Role of Hermiticity in Quantum Mechanics without exploiting the Spectral Theorem

Question

In any introduction to quantum mechanics, an explanation for why the Hamiltonian $H$ is a Hermitian operator is given. This explanation usually relies on the spectral theorem, which guarantees that $H$ has a complete orthonormal basis of eigenvectors, and real eigenvalues. Then, the eigenvectors of $H$ specify the states of well-defined energy, and the eigenvalues are the corresponding energies.

However, I am wondering if there is a simple physical explanation for why the Hamiltonian matrix elements must be symmetric (up to complex conjugation). To clarify what I mean, suppose I fix a basis $B = \{|1\rangle,\ldots, |n\rangle\}$ of my Hilbert space, and I have a Hamiltonian $H$. Suppose moreover that, when I express my Hamiltonian in the basis $B$, there is a term of the form $$c_{ij}|i\rangle \langle j|.$$ Then, by Hermiticity, there must also be a term of the form $$c_{ij}^*|j\rangle\langle i|.$$ My intuitive understanding of terms of the form $|i\rangle\langle j|$ is that it is "like a force" in the Hamiltonian which sends $| j \rangle$ to $|i \rangle$, since, when $H$ acts on $|j\rangle$, it will replace $|j\rangle$ with $|i\rangle$. Then, insisting that both $|i\rangle\langle j|$ and $|j\rangle \langle i|$ are present in the Hamiltonian is like saying that the forces must always go in both directions. Is this a correct interpretation, and if so, why must this be true?

Answer 1

I set $\hbar = 1$

Another way to come at this is to consider the Hamiltonian to be the generator of unitary time evolution. That is, say we have an initial state $|\psi(0)\rangle$. We can take it as a postulate of quantum mechanics that the state at a later time is given by

$$ |\psi(t)\rangle = U(t) |\psi(0)\rangle $$

Where $U$ is a unitary operator. By taking the time derivative of this expression we arrive at the Schrodinger equation

$$ \frac{d}{dt}|\psi(t)\rangle = -i H(t) |\psi(t)\rangle $$

Where I've defined the Hamiltonian

$$ H(t) = i \frac{dU}{dt} U^{\dagger} $$

From this definition and unitary of $U(t)$ it can be proven that

$$ H(t) = H^{\dagger}(t) $$

and, if $H(t)$ happens to be time-independent:

$$ U(t) = e^{-i H t} $$

What you seem to be gesturing at in your question is that the Hermiticity of $H$ seems to intuitively follow from the reversibility of dynamics/interactions. This reversibility is exactly encoded in the unitarity of the time evolution operator, so deriving Hermiticity of $H$ from this unitarity should then, maybe, be satisfying to you.

Answer 2

The requirement for hermicity comes from the fact that all rhe measured quantities are real. Since the results of measurements are eigenvalues of the corresponding operator, these eigenvalues must be real. Hermitian operators are the ones that have real eigenvalues.

Operators can be defined as hermitian without spectral decomposition.