22
votes
Accepted
How does canonical quantization work with Grassmann variables?
When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.
...
- 1,724
13
votes
Accepted
Path integral for complex scalar field
For simplicity, let's consider the standard complex integral (the path integral is just the limit of the product of many standard integrals).
A function defined on the complex plane $f(z)$ is the same ...
- 6,119
9
votes
Why are physical states not eigenstates of BRST charge?
For starters, the BRST charge operator is Grassmann-odd, so an eigenvalue would be Grassmann-odd as well, which is unphysical.
- 213k
8
votes
Why are Grassmann fields never classical?
Well, it depends on what is meant by the word classical.
Usually in physics, classical theories mean theories where Planck's constant $\hbar$ is zero. If that's what is meant, then there certainly ...
- 213k
8
votes
Transpose of fermion bilinears
In short, yes there's an extra minus sign when we reorder the terms. I've just encountered this myself and I found going very slowly helps to understand where the minus sign comes in. Let's use index ...
- 181
7
votes
Accepted
Derivative with respect to a spinor of the free Dirac lagrangian
If $\theta_1,\theta_2$ are a pair of Grassmann variables, then
$$
\frac{\partial}{\partial\theta_2}(\theta_1\theta_2)=-\theta_1\qquad\tag{left derivative}
$$
where the negative sign is due to the fact ...
7
votes
Accepted
What is the meaning of a Grassmann variable?
This is a quick and dirty route to Grassmann numbers. It's worth being more careful and thorough, but this is the general idea.
Start by considering a complex vector space $V$, which has a basis $\{\...
- 71.5k
7
votes
Accepted
What are self-interacting fermions?
The Dirac spinor in $d$ spacetime dimensions has $2^{[d/2]}$ complex Grassmann-odd components. Crossterms can survive in the quartic interaction term.
- 213k
7
votes
Accepted
Why must Grassmann algebras for Fermionic theories be infinite dimensional?
Ref. 1 does define a $2^N$-dimensional Grassmann algebra $\Lambda_N$ with finitely many anti-commuting generators $\xi^1,\ldots,\xi^N,$ and proceeds to write
We shall usually, though not always, deal ...
- 213k
6
votes
Accepted
Why are coherent states necessary for defining the fermionic path integral?
The integral formula that you write $\int d^N\eta d^N\bar{\eta} e^{\bar{\eta}A\eta}=\det A$ is indeed true for any Grassman numbers, and does not involve coherent states. The Grassman resolution of ...
- 290
6
votes
Accepted
Why are Grassmann fields never classical?
Let me put things in broader perspective. Bosonic fields are quantized in terms of commutators with a prefactor $\hbar$. Classical limit leads to commuting variables that may be represented by complex ...
- 309
6
votes
Accepted
Grassmann's variables under integration
I am not sure that $d(\eta^2)$ is defined at all. But if it is, then, in my opinion, you should write it in this way
$$
d(\eta^2) = d(\eta\eta) = d\eta\ \eta + \eta\ d\eta
$$
So you get not $\eta\ d\...
- 6,243
6
votes
Accepted
Is the derivative with respect to a fermion field Grassmann-odd?
Since$^1$ $$(\frac{\partial}{\partial z}z)~=~1\tag{1}$$
for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $\frac{\partial}{\partial z}$ should be the same as the ...
- 213k
6
votes
Path integral for complex scalar field
In the same way that a Grassmann-even field can be real or complex valued, a Grassmann-odd field can be real or complex valued.
Concerning P&S section 9.6, they consider a Grassmann-even complex ...
- 213k
6
votes
Grassmann numbers
The power series is the definition of exponentiation.
$e^{x+y}=e^xe^y$ is a property of that power series that can be derived for the special case when $x$ and $y$ commute.
- 6,666
6
votes
Dirac Lagrangian in Classical Field Theory with Grassmann numbers
The value of the Dirac action$^1$
$$ S~=~\int\! d^4x ~\bar{\psi}\left(\frac{i}{2}\gamma^{\mu}\underbrace{\stackrel{\leftrightarrow}{\partial}_{\!\mu}}_{=\stackrel{\rightarrow}{\partial}_{\mu}-\...
- 213k
5
votes
Accepted
Majorana Flip Relations
Recall that in an elementary Linear Algebra course, you proved that for two matrices $A,B$, $(AB)^T=B^T A^T$. Do this in component notation. For this proof you will commute the component of $A$ past ...
- 7,860
5
votes
Accepted
What is the Grassmann parameter $\epsilon$ in the BRST transformation?
That's a very good question. At the pragmatic level:
The field of complex numbers $\mathbb{C}$ has been replaced by the set of supernumbers.
Hilbert spaces and operators are not just (conjugate) ...
- 213k
5
votes
Is the fermion mass Lagrangian term imaginary instead of real?
I am going to consider the problem in $d=0+1$ dimensions to simplify the notation. For higher $d$ the discussion is identical except that you also include momentum modes, which are not relevant to the ...
5
votes
Logarithm of Grassmann numbers
I don't think we can assign a meaningful value to $\log(\theta)$. We can, however, find $\log(1+\theta)$ as follows.
Since $\theta^2=0$ then we have $\theta^k=0 \space \forall k \ge 2$. So
$\...
- 60.5k
5
votes
Why do the electron and positron creation operators anti-commute?
As far as I remember, you can choose creation operators of different fermions to commute or to anti-commute, both choices should provide correct results if used consistently.
- 27.5k
5
votes
Are Grassmann numbers always "under the hood" if we deal with fermionic ladder / field operators?
It is possible to define the antisymmetric Fock space, and related fermionic creation and annihilation operators, without any need to introduce Grassmann numbers.
Grassmann numbers are only necessary ...
- 12.2k
5
votes
Accepted
Weinberg's path integral for fermions in Volume 1
Hints:
OP's first method is correct. For clarity, it should probably be noted that it implicitly uses that the fermionic eigenvalue $q$ commutes with the bosonic state $|0\rangle$.
For OP's second ...
- 213k
5
votes
Accepted
Free fermion OPE
Yes, $:\psi(w)\psi(w):=0$ because $\psi$ is Grassmann-odd.
Yes, $:\psi(z)\psi(w):$ is only zero when $z=w$.
- 213k
5
votes
Accepted
Leibniz rule and Nakahara's definition for functional derivatives with respect to Grassmann variables
The definition given by Nakahara is correct. Specifically:
$$
\frac{\delta G[\psi(x)]}{\delta \psi(y)} \\
= \frac{1}{\epsilon}\{G[\psi(x) + \epsilon \delta(x-y)] - G[\psi(x)]\} \\
= \frac{1}{\epsilon}...
- 4,833
4
votes
Diagonalizing Faddeev-Popov Lagrangian $U(1)$
There's one serious flaw in your entire strategy. Since $\overline{c},\,c$ are fermions, they are Grassmann-number-valued. Thus any complex numbers $w,\,z$ satisfy $(w\overline{c}+zc)^2=0$. You ...
- 25.4k
4
votes
Link between the Grassmann algebra and spinors
There is no direct link between the Graßmann algebra and spinors. The reason both are connected in physics is the spin-statistic theorem: The fields/operators associated to fermionic objects, i.e. ...
- 129k
4
votes
Accepted
Plugging Majorana Spinor into Dirac Lagrangian does not give Majorana Lagrangian?
Ah. Figured it out. I want to show that
$$
i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L.
$$
Let's manipulate ...
- 11.8k
4
votes
Why is supersymmetry a continuous symmetry?
A super-Poincare algebra is a Lie superalgebra, whose elements are generators for a Lie supergroup, and therefore formally corresponds to a continuous symmetry.
Of course the elephant in the room is ...
- 213k
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