$H = \sum_{k} V(q) a_{k4}^{\dagger}b_{k3}^{\dagger}b_{k2}a_{k1}$

where $q$ is the transfer momentum, $a$ $b$ are two orbits or two sublattice sites.

Will the eigenvalues of the above Hamiltonian unchanged under the following transform (gauge transform of Bloch state):

$b_k \to e^{i\theta(k)}b_k$ where $\theta(k)$ is real

  • $\begingroup$ Are the number indices such of the momenta, i.e. $k_1, k_2, \ldots$ or such of the operators $a$ and $b$? $\endgroup$
    – Clever
    Commented Jan 15, 2015 at 11:13
  • $\begingroup$ Thanks for your comment, I just short some notation to make the question clear, here is the full one: $\sum_k := \sum_{k_4, k_3, k_2, k_1} \delta_{k_4+k_3,k_2+k_1}$, and $q=k_4-k_1$ $\endgroup$
    – Tim
    Commented Jan 15, 2015 at 16:05

1 Answer 1


The interaction term is not invariant under the gauge transformation. $b_k\rightarrow e^{i\theta_k}b_k$ is only a symmetry for the non-interacting Hamiltonian $H=\sum_k E(k)b_k^\dagger b_k$.

  • $\begingroup$ The term itself is not invariant, but the physics is invariant. Like electrons under magnetic field, different vector potential will change the Hamiltonian but not the physics. $\endgroup$
    – Tim
    Commented Jan 21, 2015 at 3:13
  • $\begingroup$ My understanding is that different gauge are different mathematical description of the same physics problem. So the eigenvalue should not changed. $\endgroup$
    – Tim
    Commented Jan 21, 2015 at 3:16
  • $\begingroup$ Or say this gauge transform is just an unitary transform on the original Hamiltonian. $\endgroup$
    – Tim
    Commented Jan 21, 2015 at 3:17
  • $\begingroup$ The physics is invariant under local gauge transformation in real space, provided you include U(1) gauge fields. Notice that the Hamiltonian of electrons in a magnetic field is actually invariant once you take into account the gauge transformations on the electron operators and the vector potential, otherwise it would not be called a symmetry. $\endgroup$
    – Meng Cheng
    Commented Jan 21, 2015 at 3:50
  • $\begingroup$ I agree the Hamiltonian is invariant. I think I mean the "form of the Hamiltonian" written in the new field operator changed. $\endgroup$
    – Tim
    Commented Jan 21, 2015 at 4:26

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