# Ground state energy

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!

• Ground state energy is the lowest eigenvalue of the Hamiltonian - not just any operator. Apr 26 at 13:16
• 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. Apr 26 at 13:52

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 $$E_i$$.