# The use in solving the Klein Gordon equation?

1. Sep 14, 2015

### space-time

I've done some reading on quantum field theory, and I went over how when Schrodinger first derived this equation, he discarded because it yielded negative energy solutions, negative probability distributions and it gave an incorrect spectrum for the hydrogen atom. The book then went on to state that it was later discovered that the main flaw with the Klein Gordon equation was its interpretation rather than the equation itself. Rather than interpreting the solution φ as a wave function, apparently φ was promoted to being an operator. Specifically, it is apparently a field responsible for the creation and annihilation of particles.

Now here is where my question comes in. Sorry if it seems too basic:

When I solve the Klein-Gordon equation am I actually solving for the field, or am I just solving an equation that is now useless and serves no other purpose due to the latter interpretation of the equation and to the latter introduction of the Dirac equation? Here is why I ask this:

Solving the Klein Gordon equation is quite similar to solving Schrodinger's equation. The solution (just like in the case with the wave function in the Schrodinger equation) just comes out to be some scalar function of x, y, z, and t. While I know that wave functions are quantized in this manner, I don't know if a field is supposed to be quantized as a scalar function or some other mathematical object such as a matrix or some higher rank tensor.

That is why I ask you all if I am really solving for the quantum field by solving the Klein Gordon equation or if the field must be calculated by some other means.

(On another note, what kind of initial and boundary conditions would the Klein Gordon equation have?)

2. Sep 14, 2015

### Staff: Mentor

3. Sep 15, 2015

### Avodyne

In QFT, a spin-zero particle is an excitation of a scalar field. When interactions are neglected, this scalar field (which is an operator that acts on states) satisfies the KG equation (in the Heisenberg picture of QM).