The free Dirac operator is the differential operator of the following form $$ T_0 = i \alpha \nabla + \beta,$$

where $\alpha$ and $\beta$ are Hermitian $4 \times 4$ matrices, and $T_0$ is selfadjoint on some domain $D(T_0)$ on infinite-dimensional Hilbert space. Dirac operator acts on $C^4-$valued functions $\psi(r) = (\psi_i(r))^4_{i=1}$ in Hilbert space $(L^2)^4.$

If we want to include the interaction in the differential equation (which is often the case, since the free Dirac operator is not so physically interesting) one could just use the following type of the interaction:

$$V(r) = \frac{Ze^2}{r^2},$$

where $Z$ and $e$ are just numbers. This is the important (Coulomb) type of the potential in physics.

Now the new 'interacting' differential operator is defined as

$$T = T_0 + V(r).$$

My questions: Could this type of interaction $V$ somehow 'destroy' the selfadjointness of the operator $T$, even if $T_0$ is selfadjoint on some domain? If yes, could this be simply avoided by choosing the new domain for $T=T_0 + V$ without the $r=0$, since $V$ is singular there? Is there a reason why would selfadjointness be ruined only for large coefficients Z, but not for small ones?

More information about $V$: $V$ is a $4 \times 4$ symmetric matrix, whose elements $V_{ij}$ are (complex valued) measurable almost everywhere finite functions on $R^3$.

  • 1
    $\begingroup$ Could you please add more details about the interaction $V(r)$, including the domain, the role of $r$ etc. In general even if both operators in the sum are self-adjoint it does not mean that so is their sum; here is an example mathoverflow.net/questions/122982/…. $\endgroup$ – m_gnacik Oct 26 '14 at 13:21
  • $\begingroup$ $V(r)$ is matrix valued measurable potential and $r$ is just the space variable $r=(r_1, r_2, r_3)$. Based on a few math papers I saw on this subject I also think that the following is true for $V$, in most of the physical situations: $$(V\psi)_i(r)=\sum_jV_{ij}(r)\psi_j(r),$$ $$D(V)=(\psi \in (L^2)^4|\sum_i \int |(V\psi)_i(r)|^2 dx < \infty). $$ $\endgroup$ – Dee Oct 26 '14 at 18:15
  • $\begingroup$ Self-adjointess of $T$ is equivalent to well-posedness of the initial value problem $(i\partial_t + T)\psi=0$. In your generality you could of course lose it. See Thaller's "The Dirac equation", §4.3: "Self-adjointess and essential spectrum". $\endgroup$ – Giuseppe Negro Oct 26 '14 at 18:31
  • $\begingroup$ This is still not enough information about V? $\endgroup$ – Dee Oct 26 '14 at 19:31
  • $\begingroup$ @Nadia: See Theorem 4.2 pag.112 of the aforementioned book by B. Thaller. This result might be interesting for you. $\endgroup$ – Giuseppe Negro Oct 29 '14 at 13:33

Your Answer

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

Browse other questions tagged or ask your own question.