# A Interpreting the K-G Annihilation Operator Expression

1. Jul 12, 2016

### bolbteppa

I can derive $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ for a free real scalar Klein-Gordon field in three ways mathematically: the usual Fourier transform way in Peskin/Srednicki, an awesome direct a = ½(2a) = ... way (exercise!), and as a by-product of a clever way of mode-expanding the Hamiltonian, but I can't tell you why $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ should be the answer we'll get before doing any mathematical work in a qualitative way.

How would you interpret $$a(k), \psi, \pi$$ in such a way so as to make the above expression, & similarly for the creation operator, in the real and complex case, obvious - without sweeping the problem under the rug by referring to the analogous expression for quantum harmonic oscillators which should also be explainable with such a description?

Thanks!

2. Jul 12, 2016

### vanhees71

Take charged Klein-Gordon particles. Then the conserved current due to global U(1) (phase) invariance is
$$j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*)=\mathrm{i} \psi^* \overleftrightarrow{\partial_{\mu}} \psi.$$
This defines an invariant pseudo-scalar product in the space of square-integrable functions
$$\langle \psi_1|\psi_2 \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathrm{i} \psi_1^* \overleftrightarrow{\partial_0} \psi_2,$$
which is Lorentz invariant and constant in time due to Noether's theorem which implies that
$$\partial_{\mu} j^{\mu}=0.$$
For the plane-wave solution
$$u_{\vec{p}}(x)=N \exp(-\mathrm{i} x_{\mu} p^{\mu}), \quad p^0=\sqrt{\vec{p}^2+m^2}=E_{\vec{p}},$$
you have
$$\langle \vec{p}|\vec{p}' \rangle =2 E_{\vec{p}} (2 \pi)^3 |N|^2 \delta^{(3)}(\vec{p}-\vec{p}').$$
Thus the correct normalization is given by
$$N=\sqrt{\frac{1}{2 E_{\vec{p}} (2 \pi)^3}}.$$
The decomposition into momentum eigenmodes thus gives (now for the field operators!)
$$\hat{\psi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [\hat{a}(\vec{p}) u_{\vec{p}}(x) + \hat{b}^{\dagger}(\vec{p}) u_{\vec{p}}^*(x)].$$
This leads from the canonical equal-time commutator relations to
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=[\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{p}')] = \delta^{(3)}(\vec{p}-\vec{p}'),$$
identifying the $\hat{a}$ and $\hat{b}$ as annihilation operators for properly normalized momentum eigenstates.

3. Jul 12, 2016

### bolbteppa

Thanks a lot, that is a nice mathematical derivation of wave functions but I don't see how that explains why

$$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ should be an equivalent way to represent a(k) before doing any mathematical work in a qualitative way. Ho does the interpretation of a(k) as an annihilation operator qualitatively imply

$$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$

4. Jul 12, 2016

### vanhees71

Well, it's the scalar product I introduced above. Note that for the KG field
$$\pi=\dot{\phi}^*.$$

5. Jul 12, 2016

### bolbteppa

Fascinating, that's something I noticed a while ago and meant to follow up on: It is a stunning mystery to me why you would take the action, apply Noether for a $U(1)$ symmetry, get the conserved current $$j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*)=\mathrm{i} \psi^* \overleftrightarrow{\partial_{\mu}} \psi$$
and then notice that since we want the functions in our function space to also be invariant under $U(1)$ symmetries via
$$<g|f> = <g|e^{-i\theta} e^{i\theta}|f>$$
the $U(1)$ symmetry of our action and of wave functions seems to derive an inner product for us in a new situation based off it's invariance in an old one Is it not absolutely shocking? Any thoughts on making that more intuitive I don't see it yet but wow thanks a lot closer!

6. Jul 12, 2016

### vanhees71

Well, the symmetry principles are the way to argue about physics in such unintuitive situations as relativistic quantum field theory is. One important point is that if you do the same analysis for non-relativistic QT you get a positive definite scalar product for the fields, and thus the wave-function interpretation as probability amplitude (aka Born's rule) and thus the "1st-quantization formalism" works, while for relativistic fields, you don't get a positive definite charge, and you thus need field quantization to be able to apply the Stueckelberg trick, i.e., writing an creation operator in front of the negative-frequency modes and thus get a local representation of the Poincare group on the field-operator algebra. All this is nicely explained by the innocent-looking symmetry analysis a la Noether! So it's worth while to study the action formalism for classical field theory first and do this analysis and then start to understand the canonical quantization as a way to get a local representation of the Poincare group with a stable ground state in terms of a relativistic QFT.