Consider a scalar field $\phi(x)$ described by an action $$ S= \frac12 \int d^4x\ \phi (F\phi) $$ where the kernel $F$ is a self-adjoint operator, in the sense that for any two complex-valued functions $\psi_1$ and $\psi_2$ supported on an open neighborhood $U$ of spacetime,

$$ \int d^4x\ [{\psi_1}^*(F\psi_2)-(F\psi_1)^*\psi_2] = 0\,.$$

Now consider a compact region $\Omega \subset U$, so that we can define a two-edged vector operator $\overleftrightarrow{f^\mu}$ corresponding to $F$ in the following manner.

$$ \int_\Omega d^4x\ [{\psi_1}^*(F\psi_2)-(F\psi_1)^*\psi_2] =: \int_{\partial\Omega} d\Sigma_\mu\ {\psi_1}^* \overleftrightarrow{f^\mu} \psi_2 \,,$$

where $d\Sigma_\mu$ is the outward directed area element of $\partial \Omega$.

Using this, we can define an inner product on the space of mode functions $\{u_i\}$ which satisfy the equation of motion for $\phi\,,$ namely that $Fu_i=0$. Given any complete Cauchy hypersurface $\Sigma$ for these equations of motion, define

$$ \langle u_i |u_j \rangle:= -i \int_\Sigma d\Sigma_\mu\ {u_1}^* \overleftrightarrow{f^\mu} u_2 $$

But, if $Fu_i=Fu_j=0$, then don't we quite clearly have that

$$ \langle u_i |u_j \rangle = 0 $$


For a contrary claim, read B. S. DeWitt, Phys. Rep. 19C, 292 (1975), section 1.1, just before Eq.(7). I am trying to understand what DeWitt really wanted to say and what I am missing.


1 Answer 1


No. Note that in the first case, the integration region is a boundary, $\partial \Omega$, and in the second case, it is an arbitrary region $\Sigma$. As the first integral vanishes, the second one is independent of deformations of $\Sigma$.

$$\int_\Omega d^4x\ [{\psi_1}^*(F\psi_2)-(F\psi_1)^*\psi_2] =: \int_{\color{red}{\partial\Omega}} d\Sigma_\mu\ {\psi_1}^* \overleftrightarrow{f^\mu} \psi_2 $$ vs. $$ \langle u_i |u_j \rangle:= -i \int_{\color{red}{\Sigma}} d\Sigma_\mu\ {u_1}^* \overleftrightarrow{f^\mu} u_2 $$ and so $$ \langle u_i |u_j \rangle_\Sigma-\langle u_i |u_j \rangle_{\Sigma'}=\langle u_i |u_j \rangle_{\partial(\Sigma-\Sigma')}=0 $$

Conclusion: the inner product does not in general vanish, but it is invariant under time-like deformations (in the flat case, it is independent of time).

For more details, see DeWitt's The Global Approach to Field Theory, chapter 4, section "The Wronskian Operator" (page 54).

  • $\begingroup$ Right. So, the first equation is not an operational definition of $\overleftrightarrow{f^\mu}$ but rather a requirement imposed on it. Correct? Otherwise, how do we compute the inner product if $\Sigma$ is not the boundary of some region, and in which case it would be trivially zero on-shell? $\endgroup$ Jan 13, 2018 at 14:54
  • $\begingroup$ BTW, what do you mean $\Sigma - \Sigma'$? $\endgroup$ Jan 13, 2018 at 14:58
  • $\begingroup$ Hi @NanashiNoGombe 1) The first equation is the statement that there must exist some $f^\mu$. But this equation does not determine this function uniquely (indeed, and as discussed in the reference, one may always add an arbitrary skew-symmetric function to it without affecting the value of the inner product). The explicit form of $f^\mu$, for some specific examples, can be found in the reference. 2) The intersection of those regions. $\endgroup$ Jan 13, 2018 at 18:45

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