# How to see that the trivial insulator is trivial?

I have been trying to better understand gapped phases of matter — which may be "topological" or "trivial" — and I have run into the problem that I don't really understand the trivial case very well.

When I say trivial, what I mean is something like the following:

A trivial state can be transformed to a product state by a finite sequence of local unitary transformations even if the system size is infinite.

or perhaps

A state if trivial if it is the ground state of a hamiltonian that can be adiabatically transformed to a hamiltonian having a product state as its ground state without closing the gap.

I'll summarize this by saying a trivial state can be adiabatically connected to a product state.

Now, when I think of a trivial band insulator, I imagine something like the following. (I neglect interactions, because I am having enough trouble thinking about the non-interacting case, but ultimately I would like to understand this with interactions turned on.) We have some local hamiltonian describing electrons in a periodic potential, and we diagonalize the hamiltonian to obtain the band structure, something like this:

(This image is taken from this question which in turn took it from this paper.) To get the ground state of the (non-interacting) many-body hamiltonian, we simply "fill up" the bands with particles. If at the end of this all the bands are either completely filled or empty, we have an insulator. So the ground state is like a Slater determinant taken over all of the states in occupied bands. My question is: how can I see that the ground state obtained in this way is adiabatically connected to a product state?

I am most interested in an answer to this question that describes (perhaps with some hand-waving) how to see that there is a finite-depth local unitary transformation from the "Slater determinant" state to a product state (If using bosons instead of fermions makes things easier, that's fine. It could also be a transformation applied to the hamiltonian rather than its ground state.)

I'm less interested in arguments that go like "We can calculate such-and-such topological invariant of the filled bands and see that it is the same for a trivial insulator and for a product state," but if you can convince me that this is the best or only way to make the argument, I'd accept that as an answer too.

I'd also like to understand why we can do this adiabatic transformation for trivial state produced by filling bands in the figure on the left above (a trivial insulator), but not for the state produced by filling up the bands on the right (a topological insulator). Again, I'd like to avoid arguments about topological invariants of bands if at all possible.

Here's one way to think about it. For the moment, let's restrict our attention to 1D, and for concreteness consider the Su-Schrieffer-Heeger model of polyacetylene which describes a bipartite chain with alternating $$A$$ and $$B$$ sites, with the following Hamiltonian:

$$H = \sum_{n} \bigg[v\ |n,A\rangle\langle n,B| + w |n,B\rangle\langle n+1,A| + h.c.\bigg]$$ where $$v,w\in\mathbb R$$ are the intra-cell and inter-cell hopping amplitudes, respectively. Performing a spatial Fourier transform yields $$H = \int_{BZ}\mathrm dk\ |k\rangle\langle k| \otimes \underbrace{\pmatrix{0 & v+we^{-ik}\\v+we^{ik} & 0 }}_{\equiv H_k}$$

where $$|k\rangle = \frac{1}{\sqrt{2\pi}}\sum_n e^{-ikn}|n\rangle$$, $$BZ:=[-\pi,\pi]$$ (with the endpoints identified), and $$H_k$$ acts on the $$A/B$$ degree of freedom. It is convenient to write $$H_k = \vec d(k) \cdot \vec \sigma$$, where $$\vec \sigma=(\sigma_1,\sigma_2,\sigma_3)$$ are the Pauli matrices. In this instance, we would have that $$\vec d(k) = \pmatrix{v+w\cos(k)\\ w\sin(k)\\0}$$ The eigenvalues and eigenvectors of $$H_k$$ are $$E_{\pm}(k) =\pm|\vec d(k)| =\pm\sqrt{v^2+w^2+2vw\cos(k)}, \qquad u^{\pm}_k = \frac{1}{\sqrt{2}} \pmatrix{\pm 1\\ e^{i\varphi(k)}}$$ where $$e^{i\varphi(k)} = (v+we^{-ik})/|v+we^{-ik}|$$. Observe that the system is gapped as long as $$d(k) \neq \vec 0$$ for all $$k$$, which in this case corresponds to $$v\neq w$$.

The principal question at play here is the following:

Given two Hamiltonians $$H_1$$ and $$H_2$$ with respective ground states $$\psi_1$$ and $$\psi_2$$, can we adiabatically deform $$H_1$$ into $$H_2$$ (and therefore $$\psi_1$$ into $$\psi_2$$, up to a phase) without closing the gap?

To visualize this more clearly, consider the image that $$\vec d$$ traces out in $$\mathbb R^3$$ as $$k$$ varies over $$BZ$$. In 1D, this would correspond to a closed loop. Maintaining a gap means that this loop must avoid the origin, so our question essentially becomes the following: Given two closed loops in $$\mathbb R^3-\{\vec 0\}$$, when is it possible to continuously deform one into the other?

From a mathematical standpoint, we are trying to find the fundamental group $$\pi_1(\mathbb R^3-\{\vec 0\})$$, whose elements are equivalence of classes of closed loops (where loops are understood to be equivalent if they can be continuously deformed into one another). It is not difficult to see that $$\mathbb R^3-\{\vec 0\}$$ is simply-connected - meaning that $$\pi_1(\mathbb R^3-\{0\})=\{0\}$$, and every loop can be deformed into every other loop. From this standpoint, every 1D Hamiltonian is topologically equivalent to every other 1D Hamiltonian.

But this is where symmetry comes into play. Observe that our toy model exhibits chiral symmetry, meaning that there exists a Hermitian and unitary operator $$\Gamma$$ such that $$\Gamma H_k \Gamma = -H_k$$. In our case, $$\Gamma= \sigma_3$$. However,

$$\sigma_3 \big(\vec d(k)\cdot \vec \sigma\big)\sigma_3 = \vec d'(k) \cdot \vec \sigma, \qquad \vec d'(k) = \pmatrix{-d_1(k)\\ -d_2(k) \\ d_3(k)}$$ Therefore, requiring chiral symmetry implies that $$d_3(k)=0$$, which constrains our loop to the $$d_3=0$$ plane. With this in mind, we may ask a variation of our original question:

Given two Hamiltonians $$H_1$$ and $$H_2$$ with respective ground states $$\psi_1$$ and $$\psi_2$$ which possess chiral symmetry, can we adiabatically deform $$H_1$$ into $$H_2$$ (and therefore $$\psi_1$$ into $$\psi_2$$, up to a phase) without closing the gap and while maintaining chiral symmetry?

The answer to this question is an emphatic no. In our intuitive picture, we are now asking about closed loops in $$\mathbb R^2-\{\vec 0\}$$, and this space is not simply connected. Instead, we may categorize our equivalence classes of closed loops by the (integer!) number of times they encircle the origin (the so-called winding number) which is reflected in the fact that $$\pi_1(\mathbb R^2-\{\vec 0\})=\mathbb Z$$.

In our toy model, observe that the curve in question is a circle of radius $$w$$ centered at $$x=v$$. As a result, the winding number is $$0$$ if $$|v|>|w|$$ and $$1$$ if $$|v|<|w|$$.

Now to address your question about the topologically trivial phase. First, we write our Hilbert space as a direct product of local Hilbert spaces, $$\mathscr H = \bigotimes_{n\in \mathbb Z} \mathscr H^{loc}_n \qquad \mathscr H^{loc}_n \simeq \mathbb C^2 \text{ where } \pmatrix{\alpha\\\beta}_n \sim \alpha|n,A\rangle + \beta |n,B\rangle$$

Given this decomposition of $$\mathscr H$$, we may now ask whether the ground state of our system may be written as a product state. Because we're interested in spatial localization, it is convenient to consider the Wannier states

$$\phi_n = \frac{1}{\sqrt{2\pi}}\int \mathrm dk \ e^{ink} |k\rangle\otimes u_k^-$$

One can show that these states form an orthonormal basis for the valence band; as a result, the ground state may be written

$$\Psi = \prod_{n\in \mathbb Z} \phi_n$$

We can evaluate these states directly:

$$\phi_n = \frac{1}{\sqrt{2\pi}} \int \mathrm dk\left[ e^{ikn} |k\rangle\otimes\pmatrix{-1\\0} + e^{ikn + \varphi(k)}|k\rangle\otimes \pmatrix{0\\1} \right]$$ $$= |n\rangle\otimes \pmatrix{-1\\0} + \sum_{m} f(n-m) |m\rangle \otimes \pmatrix{0\\1}$$

$$f(n-m) \equiv \frac{1}{\sqrt{2\pi}} \int \mathrm dk \ e^{i[(n-m) k + \varphi(k)]}$$

Observe that if $$w=0$$ then $$\varphi(k)=0$$, and so $$f(n-m)=\delta_{n,m}$$. This implies that $$\phi_n = \frac{1}{\sqrt{2}}|n\otimes\pmatrix{-1\\1}$$, the Wannier states are exactly localized to our lattice sites, and the ground state wavefunction $$\Psi$$ is a product state.

On the other hand, if $$v=0$$ then $$\varphi(k)=-ik$$, and so $$f(n-m)=\delta_{n,m+1}$$. This implies that $$\phi_n = \frac{1}{\sqrt{2}} \left[|n\rangle\otimes\pmatrix{-1\\0} + |n-1\rangle\otimes \pmatrix{0\\1}\right]$$ which means that the Wannier states are not localized to the lattice sites; there is (short-range) entanglement between neighboring sites, and so the ground state $$\Psi$$ is not a simple product state.

Here is what $$f(\Delta)$$ looks like in general:

The function $$f(\Delta)$$ suffers a discontinuity when $$v=w$$, because at that point the integrand develops a singularity. In contrast, if $$v>w$$ we can always smoothly deform $$(v,w)\rightarrow(1,0)$$ without closing the gap, and the ground state wavefunction can be adibatically connected to a the trivial product state mentioned above. This characterizes the trivial phase. On the other hand, if $$v we cannot reach this product state without closing the gap, and so this characterizes the topological phase.

It should be mentioned that this is true only as long as we impose chiral symmetry. If we add a term to the Hamiltonian such that $$d_3(k)=u$$ for some parameter $$u$$, its easy to see that by making $$u>0$$ we can switch $$v$$ and $$w$$ without closing the gap. As a result, we can always continuously deform the ground state wavefunction to a product state if we don't impose any symmetry requirements.

Summary:

Two Hamiltonians are considered to be topologically equivalent if they can be continuously deformed into one another without closing the gap. In the absence of additional symmetry requirements, all 1D (and, as it turns out, 3D) Hamiltonians are topologically trivial. With the imposition of different symmetries, we obtain a zoo of different topological invariants in all dimensions which classify different topological phases as long as the relevant symmetries remain unbroken - see Symmetry-Protected Topological Order for more. On the other hand, true (long-range) topological order can arise in 2D or when interactions between particles are considered.

• Thanks for taking the time to write this excellent answer. I do think that you've given a clear and correct answer to the principal question that you stated. Unfortunately, I realized after receiving the two answers that I was not explicit enough in my question about what I was after. Because I'm hoping for a pretty specific answer (and not because of any deficiency in your answer) I'm left a little unsatisfied. In spite of this, you've given me a lot to think about, and I hope to improve my original question in light of your post.
– d_b
Aug 4 at 22:47
• @d_b No worries - if you do clarify your question, I'll try to take another crack at it (or perhaps answer it in the context of the answer I have already written). Aug 4 at 23:48
• @d_b I thought about it a bit and revised my answer to focus more on the product state issue. Let me know if that is closer to your intent. 2 days ago
• Thank you! I think this is pretty much what I was looking for. Actually it looks simpler to write down an explicit product state wavefunction for the ground state than I was imagining. One thing is still bugging me though. Why don't we need to (anti)symmetrize the product of $\phi_n$ that you have written down for $\Psi$?
– d_b
yesterday

I can try to give a very hand-waving answer. Imagine you have a topological insulator, you found eigenstates of your hamiltonian as a function of coordinate r in real space and as a function of a k vector in a reciprocal space $$\psi_n(r, k)$$. Topological nontriviality will reveal itself if you will try to plot $$\psi$$ as a function of $$k$$, in the case of 2D periodicity it will look like vortexes or sinks that you can not adiabatically transform into the flat field which is a trivial case. Topological invariants just count the difference between the number of vortexes and sinks.