Noncommutativity and Quantum Fluctuations I have been reading a book called Quantum Phase Transitions in Transverse Field Spin Models:
From Statistical Physics to Quantum Information. Part I on this book gives an introduction to Quantum Phase Transitions (QPTs). Chapter 1 in Part I states that:


"The presence of non-commuting terms in the Hamiltonian produces a superposition
between various states. In QPTs, it is the quantum fluctuations (due to non-commuting terms in the Hamiltonian)
which are responsible for taking the system from one phase to another." (Section 1.3: page 14)


From this I understood that the claim is : non-commuting terms in the Hamiltonian leads to quantum fluctuations. To understand why non-commuting terms lead to phase transitions, I read the part of the introductory section on Transverse Ising Models, which starts from page 17. This model describes a d dimensional hypercube lattice containing N spins, either up or down. The Hamiltonian is given by equation 1.8 in the book:


$\hat{H} = -\sum_{\langle ij\rangle} J_{ij} \sigma_i^x \sigma_j^x - h\sum_i \sigma^z_i$


Here we see that that $\sigma^a_i$ is the $a^{th}$ Pauli Spin matrix at the lattice site $i$. The text also claims:


" In this
section, we assume that Jij = Jx ≥ 0, i.e., a ferromagnetic (FM) interaction between the
nearest-neighbor spins. which simplifies the Hamiltonian to the form:
$\hat{H} = -\sum_{\langle ij\rangle} J \sigma_i^z \sigma_j^z - h\sum_i \sigma^x_i$"


(Side Comment: I am not sure why the Hamiltonian in (Eq. 1.8) would transform this way; I do not see why the sum over nearest neighbors started with x Pauli matrices and ended with z Pauli matrices, but I went with the textbook). Now comes the following statement:


"The operator $\sigma_i^x$
acting on the eigenstate of $\sigma_i^z$ with eigenvalue +1 (up state)
changes it to the down state with eigenvalue −1. The non-commuting transverse field
term, therefore, introduces quantum fluctuations in the model causing a QPT" (pg 17)


I understand the first sentence completely. However, I am not sure what the second statement means. Is the noncommutative transverse field term the sum $h\sum_i \sigma^x_i$? If it is, is it noncommutative because the commutator of the first and second terms of the Hamiltonian is nonzero via the Pauli spin matrix commutator rule? Also, why does this non-commutativity lead to quantum fluctuations? For the last part, please kindly provide a conceptual and mathematical explanation.
 A: I suspect the authors are referring to the general principle that "Quantum physics reduces to classical physics when everything commutes". The usage of "noncommutative" here is fairly loose, after all it makes no sense to say "$\sigma^z$ doesn't commute" without specifying what it fails to commute with.
I believe the mental picture you're meant to have in mind is that of a perturbed "pure Ising" model, such that any "anomalous" or "non-classical" behaviour is attributable to the presence of a transverse field.
It is possible to correctly compute the partition function of the standard Ising model by evaluating the semiclassical expression $Z(\beta) = \sum_{\sigma_i = \pm 1} \exp( \beta J \sum_{\langle i j \rangle} \sigma_i \sigma_j )$.
This may be viewed either as a summation over all microstates of a discrete, classical model in which all spins point up or down, or as a trace over the $2^N$-dimensional multi-body Hilbert space. In this scenario, the eigenstates of the quantum Hamiltonian overlap with a classical idea of "spin up/spin down" precisely, so it can be said that "there are no quantum fluctuations", i.e. any fluctuations in the system are wholly attributable to classical thermodynamics. In this model, a single spin on a single site may be flipped while remaining in an eigenstate.
With a transverse field, the Hamiltonian is no longer diagonal in the standard $\sigma_z$ basis for the spin Hilbert space. As the discussion will show, this leads to a need to consider long-range quantum entanglement that does not admit such a clean classical analogue. Obviously there is still a basis for the Hamiltonian that renders it diagonal, but the key point is that this basis requires the cooperation of spins over multiple sites in a manner incompatible with classical mechanics.
"Quantum fluctuations" are only defined (albeit loosely) in the context of a particular choice of basis. For a toy example, consider the toy Hamiltonian
$$H = -\sigma_1^z \sigma_2^z + h(\sigma_1^x + \sigma_2^x)$$
Written with respect to the standard basis $\{|\uparrow \uparrow\rangle,|\downarrow \uparrow\rangle,|\uparrow \downarrow\rangle,|\downarrow \downarrow\rangle\}$, this has matrix representation
$$ [H] = \begin{bmatrix} -1 & h & h & 2h \\h & 1 & 2h &h \\ h&2h & 1 &h \\2h & h & h& -1\end{bmatrix}$$
If $h>0$, the lowest eigenvalue is $-1-2h$, corresponding to the state $\frac{1}{\sqrt{2}}\left[|\uparrow \uparrow \rangle + |\downarrow \downarrow\rangle\right]$. Viewed in the density matrix formalism, there's a big difference here between this quantum superposition and a classical ensemble of essentially independent spins. This superposition ground state has an oscillating phase that an experiment can measure by looking at spin-spin correlators (cf. larmor precession); in the $h=0$ case,  the oscillating phase is an irrelevant gauge field.
A: I'd like to complete catalogue_number's answer by elaborating on the meaning of

In QPTs, it is the quantum fluctuations (due to non-commuting terms in the Hamiltonian) which are responsible for taking the system from one phase to another.

Immediately before that, the authors state that

Each term of the Hamiltonian, when
dominant, corresponds to a specific ground state and determines a phase in the phase diagram. The presence of non-commuting terms in the Hamiltonian produces a superposition between various states.

Consider the Ising Hamiltonian
$$H = -J \sum_{<ij>}\sigma^z_i\sigma^z_j - h \sum_i \sigma^x_i$$
and its limiting case $h \to 0$
$$H_0 = -J \sum_{<ij>}\sigma^z_i\sigma^z_j.$$
Referring to the first sentence of the second quote, this can also be thought as the limiting case in which $h \neq 0$ but $J >> h$.
Denoting $\{ |\uparrow \rangle ,|\downarrow\rangle \}$ the eigenbasis of $\sigma^z$ the ground state of $H_0$ is doubly degenerate
$$\begin{align}
&|\uparrow \uparrow \dots \uparrow\rangle\\
&|\downarrow \downarrow \dots \downarrow\rangle
\end{align}$$
with energy $-NJ$, where $N$ is the number of sites on the chain and periodic boundary conditions are assumed.
Both states being eigenstates of $H_0$, when measuring the energy in any one of them the probability of getting the value $E_0 = -NJ$ is $P(-NJ) = 1$. In other words, there are no statistical fluctuations in the value of the energy.
If we turn on the magnetic field, the Hamiltonian of the system is now $H$ which is not diagonal in the $H_0$ basis (i.e. the $\sigma^z$ basis), and therefore will have a different ground state. Notice that this is a direct consequence of the non-commutativity of $H_0$ and $H_1 = H - H_0$: if the two terms commuted, $H$ and $H_0$ would share the same eigenbasis, just with different eigenvalues.
If we wait for the system to equilibrate (i.e. either we turn on $h$ very slowly or we wait until it is in the new ground state of $H$) and measure again $H_0$, we will have a probability distribution over its spectrum for the value that we will measure (i.e. no value will turn out with probability 1) and hence there will be non-zero statistical fluctuations.
At zero temperature there are no thermal fluctuations, and it is the study of how the quantum fluctuations just described change with $h$ that characterize quantum phase transitions.
