# A Canonical quantization of scalar fields

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1. Mar 19, 2016

### Higgsy

In the srednicki notes he goes from

$$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x)$$ to
$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p)$$

Where $$\tilde{a}(p) = \int \frac{d^{3}x}{(2\pi)^{\frac{3}{2}}}e^{-ipx}a(x)$$

Is this as simple as substituting or isn't there a commutator to get to
$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p)$$ ?

Do $$e^{ipx}$$ and $$P^{2}$$ commute? What about $$\nabla$$ and $$a(x)$$?

This is really bugging me any help would be greatly appreciated!

2. Mar 19, 2016

### vanhees71

This is obviously non-relativistic QFT. Then the scalar field (quantized Schrödinger field) in the Heisenberg picture has a mode decomposition
$$\hat{\psi}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^{3/2}} \exp[-\mathrm{i} (\omega_{\vec{p}}t-\vec{x} \cdot \vec{p})] \hat{a}(\vec{p}), \quad \omega_{\vec{p}}=\frac{\vec{p}^2}{2m}.$$
The free Hamiltonian is indeed given by
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \hat{\psi}^{\dagger}(t,\vec{x}) \left (-\frac{\Delta}{2m} \right ) \hat{\psi}(t,\vec{x}).$$
Now you can plug in the mode decomposition for $\hat{\psi}(t,\vec{x})$. After some simple but lengthy manipulations, using
$$\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \exp(\mathrm{i} \vec{p} \vec{x})=(2 \pi)^3 \delta^{(3)}(\vec{p})$$
you indeed find that
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_{\vec{p}} \hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
It's important, particularly in the beginning to somehow distinguish between usual real or complex numbers and operators. Above, I've written a hat above operators. Of course the operators are all linear operators on Hilbert space and thus commute with all usual real or complex numbers!

3. Mar 19, 2016

### Higgsy

But does $$P^{2}$$ commute with$$e^{ipx}$$ or how can I perform this commutation?

4. Mar 19, 2016

### vanhees71

In the above equation $P^2$ is obviously $\vec{p}^2$ and thus a real number, which commutes with all operators. Unfortunately I don't have my copy of Srednicky's book here; so I can't look at the context right now.