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In a recent paper (on an exactly solvable toy model and its dynamics), we studied such a toy model:

$$ H = \sum_{n\in \mathbb{Z}} n |n \rangle \langle n | + g \sum_{n_1,n_2 \in \mathbb{Z}} |n_1 \rangle \langle n_2 | . $$

We solved the eigenvalues and eigenstates analytically and in particular, we studied the evolution of the initial state $|\psi_0 \rangle = |0\rangle $. We calculated the matrix element of the time-evolution operator $\langle m |e^{-iHt }|0\rangle $. All is done analytically with closed expressions.

However, a few days ago, a student questioned me, is it legitimate to act $H$ on the state $|0 \rangle $? It yields the state

$$ |\phi \rangle = \sum_{n\in\mathbb{Z}} |n \rangle, $$

which is not in the Hilbert space $l_2$.

I am now a bit upset. It seems that the initial state is not in the domain of the Hamiltonian. But if we, as we did, formally decompose the initial state with respect to the eigenstates and then form a new evolved state by unitary evolution, everything seems fine.

I am not so familiar with functional analysis, but quite interested in it. Could anyone help me on this dilemma? What is the domain of $H$? Is it legitimate to have an initial state outside of the domain of $H$, but within the domain of $e^{-iHt}$? It seems that the latter is bigger than the former.

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  • $\begingroup$ The paper is also on arxiv (without paywall). $\endgroup$ – Keith McClary Mar 28 at 3:48
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    $\begingroup$ Specifying an unbounded operator means that both its action and its domain are given. So the answer to the question "what is the domain of $H$?" is ill-posed. The maximal domain of definition for $H$ is clearly $\{\psi\in \ell^2, H\psi\in \ell^2\}$. However, the relevant domain is the domain of self-adjointness (if one exists), for it is only for self-adjoint operators that the evolution $e^{-itH}$ can be defined rigorously and unambiguously. It is well-known that there are symmetric unbounded operators with no, one or infinitely many self-adjoint extensions. $\endgroup$ – yuggib Mar 28 at 15:31
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    $\begingroup$ Once a self-adjoint operator $H$ is given, with domain $D(H)$, then $e^{-itH}$ can be defined (usually by means of spectral theorem), and it is a bounded operator, that means its domain of definition is the whole Hilbert space $\mathscr{H}\supseteq D(H)$. $\endgroup$ – yuggib Mar 28 at 15:33
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The operator is unbounded. Unbounded densely defined (Hermitian) symmetric operators cannot be extended to the entire Hilbert space (see the comment by DisintegratingByParts ).

The operator in the paper is not properly defined since it does not specify a domain. They do construct vectors which are formally eigenvectors, but they do not investigate whether they form a complete orthonormal basis.

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  • $\begingroup$ It is beyond most physicists to define a domain for a hamiltonian. $\endgroup$ – Jiang-min Zhang Mar 28 at 11:10
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    $\begingroup$ @Jiang-minZhang I don't agree, well-defining the model under study should be one of the primary targets of a physicist. $\endgroup$ – yuggib Mar 28 at 15:20
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    $\begingroup$ @Jiang-minZhang, defining sensible boundary conditions is precisely the same as defining the domain of $H$ in a such a way that makes $H$ self-adjoint. It is not beyond most physicists to define boundary conditions. $\endgroup$ – Peter Kravchuk Mar 28 at 20:55
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$\def\cH{{\cal H}} \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle}$ The $H$ you have written is the sum of two terms, say $H_0$ and $H_1$. $H_0$ is unbounded, so its domain is smaller than $\cH$. I'd guess that $D(H_0)$ is dense and moreover $H_0$ is self-adjoint. Therefore no problem is expected on this side.

Much worse $H_1$. If you write $$H_1 = g \sum_{n_1n_2\in\Bbb Z} \ket{n_1}\bra{n_2}$$ the sum is not even well defined. Writing $$H_1 = g \sum_{n_1\in\Bbb Z} \ket{n_1} \sum_{n_2\in\Bbb Z}\bra{n_2}$$ we get a clearer expression, but several issues arise. It looks as a tensor product of ket $\ket K = \sum\ket{n_1}$ and of bra $\bra B=\sum\bra{n_2}$. However both expressions are undefined (this is the root of your student's objection).

$\ket K$ isn't a good ket, as $\sum|c_n|^2=\infty$. The same for $\bra B$. We could salvage the latter as follows. Don't think of it as a bra, i.e. a bounded member of the dual space $\cH^*$, but as an unbounded functional on kets, defined as follows. If $\ket X\in\cH$, with $$\ket X = \sum c_n \ket n$$ then $$\braket BX = \sum c_n.$$ Its domain is not $\cH$, but the set of kets such that $\sum|c_n|<\infty$, smaller than $\cH$ but dense.

I haven't deepened my analysis and can't answer your last question.

But if we, as we did, formally decompose the initial state with respect to the eigenstates and then form a new evolved state by unitary evolution, everything seems fine.

It might happen that by formal manipulation upon a meaningless hamiltonian a meaningful unitary evolution operator is obtained. If $H$ were self-adjoint this would be granted as @yuggib commented. But your $H$ is far from that...

Actually there is something still stranger. There is a theorem due to Stone ensuring that if there is a unitary evolution operator $U(t)$ then it can be written $$U(t) = e^{-itH}$$ with $H$ self-adjoint. But there is a catch: Stone's theorem holds if $U(t)$ is strongly continuous. I'd be tempted to anticipate that your evolution operator hasn't that property.

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  • $\begingroup$ I did some numerical simulation. In numerics, we have to truncate the Hilbert space to a finite-dimensional space. Then there is no worry any more. It turns out that everything agrees with the infinite-dimensional analytical results. So, according to you, there is really a dilemma. $\endgroup$ – Jiang-min Zhang Mar 28 at 23:29

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