Second quantization and quantum field theory

When I was an undergraduate and took the course of advanced quantum mechanics, I did not fully grasp the concept of second quantization. In my impression, second quantization was summarized as the following recipe:

1. Expand the wave function in some basis: $\psi(x, t)=\sum_k a_k(t)\phi_k(x)$. 
2. Impose the commutation relation like $[a_k, a^{\dagger}_{k'}]=\delta(k-k')$ (second quantization).
3. Replace operators $\hat{\cal{O}}$ by their expectation values $\int dx \psi(x)^{\dagger} \hat{\cal{O}} \psi(x)$.

I don't remember clearly whether the above summary was from my crash study for the final exam or exactly what I was taught on class. So I asked one of my best friends Joking for confirmation since we had that class together. Without any surprise, Joking told me that he never understand second quantization either. 

Now let me try to explain what second quantization actually is in this post. Before start, I have to say second quantization is really a misleading terminology. As shown below, there is no first or second quantization. I prefer its alias called occupation number representation. As the name suggests, it is just a representation similar to other representations in quantum mechanics.

Recap on quantum mechanics

Quantum mechanics is about states and operators. A quantum state is denoted by a ket $|q\rangle$ and an observable is associated with a Hermitian operator $\hat{\cal{O}}$.

In practice, we usually choose a basis $|\lambda\rangle$ and represent $|q\rangle$ and $\hat{\cal{O}}$ in such basis as \begin{eqnarray} |q\rangle &=& \int d\lambda\, |\lambda\rangle\,\langle \lambda |q\rangle\,.\tag{1} \\ \hat{\cal{O}}&=&\int d\lambda d\lambda' \,|\lambda\rangle\,\langle \lambda|\hat{\cal{O}}|\lambda'\rangle\,\langle \lambda'|\,.\tag{2}\end{eqnarray}

In other words, in the basis $|\lambda\rangle$, the representation of a state $|q\rangle$ is a complex column with entries $\langle\lambda | q\rangle$ and the representation of a operator $\hat{\cal{O}}$ is a complex matrix with elements $O_{\lambda\lambda'}\equiv\langle \lambda|\hat{\cal{O}}|\lambda'\rangle$. 

The above recap should look familiar to those who took a course on quantum mechanics. The occupation number representation we focus on is just a particular representation of state and operator in quantum mechanics. 

Occupation number representation

Creation and annihilation operator

Let $|q\rangle$ be the state (such as an energy level or band) occupied by a single particle with quantum numbers denoted by $q$, and $\left| q_1, q_2\right.\rangle$ be the state occupied by two identical particles, etc. The symmetrization postulate in quantum mechanics requires that \begin{equation}\langle \left. q'_2, q'_1 \right|q_1, q_2\rangle=\delta(q_1-q'_1)\,\delta(q_2-q'_2)\pm \delta(q_1-q'_2)\,\delta(q_2-q'_1)\,,\tag{3}\end{equation} with + sign for bosons and - sign for fermions.

We can define a so-called creation operator $a_q^{\dagger}$ that simply adds a particle with quantum number $q$ at the front of the particle list in the state: \begin{equation}a_{q}^{\dagger} |q_2\rangle \equiv |q, q_2\rangle\,.\tag{4}\end{equation} Since $\langle q'_2, q'_1| a^{\dagger}_{q}|q_2\rangle = \langle q_2| a_{q} |q'_1, q'_2\rangle^*$, the symmetrization postulate (1) gives that \begin{equation} a_{q}|q_1, q_2\rangle = \delta(q-q_1)\,|q_2\rangle \pm \delta(q-q_2)\, |q_1\rangle\,.\end{equation}

In other words, $a_q$ removes a particle from the state $|q_1,q_2\rangle$ and is thus called annihilation operator. The state with no particles occupied is called vaccum denoted by $|0\rangle$ with the properties $a_q|0\rangle = 0$ and $a^{\dagger}_q|0\rangle=|q\rangle$. Note that the creation and annihilation operators defined here do not relate to the real particle production and destruction. Particle number conserves in non-relativistic quantum mechanics, and particles only "jump" from one occupied state to another occupied state.

To derive the commutation relations between creation and annihilation operators, we can calculate \begin{eqnarray} a_{q_1}a^{\dagger}_{q_2}|q\rangle&=&\delta(q_1-q_2)\,|q\rangle \pm \delta(q_1-q)\,|q_2\rangle\,,\\ a^{\dagger}_{q_2}a_{q_1}|q\rangle&=&\delta(q_1-q)\,|q_2\rangle\,.\end{eqnarray} Since $|q\rangle$ is arbitrary, we have $a_{q_1}a^{\dagger}_{q_2} \mp a^{\dagger}_{q_2}a_{q_1}=\delta(q_1-q_2)$ with - sign for bosons and + sign for fermions. More formally, we have for boson\begin{equation} \left[a_{q_1},\,a^{\dagger}_{q_2}\right] = \delta(q_1-q_2)\,,\tag{5}\end{equation} and for fermion \begin{equation} \left\{a_{q_1},\,a^{\dagger}_{q_2}\right\} = \delta(q_1-q_2)\,.\tag{6}\end{equation}

In sum, a quantum state $|q_1,\cdots, q_n\rangle$ can be represented by creation (and annihilation) operators acting on the vaccum state: $a^{\dagger}_{q_1}\cdots a^{\dagger}_{q_n}|0\rangle$. Moreover, the quantum state can also be represented by creation (and annihilation) operators in another basis, for example, as \begin{equation}|q\rangle=\int d\lambda |\lambda\rangle\langle \lambda|q\rangle=\int d\lambda\, \langle \lambda|q\rangle\, a^{\dagger}_{\lambda}|0\rangle\,.\end{equation}

Observables in occupation number representation

We talked about states. What about operators of observables? Without loss of generality, take the simplest case of one boson as an example. Each operator $\hat{\cal{O}}$ in (2) has the form in the occupation number representation as \begin{equation} \hat{\cal{O}}=\int d\lambda d\lambda' \,O_{\lambda\lambda'}\, a^{\dagger}_{\lambda}\,a_{\lambda'}\,.\tag{7}\end{equation} Indeed, when writing $|\lambda\rangle$ and $\hat{\cal{O}}$ in terms of creation and annihilation operators as in (7), we can prove by the commutation relation (5) that $\langle 0 |a_{\lambda} \,\hat{\cal{O}}\, a^{\dagger}_{\lambda'}|0\rangle$ in the occupation number representation equals to the matrix elements $O_{\lambda\lambda'}$ in (2). I encourage readers to check this calculation for a sense on the algebras of creation and annihilation operators.

The single-boson Hamiltonian is of the form $\hat{\cal{H}}=\int d\lambda d\lambda' \,\langle \lambda| \hat{\cal{H}}|\lambda'\rangle\, a^{\dagger}_{\lambda}\,a_{\lambda'}$ in the occupation number representation. As a result, in Heisenberg picture, the annihilation (creation) operators evolves in time as \begin{equation} i\hbar \frac{d}{dt} a_{\lambda} = \left[a_{\lambda},\,\hat{\cal{H}}\right]=\int d\lambda' \langle \lambda| \hat{\cal{H}}| \lambda'\rangle a_{\lambda'}\,.\tag{8}\end{equation}

In sum, both quantum states and operators of observables can be "translated" to representations with creation and annihilation operators (in the Heisenberg picture), as described above. The so-called second quantization is nothing but a different representation of states and operators of observables.

The occupation number representation is widely used in many-body physics. People write down Hamiltonians in terms of creation and annihilation operators directly without "deriving" them.    

A concrete example

Consider the textbook problem of finite potential well:\begin{equation} \left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right)\psi_n(x)=E_n\,\psi_n(x)\,,\end{equation} where the quantum number is the energy level $n$. We now formulate this problem in the occupation number representation: The quantum state in the position basis is \begin{equation} |n\rangle=\int dx \langle x|n\rangle a^{\dagger}_x |0\rangle\equiv \int dx\,\psi_{n}(x)\,a_{x}^{\dagger}|0\rangle\end{equation} with $\psi_{n}(x)\equiv \langle x |n\rangle=\langle 0|a_x |n\rangle$, and the Hamiltonian is of the form \begin{equation} \hat{\cal{H}}=\int dx \,a_x^{\dagger} \left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right) a_x\,.\end{equation} Using the commutation relation (3), we can prove that $\hat{\cal{H}} |n\rangle = E_n |n\rangle$ leads to exactly the same equation on $\psi_n(x)$. Schrodinger equation is recovered!

Note that $a_x$ is the annihilation operator in the position basis. Do not confuse it with wave function. As shown in this concrete example, the wave function is actually the amplitude $\langle 0|a_x |n\rangle$.

Distinction from quantum field theory (QFT)

The above description for occupation number representation is for non-relativistic quantum mechanics such as many-body physics. The mathematical aspect is very similar, if not identical, to that in QFT. For example, my description of creation and annihilation operators is copied from Chapter 4.2 of Weinberg's QFT bible. Or in Chapter 1.2 of Mahan's many-body physics book, the creation and annihilation operators are derived from the canonical quantization in QFT. But underlying physics is different!

Such distinction is clearly pointed out at the very beginning of Srednicki's QFT book. Let me rephrase the arguments there. In non-relativistic quantum mechanics, the position is an operator: $\hat{x}|x\rangle = x |x \rangle$ while the time is only a parameter (or label) since there is no time operator. When going into the relativistic quantum theory in which space and time should be treated equally (Lorentz invariance), there only two options:

1. promote time to an operator
2. demote space to a label

For the first option, we can define a set of relativistic operators $\hat{X}^{\mu}(\tau)$ labeled by the proper time $\tau$ whose 0-component is the time operator. This option leads to the string theory when an extra label is introduced: $\hat{X}^{\mu}(\tau, \sigma)$. On the other hand, QFT formalism uses the second option: both position and time are the labels of some operator $\hat{\phi}$: $\hat{\phi}(x, t)$. Because of the label $x$, the operator $\hat{\phi}$ becomes a field, an operator-valued field. It is the "field" in the name of quantum field theory.

Remarks:

1. $\hat{\phi}(x, t)$ is the fundamental degrees of freedom in QFT. Since there is an operator in each spacetime point, QFT has an infinity number of degrees of freedom.
2. $\hat{\phi}(x, t)$ should not be confused with the annihilation operator $a_x(t)$ as described previously. $\hat{\phi}(x, t)$ represents an infinity number of operators labeled by $x$ while $a_x(t)$ is a Heisenberg operator in the basis $|x\rangle$ associated with the position operator $\hat{x}$.  
3. $\hat{\phi}(x, t)$ should NOT be confused with wave function. In analogous to non-relativistic quantum mechanics, $\hat{\phi}(x, t)$ is more like the position operator $\hat{x}$ rather than wave function:
•  As proved in Chapter 2.3.2 of Schwarz's QFT book, $\langle 0|\hat{\phi}(x,t)|\psi\rangle$ of Klein-Gordon field approximates the wave function $\langle x|\psi\rangle$ in the non-relativistic quantum mechanics.
• The wave function in QFT, if needed, should be a functional in terms of $\hat{\phi}(x, t)$, called Schrodinger functional.
4. In QFT, there are also creation and annihilation operators. A 'particle' state with momentum $p^{\mu}$ is described by creation operator: $|p\rangle=a_{p}^{\dagger}|0\rangle$. It turns out $\hat{\phi}(x, t)$ is the Fourier transform of $a_{p}$ and $a_{p}^{\dagger}$. So particles are just bundles of energy and momentum of the fields. Furthermore, 'generic' particles are truly delocalized, which is not surprising since we already touched the concept of wave-particle duality in old quantum theory.

In conclusion, although the occupation number representation shares similar math with QFT, the underlying physics is different because of the Lorentz invariance in QFT. In addition to operator-valued field, Lorentz invariance brings more consequences in QFT including antiparticles, particle production and destruction, spin-statistics theorem, etc. But that is another story.