I was wondering if there is any way to interpret the fact that the Klein Gordon equation is a 2nd order PDE in time. I mean, normally you would expect that as soon as you fix the initial wavefunction, then the evolution of your system is fixed for all further moments in time. This is true for the Schrödinger and Dirac equation but not for the Klein Gordon equation. Is there any way to see why this is still "correct"?
[Physics] Why is the Klein Gordon equation of second order in time
klein-gordon-equationquantum mechanicsquantum-field-theoryschroedinger equation
Related Solutions
I think you are mixing up two different things:
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.
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.
The vacuum Dirac equation automatically implies the Klein-Gordon equation. It means that every solution to the vacuum Dirac equation is automatically a solution to the Klein-Gordon equation.
The converse of course doesn't hold. The most basic reason is that the Klein-Gordon equation should really act on scalars, a single bosonic field, while the minimum number of components for the $d=4$ Dirac equation is four (and they should be fermionic fields). So a general (or generic) valid solution to the Klein-Gordon equation is a valid solution to the Klein-Gordon equation (this much is a tautology, but you were asking about it), but it is not a solution to the Dirac equation.
Even if you combine 4 solutions to the Klein-Gordon equation, declare that they are 4 components of a Dirac spinor, and ask whether they solve the Dirac equation, the answer is No. It's because the Dirac equation is really "stronger" than the Klein-Gordon equations for its components. Effectively, the Dirac equation is first-order while the Klein-Gordon equation is second-order. The Dirac equation implies certain correlations between the spin (up/down) of the particle and the sign of the energy (positive/negative). The quadruplet of Klein-Gordon equations allows all combinations of spin up/down and the sign of the energy.
However, the most general quadruplet of solutions to the Klein-Gordon equation may be written as a solution of the Dirac equation with a positive mass and a solution to the Dirac equation with a negative (opposite) mass.
The Dirac equation describes spin-1/2 (and therefore "fermionic") particles such as electrons, other leptons, and quarks, while the Klein-Gordon equation describes spin-0 "scalar" (and bosonic) particles such as the Higgs boson. However, before they do the proper job, the "wave functions" have to be promoted to full fields and these fields have to be quantized.
Best Answer
There's a major difference between Schrödinger/Dirac equations and Klein-Gordon one: the former are complex while the latter is real. But if you think of them a little, you'll also find a major similarity.
If you represent complex numbers of the form $a+ib$ with matrices of the form $\pmatrix{a&-b\\ b&a}$, then you can easily rewrite Schrödinger's equation like this (taking all dimensional constants equal to $1$):
$$\left\{\begin{align} \dot R&=\hat H_RI-\hat H_I I\\ -\dot I&=\hat H_RR+\hat H_I R, \end{align}\right.$$
where $R$ and $I$ are real and imaginary parts of the wavefunction $\psi=R+iI$, and Hamiltonian $\hat H=\hat H_R+i\hat H_I$.
Now Klein-Gordon equation can also be rewritten in this form:
$$\left\{\begin{align} \dot\varphi&=A\\ \dot A&=\nabla^2\varphi-\mu^2\varphi. \end{align}\right.$$
In both cases you have two simultaneous equations of the first order. In both cases you have to specify two initial conditions. For Schrödinger's equation they are real $R$ and imaginary $I$ parts of the wavefunction, and for Klein-Gordon equation they are $\varphi$ and $\dot\varphi$.