I'm trying to work through Altland and Simons' example of interacting fermions in one dimension. It's in chapter 2, page 70 (you can find it here).

They define fermionic operators $$ a_{sk}^\dagger $$ where $s=L/R$. $a_{Lk}^\dagger$ is an operator that creates an electron going to the left with momentum $(-k_F+k)$, and $a_{Rk}^\dagger$ is an operator that creates an electron going to the right with momentum $(k_F+k)$. So basically, $a_{Lk}^\dagger=a_{-k_F+k}^\dagger$, $a_{Rk}^\dagger=a_{k_F+k}^\dagger$. These operators are restricted to exist only for small $k$.

Then, they define density operators

$$ \rho_{sq}=\sum_k a^\dagger_{sk+q}a_{sk} $$

They go on to show that the commutation relations for the density operators are

$$ [\rho_{sq},\rho_{s'q'}]=\delta_{s,s'}\sum_k (a^\dagger_{sk+q}a_{sk-q'}-a^\dagger_{sk+q+q'}a_{sk}) $$

Now, here's the part I don't understand. They say they want to replace the right side of the equation with its ground state expectation value. They define the ground state of the theory by $|\Omega\rangle$. Then they claim that

$$ \langle\Omega|a^\dagger_{sk}a_{sk'}|\Omega\rangle = \delta_{kk'} $$

Why should this be true? I understand that in the noninteracting theory, $a^\dagger_{sk}a_{sk'}|\Omega\rangle$ is orthogonal to $|\Omega\rangle$ unless $k=k'$. But in the interacting theory, the ground state could be in a superposition of states that means $\langle\Omega|a^\dagger_{sk}a_{sk'}|\Omega\rangle\neq0$.

They ultimately use this to prove

$$ \langle\Omega|[\rho_{sq},\rho_{s'q'}]|\Omega\rangle = \delta_{s,s'}\delta_{q,-q'}\sum_k\langle\Omega|(a^\dagger_{sk+q}a_{sk+q}-a^\dagger_{sk}a_{sk})|\Omega\rangle $$ and I don't see any other way to prove this.

What am I missing?

  • 2
    $\begingroup$ The ground state $|\Omega\rangle$ is the filled Fermi sphere (or line in 1D). This is regarded as the vacuum and the low-energy excitations of this vacuum are the quasiparticles (charge density waves in the 1D spinless case). The ground state does not change in the presence of weak interactions. The approximation is similar to first-order perturbation theory in ordinary quantum mechanics. $\endgroup$
    – Praan
    Commented May 27, 2016 at 16:53
  • $\begingroup$ @Praan In the text, they say the only approximation they're making is that we focus only on low-lying excitations above the ground state--NOT that the ground state is close to the non-interacting ground state. Looking ahead, though, they do go on to describe $|\Omega\rangle$ as a filled fermi sea, so you might be right. Is there any way to do this calculation without this assumption? $\endgroup$ Commented May 27, 2016 at 17:09

1 Answer 1


$\def\kk{\mathbf{k}} \def\ii{\mathrm{i}} \def\qq{\mathbf{q}}$

You can prove this using translation invariance, without any other assumptions on the nature of the interacting ground state. I am going to give the argument in $D$ dimensions, which obviously holds in $D=1$ as a special case.

Spatial translations of the system as a whole are generated by the centre-of-mass momentum ($\hbar = 1$) $${\bf P} = \sum_{\kk,s} \kk a_{\kk s}^\dagger a_{\kk s},$$ where boldface type denotes a vector in $D$ dimensions. The unitary operator $T_{\bf r} = \mathrm{e}^{\ii {\bf P} \cdot {\bf r}}$ describes a translation of the coordinate system by an amount ${\bf r}$. Now, the crucial assumption is that the interacting ground state satisfies the condition ${\bf P} \lvert \Omega \rangle = 0$, i.e. $T_{\bf r} \lvert \Omega \rangle = \lvert \Omega \rangle $. This holds for any system with translation-invariant interactions, since in that case 1) the Hamiltonian $H$ commutes with ${\bf P}$, so that eigenstates of $H$ are also eigenstates of ${\bf P}$, and 2) eigenstates of ${\bf P}$ gain a positive contribution ${\bf P}^2/2M$ to their energy (in non-relativistic approximation), with $M$ the total mass of the system. Therefore, the ground state $\lvert \Omega \rangle$ lies in the ${\bf P}=0$ sector. Of course, the above formal arguments prove what should already be obvious: a system in its ground state has no overall motion in the lab frame.

The rest of the argument follows from some simple algebra. One readily proves the commutation relation $$ [\mathbf{P},a_{\kk s}] = -\kk a_{\kk s}, $$ which means physically that the ladder operator $a_{\kk s}$ reduces the total momentum of the system by an amount $\kk$, and further implies the translation property $$T_{\bf r} a_{\kk s}T_{\bf r}^\dagger = \mathrm{e}^{-\ii {\bf k} \cdot {\bf r}}a_{\kk s}.$$ It follows that \begin{align} \langle \Omega \rvert a^\dagger_{\kk s}a_{\kk' s} \lvert \Omega \rangle & = \langle \Omega \rvert \left(T_{\bf r}^\dagger T_{\bf r}\right) a^\dagger_{\kk s} \left(T_{\bf r}^\dagger T_{\bf r}\right) a_{\kk' s} \left(T_{\bf r}^\dagger T_{\bf r}\right) \lvert \Omega \rangle \\ & = \langle \Omega \rvert \left(T_{\bf r}a^\dagger_{\kk s}T_{\bf r}^\dagger \right)\left(T_{\bf r} a_{\kk' s}T_{\bf r}^\dagger \right)\lvert \Omega \rangle \\ & = \mathrm{e}^{\ii (\kk - \kk')\cdot {\bf r}}\langle \Omega \rvert a^\dagger_{\kk s}a_{\kk' s} \lvert \Omega \rangle, \end{align} where the first equality follows from unitarity of $T_{\bf r}$, the second from the translation-invariance of the interacting ground state, and the third using the translation property of the ladder operators. Noting that the above equality holds for all ${\bf r}$, we conclude that both sides must vanish unless $\kk = \kk'$.

Finally, it is worth mentioning that this argument does not rely in any way on fermionic statistics, and the same holds true in a bosonic system. Indeed, a similar property holds true for any operator which increases the momentum of the system by a definite amount (thus satisfying the translation property above). This includes the density Fourier components themselves, $$ \rho_{\qq s} = \sum_\kk a^\dagger_{\kk+\qq s}a_{\kk s},$$ which satisfy $$\langle \Omega \rvert \rho_{\qq s}^\dagger \rho_{\qq' s}\lvert \Omega \rangle \propto \delta_{\qq\qq'}.$$ Moreover, the argument also holds for other eigenstates of $\mathbf{P}$ (with non-zero eigenvalue). In fact, it generalises to expectation values with respect to any translation-invariant mixed state $\varrho$, which is any mixture of $\mathbf{P}$-eigenstates (possibly with different momenta) such that $[\varrho,\mathbf{P}]=0$.

  • $\begingroup$ Wow! This is an approach I would never have thought of. Thanks! $\endgroup$ Commented May 28, 2016 at 2:04
  • $\begingroup$ Perhaps several years late, but how are fermions supposed to have such commutation relation (instead of anti-commutation) in the first place? $\endgroup$
    – donnydm
    Commented Nov 3, 2022 at 14:59
  • $\begingroup$ Hi @donnydm, you can actually prove that commutation relation from the fundamental fermionic anti-commutation relations. That is, $[a^\dagger a,a] = -a$ is true for both fermions and bosons. $\endgroup$ Commented Nov 4, 2022 at 0:13
  • $\begingroup$ @MarkMitchison Ah, I see we can use $[A,B]=2AB-\{A,B\}$ and $a^2=0$, thanks! $\endgroup$
    – donnydm
    Commented Nov 4, 2022 at 1:29
  • 1
    $\begingroup$ @TobiasFünke Yes, I think the identity holds for any eigenstate of $\mathbf{P}$. Thanks, have edited to state this. $\endgroup$ Commented Aug 23, 2023 at 2:00

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