Does time-independence of potential energy imply time-independence of the Hamiltonian in Quantum Mechanics?

Question

Consider a quantum mechanical system for a particle with Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m}+\hat{V}$ where $\hat{V}$ is the potential energy operator. and now let us assume that $\hat{V}$ is time-independent. Does it follow that $\hat{H}$ is time-independent?

In particular, I am interested in the case of potential wells/potential barriers/$\delta$ potentials where this seems to be used in every treatment I have found so far.

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My thoughts: In general, since wavefunctions can be dependent of time, so can $\hat{p}$ by its very definition and hence $\hat{H}$ can be time-dependent. However, I wonder if $\hat{V}$ being time-independent already yields the restriction that wavefunctions do not depend on time and hence $\hat{p}$ also doesn't. This seems like a sort of circular reasoning and I'm not sure how to continue here.

Answer 1

Perhaps it would be helpful to take a more abstract view for a moment.

A linear operator $\hat A$ on a Hilbert space $\mathcal H$ is a linear map from $\mathcal H\rightarrow \mathcal H$. If $\mathcal H=L^2(\mathbb R)$, then elements of the Hilbert space essentially consist of square-integrable functions of one real variable, which we usually interpret as the position of a particle on a line.

Examples of such operators might include $\hat X$ or $\hat P$, which act on suitable vectors $f\in L^2(\mathbb R)$ to produce other vectors $$\big(\hat Xf\big)(x) = x f(x)$$ $$\big(\hat Pf\big)(x) = -i\hbar f'(x)$$

Put differently, an operator is just a rule for taking a square-integrable function and spitting out another square-integrable function.

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In contrast, consider the family of operators $\hat Q(t)$ which eat vectors $f\in L^2(\mathbb R)$ and spit out

$$\big(\hat Q(t) f\big) (x) = (x+t^2) f(x)$$

Notice that for each $t$, $\hat Q(t)$ is a different operator. $\hat Q(0)$ is just $\hat X$, while $\hat Q(1)$ multiplies the wavefunction by $(x+1)$, and $\hat Q(-17)$ multiplies the wavefunction by $(x+289)$, and so on. We typically call $\hat Q$ a time-dependent operator, which is a different way of saying that the rule by which vectors are mapped to other vectors is different for each value of $t$.

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Having made this distinction, the answer to your question becomes clear. $\hat X$, which encodes the rule "multiply the wavefunction by $x$," and $\hat P$, which encodes the rule "differentiate the wavefunction and multiply by $-i\hbar$," are both time-independent because those rules don't change with time. Similarly, if the potential energy operator is given by some $V(\hat X)$ which encodes the rule "multiply the wavefunction by $V(x)$", then it too is a time-independent operator.

In contrast, the operator (or alternatively, the family of operators) $U(\hat X,t)$ which encodes the rule "multiply the wavefunction by $x+\sin(t)$" is time-dependent because the rule is different for different values of $t$; for instance, $U(\hat X,0)$ encodes the rule "multiply the wavefunction by $x$" while $U(\hat X,\pi/2)$ encodes the rule "multiply the wavefunction by $x+1$".