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In quantum mechanics, one in principle can write down an explicit form of the corresponding wave-function. For example, $V_i$ for the $i$-th level of quantum harmonic oscillator.

In QFT, the Hilbert space which the quantum fields act on is not the space of wave-functions, but rather of functionals on the space of field configurations. Is it possible to explicitly write down those functionals and the corresponding field configurations for $0, 1, 2, \ldots$ excitations (particles) of the field?

And I don't mean things like $a^\dagger_p a^\dagger_q|\mathrm{vacuum}\rangle$, but something in coordinates, like in the case of the quantum mechanical wave-function.

Qmechanic
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stanislav-iablokov
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2 Answers2

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The correct analogy with the QM wave function can be understood from the path integral formulation. Given a theory with an action $S[\varphi]$ the path integral is denoted as $$ Z = \int [\mathrm{d}\varphi]\,e^{iS[\varphi]}\,. $$ A wave functional is, as the name says, a functional of the configuration variables of your QFT, which obviously are fields. In QM if you had, say, $N$ coordinates, your wave function would be $$ \psi(q_1,\ldots,q_N)\,. $$ QFT, on the other hand, can be though of as if there were infinitely many coordinates ($N\to\infty$). Those coordinates describe your field configuration at a certain time slice, which we can take to be $x^0 = 0$ for simplicity. Therefore the wave functional is something like $$ \Psi[\phi_0(\vec{x})]\,. $$ You can imagine discretizing your space, at which point the $q_i$'s would be the values of $\varphi$ at the point $x = (0,\vec{x}_i)$.

In the path integral language a field configuration can be thought of as a boundary condition for the integral at a certain time slice. We can therefore define a wave functional as the following path integral $$ \Psi[\phi_0(\vec{x})] = \int[\mathrm{d}\varphi]_{\varphi(0,\vec{x})\equiv\phi_0(\vec{x})}\,e^{iS[\varphi]}\,. $$ Where, as you can see, the integration is restricted only to those configurations which have $\varphi(0,\vec{x}) \equiv \phi_0(\vec{x})$. Now, this would actually the vacuum state because we are not inserting anything in the path integral. A more general state can be obtained by putting operators such as $$ \Psi_{1,\ldots,n}[\phi_0(\vec{x})] = \int[\mathrm{d}\varphi]_{\varphi(0,\vec{x})\equiv\phi_0(\vec{x})}\,\mathcal{O}_1(x_1)\cdots\mathcal{O}_n(x_n)\,e^{iS[\varphi]}\,. $$

MannyC
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Yes, you can generally map between second quantization picture (as in QFT) to first quantization (each particle has a coordinate and you get the wave functions). For more than one particle, the QFT version is guaranteed to contain the correct symmetrization or anti-symmetrization of the wave-function.

The wave-function for a two-particle state, for example, is given by $$ \Phi(r_1, r_2) = \langle r_1, r_2 | a^{\dagger}_p a^{\dagger}_q | {\rm{vac}}\rangle = \langle {\rm{vac}}| a(r_2)a(r_1) a^{\dagger}_p a^{\dagger}_q | {\rm{vac}}\rangle$$ where $a(r)$ is the field annihilating a particle at position $r$, which you can get from a Fourier-transform of $a_p$

  • By the explicit field configuration for 2 particles I mean something of the form $\phi(x)=\phi_1(x) + \phi_2(x)$. For the harmonic oscillator in QM we have the wave-function of variable x (vector with finite index). No matter what the oscillator excitation state is, it depends just on one vector variable. I want the same thing for quantum fields, i.e., the wave-functional as a map from the field configuration (which is basically a vector with continuous index) to numbers. So no matter how many particles are in the system, I want it still to be a functional of the field configuration. – stanislav-iablokov Mar 13 '20 at 12:07
  • And the corresponding field configuration should also be a function of one vector variable: $\phi(x)$, not the 2-particle wave-function $\Phi(x1,x2)$. – stanislav-iablokov Mar 13 '20 at 12:08
  • I should rephrase it more explicitly. For QM you give me 3 numbers (coordinates) - I give you the amplitude to find a particle in the designated point in space. For QFT you give me a field configuration (infinite bunch of numbers) - I give you the amplitude to find the field in this configuration. So what this infinite bunch of numbers is for 2-particle state? And how does the corresponding wave-functional which extracts the amplitude looks like? – stanislav-iablokov Mar 13 '20 at 12:34
  • You are correct, this is not the right analogy for a wave functional in QFT. It is still called a wave function because us physicists are bad at naming things. – MannyC Mar 13 '20 at 12:42
  • @pgr I'm really not sure that I understand. In QM, only when you have a one particle problem your wave function is described by a set of $3$ numbers. When you, for example, solve the Helium-atom problem you have in your Hamiltonian 2 vectors, for the potential $-e^2/|r_1-r_2|$, and the wave-function similarly depends on $6$ coordinates. Basically, in 1st quantization we restrict the number of particles while in 2nd quantization we also quantize it and allow it to change. But for a fixed number of particles the analogy is as I wrote it. Are you interested in a single-particle description? –  Mar 13 '20 at 12:50