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# 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Second quantization

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## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Second quantization

Since the modes of an electromagnetic field have the same classical equations as a simple harmonic oscillator, we can quantize them in the same way. We introduce an annihilation operator $a_p$ and its conjugate creation operator $a_p^{\dagger}$ for each wavenumber $\vec{p}$ and integrate over them to get the Hamiltonian for the free theory:
$$H_0=\int \frac{d^3 p}{(2 \pi)^3} \omega_p\left(a_p^{\dagger} a_p+\frac{1}{2}\right),$$
with
$$\omega_p=|\vec{p}|$$
This is known as second quantization. At the risk of oversimplifying things a little, that is all there is to quantum field theory. The rest is just quantum mechanics.

First quantization refers to the discrete modes, for example, of a particle in a box. Second quantization refers to the integer numbers of excitations of each of these modes. However, this is somewhat misleading – the fact that there are discrete modes is a classical phenomenon. The two steps really are (1) interpret these modes as having energy $E=\hbar \omega$ and (2) quantize each mode as a harmonic oscillator. In that sense we are only quantizing once. Whether second quantization is a good name for this procedure is semantics, not physics.
There are two new features in second quantization:

1. We have many quantum mechanical systems – one for each $\vec{p}$ – all at the same time.
2. We interpret the $n$th excitation of the $\vec{p}$ harmonic oscillator as having $n$ particles.
Let us take a moment to appreciate this second point. Recall the old simple harmonic oscillator: the electron in a quadratic potential. We would never interpret the states $|n\rangle$ of this system as having $n$ electrons. The fact that a pointlike electron in a quadratic potential has analogous equations of motion to a Fourier component of the electromagnetic field is just a coincidence. Do not let it confuse you. Both are just the simplest possible dynamical systems, with linear restoring forces. ${ }^1$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Field expansion

Now let us get a little more precise about what the Hamiltonian in Eq. (2.65) means. The natural generalizations of
$$\left[a, a^{\dagger}\right]=1$$
are the equal-time commutation relations
$$\left[a_k, a_p^{\dagger}\right]=(2 \pi)^3 \delta^3(\vec{p}-\vec{k})$$
The factors of $2 \pi$ are a convention, stemming from our convention for Fourier transforms (see Appendix A). These $a_p^{\dagger}$ operators create particles with momentum $p$ :
$$a_p^{\dagger}|0\rangle=\frac{1}{\sqrt{2 \omega_p}}|\vec{p}\rangle,$$
where $|\vec{p}\rangle$ is a state with a single particle of momentum $\vec{p}$. This factor of $\sqrt{2 \omega_p}$ is just another convention, but it will make some calculations easier. Its nice Lorentz transformation properties are studied in Problem 2.6.
$$\langle 0 \mid 0\rangle=1$$
$$\langle\vec{p} \mid \vec{k}\rangle=2 \sqrt{\omega_p \omega_k}\left\langle 0\left|a_p a_k^{\dagger}\right| 0\right\rangle=2 \omega_p(2 \pi)^3 \delta^3(\vec{p}-\vec{k}) .$$
The identity operator for one-particle states is
$$\mathbb{1}=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p}|\vec{p}\rangle\langle\vec{p}|,$$
which we can check with
$$|\vec{k}\rangle=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p}|\vec{p}\rangle\langle\vec{p} \mid \vec{k}\rangle=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p} 2 \omega_p(2 \pi)^3 \delta^3(\vec{p}-\vec{k})|\vec{p}\rangle=|\vec{k}\rangle$$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Second quantization

$$H_0=\int \frac{d^3 p}{(2 \pi)^3} \omega_p\left(a_p^{\dagger} a_p+\frac{1}{2}\right),$$

$$\omega_p=|\vec{p}|$$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Field expansion

$$\left[a, a^{\dagger}\right]=1$$

$$\left[a_k, a_p^{\dagger}\right]=(2 \pi)^3 \delta^3(\vec{p}-\vec{k})$$
$2 \pi$的因子是一种惯例，源于我们对傅里叶变换的惯例(见附录a)。这些$a_p^{\dagger}$算子产生动量为$p$的粒子:
$$a_p^{\dagger}|0\rangle=\frac{1}{\sqrt{2 \omega_p}}|\vec{p}\rangle,$$

$$\langle 0 \mid 0\rangle=1$$

$$\langle\vec{p} \mid \vec{k}\rangle=2 \sqrt{\omega_p \omega_k}\left\langle 0\left|a_p a_k^{\dagger}\right| 0\right\rangle=2 \omega_p(2 \pi)^3 \delta^3(\vec{p}-\vec{k}) .$$

$$\mathbb{1}=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p}|\vec{p}\rangle\langle\vec{p}|,$$

$$|\vec{k}\rangle=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p}|\vec{p}\rangle\langle\vec{p} \mid \vec{k}\rangle=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 \omega_p} 2 \omega_p(2 \pi)^3 \delta^3(\vec{p}-\vec{k})|\vec{p}\rangle=|\vec{k}\rangle$$

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