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I am trying to compute the $2$-point Green function $\tau_2(x,y)$ for free Dirac fields. The corresponding formula for $\tau_2(x,y)$ is given by

$$\tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta_w, \bar{\eta}_z],$$

where $Z_0$ is the generating functional for free Dirac fields given by $$ Z_0[\eta_w, \bar{\eta}_z] = \exp\left(-i\int \bar{\eta}_z \, S(z-w) \, \eta_w \, dz \, dw\right). $$

Here, $\eta$ and $\bar{\eta}$ are source terms. Also, $S^{-1} = i\gamma\cdot\partial - m$ is the operator appearing in the quadratic term of the lagrangian.

Notation $\eta_x \equiv \eta(x)$ etc.

At first, I determine $\frac{\delta Z_0}{\delta\bar{\eta}_y}$ as

$$ \frac{\delta Z_0}{\delta\bar{\eta}_y} = -iZ_0 \int S(y-w) \eta_w \, dw. \label{a}\tag{1} $$

Then I try to compute

\begin{align} -\frac{\delta^2 Z_0}{\delta\eta_x\delta\bar{\eta}_y} &= -\frac{\delta}{\delta\eta_x} \left[-iZ_0 \int S(y-w) \eta_w \, dw \right] \\ &= i\frac{\delta}{\delta\eta_x} \left[Z_0[\eta_w, \bar{\eta}_z] \int S(y-w) \eta_w \, dw \right] \label{b}\tag{2} \end{align}

Question

  1. How to proceed from this step (eq. (\ref{b}))? I have to take the functional derivative of the product of two Grassmann functionals. What is the relevant formula for it? If you also mention any reference, that would be great.

  2. In eq. (\ref{a}) I have written $Z_0$ before the functional derivative part. Should I write it after the functional derivative term? In other words, what is the chain rule for Grassmann functionals?

  3. In the Appendix, I have mentioned the formula to take the functional derivative of a product of Grassmann functionals. Let's say, I have a product of some Grassmann functionals and an ordinary function $f(x) \in \mathbb{C} \forall x$. Then how to evaluate this functional derivative? That is,

$$ \frac{\delta}{\delta\psi(x)} [\psi(y_1) f(y_2) \psi(y_3)] = ? $$ where $\psi$ is a Grassmann field.

Appendix

The formula used to compute the eq. (\ref{a}) is given below.

$$ \frac{\delta}{\delta\psi(x)} [\psi(y_1) \cdots \psi(y_n)] = \delta(y_1-x) \psi(y_2) \cdots \psi(y_n) + (-1) \delta(y_2-x) \psi(y_1) \cdots \psi(y_n) + \cdots \cdots + (-1)^{n-1} \delta(y_n-x) \psi(y_1) \cdots \psi(y_{n-1}). $$

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    $\begingroup$ The answer is obviously $S$. Are you sure you’re not overthinking this? $\endgroup$ – Prof. Legolasov Sep 30 '18 at 4:43
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First of all, the functional integral $Z_0$ is a real number, since it is defined as an vacuum expectation value: $$Z_0[\zeta,\bar{\zeta}]:=\langle0|T\,e^{i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}|0\rangle$$ where $T$ is the time ordering operato and: $$\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle:=\int d^4x(\bar{\zeta}(x)\psi(x)+\bar{\psi}(x)\zeta(x))$$ Now, after some analytical steps, it is found that this object must satisfy the Symanzik equation: $$\left[(i\gamma^\mu\partial_\mu-m)\frac{\delta}{i\delta\bar{\zeta}_z}-\zeta_z\right]Z_0[\zeta,\bar{\zeta}]=0$$ It is found that a solution is readly obtained by putting (as an ansatz): $$Z_0[\zeta,\bar{\zeta}]=e^{-\int d^4x\int d^4y\,(\bar{\zeta}(x)S_F(x-y)\zeta(y))}=e^{-\langle\bar{\zeta}_xS^F_{xy}\zeta_y\rangle}$$ But, in general, a solution for a linear differential equation can be searched by means of a fourier transform. In this way we define the functional Fourier transform of $Z_0$ as: $$Z_0[\zeta,\bar{\zeta}]=\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,\tilde{Z}[\psi,\bar{\psi}]e^{i\int d^4x(\bar{\zeta}(x)\psi(x)+\bar{\psi}(x)\zeta(x))}=\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,\tilde{Z}[\psi,\bar{\psi}]e^{i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}$$ By putting this functional fourier transform in the Symanzik equation, we can identify: $$\tilde{Z}[\psi,\bar{\psi}]:=\mathcal{N}e^{i\int d^4x\,\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi}=\mathcal{N}e^{iS_D[\psi,\bar{\psi}]}$$ where $\mathcal{N}$ is a constant, and obtain: $$Z_0[\zeta,\bar{\zeta}]=\mathcal{N}\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,e^{iS_D+i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}$$ Now (in analogy with the bosonic case), the 2n-point Green Function can be written as: $$S^{(2n)}_0(x_1,...,x_n;y_1,...,y_n)=\langle 0|\psi(x_1)\cdots\psi(x_n)\bar{\psi}(y_1)\cdots\bar{\psi}(y_n)|0\rangle=\frac{\delta^{(2n)}Z_0[\zeta,\bar{\zeta}]}{\delta\bar{\zeta}(x_1)\cdots\delta\bar{\zeta}(x_n)\delta\zeta(y_1)\cdots\delta\zeta(y_n)}\bigg|_{\zeta=0,\bar{\zeta}=0}$$ And if we want to evaluate the 2-point Green function, we first evaluate: $$\frac{\delta Z_0}{\delta\zeta(x_2)}=\frac{\delta}{\delta\zeta(x_2)}e^{-\int d^4x\int d^4y (\bar{\zeta}(x)S_F(x-y)\zeta(y)}=-\int d^4x\, (-1)\bar{\zeta}(x)S_F(x-x_2)\cdot Z_0[\zeta,\bar{\zeta}]$$ where the $(-1)$ is due to the fact the grassman functional derivative has to jump over the $\bar{\zeta}$, which is a grassman valued function. Then: $$\frac{\delta}{\delta\bar{\zeta}(x_1)}\left(\frac{\delta Z_0}{\delta\zeta(x_2)}\right)=\frac{\delta}{\delta\bar{\zeta}(x_1)}\left(\int d^4x\, \bar{\zeta}(x)S_F(x-x_2)\cdot Z_0[\zeta,\bar{\zeta}]\right)=S_F(x_1-x_2)\cdot Z_0+\left(\int d^4x\, \bar{\zeta}(x)S_F(x-x_2)\cdot (-1)\int d^4 y\, S_F(x_1-y)\zeta(y)\cdot Z_0\right)$$ By putting $\zeta=0,\bar{\zeta}=0$, the $Z_0$ is $1$ and the second term goes to zero, obtaining: $$S^{(2)}_0(x_1,x_2)=S_F(x_1-x_2)$$

The functional integral is a real Number, it is not Grassman valued, so you don't have to worry about the order of your equation (2).

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  • $\begingroup$ "the functional integral $Z_0[\zeta, \bar{\zeta}]$ is a real number". NO, $Z_0$ is NOT a real number. It's instead EVEN-graded in Grassmann numbers $\zeta$ and $\bar{\zeta}$. The n-point Green functions are real/complex though. $\endgroup$ – MadMax Oct 15 '18 at 16:45
  • $\begingroup$ In this non-supersymmetric scheme, does it even matter this dinstinction? :) $\endgroup$ – Kevin De Notariis Oct 15 '18 at 19:21
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I am trying to compute the 2-point Green function $\tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta, \bar{\eta}]$

You have to take the $\eta=0$/$\bar{\eta}=0 $ limit as the final step $$ \tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta, \bar{\eta}] \bigg|_{\eta=0,\bar{\eta}=0}.$$ Green functions (as opposed to the functional integral $Z_0[\eta, \bar{\eta}]$) should not be explicitly $\eta$/$\bar{\eta}$ dependent.

  1. How to proceed from this step (eq. (2)?

The $\frac{\delta}{\delta\eta_x} \left[Z_0[\eta, \bar{\eta}] \right]$ part in eq.(2) will drop out after taking the $\eta=0$/$\bar{\eta}=0$ limit at the final step. Thus you only care about the $\frac{\delta}{\delta\eta_x} \left[\int S(y-w) \eta_w \, dw \right]$ part.

  1. In eq. (1) I have written $Z_0$ before the functional derivative part. Should I write it after the functional derivative term?

The key point is that the functional integral $Z_0[\eta, \bar{\eta}] $ is EVEN-graded in Grassmann numbers $\eta$/$\bar{\eta}$. The order of $Z_0$ in eq. (1) does NOT matter.

  1. Then how to evaluate this functional derivative?

Just treat the ordinary function $f(x)$ as a constant real/complex number. It does not interfere with the functional derivative.

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