HOWTO /
## SMAGiven a translationally invariant iMPS, we can construct an excited state as a sum of local perturbations. This has the general form {$$\vert A \rangle = \sum_i A_i \vert 0 \rangle$$} where {$A_i$} is some operator that has finite support in the vicinity of site {$i$}. We can always choose {$A$} such that the state is orthogonal to {$\vert 0 \rangle$} and normalized such that {$\langle A \rangle = 0$} and {$\langle A^\dagger A \rangle = N$}. Now consider the energy of this state, {$\langle A^\dagger H A\rangle$}. Decomposing this operator, {$$\langle A^\dagger H A\rangle = \langle A^\dagger A \rangle \langle H \rangle + \Delta \langle A^\dagger A \rangle = EN^2 + \Delta N$$} Divide through by {$\langle A^\dagger A\rangle$} to normalize the expectation value, and we have {$E_A = EN + \Delta$}, hence this wavefunction represents an excitation with energy {$\Delta$} above (or below, if {$\Delta$} is negative) the energy of the original state. What is the variance of this state? To obtain this, we consider the expectation value of {$H^2$} on this state, {$\langle A^\dagger H^2 A\rangle$}. This decomposes as - {$N^4$} part: this would be {$\langle A^\dagger \rangle \langle H \rangle^2 \langle A \rangle$} but this is zero.
- {$N^3$} part: the only non-zero 3rd order part is {$\langle A^\dagger A\rangle \langle H \rangle^2 = E^2 N^3$}
- {$N^2$} part: There are several contributions. {$\langle A^\dagger A\rangle (\langle H^2 \rangle - \langle H \rangle^2) = \sigma^2 N^2$} (where {$\sigma^2$} is the variance of {$\vert 0 \rangle$}).
{$2 (\langle A^\dagger H A\rangle - \langle H \rangle \langle A^\dagger A \rangle) \langle H \rangle = 2E\Delta N^2$} {$2 \langle A^\dagger H \rangle \langle H A \rangle = 2 |k|^2 N^2$}, where {$k = \langle H A\rangle / \langle A^\dagger A \rangle$}. This term appears with a prefactor 2 because there are two ways to group the {$H^2$} term. This term is interesting -- if the operator {$A$} is not a scalar then $\vert A \rangle$ is in a different symmetry sector to {$\vert 0 \rangle$}, and {$k=0$} identically. However for symmetry-broken states we will in general have $k \neq 0$. - {$N$} part: The remaining connected part. There will be a contribution {$\Delta^2 N$}, and a part that is left over, call that {$\sigma_A^2 N$}. This is the variance due to the excitation. But it is not a priori clear that {$\sigma_A^2$} is necessarily positive.
To verify whether {$\sigma_A^2$} is positive, we attempt to isolate this term as the expectation value of a positive operator. The variance of {$\vert A \rangle$} is is obtained from {$\langle A^\dagger (H - EN - \Delta)^2 A \rangle$}. The operator EN is simply the MPO that has a constant E per site. This is simply an energy shift -- we might as well assume that this is incorporated into {$H$} itself and set {$E=0$}. The {$\Delta$} term doesn't formally exist as an MPO representation, but it does exist as a multiplier on the operator {$A$}. Hence we can form the expression {$\langle ( A^\dagger H - \Delta A^\dagger ) ( H A - \Delta A) \rangle$}. Expanding, {$$\langle A^\dagger H^2 A \rangle - 2 \Delta \langle A^\dagger H A \rangle + \Delta^2 N$$} After normalizing by {$\langle A^\dagger A \rangle$}, this gives: {$\sigma^2 N + 2|k|^2 N + \sigma_A^2$}. The extensive part of the variance has increased by {$2|k|^2$}, which is positive-definite. What about the constant part? To determine whether this is positive, we can attempt to remove the extensive part. Keeping the convention that {$E=0$}, let {$P = (H - \Delta)A - sI$}, where {$s$} is some constant (we allow it to be complex), and by {$sI$} we mean the operator that is a constant shift {$s$} per site. Consider the positive operator {$$\langle P^\dagger P \rangle = \langle A^\dagger (H - \Delta)^2 A \rangle - s \langle (H-\Delta)A \rangle - s^* \langle A^\dagger (H-\Delta)\rangle)$$} {$$= \sigma^2 N^2 + 2|k|^2 N^2 - (s^*k + k^*s)N^2 + |s|^2 N^2 + \sigma_A^2 N$$} It seems that this never cancels, there is no solution for {$s$} such that the {$N^2$} terms cancel, even in the case {$k=0$}. So it seems that there is no constraint on the sign of {$\sigma_A^2$}. ## Part 2, variational energyConsider the matrix elements of {$H$}, for our two candidate states {$\vert 0 \rangle$} and {$\vert A \rangle$}. Firstly, the norm matrix, {$\langle 0 \vert 0 \rangle = 1$} {$\langle A^\dagger \vert 0 \rangle = \langle 0 \vert A \rangle = 0$} {$\langle A^\dagger \vert A \rangle = N$} This is diagonal, and has inverse diag(1,1/N). {$\langle 0 \vert H \vert 0 \rangle = EN$} {$\langle 0 \vert H \vert A \rangle = kN$} {$\langle A^\dagger \vert H \vert 0 \rangle = k^\dagger N$} {$\langle A^\dagger \vert H \vert A \rangle = EN^2 + \Delta N$} We now solve the generalized eigenvalue problem. {$(EN - \lambda) (EN^2 + \Delta N - \lambda N) - |k|^2 N^2 = 0$} {$ E^2 N^3 + E \Delta N^2 - E \lambda N^2 - \lambda EN^2 - \lambda \Delta N + \lambda^2 N - |k|^2N^2 = 0$} Divide through by {$N$}. {$ E^2 N^2 + E \Delta N - E \lambda N - \lambda EN - \lambda \Delta + \lambda^2 - |k|^2N = 0$} {$ E^2 N^2 + E \Delta N - \lambda (2EN + \Delta) + \lambda^2 - |k|^2 N = 0$} Consider the large {$N$} limit, by dropping the non-leading terms. Then we have {$ E^2 N^2 - 2 \lambda E N + \lambda^2 = 0$} This has the solution {$\lambda = EN$}. Let {$\lambda = EN + \alpha$} and substitute. {$ 2 E^2 N^2 + E \Delta N - (EN + \alpha) (2EN + \Delta) + 2 EN \alpha + \alpha^2 - |k|^2 N = 0$} {$ -\alpha \Delta + \alpha^2 - |k|^2 N = 0$} {$ \alpha = \Delta/2 \pm \frac{\sqrt{\Delta^2 + 4 |k|^2 N}}{2}$} This is essentially the same result obtained for the thermodynamic limit of PT-DMRG. If {$k=0$}, we obtain the energies {$EN$} and {$EN+\Delta$}, as we expect. If {$k \neq 0$} however, there is a {$\sqrt N$} correction -- in the large {$N$} limit the energies of the two states are {$EN \pm |k| \sqrt N + O(1)$}. Thus the energy per site is {$E \pm |k| / \sqrt N$}. This explains the success of PT-DMRG for finite systems: if {$|k|$} is O(1), then even for {$N \sim 100$} or so the correction to the energy is quite big. |

Page last modified on August 03, 2017, at 10:18 AM