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My book about quantum mechanics states that the hamiltonian, defined as $H=i\hbar\frac{\partial}{\partial t}$ is a hermitian operator. But i don't really see how I have to interpret this. First of all: from which to which space is this operator working? They are defining a vectorspace called the "wavefunctionspace $F$" which contains all square-integrable functions that are continious and infinite differentiable (and everywhere defined). But it looks to me, that if the hamiltonian acts on this space, it's not necessary true that the image of a random vector of $F$ is again in $F$.

I think in fact, that there are some vectors of $F$ so that the hamiltonian of those vectors is not an element of $F$ (so that it's not an endomorphism on $F$). And if the hamiltonian has to be hermitian, it has to be an endomorphism on some space.

If we define instead the vectorspace $V$, which is the same space as $F$ but where functions don't have to be square-integrable, the hamiltonian will be an endomorphism (so at first I thought this was the solution). But now the inner product on functions $<f,g> = \int_{-\infty}^\infty{f^*g}$ which was defined well on $F$ because the integral will always exist if $f$ and $g$ are function of $F$, is no longer properly defined.

I hope someone can clarify how I have to interpet this operator (the same question holds in fact for some other operators).

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Comment to the question (v1): The formula $H=i\hbar\frac{\partial}{\partial t}$ is not generally valid, and in particular, not a definition, cf. e.g. this Phys.SE post. –  Qmechanic Apr 29 '13 at 18:12
For a mathematically rigorous formulation of QM one needs notion of rigged Hilbert spaces. However most of the physicists who use QM as a tool to understand nature usually don't try to be too rigorous(I myself have never read the wiki page I referred to:). –  user10001 Apr 29 '13 at 20:17
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2 Answers 2

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As you said, the Equation you give works only in "wavefunctionspace". As long as you deal with wave functions, you can always decompose the object your derivative acts on as $$\int \frac{\mathrm d\omega }{\sqrt{2\pi}} \tilde f(\omega) e^{i \omega t}$$ so your derivative will draw a factor of $i \omega$ from the exponential. The general shape still is a superposition of plane waves and therefore still in $\mathbf F$, and the operator $i\hbar\frac{\mathrm d}{\mathrm dt}$ is self-adjoint.

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I'd like to add that I'm not particularly fond of such argumentations but I can see why many authors and lecturers employ them first before introducing the more abstract yet clearer Dirac notation. –  Neuneck Apr 30 '13 at 6:32
Thank you for your answer :). –  yarnamc May 1 '13 at 8:53
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You may be able to find some pathological examples for why operators in quantum mechanics aren't hermitian, but these aren't physical. This is physics, there is bound to be some sketchiness to the math. You might be interested in reading this for some interesting mathematical problems/surprises with the mathematical formulation of quantum mechanics: http://lanl.arxiv.org/pdf/quant-ph/9907069v2.pdf

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Aha, that link helped me indeed. Thank you for your answer. –  yarnamc May 1 '13 at 8:53
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