Recently I am reading some material regarding the Lieb-Schultz-Mattis(LSM) theorem and I do not understand the proof of LSM theorem. The proof is from Ian Affleck's paper. Let's start with the spin-1/2 Heisenberg Hamiltonian: \begin{equation} H = \sum_{i} \vec{S}_{i} \cdot \vec{S}_{i+1} \end{equation} We wish to prove that there exists a low energy excitation of $\mathcal{O}(1/L)$, where $L$ is the size of periodic chain which is even. We may assume that the ground state $| \psi_{0} \rangle$ is unique. To prove there is a low energy excitation, we need to construct an excited state $| \psi_{1} \rangle $ such that $ \langle \psi_{1} | H - E_{0} | \psi_{1} \rangle = \mathcal{O}(1/L) $. We construct this $| \psi_{1} \rangle $ by applying a unitary transformation to the ground state $| \psi_{0} \rangle$. \begin{equation} | \psi_{1} \rangle = U |\psi_{0} \rangle ~~,~~ U = \exp \Big(i \frac{2 \pi}{L} \sum_{n} nS^{z}_{n} \Big) \end{equation}

My questions are why we need to apply a unitary transformation to construct the excited state $| \psi_{1} \rangle $ and why the unitary transformation $U$ has this form( $\sum_{n} n S^{z}_{n}$ instead of $ \sum_{n}S^{z}_{n}$)? I appreciate any comment.


1 Answer 1


We want to calculate the expectation value $\langle H - E\rangle$ on the excited state we are constructing. This is going to be much easier if the state we construct is correctly normalized. We could explicitly normalize the state but calculating the normalization constant for many-body states can be painful. Applying a unitary operator to an already normalized state is guaranteed to give us another normalized state. It would be enough for the operator to give us a normalized state when applied to the ground state (because that is what we are going to do with it), but the easiest way to achieve this is often to just ensure that the operator always preserves normalization.

For the factor of $n$, well if we think about the ferromagnetic case we know that the low energy excitations are spin waves. The antiferromagnetic case is more difficult to picture due to the structure of ground state, but we would again expect some sort of wavelike pattern on top of the ground state. A wave like structure will have the form $e^{i \omega n}$. To construct our unitary operator we promote $\omega$ to an operator (with the smallest frequency allowed by the size of the system, as we expect this will be the lowest energy). The factor of $n$ plays the same roll of increasing the phase of the wave as you move through the system.


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