# Showing that an operator is Hermitian

Consider the operator $$T=pq^3+q^3p=-i\frac{d}{dq}q^3-iq^3\frac{d}{dq}$$

defined to act on the Hilbert Space $H=L^2(\mathbb{R},dq)$ with the common dense domain $S(\mathbb{R})$. Here $S(\mathbb{R})$ denotes Schwartz space.

How would I show that this operator is hermitian?

I understand the procedure of using the scalar products, i.e. $\forall f,g\in S(\mathbb{R})$ $$<f,Tg>=<T^*f,g>=<Tf,g>$$

But, I am unsure if I could just apply complex conjuagtion right to $T$ to test the operator. It seems if I do this that the operator is not hermitian as the minus sign switches to positive. This leads me to believe this method is flawed and a valid test of the operator being hermitian.

The generalization of this operator to $$T_n=pq^n+q^np$$ and its importance as an example of the subtle difference between hermitian and self-adjoint matrices defined on infinite dimensional Hilbert Spaces $H$ and usage with deficiency indices led me bring the question up.

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If I recall correctly, the inner product is an integral, hence do that integral with a test function (use integration by parts). –  Love Learning Jan 29 at 0:38
The inner product equals an integral and $S(\mathbb{R})$ allows us to throw away the boundary terms. I wanted to know if there were other ways to check hermicity. –  sunspots Jan 29 at 0:55
Since p and q are hermitian, so is $p q^n+q^n p$. You can verify it explicitly, it's basically the same as doing it for $p$. –  fqq Jan 29 at 0:56
–  Qmechanic Jan 29 at 1:00
Echoing what "fqq" wrote, the composition or sum of two Hermitian operators is also Hermitian, so it's automatically true (unless I'm leaving something out?). –  DumpsterDoofus Jan 29 at 1:03
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## 1 Answer

In a complex Hilbert space ${\cal H}$, an operator $A: D(A) \to {\cal H}$, with $D(A)\subset {\cal H}$ a (not necessarily dense) subspace, is said to be Hermitian if $$\langle A \psi | \phi \rangle = \langle \psi| A \phi\rangle \quad \forall \psi, \phi \in D(A)\:. \quad (1)$$ It seems to be worth stressing that, to check (1), it is not necessary to exploit the definition of adjoint operator, $A^\dagger$ that, generally, does not exist when $D(A)$ is not dense.

If $D(A)$ is dense, the Hermitian operator $A$ is said to be symmetric.

In your case(s) $A:=T_n$ and $D(T_n)= S({\mathbb R})$, the Hilbert space ${\cal H}$ being $L^2(\mathbb R)$.

Just using (a) integration by parts, (b) the definition of scalar product in $L^2(\mathbb R)$ and (c) the fact that the elements of $S(\mathbb R)$ rapidly vanishes for $|x|\to \infty$ with all derivatives, you immediately establish that (1) is valid for $T_n$.

Actually these operators are symmetric because $S(\mathbb R)$ is dense in $L^2(\mathbb R)$.

Existence of self-adjoint extensions can be studied examining deficiency indices of symmetric operators. It is not necessary to prove that $T_n$ is closed, even if this condition is assumed in several textbooks as Reed and Simon, as stated in Dunford Schwartz books (vol II Corollary 13 ch. XII.4.13). However, if $n$ is even, self-adjoint extensions do exist in view of the fact that $T_n$ commute with the anti unitary involutive operator given by the complex conjugation of Fourier transforms (exploiting Theorem 18 ch. XII.4.13) which establishes that the deficiency indices coincide.

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