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

$$\left<a\right> = \left<a^\dagger\right> = 0,$$

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:

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

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

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

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.

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

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

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

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

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

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?


OP asks about the path integral formalism.

  • More generally, one may prove that

    1. if the action $S[\phi]$ and the path integral measure ${\cal D}\phi$ have a global $U(1)$ symmetry $$\phi~\longrightarrow~ \phi^{\prime}~=~e^{iq\theta}\phi;\tag{A}$$

    2. and if the quantity $$F[\phi]~\longrightarrow~ F[\phi^{\prime}]~=~e^{iQ\theta}F[\phi]\tag{B}$$ has non-zero $U(1)$-charge $Q\neq 0$;

    then by changing the integration variables in the path integral one may show that the corresponding correlator transforms as

    $$\begin{align} \langle F \rangle_J ~:=~~& \frac{1}{Z[J]}\int \! {\cal D}\phi~\exp\left(\frac{i}{\hbar}\left(S[\phi] +J_k\phi^k \right)\right)F[\phi] \cr ~=~~&\frac{1}{Z[J]}\int \! {\cal D}\phi^{\prime}~\exp\left(\frac{i}{\hbar}\left(S[\phi^{\prime}] +J_k\phi^{\prime k} \right)\right)F[\phi^{\prime}] \cr ~\stackrel{(A)+(B)}{=}& \frac{1}{Z[e^{iq\theta}J]}\int \! {\cal D}\phi~\exp\left(\frac{i}{\hbar}\left(S[\phi] +J_ke^{iq_k\theta}\phi^k \right)\right)e^{iQ\theta}F[\phi] \cr ~=~~&e^{iQ\theta}\langle F \rangle_{e^{iq\theta}J}. \end{align} \tag{C}$$

  • In particular

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

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

  • We can say more. Such symmetry implies $U(1)$-invariance of

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

    2. the generator $$W_c[J]~=~W_c[e^{iq\theta}J]\tag{F}$$ of connected Feynman diagrams;

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

  • In particular

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

    • Similarly, the $n$-point 1PI-coefficient functions $$\Gamma_{n,k_1\ldots k_n}~=~\left.\frac{\delta^n \Gamma[\phi_{\rm cl}]}{\delta\phi^{k_1}_{\rm cl}\ldots \delta\phi^{k_n}_{\rm cl}} \right|_{\phi_{\rm cl}=0}\tag{I}$$ in $\Gamma[\phi_{\rm cl}]$ must preserve the $U(1)$-symmetry, cf. e.g. my Phys.SE answer here.

    • In other words, Feynman diagrams must respect the $U(1)$ symmetry. This can e.g. explain why a fermion line cannot break.


$^1$ Eq. (G) can be seen from the fact that if the sources transform as $$J~\longrightarrow~ J^{\prime}~=~e^{-iq\theta}J,\tag{J}$$ then the classical fields $\phi_{\rm cl}=\frac{\delta W_c[J]}{\delta J}$ transform as $$\phi_{\rm cl}~\longrightarrow~ \phi^{\prime}_{\rm cl}~=~e^{iq\theta}\phi_{\rm cl}.\tag{K}$$


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 Mar 13 '20 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 Mar 13 '20 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 Mar 14 '20 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 Mar 13 '20 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 Vadim Mar 13 '20 at 21:11
  • $\begingroup$ In fact, for superconductor the particle number should change by 2 - a Cooper pair. $\endgroup$ – Roger Vadim Mar 13 '20 at 21:13
  • $\begingroup$ Oh yeah you are right. I was thinking way too hard in the path integral direction! Thank you! $\endgroup$ – yanscha Mar 13 '20 at 21:19

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