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The Hellmann-Feynman Theorem says $$\tag{1} \frac{d E_\lambda}{d \lambda} ~=~ \bigg\langle \psi(\lambda) \bigg| \frac{d H_\lambda}{d \lambda} \bigg| \psi(\lambda) \bigg\rangle$$ where $H_\lambda$ is a Hamiltonian parametrized by $\lambda$. Here, $| \psi(\lambda) \rangle$ is an eigenstate of $H_\lambda$ with energy $E_\lambda$.

Naively, if $\frac{d E_\lambda}{d \lambda} = 0$ for all $\lambda$ and $| \psi(\lambda) \rangle$, I would suppose that $\frac{d H_\lambda}{d \lambda} = 0$ because $| \psi(\lambda) \rangle$ is a complete set of states. However, I can construct a Hamiltonian where this is not true: $$\tag{2} H_\lambda ~=~ \frac{1}{2} \begin{pmatrix} E_1+E_2 & (E_2-E_1) e^{-i \lambda} \\ (E_2-E_1) e^{i \lambda} & E_1+E_2 \\ \end{pmatrix}$$ The eigenvalues are $E_1$ and $E_2$, independent of $\lambda$, so while $\frac{d E_\lambda}{d \lambda} = 0$, it's clear that $\frac{d H_\lambda}{d \lambda} \neq 0$. What is the flaw in my logic?

For those who are interested, the motivation for this question is the Dirac Hamiltonian for graphene (and specifically, the quantized conductivity in the Integer Quantum Hall Effect and all that good stuff). If $\vec{k}=(k_x,k_y)=k(\cos(\lambda),\sin(\lambda))$, then the Hamiltonian for graphene is $$\tag{3} H(k,\lambda) ~=~ \hbar v_F k \begin{pmatrix} 0 & e^{-i \lambda} \\ e^{i \lambda} & 0 \\ \end{pmatrix}$$ Then I replaced the linear band structure with two flat bands of energy $E_1$ and $E_2$ to get equation (2).

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I don't think one can conclude $ \langle | dH_{\lambda}/ d \lambda | \rangle =0 $ hence $dH_{\lambda}/ d \lambda=0$ from the zero values of the HF theorem, here $|\rangle$ is any state. Since $ \langle | dH_{\lambda}/ d \lambda | \rangle = \sum_{ij} c_i^* c_j \langle i | dH_{\lambda}/ d \lambda | j \rangle $, ($|i \rangle$ is the $i$-th eigenstate of the Hamiltonian $H_{\lambda}$) where the zero value of the HF theorem only guarantees the diagonal part of this equation. – user26143 Nov 15 '13 at 6:34
@user26143 Ah! How could I have made such a silly mistake?! Just because the diagonal terms of $dH_\lambda/d\lambda$ in the basis of $H_\lambda$ eigenstates are zero doesn't imply the off-diagonal terms are zero. Then, can we conclude that even if all $dE_\lambda/d\lambda=0$, $dH_\lambda/d\lambda$ can still be nonzero? If you think so, please post it as an answer so that I can accept it! – ChickenGod Nov 15 '13 at 9:27
And just a little more commentary: This would mean a hypothetical solid with flat bands and thus zero group velocity could still have $\nabla_{\vec{k}} H \neq 0$. Counterintuitive in my opinion, but very interesting! – ChickenGod Nov 15 '13 at 9:43
I think if $dE_{\lambda}/d\lambda=0,\forall \lambda$, $dH_{\lambda}/d \lambda$ can still be non-zero. I will post an answer, although I know nothing about graphene, could not share your interest :( – user26143 Nov 15 '13 at 12:17
up vote 5 down vote accepted

I don't think one can conclude $\langle\; | \frac{d H_{\lambda}}{d \lambda} |\;\rangle=0$, hence $\frac{dH_{\lambda}}{d\lambda}=0$, from the zero values of the HF theorem, here $|\rangle$ is any state.

Since $\langle\;| \frac{d H_{\lambda}}{d\lambda}|\;\rangle = \sum_{ij} c^∗_i c_j \langle i | \frac{dH_{\lambda}}{d\lambda}|j⟩$, ($|i\rangle$ is the $i$-th eigenstate of the Hamiltonian $H_{\lambda}$) the zero values of the HF theorem only guarantee the diagonal part of this equation.

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