People define field operators as

$\bf{\Psi}(x)$ = $\sum_k$ $\phi_k(x)$ $c_k$
$\bf{\Psi}^{\dagger}(x)$ = $\sum_k$ $\phi^*_k(x)$ $c^{\dagger}_k$

where $\phi_k(x)$ is a single particle basis.

My question is: Are these operators parametrized by "x"? In the sense that for a fixed number $x_0 \in R^n$ $\rightarrow$ $\bf{\Psi}_{x_0}$ $\equiv$ $\bf{\Psi}(x_0)$ = $\sum_k$ $\phi_k(x_0)$ $a_k$. One analogy that comes to my mind is the one-parameter family of evolution operators, for fixed $t_0$ $\rightarrow$ $U_{t_0}$. I am asking this because when one computes the zero temperature Green's function for a fermionic gas one has the following derivation

enter image description here And enter image description here

Where $\hat{\psi}$(x)= $\frac{1}{\sqrt{V}}$ $\sum_k$ $e^{ikx}$ $c_k$
In Eq. 1.8 one can see $\frac{e^{i(kx-k'x')}}{V}$,where $e^{ikx}$ was pulled out of the inner product. This can happen only if $e^{ikx}$ is a complex number in the field operator $\hat{\psi(x)}$ and not a function of position,I guess.


1 Answer 1


Yes it is an operator on the hilbert space that is parameterized by $x$. In other words, each value of $x$ gives a different operator $\hat \psi(x): {\mathcal H}\to {\mathcal H}$. You can pull the $e^{ikx}$ out of the integral because for each chosen $x$ it is just a number --- an element of the field over which the Hilbert space is defined. Do not confuse the Hilbert space on which the $c$ and $c^\dagger$ operators act with the one we use in quantum mechanics. The inner product here does not involve integrals over $x$.

When we quantize a real scalar field $\varphi(x)$ and expand $$ \hat \varphi(x,t)= \sum_k \left\{a_k e^{i(kx-\omega_k t)}$ + a^\dagger_k e^{-i(x-\omega_kt)}\right\} $$ the $a_k$ and $a_k^\dagger$ are linear combinations of operators that act of fucntionals of $\varphi$. For example the "momentum" conjugate to $\varphi(x)$ is $$ \hat \Pi(x) = -i \hbar \frac{\delta}{\delta \varphi(x)} $$ (a functional derivative) so the inner product in this space is something like $$ <F|G>= \int d[\varphi] F^*(\varphi) G(\varphi) $$ You should think of $\varphi(x)$ as the displacement of a rubber sheet at position $x$ and the field expansion as one in the normal modes of the sheet that behave like harmonic oscillators. The Hilbert space in which the modes themselves are treated as elements of $L^2({\mathbb R}^N)$ is something entirely different --- you would have this normal mode space even for a classical syste, After you quantize your rubber sheet, you later discover that these classical normal modes can also be regarded as the wavefunctions of the resuting phonon particles, but it is still a different space than that acted on by the $a_k$ and $a^\dagger_k$

  • $\begingroup$ "Do not confuse the Hilbert space on which the c and c† operators act with the one we use in quantum mechanics. The inner product here does not involve integrals over x." Can you please explain. I am under the impression that for a single particle Hamiltonian you have say $\hat{H}$ $\psi_k$=$\epsilon_k$ $\psi_k$. Now,I can build the Hilbert space $H^N$ as the direct sum of N copies of H. After that I can define,for example,$c^{\dagger}$ :$ H^N$ $\rightarrow$ $H^{N+1}$. Say my ground state is $\phi_0$ and <$\phi_0$ | $c_k$ $c^{\dagger}$|$\phi_0$ > reduces to a inner product over $H^N$. Right? $\endgroup$
    – Small Pole
    Feb 20, 2017 at 19:34
  • $\begingroup$ Sorry,as the direct product not sum. $\endgroup$
    – Small Pole
    Feb 20, 2017 at 19:37
  • $\begingroup$ I am trying to make sense of the spaces people use. One one hand, we have the parametrization of the field operators x $\in$ R $\rightarrow$ L($H^N$,$H^{N+1}$} and all the integrals and derivatives with respect to x are defined on this space. On the other hand,we have the spatial coordinates of the domains on which the Hilbert spaces $H^N$ exists and these go into the inner products we talked above,and these we can defines as ($x_1$,$x_2$,..,$x_n$} . Here,x and ($x_1$,$x_2$,..,$x_n$} will be different coordinates. This is what you mean above? $\endgroup$
    – Small Pole
    Feb 20, 2017 at 20:06
  • $\begingroup$ I expanded my answer to address these points. $\endgroup$
    – mike stone
    Feb 20, 2017 at 23:39
  • $\begingroup$ Thank you very much for the explanation. I would like to ask you if you can give me some reference ( i.e books or articles) that go deep into the mathematics of this stuff ( i.e. the spaces they use,etc). Thank you $\endgroup$
    – Small Pole
    Feb 22, 2017 at 7:18

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