In my lecture notes it says, that for an essentially self-adjoint operator $(H,D(H))$, that is bounded from below, and its self-adjoint closure $(\bar{H},D(\bar{H}))$ the ground state energy $E_0$ of $H$ agrees with $\bar{E_0}$ of $\bar{H}$.

I don't really know why this statement is true, maybe because I did not understand the concept of ground state energy yet... We defined the ground state enery $E_0$ as $$E_0= \inf_{\psi\in D(H)\setminus\{0\}} \frac{\langle \psi, H\psi\rangle}{||\psi||^2}.$$

But why does ist make sense and how would i see that $H$ and its closure have the same ground state energy?

  • $\begingroup$ I think this question is more likely to be answered quickly on math.se $\endgroup$ Apr 27, 2021 at 7:46
  • 1
    $\begingroup$ I think that there are several physically-minded facets of this issue which cannot be properly treated by pure mathematicians. So also here is a good place to ask this question. $\endgroup$ Apr 27, 2021 at 9:02

1 Answer 1


Since the domain of the closure $\overline{H}$ is larger than the domain of $H$ and the two operators coincide on $D(H)$, we have $$\overline{E}_0 \leq E_0\:.\tag{1}$$ From the definition of $\overline{E}_0$, there is a sequence of vectors $D(\overline{H}) \ni \psi_n$ such that $$\frac{\langle \psi_n, \overline{H} \psi_n\rangle}{||\psi_n||^2} \to \overline{E}_0\:.$$
On the other hand, for the definition of closure of an operator, for every $\psi_n\in D(\overline{H})$ there is a sequence $D(H) \ni \phi^{(n)}_m \to \psi_n$, for $m\to +\infty$, such that $H\phi^{(n)}_m \to \overline{H} \psi_n$. At this point it is easy to construct a sequence of vectors $\phi_m \in D(H)$ such that $$\frac{\langle \phi_m, H \phi_m\rangle}{||\phi_m||^2} \to \overline{E}_0\:,$$ so that $$\overline{E}_0 \geq E_0\:.\tag{2}$$ (1) and (2) implies $E_0 = \overline{E}_0$.

Regarding the definition of $\overline{E}_0$, from the spectral theorem it turns out that it is the minimum value of the spectrum of the selfadjoint operator $\overline{H}$. Not only the infimum, but exactly the minimum, as the spectrum is closed (see ADDENDUM below).

Assuming that $\overline{H}$ is the energy observable of a stationary quantum system, there are two possibilities then: $\overline{E}_0$ is an eigenvalue, i.e. there is (at least) a vector in $\psi_0 \in D(\overline{H})$ -- the/a ground state -- representing a stationary state of minimal energy, such that $\overline{H}\psi_0 = \overline{E}_0 \psi_0$ or there is not such a vector.

In the second case the physical relevance of $\overline{E}_0$ is disputable...

Regarding instead $E_0$, it has no evident physical meaning as $H$ is not assumed to be selfadjoint but only symmetric. I do not think that it is the infimum of the real part of the spectrum since there is no spectral decomposition (in infinite dimensional Hilbert spaces) for $H$ in terms of a PVM (there is in terms of a POVM, but the support of the latter does not coincide with the spectrum of the operator). However the found identity is quite useful because usually and differently form $\overline{H}$, the operator $H$ is a differential operator and there are chances to compute the infimum $E_0$ of the corresponding quadratic form.

ADDENDUM. Suppose that $A$ is selfadjoint and bounded below (so that the spectrum is bounded below), then $$\langle x,A x\rangle= \int_{\mathbb{R}} \lambda d \mu_{xx}(\lambda)= \int_{\sigma(A)} \lambda d \mu_{xx}(\lambda)$$ for every normalized $x\in D(A)$. Therefore $$\langle x,A x\rangle \geq \inf \sigma(A) \int 1 d \mu_{xx}(\lambda) = \inf \sigma(A) ||x||^2 = \inf \sigma(A) = \min \sigma(A) \quad (>-\infty)\:,\tag{A1}$$ where I used the fact that $\sigma(A)$ is closed. To conclude observe that, if $\lambda_0 = \min \sigma(A)$ is an eigenvalue with normalized eigenvector $\psi_0$, then $$\langle \psi_0, A \psi_0 \rangle = \lambda_0 = \min \sigma(A)\:,$$ so that $$\min \sigma(A) = \inf \sigma(A) = \inf_{\psi \in D(A), ||\psi||=1} \langle \psi, A \psi\rangle\:.$$ If $\lambda_0 = \min \sigma(A)$ is not an eigenvalue, then it must be part of the continuous spectrum $\sigma_c(A)$ (the residual spectrum is empty as $A$ being selfadjoint). In this case $P_{(\lambda_0-1/n, \lambda_0+1/n)} \neq 0$ where $P$ is the spectral measure (PVM) of $A$. Hence, take $\psi_n \in P_{(\lambda_0-1/n, \lambda_0+1/n)} ({\cal H}) \neq \{0\}$ with $||\psi_n|| =1$ (notice that it implies $\psi_n \in D(A)$) so that $$|\langle \psi_n,A \psi_n \rangle - \lambda_0| = \left|\int_{(\lambda_0-1/n, \lambda_0+1/n)} (\lambda - \lambda_0) d\mu_{\psi_n,\psi_n}(\lambda) \right|\leq \int_{(\lambda_0-1/n, \lambda_0+1/n)} |\lambda - \lambda_0| d\mu_{\psi_n,\psi_n}(\lambda)$$ $$ \leq \sup_{\lambda \in (\lambda_0-1/n, \lambda_0+1/n)} |\lambda - \lambda_0| ||\psi_n||^2 = \sup_{\lambda \in (\lambda_0-1/n, \lambda_0+1/n)} |\lambda - \lambda_0| \to 0$$ as $n\to +\infty$. This result together with (A1) yields the thesis again $$\min \sigma(A) = \inf \sigma(A) = \inf_{\psi \in D(A), ||\psi||=1} \langle \psi, A \psi\rangle\:.$$

  • $\begingroup$ When you say that from the spectral theorem it follows, that $E_0$ is the minimum value of the spectrum of the selfadjoint operator, do you mean we can see that from the identity $$\langle \psi, \bar{H}\psi \rangle = \int_{\sigma(\bar{H})}\lambda \text{d}\mu_{\psi,\psi}$$ ? Or how else would I see that? $\endgroup$
    – uzizi_1
    Apr 27, 2021 at 11:06
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    $\begingroup$ I added a proof.... $\endgroup$ Apr 27, 2021 at 11:28

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