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In addition to other problems described here, is there also a problem of hermiticity of the Hamiltonian operator $\hat{H}$, if someone erroneously writes $$\hat{H}:=i\hbar\frac{\partial}{\partial t}?$$

If yes, what is the origin of non-hermiticity? I guess this because $t$ is usually taken to be in the domain $0\leq t\leq +\infty$ whereas $-\infty\leq x,p\leq+\infty$. But I'm not sure because very often by the choice of origin $t$ can be made to lie in the domain $-\infty\leq t\leq+\infty$.

Qmechanic
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SRS
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    hermiticity with respect to which scalar product? – AccidentalFourierTransform Sep 04 '17 at 13:12
  • @AccidentalFourierTransform the sense in which $\hat{x}=x$, $\hat{p}=-i\hbar\frac{d}{dx}$ is hermitian in position basis? – SRS Sep 04 '17 at 13:14
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    To expand the comment of @AccidentalFourierTransform : what is the vector space your operator acts on? Since you are taking the derivative with respect to the variable $t$, and you want it to be a linear operator, $H$ should probably act on a suitable vector space of functions in the variable $t$ (with values in another vector space, e.g. the vector space $L^2(\mathbb{R}^d)$ of wavefunctions I guess). – yuggib Sep 04 '17 at 14:24
  • Consider, for example, the problem of a particle in a box. @yuggib – SRS Sep 04 '17 at 14:26
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    Which properties do you want the map $t\mapsto \psi(t)$ to have? (where $\psi(t)$ is a wavefunction describing a particle in a box) If you want to give meaning to $i\partial_t$ as a linear operator (and this is really the point zero of your question, even before eventual questions about being hermitian), you have to specify the space it acts upon. This amounts, as I said above more concretely, to the specification of the properties of the maps $t\mapsto \psi(t)$ that you are willing to consider. If they do not form a vector space, then the question is already ill-defined. – yuggib Sep 04 '17 at 14:40

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The real question, as others have noted, is what are you going to apply this operator to exactly? Let's just elaborate on that to show you what goes wrong.

Let $\Psi(x,t)$ be a generic solution to the 1-D Schrodinger Equation. The conventional interpretation of this solution is that at each time $t$, the single-variable function $$\psi_t(x):=\Psi(x,t)$$denotes the state of the system at time $t$, and the totality of such states form a Hilbert Space, replete with Hermitian operators and all that jazz.

But since these wavefunctions are functions of $x$, how do we apply an operator like $i\hbar\partial_t $ to them? For this we need functions of time $t$.

One obvious strategy to try is to simply switch $t$ and $x$ around in the above definition to get $$\psi_x(t):=\Psi(x,t)$$which presumably describes the full temporal behavior of the system at a fixed point $x$.

However a quick check shows that these functions do not form a Hilbert Space, since they are not square integrable. To see this, just consider a stationary state of the form $$\psi_x(t):= e^{-iEt}\phi(x)$$and integrate over time (remembering that $x$ is fixed) to get$$\parallel\psi_x(t)\parallel^2=\int_{\mathbb R}|e^{-iEt}\phi(x)|^2dt=|\phi(x)|^2\int_{\mathbb R}|e^{-iEt}|^2dt=\infty$$ In other words, when you try to interpret the solutions of Schrodinger's Equation as functions of time, in order to make the action of $i\hbar\partial_t $ meaningful, you don't get a Hilbert Space, which makes the entire subject of Hermitian operators and spectral theory meaningless.

Physics Footnotes
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