Definition of ground state?

Question

In quantum mechanics, a ground state is an eigenstate of the hamiltonian with the minimal eigenvalue and its existence is guaranteed by appropriate theorems.

At least that's how it's defined in undergraduate courses.

In the formalism of operator algebras, the definition of ground state is equally clear but so different that I cannot relate it to the classic definition.

Given a suitable operator algebra $U$ and a continuous group of automorphisms $\tau$ defineing a dynamical system on $U$, and calling $\tau$ the infinitesimal generator of the system (for a quantum system this would be the Hamiltonian) a ground state $\omega$ is one that for every operator $A$ of the algebra with adjoint $A*$ satisfies

$$- i \omega(A^* \delta(A)) \geq 0 \ \ \ \ \ (1) $$

Eq. (1) follows from the KMS condition, I will paste a screenshot from one source here:

I wonder if anybody has an intuition on how to bridge this definition to the classic one.

Answer 1

In finite dimensions the definition of a ground state boils down to the following: the closed system (represented by $\mathbb C^n$) admits internal dynamics $\tau$, i.e. $\tau(\rho)=e^{itH}\rho e^{-itH}$ for some $H\in\mathbb C^{n\times n}$ Hermitian. Taking this as a map from $\mathbb R\to\mathcal L(\mathbb C^{n\times n})$, $t\mapsto e^{itH}(\cdot)e^{-itH}$ we obtain a one-parameter group with generator $$ \boxed{\delta=i[H,\cdot]=iH(\cdot)-i(\cdot)H} $$ This is due to the identity $$ e^{t\delta}(\rho)=e^{it[H,\,\cdot\,]}\rho=(e^{itH}(\cdot)e^{-itH})\rho=e^{itH}\rho e^{-itH}=\tau(\rho)\,. $$ Moreover, in finite dimensions there is a one-to-one correspondence between states $\omega:\mathbb C^{n\times n}\to\mathbb C$ (in the $C^*$-algebraic sence) and positive semi-definite matrices of trace one. With this in mind, a state $\rho\in\mathbb C^{n\times n}$ is a ground state (with respect to $\tau$) iff $$ -i\operatorname{tr}(\rho A^* i[H,A])\geq 0\qquad\Longleftrightarrow\qquad \operatorname{tr}(\rho A^*HA)\geq\operatorname{tr}(\rho A^*AH) \tag{1} $$ for all $A\in\mathbb C^{n\times n}$.With this we can show that (in the finite-dimensional setting) a ground state in the sense of (1) is indeed what one would expect:

Proposition. Given any $H\in\mathbb C^{n\times n}$ Hermitian the following statements hold.

1. $\rho$ is a ground state in the sense of (1) if and only if $\operatorname{tr}(H\rho)=\min\sigma(H)$ with $\sigma(\cdot)$ the spectrum
2. If $H$ has unique zero-point energy (with corresponding ground state $|\psi\rangle$), then $|\psi\rangle\langle\psi|$ is the unique ground state.

Proof. Our main tool will be the spectral decomposition $H=\sum_{j=1}^n E_j|j\rangle\langle j|$ where $E_1\leq E_2\leq\ldots\leq E_n$. With this (1) becomes $$ \operatorname{tr}(\rho A^*HA-\rho A^*AH)=\sum_{j=1}^nE_j\big(\langle A^*j|\rho|A^*j\rangle-\langle j|\rho A^*A|j\rangle\big)\geq 0\tag{2} $$ for all $A\in\mathbb C^{n\times n}$. Moreover, assume that $k\in\{1,\ldots,n\}$ is the smallest number such that $\min\sigma(H)=E_k<E_{k+1}$ (i.e. the zero-point energy of $H$ is $k$-fold degenerate).

1. First assume that $\rho$ satisfies $\operatorname{tr}(H\rho)=\min\sigma(H)$; we have to show that $\rho$ satisfies (1). One readily verifies $\langle j|\rho|j\rangle=0$ for all $j>k$. Because $\rho$ is positive semi-definite this implies $\langle j|\rho|j'\rangle\neq 0$ only if $j,j'\in\{1,\ldots,k\}$. We compute \begin{align*} \operatorname{tr}(\rho A^*AH)=\sum_{j=1}^n E_j\langle j|\rho A^*A|j\rangle&=\sum_{j=1}^{\bf k} E_j\langle j|\rho A^*A|j\rangle\\ &=\min\sigma(H)\sum_{j=1}^k\langle j|\rho A^*A|j\rangle=\min\sigma(H)\operatorname{tr}(\rho A^*A) \end{align*} which implies \begin{align*} \operatorname{tr}(\rho A^*HA)-\operatorname{tr}(\rho A^*AH) &=\operatorname{tr}(\rho A^*HA)-\min\sigma(H)\operatorname{tr}(\rho A^*A)\\ &=\operatorname{tr}\big(A\rho A^*(\underbrace{H-\min\sigma(H)\cdot{\bf 1}}_{\geq 0})\big) \geq 0 \end{align*} because $\rho$ (and thus $A\rho A^*$ for all $A$) is positive semi-definite; hence (1) holds. Conversely, assuming (1) (resp. (2)) holds let us evaluate this for $A=|1\rangle\langle\beta|$ with $\beta\in\{k+1,\ldots,n\}$ arbitrary. This yields $ \langle\beta|\rho|\beta\rangle(\min\sigma(H)-E_\beta)\geq 0 $; but $\min\sigma(H)-E_\beta<0$ because of how we chose $\beta$ so---because $\langle\beta|\rho|\beta\geq 0$ ($\rho$ is a state)---we find that $\langle\beta|\rho|\beta\rangle$ has to be zero. Therefore \begin{align*} \operatorname{tr}(\rho H)=\sum_{j=1}^n E_j\langle j|\rho|j\rangle=\sum_{j=1}^{\bf k} E_j\langle j|\rho|j\rangle&=\min\sigma(H)\sum_{j=1}^k \langle j|\rho|j\rangle\\ &=\min\sigma(H)\underbrace{\sum_{j=1}^n \langle j|\rho|j\rangle}_{=\operatorname{tr}(\rho)=1}=\min\sigma(H)\,. \end{align*}
2. Now this is easy: by (1) $\rho$ is a ground state iff $\operatorname{tr}(\rho H)=\min\sigma(H)$. If we write $H=\min\sigma(H)|\psi\rangle\langle\psi|+\sum_{j=2}^n E_j|j\rangle\langle j|$ (with $E_j>\min\sigma(H)$; this is possible by assumption), then $\operatorname{tr}(\rho H)=\min\sigma(H)$ is equivalent to $\langle\psi|\rho|\psi\rangle=1$; but this holds if and only if to $\rho=|\psi\rangle\langle\psi|$: decomposing $\rho=\sum_j r_j|j'\rangle\langle j'|$ and using that $r_j,|\langle\psi|j'\rangle|\in[0,1]$ as well as Cauchy-Schwarz we find \begin{align*} 1=\langle\psi|\rho|\psi\rangle=\sum_j r_j|\langle j'|\psi\rangle|^2&\overset{\text{CS}}\leq\Big( \sum_j r_j^2 \Big)^{1/2}\Big( \sum_j |\langle j'|\psi\rangle|^4\Big)^{1/2}\\ &\leq\Big( \sum_j r_j \Big)^{1/2}\Big( \sum_j |\langle j'|\psi\rangle|^2\Big)^{1/2}=1\,. \end{align*} In particular $r_j,|\langle j'|\psi\rangle|\in\{0,1\}$ in order for equality to hold. From this we quickly arrive at $\rho=|\psi\rangle\langle\psi|$ as desired. $\quad\square$