Selfadjointness of the Dirac operator on the infinite-dimensional Hilbert space

Question

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?

Answer 1

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!

Answer 2

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".

Answer 3

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.