I am trying to get familiar with the ground state energy of an operator. In my lecture we defined the ground state energy of a self-adjoint operator $H$ that is bounded from below as $$ E_0= \inf_{\psi\in D(H)\setminus\{0\}} \frac{\langle \psi, H\psi\rangle}{||\psi||^2}= \inf \sigma(H)$$

What I quite dont understand is, why second equality. Why is this the same as the spectrum of the operator? Does it have to do with the spectral theorem? I tried to show the equality in the following way:

$``\geq"$ \begin{align} \langle \psi, H\psi\rangle&=\int_{\sigma(H)}\lambda ~\text{d}\mu_{\psi,\psi}\\ &\geq \inf_{\lambda\in \sigma(H)} \lambda \int_{\sigma(H)} \text{d}\mu_{\psi,\psi}\\ &= \left( \inf \sigma(H)\right)||\psi||^2\\ \Rightarrow E_0 \geq \inf \sigma(H) \end{align}

$``\leq"$ Let $\varepsilon>0$. Find $\psi\in D(H)$ with $||\psi||=1$, such that \begin{align} \langle \psi, H\psi\rangle&\leq \inf\sigma(H)+\varepsilon \end{align} Since $\varepsilon$ was chosen arbitrarily, we find that $E_0\leq \inf \sigma(H)$.

Does this make any sense?

I also have to show that for an essentially self-adjoint operator $(H,D(H))$ 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... I would be really grateful for any help!

  • $\begingroup$ Ground state energy is the lowest eigenvalue of the Hamiltonian - not just any operator. $\endgroup$
    – Roger V.
    Apr 26, 2021 at 13:16
  • $\begingroup$ The "ground state energy" is by definition the lowest spectral value of the self-adjoint operator called Hamiltonian, denoted by $H$. Therefore your initial formula should be: $$ E_0 := \inf \sigma (H) = \inf_{\psi\in D(H)\setminus\{0\}} \frac{\langle \psi, H\psi\rangle}{\langle \psi,\psi\rangle}$$ So the equality you question is why is the third member equal to the second. $\endgroup$
    – DanielC
    Apr 26, 2021 at 13:52

1 Answer 1


Not sure if this is what you're looking for or if you're looking for a more mathematically sophisticated answer. The physicist `proof' would just be to say that because $H$ is self adjoint, it breaks up the state space into a complete basis orthonormal of eigenfunctions $\psi_{E_i}$ satifying $H \psi_{E_i} = E_i \psi_{E_i}$. Therefore, if you have a general state $$ \psi = \sum_i c_i \psi_{E_i} $$ then, assuming $||\psi||^2 = 1$, this gives $$ \langle \psi, H \psi \rangle = \sum_i |c_i|^2 E_i. $$ Therefore choosing the minimum state $\psi$ amounts to choosing the coefficients $c_i$ such that you only pick up the smallest energy eigenvalue $E_i$.


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