Why is the Klein Gordon equation of second order in time?

Question

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"?

Answer 1

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 $\dot\varphi$.

Answer 2

In non-relativistic theories, you have $$ \frac{\mathbf p^2}{2m}=E\to\frac{\hat{\mathbf p}^2}{2m}=\hat E $$ which gives us the normal $\nabla^2$ and $\partial_t$ terms we find in the Schroedinger equation. When you account for relativity, the above energy-momentum relationship becomes $$ \sqrt{\mathbf p^2c^2+m^2c^4}=E\tag{1} $$ When you turn those into operators, you get some wonky second-derivative under the square root term: $$ \sqrt{(-i\hbar\nabla)^2 c^2+m^2c^4}=i\hbar\frac{\partial}{\partial t} $$ which is difficult to work with. So instead of (1), you just use $$ \mathbf p^2c^2+m^2c^4=E^2\tag{2} $$ as the start. This naturally leads to the second order in time.

Answer 3

The time evolution of the wavefunction is always described by the Schroedinger equation $$ \dot{\psi} = -i \hat{H} \psi,$$ which is linear in time. The Klein-Gordon equation is a classical relativistic equation describing the propagation of disturbances in a field $\phi$ carrying a mass $m$. This equation is second order in time, just like the majority of classical equations of motion one could think of (e.g. Newton's second law). When you quantise such a field theory, the wavefunction is now a functional of the field, i.e. $\psi[\phi(x)]$ describes the probability amplitude for a particular field configuration taking the values $\phi(x)$ at all possible space-time points $x$. Like always in quantum theory, the dynamical variables are represented as non-commuting operators. In this case, the field $\phi$ therefore becomes an operator $\hat{\phi}$, and the Hamiltonian $\hat{H}$ is expressed in terms of these field operators.