You have to understand, most of these people are not ready to be unplugged. And many of them are so inured, so hopelessly dependent on the system, that they will fight to protect it.
You have to understand, most of these people are not ready to be unplugged. And many of them are so inured, so hopelessly dependent on the system, that they will fight to protect it.
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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
{$2 (\langle A^\dagger H A\rangle - \langle H \rangle \langle A^\dagger A \rangle) \langle H \rangle = 2E\Delta N^2$} {$2 \langle A^\dagger H \rangle \langle H A \rangle = 2 |k|^2 N^2$}, 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 $k \neq 0$.
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. |