Schrödinger equation from Klein-Gordon equation?

Question

One can view QM as a 0+1 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:

$$i\partial_t\phi~=~-\frac{\Delta}{2m}\phi.$$

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.

Answer 1

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.

Answer 2

In this answer we include for clarity the correct factors$^1$ of $\hbar$ and $c$ in the calculation in chapter III.5 of Zee's QFT in a Nutshell.

1. The massive complex Klein-Gordon (KG) field $$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\exp\left(-\frac{imc^2t}{\hbar}\right)\phi(\vec{x},t).\tag{4} $$ is expanded around field configurations $\phi$ of energy $\approx mc^2$. (In particular, the expansion (4) suppresses negative energy/frequency components of the KG field.)

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

3. Since the correspondence (4) works at the level of Lagrangians (5), it also works at the level of equations of motion.

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$^1$ Since the Lagrangian density $\cal L$ has dimension of an energy density, the dimensions of the fields are $$ \dim|\Phi|^2~=~\frac{\text{Energy}}{(\text{Length})^{d-2}}\qquad\text{and}\qquad \dim|\phi|^2~=~(\text{Length})^{-d},$$ where $d$ is the number of spatial dimensions.

$^2$ In case of a background EM gauge potential $A_{\mu}$, make a minimal coupling substitution $$ \vec{\nabla}~\to~\vec{D}~=~\vec{\nabla}\mp\frac{iq}{\hbar}\vec{A}, \qquad\partial_t~\to~D_t~=~\partial_t\mp\frac{iq}{\hbar}A_t,$$ where the Minkowski signature is $(\mp,\pm,\ldots,\pm)$, respectively. Here the electric potential is $$\mp A_t~=~\mp c A_0 ~=~c A^0 ~=~c^2 A^t,$$ and the electric potential energy is $V=\mp q A_t$.