Definition of Hermiticity for Time-Dependent Operators

Question

I am learning quantum mechanics and am reading about Hermitian operators, which are defined to satisfy $$\int f^* Ag d\tau = \left(\int g^* A fd\tau\right)^*.$$ for all (square integrable) functions $f$ and $g$. The limits of integration are often vaguely stated as "over all space" which I assume to mean over all spatial coordinates. Thus, when we want to prove that the momentum operator $\hat{p_x}$ is Hermitian, we must show that $$\int_{-\infty}^\infty f^* \hat{p_x}g dx = \left(\int_{-\infty}^\infty g^* \hat{p_x} fdx\right)^*$$ because the operator only acts on the $x$ component of a function so any other spatial dimensions are irrelavant to this computation. However, I am confused about how to define Hermiticity for operators involving time. For example, in the time-dependent Schrodinger equation (in one dimension) $$-\frac{\hbar}{i}\frac{\partial \psi(x,t)}{\partial t} = \hat{H}\psi(x,t)$$ the Hamiltonian operator acts on a wavefunction that is both a function of spatial and time coordinates. I know that the Hamiltonian operator is Hermitian (as it corresponds to energy, a physical observable), but I don't quite see how to correctly formulate the definition of Hermiticity to involve time as well. What immediately comes to mind is: $$\int_{-\infty}^\infty\int_0^\infty f^* \hat{p_x}g dtdx = \left(\int_{-\infty}^\infty \int_0^\infty g^* \hat{p_x} fdtdx\right)^*$$ but this doesn't quite seem right. Furthermore, when we then talk about the time-independent Schrodinger equation (again in one dimension), we make the assumption that the Hamiltonian is independent of time and we get $$\hat{H}\psi(x) = E\psi(x)$$ where we consider $\hat{H}$, the Hamiltonian, to only act on complex-valued square integrable functions of a single real variable. Technically speaking, it seems that the Hamiltonian appearing in the time dependent and time independent Schrodinger equations are slightly different (because they act on different function spaces), although very related, and I feel like the way Hermiticity is defined for the time-dependent Hamiltonian should be such that when the Hamiltonian does not explicitly depend on time, hermiticity of its time-independent counterpart should follow. I apologize in advance if this question is poorly phrased, but I was hoping someone could clear this up.

Answer 1

The time here is simply a parameter that should have the same value for the operator and the functions, for the definition of hermicity to be meaningful: $$\int f^*(x,t) A(x,t)g(x,t) dx = \left(\int g^*(x,t) A(x,t) f(x,t)dx\right)^*.$$ (Admittedly, I am being a bit schematic here - more appropriately an operator should be $A(x,x',t)$ with integration over $dxdx'$, but the Q. seems to follow the presentation in L&L or a book, inspired by it.)

Answer 2

You're right in noticing that time and space are not treated on the same footing in quantum mechanics (if you read quantum field theory in the future, this will be rectified). Generally, part of defining an operator is deciding what vector space it acts on (In QM it's more generally a Hilbert space). Let's label the elements of that vector space as different kets $|\psi\rangle$, and the operator as $\hat{O}$. Then an operator's Hermitian conjugate is defined as follows (the "bras" $\langle \phi|$ are vectors in the dual space): $$\langle \phi|\hat{O}\psi\rangle=\langle \hat{O}^\dagger \phi|\psi\rangle$$ An operator is Hermitian if $\hat{O}^\dagger = \hat{O}$ satisfies the above relation. Checking if $\hat{H}$ is Hermitian should be clear from this definition.

This isn't in the language of square-integrable functions that you used in your question, but you can get to there from here with a few tools:

1. eigenstates of observables (even with continuous spectra) are complete: for example the position operator $\hat{x}$ has eigenvalues $|x\rangle$ such that: $I (\text{identity operator})=\int dx |x\rangle \langle x|$, so you can insert this wherever you want.
2. The wavefunctions of QM are defined by taking the inner product of a state with an eigenstate of some relevant operator. For example the wavefunction in position space is just $\psi(x) = \langle x|\psi \rangle$.

The issue you were running into that didn't "seem quite right" is caused by there being no time operator $\hat{t}$ in quantum mechanics that could give rise to an integral like that. Time is just a parameter in regular QM (unlike position in QM, which is an observable), so although integrating over time is valid depending on what you're calculating, it has nothing to do with the Hilbert space and thus nothing to do with hermiticity.