4
$\begingroup$

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

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

identically?

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.

$\endgroup$
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nanashi No Gombe Jan 13 '18 at 14:54
  • $\begingroup$ BTW, what do you mean $\Sigma - \Sigma'$? $\endgroup$ – Nanashi No Gombe Jan 13 '18 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. $\endgroup$ – AccidentalFourierTransform Jan 13 '18 at 18:45

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.