Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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).

share|improve this question
1  
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

4 Answers 4

up vote -1 down vote accepted

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.

share|improve this answer
    
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
    
Are you sure that this proves that $ih\frac{d}{dt}$ is Hermitian? I know that the Hamiltonian can be pseudo-hermitian and $H=ih\frac{d}{dt}$ is still valid from Schrodinger equation. –  Bob Aug 22 at 13:38
1  
Sorry, completely wrong. $id/dt$ is not an operator on the Hilbert space of the system. The Hamiltonian operator is not $i\hbar \partial /\partial t$. –  Valter Moretti Aug 22 at 15:11

The general situation is the following one. There is a self-adjoint operator $H :D(H) \to \cal H$, with $D(H) \subset \cal H$ a dense linear subspace of the Hilbert space $\cal H$. (An elementary case is ${\cal H} = L^2(\mathbb R, dx)$, but what follows is valid in general for every complex Hilbert space $\cal H$ associated to a quantum physical system.)

It turns out that $D(H) = \cal H$ if and only if $H$ is bounded (it happens, in particular, when $\cal H$ is finite dimensional).

Physically speaking $H$ is bounded if and only if the values the corresponding observable (the energy of the system) attain form a bounded set, so it hardly happens in concrete physical cases. $D(H)$ is almost always a proper subset of $\cal H$.

If $\psi \in \cal H$ represents a (pure) state of the system, its time evolution is given by $$\psi_t = e^{-i \frac{t}{\hbar} H}\psi \tag{1}\:.$$ The exponential is defined via spectral theorem. The map $\mathbb R \ni t \mapsto e^{-i \frac{t}{\hbar} H}\psi$ is always continuous referring to the topology of $\cal H$. Moreover it is also differentiable if and only if $\psi_t \in D(H)$ (it is equivalent to say that $\psi \in D(H)$). In this case one proves that (Stone theorem) $$\frac{d\psi_t}{dt} = -i \frac{1}{\hbar} H e^{-i \frac{t}{\hbar} H}\psi= -\frac{i}{\hbar} H\psi_t\:.$$ In other words, $$i \hbar \frac{d\psi_t}{dt} = H\psi_t\:.\tag{2}$$ It should be clear that $\frac{d}{dt}$ is not an operator on $\cal H$, as it acts on curves $\mathbb R \ni t \mapsto\psi_t$ instead of vectors. $$\frac{d\psi_t}{dt} = \lim_{s \to 0} \frac{1}{s} \left(\psi_{t+s}-\psi_t \right)$$ and the limit is computed with respect to the Hilbert space norm. Identity (2) holds if and only if $\psi \in D(H)$ and not in general.

ADDENDUM.

Identities or even definitions (!) like $$H = i \hbar \frac{d}{dt}\:.\tag{3}$$ have no sense. An observable in QM first of all is an operator (self-adjoint) on the Hilbert space $\cal H$ of the theory. In other words it is a linear map $A$ associating any given vector $\psi \in \cal H$ (or some suitable domain) to another vector $A\psi$. If $\psi$ is a given single vector of $\cal H$ - and not a curve $t \mapsto \psi_t$ - the formal object $$\frac{d}{dt}\psi$$ has no meaning at all as it cannot be computed! Thus, wondering whether or not $H$, "defined" by means of (3), is Hermitian does not make sense in turn, because the RHS of (3) is not an operator in $\cal H$.

The concrete definition of $H$ can be given as soon as the physical system is known and taking advantage of some further physical principles like some supposed correspondence between classical observables and quantum ones, or group theoretical assumptions about the symmetries of the system.

For non-relativistic elementary systems described in $L^2(\mathbb R^3)$, the Hamiltonian operator has the form of the (hopefully unique) self-adjoint extension of the symmetric operator $$H := -\frac{\hbar^2}{2m}\Delta + V(\vec{x})$$ That is the definition of $H$.

Nevertheless, Schroedinger equation (2) is always valid, no matter the specific features of the quantum (even relativistic) system, when $\psi \in D(H)$. Time evolution is however always described by (1) regardless any domain problem.

share|improve this answer
    
In short, you have derived the form of the Hamiltonian from equation 1, instead of the other way round, which is more conventional. Either way, it's not very well motivated physically. –  akrasia Aug 22 at 16:41
1  
No, I did not. There is no explicit form of the Hamiltonian in my answer. There is no definition of the Hamiltonian at all. I insist on the fact that $H = i d/dt$ is a non-sense both in physics and in mathematics. It is simply a confused idea. The explicit form of the Hamiltonian is obtained adding details to the specific physical system one is considering. For instance, in non relativistic QM, for a single particle in $\mathbb R^3$, $H$ is the (hopefully unique) self-adjoint extension of $-\frac{\hbar^2}{2m} \Delta_{\vec{x}} + V(\vec{x})$. –  Valter Moretti Aug 22 at 17:58

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

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

If the hamiltonian is hermitian, the eigenfunction make a base of the space of squared integrable function. So the action of H on any function stays on the squared integrable function space.

share|improve this answer

protected by Qmechanic Aug 22 at 12:48

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.