David Z Jan 19 '11 at 21:29
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.