I could not find anything on that on google, or here on physics stack exchange, which surprises me. My problem is, that I do not see, why exactly

$$\left<a\right> = \left<a^\dagger\right> = 0,\tag{1}$$

where $<...>$ is defined via the path integral, with $A$, $B$ being operators, $a$,$b$ their coherent states and S the action of the physical system:

$$\left<AB\right> = \frac{\int\mathcal{D}(a,b) ~a, b ~\mathrm{exp}(-S)}{\int\mathcal{D}(a,b)~ \mathrm{exp}(-S)},\tag{2}$$

e.g. used to calculate the correlation function / Green's function, e.g. in imaginary time:

$$G(\tau)=\left<a(\tau) a^\dagger(0)\right>.\tag{3}$$

Now I know, that $\left<a\right>$ and $\left<a^\dagger\right>$ have to vanish, as long as the Hamiltonian in S conserves the particle number. But I can't see, why it is zero on mathematical grounds.

Note: Usually I compute the correlation function as follows. $\mathcal{Z}$ is the partition function, $\bar \phi, \phi$ are coherent states, defined via $a \left|\phi\right> = \phi \left|\phi\right>$ and $H$ is the Hamiltonian.

$$S=\int\mathrm{d}\tau ~ \bar\phi ~ (\partial_\tau-H) ~\phi.\tag{4}$$

Lets assume a quadratic Hamiltonian and $\phi$, $\bar \phi$, $u$ and $\bar u$ all are functions of $\tau$.

$$\left<a(\tau) a^\dagger(0)\right> = \frac{1}{\mathcal{Z_0}} \int\mathcal{D}(\bar \phi, \phi) ~\phi \bar \phi ~\mathrm{exp}(-S) $$

$$= \frac{1}{\mathcal{Z_0}} \frac{\delta^2}{\delta u \delta \bar u}\Big|_{u=\bar u=0} \int\mathcal{D}(\bar \phi, \phi) ~\mathrm{exp}\Big[-\int\mathrm{d}\tau ~ \big\{\bar\phi ~ (\partial_\tau-H) ~\phi + \bar\phi u + \bar u \phi\big\}\Big] $$

$$= \frac{\delta^2}{\delta u \delta \bar u}\Big|_{u=\bar u=0} ~ ~\mathrm{exp}\Big[\int\mathrm{d}\tau ~ \bar u (\partial_\tau - H)^{-1} u\Big]= \mathrm{det}\big(\partial_\tau-H\big)^{-1} .\tag{5}$$

Now I cannot see, where this goes wrong, if I take $\left<a\right>$ or $\left<a^\dagger\right>$. Especially (!) because for a superconductor, $\left<a^\dagger a^\dagger\right>$ can be non-zero, so it depends on the Hamiltonian, which correlators vanish and which do not, right?


3 Answers 3


OP asks about the path integral formalism.

  • More generally, one may prove that

    1. if $$G~=~U(1)^m\times \mathbb{Z}_{n_1}\times\ldots\times \mathbb{Z}_{n_r}\tag{A}$$ is an abelian group$^1$;

    2. if the $k$th field of the theory $$\phi^k~\longrightarrow~ \phi^{\prime k}~=~g^{q_k}\phi^k, \qquad g~\in~G,\tag{B}$$ transforms in an 1-dimensional representation$^1$ characterized by an integer charge tuple $q_k\in\mathbb{Z}^{m+r}$;

    3. if the action $S[\phi]$ and the path integral measure ${\cal D}\phi$ have a global $G$-symmetry;

    4. and if the quantity $$F[\phi]~\longrightarrow~ F[\phi^{\prime}]~=~g^QF[\phi], \qquad g~\in~G,\tag{C}$$ has a non-trivial charge $Q\in\mathbb{Z}^{m+r}$: $$ \exists g\in G:~~ g^Q~\neq~1~\in~\mathbb{C}; \tag{D}$$

    then by changing the integration variables in the path integral one may show that the corresponding correlator transforms as $$\begin{align} \langle F \rangle_J ~:=~~& \frac{1}{Z[J]}\int \! {\cal D}\phi~\exp\left(\frac{i}{\hbar}\left(S[\phi] +J_k\phi^k \right)\right)F[\phi] \cr ~=~~&\frac{1}{Z[J]}\int \! {\cal D}\phi^{\prime}~\exp\left(\frac{i}{\hbar}\left(S[\phi^{\prime}] +J_k\phi^{\prime k} \right)\right)F[\phi^{\prime}] \cr ~\stackrel{(B)+(C)}{=}& \frac{1}{Z[Jg^q]}\int \! {\cal D}\phi~\exp\left(\frac{i}{\hbar}\left(S[\phi] +J_kg^{q_k}\phi^k \right)\right)g^QF[\phi] \cr ~=~~&g^Q\langle F \rangle_{Jg^q}. \end{align} \tag{E}$$

  • In particular then

    $$ \langle F \rangle_{J=0} ~\stackrel{(D)+(E)}{=}~0.\tag{F} $$

  • The $G$ symmetry could e.g. be global gauge symmetry, flavor symmetry, particle number symmetry, ghost number symmetry, parity symmetry, etc. OP's scenario (1) can be viewed as a special case of eq. (F).

  • We can say more. Such symmetry implies $G$-invariance of

    1. the path integral/partition function $$Z[J]~=~Z[Jg^q],\tag{G}$$ i.e. the generator of all Feynman diagrams;

    2. the generator $$W_c[J]~=~W_c[Jg^q]\tag{H}$$ of connected Feynman diagrams;

    3. and the effective/proper action $$\Gamma[\phi_{\rm cl}]~=~\Gamma[g^q\phi_{\rm cl}],\tag{I}$$ i.e. the generator of 1PI Feynman diagrams.$^2$

  • In particular, the following quantities must respect the $G$-symmetry, i.e. be $G$-invariant or vanish:

    • The connected $n$-point correlation functions $$\langle \phi^{k_1}\ldots \phi^{k_n}\rangle^c_{J=0}~=~\left(\frac{\hbar}{i}\right)^{n-1}W_{c,n}^{k_1\ldots k_n}.\tag{J}$$

    • The $n$-point 1PI-correlation functions $$\Gamma_{n,k_1\ldots k_n}~=~\left.\frac{\delta^n \Gamma[\phi_{\rm cl}]}{\delta\phi^{k_1}_{\rm cl}\ldots \delta\phi^{k_n}_{\rm cl}} \right|_{\phi_{\rm cl}=0}.\tag{K}$$

    • Feynman diagrams. This e.g. explains why a fermion line cannot break in a Feynman diagram.

$^1$ Recall that an irreducible representation $\rho$ of the abelian group (A) is 1-dimensional and classified by an integer charge tuple $q=(q_1,\ldots,q_{m+r})\in\mathbb{Z}^{m+r}$: $$G~\ni~g~=~(g_1,\ldots,g_{m+r})\quad\stackrel{\rho}{\mapsto}\quad \rho(g)~=~g^q~:=~g_1^{q_1}\ldots g_{m+r}^{q_{m+r}}~\in~\mathbb{C}. \tag{L}$$

$^2$ Eq. (I) can be seen from the fact that if we introduce a counter-transformation of the sources
$$J~\longrightarrow~ J^{\prime}_k~=~J_kg^{-{q_k}},\tag{M}$$ then the classical fields $\phi_{\rm cl}=\frac{\delta W_c[J]}{\delta J}$ also transform as $$\phi_{\rm cl}~\longrightarrow~ \phi^{\prime k}_{\rm cl}~=~g^{q_k}\phi^k_{\rm cl}.\tag{N}$$


Yes, it depends on the Hamiltonian. Specifically, if we add a linear term $H\to H+\lambda (a+a^{\dagger})$ this will not vanish. A simple mathematical way the expectation value $\langle a \rangle$ must vanish for a standard Hamiltonian quadratic in $a$ is that while the Hamiltonian is even under $a\to -a$ this expectation value is odd, therefore it must vanish (unless we take into account a spontaneous symmetry breaking. Then it can get a non-zero VEV)

  • $\begingroup$ Ah that's how you can show it, thank you!! What about $<aa>$ though? It is also even under $a \rightarrow -a$, right? $\endgroup$
    – yanscha
    Commented Mar 13, 2020 at 20:02
  • $\begingroup$ Oh wait, I think, I misunderstood you. With "quadratic in a" you mean quadratic in the annihilation operator, not like $a a^\dagger$, right? Then I see your argument! Thank you. But then still, I have no idea how to see, that $<a>$ is zero for example for a tight binding Hamiltonian $H=-t \sum_{<ij>}a_i^\dagger a_j$. $\endgroup$
    – yanscha
    Commented Mar 13, 2020 at 21:12
  • $\begingroup$ No, I mean a standard Hamiltonian like the tight-binding one you wrote. Note that $a$ and $a^{\dagger}$ are not independent. If you take $a\to -a$ you must do the same for its hermitian conjugate, in order for it to remain the hermitian conjugate. In fact, for tight-binding Hamiltonians the symmetry is more general $a\to \exp(i\theta)a$ and this $U(1)$ symmetry reflects the conservation of number of particles. For a SC Hamiltonian this $U(1)$ symmetry breaks and only the $Z_2$ remains (multiplication by $-1$) as the number of particles if not conserved, but the parity of it is. $\endgroup$
    – user245141
    Commented Mar 14, 2020 at 8:11

The average is taken over the ground state or it is a thermodynamic average (not average over the coherent states), i.e. \begin{equation} \langle A \rangle = \sum_n e^{-\beta E_n} \langle n | A| n\rangle . \end{equation} The quantum numbers implied by $n$ include the number of particles. If operator $A$ changes this number, the matrix elements are zero, because the wave functions corresponding to different quantum numbers are orthogonal.

  • $\begingroup$ Thank you! Your right, maybe I should have made it more clear, that it's a thermodynamic average! But still, this average (at least in QM) depends on the Hamiltonian, as can be seen for conventional superconductors, where $<a^\dagger a^\dagger>$ does not vanish, because the ground state is a superposition of the vacuum and $a_k^\dagger a_{-k}^\dagger |0>$. So there must be another way to argue, that it vanishes, depending on the Hamiltonian, right? $\endgroup$
    – yanscha
    Commented Mar 13, 2020 at 21:06
  • $\begingroup$ I don't think that it vanishes in the case of a superconductor... but this could be checked by direct calculation. $\endgroup$
    – Roger V.
    Commented Mar 13, 2020 at 21:11
  • $\begingroup$ In fact, for superconductor the particle number should change by 2 - a Cooper pair. $\endgroup$
    – Roger V.
    Commented Mar 13, 2020 at 21:13
  • $\begingroup$ Oh yeah you are right. I was thinking way too hard in the path integral direction! Thank you! $\endgroup$
    – yanscha
    Commented Mar 13, 2020 at 21:19

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