2
$\begingroup$

For the free scalar field theory, the generating functional is given by

$$Z[J] = N \int D\phi e^{i \int d^4x \left( \frac{1}{2} \phi(x) A\phi(x)+J(x)\phi(x)\right)} .$$

with the differential operator $A = -(\Box + m^2)$. The way this is calculated in my book is by analogy with the following multidimensional integral:

$$\left(\prod_k\int_{-\infty}^{\infty} q_k \right)e^{[(i/2)\sum_{ij} q_i A_{ij} q_j ]+ [i\sum_i J_iq_i]} = Ce^{(-i/2)\sum_{ij}J_i (A^{-1})_{ij}J_j}$$

The sums are replaced by integrals, vectors by functions and matrices by operators. So that:

$$Z[J] = Z_0 \exp(-i/2)\iint d^4x d^4y J(x) D_F(x-y)J(y)$$

where $D_F(x-y)$ is the inverse of $A$.

My question is this: Why is there an equivalence between the discrete and continuous case - in the first expression for $Z[J]$ there is a single integral while the discrete 'equivalent' contains one double and one single sum. Wouldn't the equivalence only hold if the first expression for $Z[J]$ had something like $$\iint d^4xd^4y \phi(x)A\phi(y) + \int d^4x J(x)\phi(x)$$ in the exponent?

$\endgroup$
  • $\begingroup$ This is from Zee's QFT book, chapter I.3 (page 22 in my edition). $\endgroup$ – Javier Oct 1 '16 at 16:43
3
$\begingroup$

Essentially, the sum over $j$ is hidden inside $A\phi$, which is a linear combination of the "components" of $\phi$. Let me elaborate:

The "equivalence" between vectors and functions means that we break up spacetime into discrete points $x_i$, and $q_i = \phi(x_i)$. The integral

$$\int dx\ \phi(x)A \phi(x)$$

is discretized to

$$\sum_i \phi(x_i) (A\phi)(x_i) = \sum_i \phi(x_i) \sum_j A_{ij} \phi(x_j) = \sum_{ij}q_i A_{ij} q_j$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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