this might be a "standard trick" for many solid state physicists, however it's one that I'm not familiar with so maybe you can help me. Here's the Problem:

Suppose we're given a Hamiltonian of the form $H=\sum_{k} \epsilon_{k} c^{\dagger}_{k}c_{k} + \sum_{k} U c^{\dagger}_{k+Q} c_{k}$. Here U is some complex (!) number, k is a 2 dimensional wavevector and Q=$(\pi,\pi)$. Furthermore we impose $\epsilon_k = - \epsilon_{k+Q}$. The first Brillouin Zone is the set $\{ (k_x,k_y) ; |k_x|+|k_y|<\pi \} $ in 2D-k-space.

Now define the 2D-vector $\Psi_k = (c_k,c_{k+Q})$. Then the Hamiltonian be written as: $H= \sum_{k}' \Psi_k^{\dagger} A_k \Psi_k $ with the k-dependent 2x2 Matrix $A_k$ defined by:

$$ \left[ \begin{array}{ c c } \epsilon_k & U \\ U^* & \epsilon_{k+Q} \end{array} \right] $$

The prime (') in the sum denotes that it has to be taken over wavevectors in the frist Brillouin zone only!

Now multiplying this quadratic form out "in reverse" I obtain something like: $H=\sum_{k} \epsilon_{k} c^{\dagger}_{k}c_{k} + \sum_{k} U c^{\dagger}_{k+Q} c_{k} + \sum_{k} \epsilon_{k+Q} c^{\dagger}_{k+Q}c_{k+Q} + \sum_{k} U c^{\dagger}_{k} c_{k+Q} $

It's not clear to me why the third and fourth term are supposed to vanish in case I'm restricting my sum to the frist Brillouin zone.

I hope someone can help. It should be rather technical, but still important I think.

Thanks in advance.

  • $\begingroup$ I don't really get the question. Vector $k+Q$ does not lie in first Brillouin zone and equivalent to $k$. You should either replace it with $k$ or consider that Brillouin zone too. Are you trying to fold the energy spectrum? $\endgroup$
    – Misha
    Commented Feb 23, 2012 at 17:56
  • $\begingroup$ Note that the sum in the original hamiltonian is NOT restricted to the first Brillouin zone, while the sum in the Hamiltonian given in form a "quadratic form" is just over the first Brillouin zone. The model describes a charge density wave. The condition on the energies is the so-called nesting condition. $\endgroup$
    – MrLee
    Commented Feb 23, 2012 at 18:30
  • $\begingroup$ Then your original Hamiltonian should be non-Hermitian. Otherwise you've had terms like $U^*c^{\dagger}_{k}c_{k+Q}$ there. $\endgroup$
    – Misha
    Commented Feb 23, 2012 at 18:35
  • $\begingroup$ Ok. To let's assume for simplicity that U is real. So that we don't have to worry about the difficulties you are describing. So replace U* by U everywhere $\endgroup$
    – MrLee
    Commented Feb 23, 2012 at 19:01
  • $\begingroup$ Still, you may think of this term (with $U$) as of amplitude for an electron to from $k$ to $k+Q$. It is natural to assume (to get hermitian Hamiltonian, to be more precise) that amplitude to go back is the same. Thus there should be also term for transition $k+Q\to k$. $\endgroup$
    – Misha
    Commented Feb 23, 2012 at 20:01

2 Answers 2


The terms don't disappear--- they are the same as previous terms when you shift the dummy summation variable $k\rightarrow k+Q$.


I will formulate as an answer, because the discussion in comments is too long: during the following discussion I will be able to add more details here.

Third term indeed disappears, but by slightly different reasoning. Obviously $$ \varepsilon_k c^+_{k\sigma}c_{k\sigma} = \varepsilon_{k+Q}c^+_{k+q\sigma} c_{k+Q\sigma} $$ due to the fact that these states are equal.

It is not that evident but probably true that $$ c^+_{k\sigma} W c_{k+Q\sigma} = c^+_{k+Q\sigma} W c_{k\sigma} $$ as probability go "there" and "back" usually more or less equal.

However, it is not general rule or something. There could be interaction which makes them non-equal.

Probably, it can be derived from momentum conservation (thus assuming momentum concerving interaction), I am not sure.

I suggest consider this fact a part of a model.

  • $\begingroup$ There are several confusions here: Q is $(\pi,\pi)$, not $(2\pi,2\pi)$, so the shift is nontrivial--- it is relating the field and its complex conjugate. The probability amplitudes to "go there" and "back" are not equal, they are complex conjugate, and this is Hermiticity. See my answer--- it's just a shift of dummy variables. $\endgroup$
    – Ron Maimon
    Commented May 25, 2012 at 3:57

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