I believe this question may seem silly, every student who has studied quantum mechanics in school must has been told that Hamiltonian is a linear operator on Hilbert space.

However, today I think this statement doesn't make sense, because when we say something is a linear operator on vector space, we mean it acts on a vector and send it to a vector in the same space. But given a vector in Hilbert space, that means it's square integrable, then clearly Hamiltonian does not necessary to send it to another square integrable vector. Then what do we actually mean by saying Hamiltonian is a linear operator on Hilbert space?


1 Answer 1


For a quantum system corresponding to a Hilbert space $\mathcal H$, a physical observable corresponds to a (possibly unbounded) self-adjoint operator $A$ on $\mathcal H$. This means in particular that $A$ is a linear map from some dense subspace $\mathcal D_A\subseteq \mathcal H$ to $\mathcal H$.

The Hamiltonian $H$ is just such an operator. If we decide on a formula for $H$, e.g.

$$H\psi := -\frac{\hbar^2}{2m}\psi'' + V(x)\psi$$ then $D_H$ can only consist of those elements $\psi\in\mathcal H$ such that $H\psi\in\mathcal H$ as well. The domain on which $H$ is self-adjoint may in fact be smaller than this, but it is never larger.

For the case of the free particle, where $H:=-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$, this means that if you have some square-integrable function $\phi(x)$ whose second derivative is not square-integrable, then $\phi\notin D_H$.

  • $\begingroup$ So what you mean is that we make it to be an axiom? $\endgroup$
    – GK1202
    Jul 12, 2020 at 22:37
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    $\begingroup$ @y255yan The Hamiltonian is (generally unbounded) linear operator on a Hilbert space $\mathcal H$. That means that it is a linear map from some domain $D_H\subseteq \mathcal H$ to $\mathcal H$. $\endgroup$
    – J. Murray
    Jul 12, 2020 at 22:40
  • $\begingroup$ @y255yan I have expanded my answer a bit. $\endgroup$
    – J. Murray
    Jul 12, 2020 at 23:00
  • $\begingroup$ So in a word, that's our construction (requirement) for Hamiltonian? Now what if we know $(H\psi)\in \mathcal H$? can we say $\psi$ is in $\mathcal H$? $\endgroup$
    – GK1202
    Jul 12, 2020 at 23:01
  • $\begingroup$ @y255yan No. For example, let $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ and $\phi\in\mathcal H$ be such that $H\phi\in\mathcal H$. Then $\psi(x) = x + \phi(x)$ is not in $\mathcal H$, but $H\psi$ is. $\endgroup$
    – J. Murray
    Jul 12, 2020 at 23:03

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