I am a physicist, so my background in functional analysis is limited only to basics. However, I would like to prove that the free Dirac operator is selfadjoint (or Hermitian, or neither). The free Dirac operator is a differential operator of the following form:

$D = -i\alpha \nabla + \beta$,

where $\alpha$ and $\beta$ are just Hermitian $4 \times 4$ matrices. This operator acts on a state of 4-component' functions from $\mathbb{R}^3$ to $\mathbb{C}^4$.

The inner product is defined as integral of the product of two functions from this space (one of them being a complex conjugate).

I suppose that these functions should also be square-integrable, i.e. from $L^2$. (If they should also be defined on some bounded interval, then the boundary conditions could just be: $f(0) = f(1) = 0$.)

From a mathematical point of view, what else is needed to formally prove that this operator is self adjoint?

  • $\begingroup$ I'm confused by your reference to boundary conditions. I thought you wanted to your operator to act on functions from $\mathbb{R}^3$ to $\mathbb{C}^4$. $\mathbb{R}^3$ has no boundary so there are no boundary conditions. (It might in general be necessary to impose growth or integrability conditions, however.) I don't see how it would make sense for $D$ to act on functions defined on a bounded (one-dimensional) interval. $\endgroup$ – Nate Eldredge Oct 19 '14 at 14:11
  • $\begingroup$ I wasn't sure how to pick a domain for this operator. This was just a speculation. I hoped that someone with more math background will know what is the underlying problem with this operator and/or how to specify a domain. As I understood, this is related to our choice of 'boundary conditions' in physics, except that boundary conditions we use are usually just in the form: f=something, f'=something, without much details about domains. $\endgroup$ – Dee Oct 21 '14 at 11:58

It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function.

There are many possible approaches, one which is sometimes useful is the following

  • Show that $D$ is symmetric, i.e. $\langle Du,v\rangle=\langle u,Dv\rangle$ for all $u,v\in\mathcal D(D)$.
  • Show that the ranges of $D\pm iI$ are dense (that is, you need to show it for both $D+i$ and $D-i$). Equivalently, you can show that $D^*\pm iI$ are injective.

For differential operators, the first bullet point usually reduces to an integration by parts. It is typically the second bullet point which is most difficult, because you actually need to show that the eigenvalue equations $D^*u=\pm iu$ have no solutions (in a sense, this is the input that you need to feed the machinery before you can take advantage of the huge amount of strong results for self-adjoint operators).

This proves that $D$ is essentially self-adjoint, i.e. the closure of $D$ is self-adjoint. If you want to show the stronger result that $D$ is self-adjoint, you now

  • Show that D is closed. Equivalently, show that $D+i$ or $D-i$ is onto.

Alternatively, you just circumvent this problem by simply working with closure of $D$.

Another approach, which perhaps more clearly shows how 'difficult' it is to show that an operator is self-adjoint (and also showcases the physical relevance of self-adjointedness) is the following: Show that the Cauchy problem for the Schröding equation $$ \partial_t u=-iDu $$ admits a solution in the sense that for any $u_0\in\mathcal D(D)$ there is a unique family $u(t)$ such that

  • $u(t)$ solves the Schrödinger equation
  • the family of operators $U(t)$ defined by $U(t)u_0=u(t)$ is a strongly continuous unitary group.

By Stone's Theorem, this suffices to show that $D$ has a self-adjoint extension, namely the generator of $U(t)$. Since any self-adjoint extension would give rise to such a unitary family, it actually follows that $D$ has a unique self-adjoint extension, which shows that $D$ is essentially self-adjoint, with closure equal to the generator of $U(t)$ (the last bit is a consequence of Von Neumann's extension theory).

Also, the following blog entry by Terence Tao probably explains the problem better than I ever could!

  • $\begingroup$ Do I have to actually solve the differential equation with some boundary conditions, i.e. do I have to work with some concrete set of solutions for D? $\endgroup$ – Dee Oct 19 '14 at 11:51
  • $\begingroup$ Yes, unless you operator is bounded you have to pick a domain for it (I denote the domain by $\mathcal D(D)$ above). This is the subtle part, you have to pick the domain correctly, so that you operator becomes self-adjoint. Commonly for differential operators, picking a domain corresponds to ascribing boundary conditions. $\endgroup$ – Jonas Dahlbæk Oct 19 '14 at 12:06
  • 1
    $\begingroup$ Also, for many differential operators with constant coefficients defined on all of space, the space of compactly supported smooth functions turns out to be a core for the operator (so that, if the operator is self-adjoint, it is essentially self-adjoint on the space of compactly supported smooth functions). So that might be a reasonable 'boundary condition' for you to try out, if you are satisfied with defining your operator simply as the closure of the resulting operator. $\endgroup$ – Jonas Dahlbæk Oct 19 '14 at 12:22
  • $\begingroup$ Further, if you are trying to show that $D^*$ has no imaginary eigenvalues, make sure that you are not simply using the formal definition of the adjoint. If you consider the space of smooth compactly supported functions as domain for $D$, then the domain of $D^*$ will be larger. If this comment seems strange to you, read up on the definition of the adjoint of an unbounded operator. $\endgroup$ – Jonas Dahlbæk Oct 19 '14 at 13:52

When you are on the free space and your operators have constant coefficients, you can diagonalize them via the Fourier transform. (To "diagonalize" a linear operator on a Hilbert space here means finding a unitary transformation that turns it into a multiplication operator on $L^2$-space). This solves the self-adjointness issue, since it is easy to tell when a multiplication operator is self-adjoint. For the case of the Dirac operator $$ D_0=\boldsymbol{\alpha}\cdot (-i\nabla) + \beta,$$ the Fourier transformation gives $$ \hat{D}_0 \hat{\psi}(\boldsymbol{p})=(\boldsymbol{\alpha}\cdot \boldsymbol{p}) \hat{\psi}(\boldsymbol{p}) + \beta \hat{\psi}(\boldsymbol{p}),$$ and the biggest domain of self-adjointness is therefore $$ \text{Dom}(D_0) =\left\{ \psi\in L^2(\mathbb{R}^3)\otimes \mathbb{C}^4\ :\ \lvert\boldsymbol{p}\rvert\hat{\psi}(\boldsymbol{p})\in L^2(\mathbb{R}^3)\otimes\mathbb{C}^4 \right\}. $$ This domain is exactly the first Sobolev space $H^1(\mathbb{R}^3)\otimes \mathbb{C}^4$.

For more information see Thaller's book "The Dirac equation", §1.4: "The free Dirac operator".


Let $H = {\left( {{L^2}\left( {{\mathbb{R}^3},{\mathbb{C}^4}} \right)} \right)^4}$ be the Hilbert space in question and for $u = \left( {{u_1},{u_2},{u_3},{u_4}} \right),v = \left( {{v_1},{v_2},{v_3},{v_4}} \right) \in H$ let the inner product on $H$ be defined by $\left\langle {u,v} \right\rangle = \sum\limits_{i = 1}^4 {\int_{ - \infty }^{ + \infty } {{u_i}\left( {{x_i}} \right)\overline {{v_i}\left( {{x_i}} \right)} d{x_i}} } $.

${\left[ {\nabla u} \right]_{i,j}} = {\partial _j}{u_i}$ so

${\left[ {Du} \right]_{i,j}} = - i\sum\limits_{k = 1}^4 {{\alpha _{i,k}}{{\left[ {\nabla u} \right]}_{k,j}}} + {\beta _{i,j}} = - i\sum\limits_{k = 1}^4 {{\alpha _{i,k}}{\partial _j}{u_k}} + {\beta _{i,j}}$

$D:H \to {\left( {{M_{4 \times 4}}\left( \mathbb{C} \right)} \right)^H}$, where ${\left( {{M_{4 \times 4}}\left( \mathbb{C} \right)} \right)^H}$ is a vector space of functions from $H$ to $4 \times 4$ matrices with complex-valued entries.

This means that $\left\langle {Du,v} \right\rangle $ is ill-defined, since $\left\langle { \cdot \,,\, \cdot } \right\rangle :H \times H \to \mathbb{C}$.

Suppose that we somehow redefine $D$ or the inner product so that $\left\langle {Du,v} \right\rangle $ becomes well-defined.

In that case, it would suffice to show that

1) $\left\langle {Du,v} \right\rangle - \left\langle {u,Dv} \right\rangle = 0$ (integration by parts combined with boundary conditions $u\left( { \pm \infty } \right) = v\left( { \pm \infty } \right) = 0$ - which must hold since $u,v \in H$ - should provide the needed equality). We conclude that $D$ is symmetric.

2) $D$ is well-defined on its domain. There should be no problems verifying this

It's easy to show that a symmetric operator is linear, so Hellinger-Toeplitz theorem then implies that $D$ is bounded and hence hermitian.

Note: In the text above, the definitions in the accepted answer here are used, meaning that hermitian implies self-adjoined implies symmetric and converse implications don't necessarily hold.

  • 1
    $\begingroup$ This is true on its face but I think it misses the point. $D$ will certainly be an unbounded operator so we should not expect it to be defined everywhere on the Hilbert space $H$, but instead on some dense subspace (its domain). Part of the question generally is making an appropriate choice for the domain, but in this case I think the natural choice is going to be the Sobolev space $H^1(\mathbb{R}^3; \mathbb{C}^4)$. To say it in the terms I used in the answer you link, this operator will be self-adjoint but not Hermitian. We should not expect to use Hellinger-Toeplitz. $\endgroup$ – Nate Eldredge Oct 19 '14 at 14:08

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