second quantization

Occupation number representation

In the study of quantum mechanics of many particles it is convenient to have at our disposal a compact representation of multiparticle states. For this it is useful to recall the raising and lowering operators of the harmonic oscillator, $a^\dagger\,$ and $a\,$, with which we can build arbitrary excited states by acting on the ground state $\left|0\right\rangle\,$ a number of times with the raising operator $a^\dagger\,$. For instance, the $k^\mathrm{th}\,$ excited state can be expressed as

$\left|n\right\rangle= \frac{{a^\dagger}^k}{\sqrt{k!}}\left|0\right\rangle$.

Without further justification, let us introduce a mapping between conventional multiparticle states of bosons, and states generated by generalized raising and lowering operators, which we will call creation and annihilation operators. Starting with a vacuum state $\left|0\right\rangle\,$, write the single particle state $\left|\psi_m\right\rangle_1\,$ as

$a^\dagger (\psi_m) \left|0\right\rangle\ \mapsto \left|\psi_m\right\rangle_1\,$.

Here the index $m\,$ labels a single-particle energy eigenstate. The two-particle state is simply

$\frac{1}{\sqrt{2!}}a^\dagger(\psi_n) a^\dagger(\psi_m)\left|0\right\rangle \mapsto \frac{1}{\sqrt{2}}\left( \left|\psi_n\right\rangle_1 \left|\psi_m\right\rangle_2 + \left|\psi_m\right\rangle_1 \left|\psi_n\right\rangle_2 \right)\,$,

while the three-particle state is

$\frac{1}{\sqrt{3!}} a^\dagger(\psi_p) a^\dagger(\psi_n) a^\dagger(\psi_m) \left|0\right\rangle \mapsto \frac{1}{\sqrt{3!} }\left( \left|\psi_m\right\rangle_1 \left|\psi_n\right\rangle_2 \left|\psi_p\right\rangle_3 +\left|\psi_n\right\rangle_1 \left|\psi_m\right\rangle_2 \left|\psi_p\right\rangle_3 +\left|\psi_p\right\rangle_1 \left|\psi_m\right\rangle_2 \left|\psi_n\right\rangle_3 +\left|\psi_n\right\rangle_1 \left|\psi_p\right\rangle_2 \left|\psi_m\right\rangle_3 +\left|\psi_m\right\rangle_1 \left|\psi_p\right\rangle_2 \left|\psi_n\right\rangle_3 +\left|\psi_p\right\rangle_1 \left|\psi_n\right\rangle_2 \left|\psi_m\right\rangle_3 \right)\,$,

etc. We therefore say that $a^\dagger(\psi_m)\,$ creates a particle in the $n^\mathrm{th}\,$ excited state. Note that our states are automatically symmetric under particle exchange, as a consequence of the fact that $[a^\dagger(\psi_m), a^\dagger(\psi_n)] = 0\,$. Such states are called Fock states, and their Hilbert space is called a Fock space.

We may also write the Hamiltonian as

$H = \sum_m E_m \, a^\dagger(\psi_m) a^\dagger(\psi_m)\,$.

Let us emphasize that whether or not we are truly allowed to create or destroy particles will depend on the application, and is properly the subject of quantum field theory. For now, we will merely use this technique as a formal tool, the utility of which should already be evident.

We can also treat multiparticle states of fermions in this way. Start with the raising and lowering operators of the fermionic harmonic oscillator, which satisfy

$\left\{c, c^\dagger\right\}= c c^\dagger + c^\dagger c= 1\,$,
$\left\{c, c\right\} = \left\{c^\dagger, c^\dagger\right\}= 0\,$.

Again, starting with the vacuum state $\left|0\right\rangle\,$, write the single particle state $\left|\psi_m\right\rangle_1\,$ as

$c^\dagger (\psi_m) \left|0\right\rangle\ \mapsto \left|\psi_m\right\rangle_1\,$.

The two-particle state is

$\frac{1}{\sqrt{2!}}c^\dagger(\psi_n) c^\dagger(\psi_m)\left|0\right\rangle \mapsto \frac{1}{\sqrt{2}}\left( \left|\psi_n\right\rangle_1 \left|\psi_m\right\rangle_2 - \left|\psi_m\right\rangle_1 \left|\psi_n\right\rangle_2 \right)\,$,

while the three-particle state is

$\frac{1}{\sqrt{3!}} c^\dagger(\psi_p) c^\dagger(\psi_n) c^\dagger(\psi_m) \left|0\right\rangle \mapsto \frac{1}{\sqrt{3!} }\left( \left|\psi_m\right\rangle_1 \left|\psi_n\right\rangle_2 \left|\psi_p\right\rangle_3 -\left|\psi_n\right\rangle_1 \left|\psi_m\right\rangle_2 \left|\psi_p\right\rangle_3 +\left|\psi_p\right\rangle_1 \left|\psi_m\right\rangle_2 \left|\psi_n\right\rangle_3 -\left|\psi_m\right\rangle_1 \left|\psi_p\right\rangle_2 \left|\psi_n\right\rangle_3 +\left|\psi_n\right\rangle_1 \left|\psi_p\right\rangle_2 \left|\psi_m\right\rangle_3 -\left|\psi_p\right\rangle_1 \left|\psi_n\right\rangle_2 \left|\psi_m\right\rangle_3 \right)\,$,

etc. Note that antisymmetry under particle exchange follows if we have the following anticommutation relations:

$\left\{c(\psi_m), c^\dagger(\psi_n) \right\} = \delta_{mn}\,$
$\left\{c(\psi_m), c(\psi_n) \right\} = \left\{c^\dagger(\psi_m), c^\dagger(\psi_n) \right\} = 0\,$

Non-relativistic field theory

It is not essential to express states in terms of stationary states of the single particle Hamiltonian, and we may instead consider eigenstates of other operators. For example, let

$a^\dagger(\mathbf{r}) \left|0\right\rangle \mapsto \left|\mathbf{r}\right\rangle_1\,$

represent a single-particle eigenstate of the position operator $\mathbf{r}\,$. Similarly, let

$a^\dagger(\mathbf{k}) \left|0\right\rangle \mapsto \left|\mathbf{k}\right\rangle_1\,$

represent a single-particle eigenstate of the momentum operator $\mathbf{p}\,$ with eigenvalue $\hbar \mathbf{k}\,$. Furthermore, we will impose the commutation relations $[a_\mathbf{k}, a^\dagger_{\mathbf{k}'}] = (2\pi)^3 \delta^{3}(\mathbf{k}-\mathbf{k}')\,$. Up to now, however, we have only heuristically introduced the operators $a\,$ and $a^\dagger\,$. Let us therefore ask: Is there a system, that when quantized, gives the correct multiparticle Hamiltonian and spectrum of states?

Suppose we have a non-interacting system of bosons of mass $m\,$. Then

$H = \int\!\frac{d^3k}{(2\pi)^3} \frac{\hbar^2\mathbf{k}^2}{2m} a^\dagger(\mathbf{k}) a(\mathbf{k})\,$.

Define the Heisenberg picture operator $\Psi\,$ given at some fixed time by

$\Psi(\mathbf{r})= \int\!\frac{d^3k}{(2\pi)^3} a(\mathbf{k}) e^{i \mathbf{k}\cdot\mathbf{r}}\,$.

Note that

 $i\hbar \frac{d \Psi}{d t}\,$ $= \left[ \Psi, H \right]\,$, $= \int\! \frac{d^3k}{(2\pi)^3} \frac{d^3k'}{(2\pi)^3} \frac{\hbar^2\mathbf{k}^2}{2m} \left[ a(\mathbf{k}'), a^\dagger(\mathbf{k}) a(\mathbf{k}) \right] e^{\frac{i}{\hbar} \mathbf{p}'\cdot\mathbf{r}}\,$, $= \int\! \frac{d^3k}{(2\pi)^3} \frac{\hbar^2\mathbf{k}^2}{2m} a(\mathbf{p}) e^{\frac{i}{\hbar} \mathbf{k}\cdot\mathbf{r}}\,$, $= -\frac{\hbar^2 }{2m} \nabla^2 \int\! \frac{d^3k}{(2\pi)^3} a(\mathbf{k}) e^{\frac{i}{\hbar} \mathbf{k}\cdot\mathbf{r}}\,$, $= -\frac{\hbar^2 }{2m} \nabla^2 \Psi\,$.

We see therefore that the operator $\Psi(\mathbf{r},t)\,$ obeys the Schrödinger equation. Therefore, if we treat $\Psi(\mathbf{r},t)\,$ as a classical field, and treat the Schrödinger equation as a classical wave equation, then upon quantizing this field we obtain a quantum mechanical description of multiple particles. This process has historically been called second quantization, although it should be emphasized that the system has only been quantized once. Any confusion lets up once we consider other fields such as the electromagnetic potential $A_\mu\,$.

The above dynamics follow from a Lagrangian such as

$L = \int\!d^3x\, \left( i \hbar \Psi^* \frac{\partial \Psi}{\partial t} + \frac{\hbar^2}{2m} \Psi^* \nabla^2 \Psi - V(\mathbf{r}) \Psi^* \Psi \right)\,$,

if we treat $\Psi^*\,$ and $\Psi\,$ as independent[1]. Properly, we should require that the Lagrangian be real, so we integrate by parts:

$L = \int\!d^3x\, \left( \frac{i \hbar}{2}\Psi^* \frac{\partial \Psi}{\partial t} - \frac{i\hbar}{2}\frac{\partial \Psi^*}{\partial t} \Psi - \frac{\hbar^2}{2m} \boldsymbol\nabla \Psi^* \cdot \boldsymbol \nabla \Psi - V(\mathbf{r}) \Psi^* \Psi \right)\,$.

In any case, the momentum conjugate to $\Psi\,$ is $i \hbar \Psi^*\,$. Following the general canonical quantization procedure, we should impose

$[\Psi(\mathbf{r}), i \hbar \Psi^\dagger (\mathbf{r}')] = i \hbar \delta^{3}(\mathbf{r}-\mathbf{r}')\,$,

i.e.,

 $[\Psi(\mathbf{r}), \Psi^\dagger(\mathbf{r}') ] =\delta^{3}(\mathbf{r}-\mathbf{r}')\,$ for bosons.

We may therefore interpret $\Psi^\dagger(\mathbf{r})\,$ and $\Psi(\mathbf{r})\,$ as creating or destroying an excitation located at $\mathbf{r}\,$, respectively, and refer to these as creation and annihilation operators. We may also interpret $\Psi^\dagger(\mathbf{r}) \Psi(\mathbf{r})\,$ as the number density of particles at the point $\mathbf{r}\,$.

Because the Schrödinger equation is dispersive, the position basis is not the most suited. In momentum space,

 $\int\!d^3x\, \Psi^\dagger(\mathbf{r})\Psi (\mathbf{r})\,$ $=\int\!d^3x\, \int\!\frac{d^3k}{(2\pi)^3} \frac{d^3k'}{(2\pi)^3} a^\dagger(\mathbf{k}) a(\mathbf{k}') e^{-i (\mathbf{k}-\mathbf{k}')\cdot \mathbf{r}}\,$, $=\int\!\frac{d^3k}{(2\pi)^3} \frac{d^3k'}{(2\pi)^3} a^\dagger(\mathbf{k}) a(\mathbf{k}') \delta^{3}(\mathbf{k}-\mathbf{k}')\,$, $=\int\!\frac{d^3k}{(2\pi)^3} a^\dagger(\mathbf{k}) a(\mathbf{k})\,$.

References

1. Alternatively, $Re\, \Psi\,$ and $Im\, \Psi\,$.