# Lowest energy product state of a local hamiltonian

Lets consider one-dimensional spin chain with periodic boundary condition and $N$ sites. We are given a translationally-invariant local hamiltonian $H$ which is defined as $H=\sum_{i=1}^N h_{i,i+1}$, where each $h_{i,i+1}$ acts non-trivially only on sites $i,i+1$. More specifically $h_{i,i+1} = I_1\otimes I_2 \otimes\ldots I_{i-1}\otimes G\otimes I_{i+2}\ldots I_N$. The interaction $G$ is same for all $h_{i,i+1}$ which justifies our assumption that $H$ is translationally invariant.

A ground state of $H$ is one of its lowest energy eigenstates. It is easy to see that there is a ground state of $H$ which is translationally invariant. Argument goes as follows.

Let $\mathcal{T}$ be the operator that translates each spin by one lattice site. More specifically, given a vector $\mid i_1\rangle\mid i_2\rangle\ldots \mid i_N\rangle$, the action of $\mathcal{T}$ is

$\mathcal{T}\mid i_1\rangle\mid i_2\rangle\ldots \mid i_N\rangle = \mid i_N\rangle\mid i_1\rangle\ldots \mid i_{N-1}\rangle$.

Clearly $\mathcal{T} H\mathcal{T}^{\dagger} = H$. Now let $\mid \Psi\rangle$ be any ground state of $H$ with energy $e_0$. Consider the translationally invariant state (which is not a unit vector, but can be normalized without changing the argument)

$\mathcal{T}\mid\Psi\rangle + \mathcal{T}^2\mid\Psi\rangle+\ldots \mathcal{T}^N\mid\Psi\rangle$.

Act with $H$ on left hand side and follow the equation

$H\mathcal{T}\mid\Psi\rangle + H\mathcal{T}^2\mid\Psi\rangle+\ldots H\mathcal{T}^N\mid\Psi\rangle = \mathcal{T}H\mid\Psi\rangle + \mathcal{T}^2H\mid\Psi\rangle+\ldots \mathcal{T}^NH\mid\Psi\rangle = e_0(\mathcal{T}\mid\Psi\rangle + \mathcal{T}^2\mid\Psi\rangle+\ldots \mathcal{T}^N\mid\Psi\rangle)$.

Thus it is a ground state which is translationally invariant.

QUESTION: My question is about lowest energy product state of $H$. Product states have often been used to find upper bounds on ground energy of $H$. They are efficient to work with.

A product state is a state of the form $\mid \phi_1\rangle\mid \phi_2\rangle\ldots \mid \phi_N\rangle$. Suppose I want to find the product state that minimizes the energy

$\langle \phi_1\mid\langle \phi_2\mid\ldots \langle\phi_N\mid H\mid \phi_1\rangle\mid \phi_2\rangle\ldots \mid \phi_N\rangle$.

Note that $H$ may not have any product state as its eigenstate . Still, is it true that this state must be translationally invariant as well? Above argument about translationally invariant ground state clearly breaks down for product states.

If the answer is no, then I am curious where does the breaking of symmetry really take place. For example, lets put a further restriction on the two-site interaction $G$. Lets assume that $G$ remains unchanged even if we swap the two sites. More specifically, given swap operator $\mathcal{S}$ defined as $\mathcal{S} \mid i \rangle \mid j\rangle = \mid j \rangle\mid i \rangle$, we have $\mathcal{S} G \mathcal{S}^{\dagger} = G$. This system is now more symmetric: it is translationally invariant and also left-right symmetric. Does $H$ now have a translationally invariant product state which minimizes the energy of H over all product states?

Consider the Ising antiferromagnet $$H=\sum Z_i Z_{i+1}$$ on a ring. It even has product ground states $|0101\cdots\rangle$ and $|1010\cdots\rangle$ (and only those product ground states). Thus, the lowest energy product states are not translational invariant.

I don't think this is true. Let us consider a 2 site hamiltonian:

$H= - \mu_1^z \cdot \mu_2^z + \epsilon( \mu_1^x + \mu_2^x )$

So the ground state manifold is nondegenerate and it spanned by vectors (-,+) and (+,-) for $\epsilon=0$ for $\epsilon>0$ but small, it is still nondegenerate and close a vector $\frac{1}{\sqrt{2}}((+,-) + (-,+))$ . For exapmple for $\epsilon=0.2$ the ground state energy is: -1.07703. The GSvector in ++,-+,+-,-- basis is: $(0.133719, -0.694348, -0.694348, 0.133719)$

Now if we apply the product state approach, it fails miserably with translational invariance assumed as the true ground state is a highly entangled state and it cannot be approximated using such a state (translational invariance of a product state implies zero entanglement entropy). Indeed the energy is 0.4 fora vector: $(0.61685, 0.486154, 0.486154, 0.38315)$

If we drop the translational invariance, then it starts to work better: the minimum is 2-fold degenerate it corresponds to a vector of $(0.1, -0.989898, -0.0101021, 0.1)$ (and a second one symmetric) and the energy: -1.04.

• Hi. Thanks for your reply. I do not quite understand the example though. In your case, consider the translationally invariant product state: (a^2,ab,ab,b^2). If we compute the energy of this product state and optimize over all a,b (with $\varepsilon = 0.2$), we get lowest energy -1.04. So even without dropping translational invariance, we get lowest value -1.04. So your example is not really a counterexample: It does not exhibit a non-translationally invariant product state that has lower energy than all translationally invariant product states. – anurag anshu Apr 12 '16 at 5:00