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.

You're final equation is false. It is **not** true that,
$$
\left\langle ... n _i ... \right| a = \sqrt{ n _i } \left\langle ... n _i - 1 ... \right|
$$
The correct result is
$$
\left\langle ... n _i ... \right| a = \left\langle ... n _i + 1 ... \right| \sqrt{ n _i +1}
$$
as you have just proven.

A non-Hermitian operator doesn't act in the same way to the left as it does to the right. The rules of working with bra's and ket's are if you are acting backwards with the operator then you need to use the Hermitian adjoint (see for example Wikipedia). Otherwise you end up with inconsistencies as you did above. More formally, for any operator $ A$ and states $ \left| \alpha \right\rangle $ and $ \left| \beta \right\rangle $ you can write,
\begin{equation}
\left\langle \alpha \right| A \left| \beta \right\rangle \equiv \left\langle \alpha |A \beta \right\rangle = \langle A ^\dagger \alpha | \beta \rangle
\end{equation}

and only for a Hermitian operator can you omit the dagger.

## Best Answer

The correct answer is: they're not, necessarily. There are two things that happen when you take the Hermitian conjugate: a transpose, and a complex conjugation. If you break down the Hermitian operators into symmetric and anti-symmetric parts, the symmetric part is real and the anti-symmetric part is imaginary. In equations: $$\begin{align} \mathrm{let:}&\ A^\dagger = A,\ A_s \equiv \frac{A+A^T}{2},\ \mathrm{and}\ A_a \equiv \frac{A - A^T}{2} \Rightarrow \\ A_s^T = A_s & \Rightarrow A_s^\star = A_s \\ A_a^T = -A_a & \Rightarrow A_a^\star = -A_a. \end{align}$$ You can also prioritize real and imaginary parts of $A$ to get: \begin{align} \mathrm{let:}&\ A_r \equiv \frac{A + A^\star}{2},\ \mathrm{and}\ A_i \equiv \frac{A - A^\star}{2} \Rightarrow \\ A_r^\star = A_r & \Rightarrow A_r^T = A_r \\ A_i^\star = -A_i & \Rightarrow A_i^T = -A_i. \end{align}

You can add the $\langle y|$ and $|x\rangle$ to both sides of the above equations, and the conclusion won't be affected - that's just working in terms of the operator's matrix elements.