One can view QM as a 1+0 dimensional QFT, fields are only depending on time and so are only called operators, and I know a way to derive Schrödinger's equation from Klein-Gordon's one.

Assuming a field $\Phi$ with a low energy $ E \approx m $ with $m$ the mass of the particle, by defining $\phi$ such as $\Phi(x,t) = e^{-imt}\phi(x,t)$ and developing the equation

$$(\partial^2 + m^2)\Phi~=~0$$

neglecting the $\partial_t^2 \phi$ then one finds the familiar Schrödinger equation:


Still, I am not fully satisfied about the transition field $\rightarrow$ wave function, even if we suppose that the number of particle is fixed, and the field now acts on a finite dimensional Hilbert Space (a subpart of the complete first Fock Space for a specific number of particles). Does someone has a another proposition/argument for this derivation?

Edit: for reference, the previous calculations are taken from Zee's book, QFT in a Nutshell, first page in Chapter III.5. Equivalently, see Wikipedia.


2 Answers 2


I think you are mixing up two different things:

  1. First, you can see QM as $0+1$ (one temporal dimension) QFT, in which the position operators (and their conjugate momenta) in the Heisenberg picture play the role of the fields (and their conjugate momenta) in QFT. You can check, for instance, that spatial rotational symmetry in the quantum mechanical theory is translated into an internal symmetry in QFT.

  2. Secondly, you can take the "non-relativistic limit" (by the way, ugly name because Galilean relativity is as relativistic as Special relativity) of Klein-Gordon or Dirac theory to get "non-relativistic" Schrödinger QFT, where $\phi$ (in your notation) is a quantum field instead of a wave function. There is a chapter in Srednicki's book where this issue is raised in a simple and nice way. There, you can also read about spin-statistic theorem and the wave function of multi-particle states. Let me add some equations that hopefully clarify that (I'm using your notation and of course there may be wrong factors, units, etc.):

The quantum field is: $$\phi \sim \int d^3p \, a_p e^{-i(p^2/(2m) \cdot t - p \cdot x)}$$

The Hamiltonian is:

$$H \sim i\int d^3x \left( \phi^{\dagger}\partial_t \phi - \frac{1}{2m}\partial _i \phi ^{\dagger} \partial ^i \phi \right) \, \sim \int d^3p \, \frac{p^2}{2m} \,a^{\dagger}_p a_p$$

The evolution of the quantum field is given by:

$$i\partial _t \phi \sim [\phi, H] \sim -\frac{\nabla ^2 \phi}{2m}$$

1-particle states are given by:

$$|1p\rangle \sim \int d^3p \, \tilde f(t,p) \, a^{\dagger}_p \, |0\rangle $$

(one can analogously define multi-particle states)

This state verifies the Schrödinger equation:

$$H \, |1p\rangle=i\partial _t \, |1p\rangle$$ iff

$$i\partial _t \, f(t,x) \sim -\frac{\nabla ^2 f(t,x)}{2m}$$

where $f(t,x)$ is the spatial Fourier transformed of $\tilde f (t,p)$.

$f(t,x)$ is a wave function, while $\phi (t, x)$ is a quantum field.

This is the free theory, one can add interaction in a similar way.

  • 2
    $\begingroup$ Thanks for the update, but I am specifically under a limit operation that would lead me to a "first quantization scheme" from a "second quantization scheme", alias, is it enough from the fact that I recover a Schrodinger equation and can then construct a conserved probability current with a positive density component ($j^0$) to reinterpret the field as a wave function whose modulus square gives probabilities? $\endgroup$
    – toot
    Commented Jul 19, 2012 at 17:41
  • $\begingroup$ I'm not sure what you are looking for. $f$ is a wave function that verifies the Schr. equation. The expectation value of $\phi$ is also a function that verifies the Schr. equation. So, as long as you can normalize them you get a quantum mechanical probabilistic interpretation. Have I answered your question? $\endgroup$ Commented Jul 19, 2012 at 22:48
  • 3
    $\begingroup$ @drake: The expectation value of $\phi$ is not the correct way to extract wavefunction from field. The right way is to consider the state $\int \psi(x)\phi^\dagger(x)$ where $\psi$ is a number and $\phi^\dagger(x)$ is the nonrelativistic field. $\endgroup$
    – Ron Maimon
    Commented Jul 27, 2012 at 4:25
  • 1
    $\begingroup$ @toot: The comment above is the correspondence between nonrelativistic fields and wavefunctions. If you smear a nonrelativistic creation field with a function, you produce a particle with wavefunction $\psi$. $\endgroup$
    – Ron Maimon
    Commented Jul 27, 2012 at 4:26
  • $\begingroup$ Thanks. You are just changing my notation: your $\psi$ is my $f$. The expectation value is a solution of the equation, it is usually called the classical field. $\endgroup$ Commented Jul 27, 2012 at 6:57

In this answer we include for clarity the correct factors of $\hbar$ and $c$ in the calculation in chapter III.5 of Zee's QFT in a Nutshell.

$$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\exp\left(-\frac{imc^2t}{\hbar}\right)\phi(\vec{x},t).\tag{4} $$

The free Klein-Gordon Lagrangian density is

$$\begin{align} {\cal L} ~=~&\left|\frac{1}{c}\partial_t \Phi\right|^2 - |\nabla \Phi|^2 -|\frac{mc}{\hbar}\Phi|^2\cr ~\stackrel{(4)}{=}~&\left|\left(\frac{\hbar}{c\sqrt{2m}}\partial_t-ic\sqrt{\frac{m}{2}}\right) \phi\right|^2 - \frac{\hbar^2}{2m}|\nabla \phi|^2 -\frac{mc^2}{2}|\phi|^2\cr ~=~&\left|\frac{\hbar}{c\sqrt{2m}}\partial_t\phi\right|^2 +\frac{i\hbar}{2}\left(\phi^{\dagger}\partial_t\phi-\partial_t \phi^{\dagger}\phi \right)- \frac{\hbar^2}{2m}|\nabla \phi|^2 \cr ~\longrightarrow& \frac{i\hbar}{2}\left(\phi^{\dagger}\partial_t\phi-\partial_t \phi^{\dagger}\phi \right)- \frac{\hbar^2}{2m}|\nabla \phi|^2\quad {\rm for }\quad c~\to~\infty. \end{align} \tag{5}$$ The last expression is the free Schrödinger Lagrangian density (up to total spacetime derivative terms), cf. e.g. this Phys.SE post.


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