# A Quantum fields and the harmonic oscillator

#### Higgsono

When defining quantum fields as a sum of creation and annihilation operators for each momenta, we do it in analogy with the simple example of the harmonic oscillator in quantum mechanics. But why do we assume that the coefficients in the expansion can be interpreted in the same way as in the case of the harmonic oscillator? Is this just an assumption that seems to work, or is it a good explanation for it? For in the case of the harmonic oscillator we had an Hamiltonian and so could easily prove that a and a* could be interpreted as creation and annihilation operators respectively. But when defining the quantum field, we are not given an Hamiltonian, but still assumes that we could expand or solution in terms of those operators.

Sorry if my explanation is poorly written, by I think you get the point.

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#### king vitamin

Science Advisor
Gold Member
We only do so for Gaussian quantum field theories where the expansion you mention diagonalizes the Hamiltonian. In that case, it is as justified as it is for the harmonic oscillator - if the decomposition manages to diagonalize our Hamiltonian, we are done! Of course, in both QFT and the simple harmonic oscillator we are also interested in adding more complicated terms and using perturbation theory, but then the degrees of freedom which diagonalize our unperturbed Hamiltonian are still important.

Outside of Gaussian theories and perturbation theory around them, an expansion in creation and annihilation operators does not make sense. For a generic interacting QFT, there is no easy way forward.

#### Higgsono

We only do so for Gaussian quantum field theories where the expansion you mention diagonalizes the Hamiltonian. In that case, it is as justified as it is for the harmonic oscillator - if the decomposition manages to diagonalize our Hamiltonian, we are done! Of course, in both QFT and the simple harmonic oscillator we are also interested in adding more complicated terms and using perturbation theory, but then the degrees of freedom which diagonalize our unperturbed Hamiltonian are still important.

Outside of Gaussian theories and perturbation theory around them, an expansion in creation and annihilation operators does not make sense. For a generic interacting QFT, there is no easy way forward.
ok. But why do we do it? Why do we expand the field in creation and annihilation operators? Is it because we amagine an oscillator at each point in spacetime? But then why so?

#### Paul Colby

Gold Member
ok. But why do we do it?
Once the plane wave coefficients are taken as canonical coordinates, the operator algebra follows from canonical quantization. Plane waves are chosen because the problem is isotropic and homogeneous in space and time. Other than plane wave modes may be chosen.

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
Consider a free classical scalar field $\phi \left( x, t \right)$ that satisfies the Klein-Gordon equation
$$\frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} + m^2 \phi = 0.$$
The Fourier transform of $\phi \left( x, t \right)$ is
$$\tilde{\phi} \left( k, t \right) = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx} \phi \left( x, t \right) dx.$$
Then,
$$\phi \left( x, t \right) = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{ikx} \tilde{\phi} \left( k, t \right) dk.$$
Substituting this into the Klein-Gordon equation gives ...

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