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.

In canonical quantization, we view the Dirac field $\psi$ as a $4\times1$ matrix of complex number. While in path integral quantization, we view the Dirac field $\psi$ as a Grassmann number.

For two Grassmann number $\psi_1$, $\psi_2$, we can calculate $\psi_1\cdot\psi_2$ and it has the properity $\psi_1\cdot\psi_2=-\psi_2\cdot\psi_1$. While for two column vectors, we can't even define $\psi_1\cdot\psi_2$. So what's the correspondence between Grassmann number and column vector?

share|improve this question
    
I recall that the fact that a Dirac spinor consists of Grassmann numbers is important when one tries to find that a bilinear covariant is invariant under charge conjugation, otherwise you get spurious minus signs. –  auxsvr Apr 21 at 17:51

1 Answer 1

One thing to note is that, even in the path-integral formulation, we do not actually use products of the form $\psi_1 \cdot \psi_2$, but of the form $\bar{\psi}_1 \cdot \psi_2$ (usually with $\psi_1 \equiv \psi_2$). The bar is important since it involves taking the transpose (and thus giving a row vector, rather than a column). Since, as you point out, the product would be undefined otherwise, the bar must appear REGARDLESS of whether we are in the canonical or path-integral quantization.

I think the conceptual piece you are missing is that, even in the path-integral formulation, $\psi$ is still a 4-component entity (it HAS to be, if we want it to encode the space-time behavior of a Dirac field). The way it was introduced historically may obscure this fact, but there is not anything inherently quantum mechanical about the Dirac equation. Rather, it was derived by demanding consistency with SPECIAL RELATIVITY, and only then, once derived, was it quantized via the canonical prescription borrowed from non-relavistic quantum mechanics. One could perfectly well view it as describing a completely classical field $\psi$ (though it might be hard to come up with physically-realized observable that would correspond to a complex-valued field), whose 4-component nature was the direct mathematical result of demanding a LINEAR field equation consistent with $E^2= p^2 + m^2$. All this is to say that the space time nature of $\psi$ is fixed BEFORE quantization is even considered, so it certainly cannot depend on the method of quantization we choose.

So back to your over-arching question: what is the correspondance between the two pictures? Well, when you integrate over $\psi$, you are REALLY integrating over the 4 complex-valued components of $\psi$ independently. The integrals are often written in a way that obscure this fact for ease of manipulation (particularly since one is rarely, if ever, carrying out the integrals explicitly, and is more often using them formally, in derivations, where this kind of detail can be glossed over), but it should be kept in the back of the reader's mind. Even if they are called Grassmann NUMBERS in some contexts, they really represent 4-component VECTORS of (complex) Grassmann numbers. The integration variables, which in the canonical quantization would be viewed as operators, must, in the path-integral formulation, be viewed as ordinary numbers, so that, for example, $\psi$ and its canonical momentum operator, $\pi$, commute in the integrand. Basically, one can either have non-commuting operators acting on a Hilbert space which obey the Lagrangian/Hamiltonian equations of motion (canonical picture), or commuting variables with are integrated over ALL possible values (even those which violate the equations of motion-- the path-integral formulation). I won't discuss here why those two pictures are equivalent, since I think the text you are following will show that, but will say one last thing that touches on why one must use Grassmann variables.

It is related to the fact that Dirac particles are Fermions. EVEN IN THE CANONICAL PICTURE, the fact that fermionic fields anti-commute must be put in "by hand". That is, it is not a direct consequence of the Dirac equation. Rather, one finds that without it, there are negative-energy states that would cause the vacuum to be unstable (a problem that was famously solved by Dirac postulating an electron "sea", leading to his prediction of anti-matter), and also without which causality would be violated (see page 56, and section 3.5 of Peskin and Schroeder). So, to solve these problems, one posits that $\bar \psi$ and $\psi$ anti-commute. With this modification, the canonical picture procedes more or less as it does for bosons (with the consequences of the anti-commutation being things such as Pauli exclusion, and the extra minus sign one gets for fermion loops in perturbation theory). But we must also account for this modification in the transition to the path-integral picture. This is exactly the reason for taking $\psi$ and $\bar \psi$ to be Grassmann variables there. By viewing them as ordinary numbers EXCEPT that they anti-commute (which is all Grassmann variables really are), we incorporate the proper "spin-statistics" into the path integral description of fermions.

Of course, as I like to say, I know just enough field theory to make me dangerous, so I apologize in advance if anything I said was unclear :) (and hopefully others will give their two cents to clarify where they see fit). Hope this was helpful!

share|improve this answer
    
Was this answer useful at all?? If so, I'd appreciate an upvote :). –  rdjain1 Apr 28 at 17:09

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.