# Ground state energy of the closure of an essentially self-adjoint operator

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?

• I think this question is more likely to be answered quickly on math.se – user2723984 Apr 27 at 7:46
• 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. – Valter Moretti Apr 27 at 9:02

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 $$\min \sigma(A) = \inf \sigma(A) = \inf_{\psi \in D(A), ||\psi||=1} \langle \psi, A \psi\rangle\:.$$

• 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? – uzizi_1 Apr 27 at 11:06
• I added a proof.... – Valter Moretti Apr 27 at 11:28