# Non-hermitian hamiltonians and symmetry properties of $\partial_t$ [duplicate]

From this question another one came to my mind. Consider the Hilbert space $\mathcal{H}$ of function square integrables $\psi:\mathbb{R}^n\rightarrow\mathbb{C}$ with the usual inner product:

$$\langle\psi,\chi\rangle = \int_\mathbb{R} d^n\!x\psi^*\chi\quad\mbox{ and }\quad\psi\in\mathcal{H}\Rightarrow\langle\psi,\psi\rangle\in\mathbb{R},$$

where $\psi^*$ is the complex conjugate of $\psi$. I am thinking in particular in two-variable functions $\psi(x,t)$, but the generalisation to more spatial dimensions is straightforward. Given an operator $A$, the hermitian conjugate $A^\dagger$ saisfies $\langle\psi,A\chi\rangle = \langle A^\dagger\psi,\chi\rangle\,\forall\psi\chi\in\mathcal{H}$.

Take $A=i\partial_t$, hence:

$$\langle\psi,A\chi\rangle = i\int dx\int dt\psi^*\partial_t\chi = -i\int dx\int dt\partial_t(\psi^*)\chi = \int dx\int dt(i\partial_t\psi)^*\chi = \langle A\psi,\chi\rangle$$ iwhere in the second steps I integrated by parts and dropped the global derivative assuming suitable boundary conditions, and in the third equality I used $(\partial_t\psi)^*=\partial_t(\psi^*)$. Then $A = A^\dagger$.

But since $A\psi =i\partial_t\psi = H\psi$ from the Schrödinger equation, doesn't it imply that the Hamiltonian must be always Hermitic?.

## marked as duplicate by AccidentalFourierTransform, Community♦Mar 1 '17 at 8:41

• $\langle \cdot|\cdot\rangle$ does not include integration over $\mathrm dt$ – AccidentalFourierTransform Feb 28 '17 at 20:17