The Origins of the Second Quantization

Question

I've been studying quantum theory for a while now and have a number of closely related questions that are not giving me any peace. I am not sure if such a long format is appropriate here, but I'd like to seize this opportunity and share my questions on this wonderful website.

One thing I'm trying to do in order to have a better understanding of the QFT, is to apply the formalism of occupation numbers to the lowest possible numbers of identical particles in the system. Such formalism is based on the usage of the wave vectors ('Fock states', 'second-quantised states') of the form \begin{equation} |n_{1},\,n_{2},\,... ,\,n_{k},\,... \rangle \quad, \end{equation} where $n_{k}$ stands for the number of particles in a symmetrised wave function in a state $k$ (I consider the Bose case only) . For example, if the total number of particles in the system is $3$, then \begin{equation} |\underset{k_1}{2},\,\underset{k_2}{1}\rangle = \dfrac{\sqrt{1!\,1!\,2!}}{\sqrt{3!}} ( |k_1\rangle |k_1\rangle |k_2\rangle + |k_1\rangle |k_2\rangle |k_1\rangle + |k_2\rangle |k_1\rangle |k_2\rangle ) \quad, \end{equation} where $|k_1\rangle$ is the one-particle state. Let me begin with demonstrating how this formalism works for a single harmonic oscillator.

If we set $\hbar = \omega = 1$ and ignore the vacuum energy $\hbar \omega /2 $, the Hamiltonian takes the form of \begin{equation} \hat{H} = \hat{a}^\dagger \hat{a} \quad, \end{equation} $\hat{a}^\dagger$ and $\hat{a}$ being the usual ladder operators (please refrain from calling them 'creation' and 'annihilation' operators, since $\hat{a}$ creates a single-particle state when acts on vacuum; otherwise it just changes the energy of the state; see the discussion later).

The correspondence between the second-quantised states and the eigenfunctions of the Hamiltonian is: \begin{equation} \begin{alignedat}{6} |0\rangle&\equiv|0,\,0,\,0,\,...\rangle \quad,\\ |1\rangle&\equiv|\underset{1}{1},\,0,\,0,\,...\rangle \quad,\\ |2\rangle&\equiv|0,\,\underset{2}{1},\,0,\,...\rangle \quad,\\ |3\rangle&\equiv|0,\,0,\,\underset{3}{1},\,...\rangle \quad. \end{alignedat} \end{equation} By definition, the operator $\hat{a}^\dagger_k$ creates a particle in the state $k$, while $\hat{a}_k$ annihilates the state with no particles of type $k$: \begin{equation}\begin{alignedat}{4} \hat{a}^\dagger_k\,|0\rangle = |...,\,\underset{k}{1},...\rangle \quad,\qquad \hat{a}_k \, |...,\,\underset{k}{0},...\rangle = 0 \quad. \end{alignedat}\end{equation} From this definition, it is clear that we cannot define the operators $\hat{a}_k^\dagger$ in the following way: \begin{equation} \text{(wrong!)}\quad{\hat{a}_k^\dagger \propto (\hat{a}^\dagger)^k} \quad. \end{equation} Indeed, the requirement of annihilation would not be satisfied in this case: \begin{equation}\begin{alignedat}{4} (&\hat{a})^2\, &&|...,\,\underset{5}{1},...\rangle = (\hat{a})^2\, |5\rangle \propto |3\rangle \quad,\\ &\hat{a}_2\, &&|...,\,\underset{5}{1},...\rangle = 0 \quad. \end{alignedat}\end{equation}

This construction may look pretty unusual. To better understand the rules of the game, employ the matrix formalism: \begin{equation}\begin{gathered} \hat{a}^\dagger \cong \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots \\ \sqrt{1} & 0 & 0 & 0 & \cdots \\ 0 & \sqrt{2} & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}\quad,\qquad \hat{a} \cong \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{2} & 0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}\quad,\\ |0\rangle \cong \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix} \quad. \end{gathered}\end{equation}

Now one can easily guess the form of the operators which obey the desired requirements: \begin{equation}\begin{gathered} \hat{a}^\dagger_1 \cong \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots \\ 1 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}\quad,\qquad \hat{a}^\dagger_2 \cong \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}\quad,\\ \hat{a}^\dagger_3 \cong \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}\quad. \end{gathered}\end{equation}

With the use of those operators, one can immediately write the Hamiltonian of a single harmonic oscillator in the 'QFT form': \begin{equation} \hat{H} = \sum \limits_{k=1}^\infty\, \omega_k\, \hat{a}_k^\dagger\,\hat{a}_k \quad, \qquad \omega_k = k\quad. \end{equation}

Regarding the operators $\hat{a}_k^\dagger$. First of all, they turn out to be nilpotent. Second, we have now some issues with the commutation relations which we have never discussed so far. Of course, we would prefer the newly constructed operators to obey the standard Heisenberg commutation relations \begin{equation} [\hat{a},\,\hat{a}^\dagger] = 1 \quad. \end{equation} However, the matrix form tells us that each operator $\hat{a}_k^\dagger$, together with its partner, gives birth to the $\mathfrak{su}(2)$ algebra. \begin{equation} [\hat{a}_k,\,\hat{a}^\dagger_k] \cong \begin{pmatrix} 1 & \cdots & 0 & \cdots \\ \vdots & \ddots & 0 & \cdots \\ 0 & 0 & -1 & \cdots \\ \vdots & \vdots & 0 & \ddots \end{pmatrix} \end{equation} Let's summarise these sad results in a slightly different form. Being sandwiched by the vacuum, the commutator $[a_k,\,\hat{a}^\dagger_k]$ behaves decently: \begin{equation}\begin{alignedat}{4} \langle 0| \, [\hat{a}_k,\,\hat{a}^\dagger_k] \,|0\rangle = \langle 0| \, \hat{a}_k\,\hat{a}^\dagger_k \,|0\rangle - \langle 0| \, \hat{a}^\dagger_k\,\hat{a}_k \,|0\rangle = 1 - 0 = \phantom{-} 1 \quad. \end{alignedat}\end{equation} However, we get into trouble as we apply the creation operator to the particle which was already created: \begin{equation}\begin{alignedat}{4} \langle \underset{k}{1}| \, [\hat{a}_k,\,\hat{a}^\dagger_k] \,|\underset{k}{1}\rangle = \langle \underset{k}{1}| \, \hat{a}_k\,\hat{a}^\dagger_k \,|\underset{k}{1}\rangle - \langle \underset{k}{1}| \, \hat{a}^\dagger_k\,\hat{a}_k \,|\underset{k}{1}\rangle = 0 - 1 = -1\quad. \end{alignedat}\end{equation} In the second line we were trying to apply the creation operator $\hat{a}^\dagger_k$ to the state $|\underset{k}{1}\rangle$, i.e. create another particle. This, however, does not make much sense within the single-particle formalism. Therefore, it seems natural to suggest that in the case of two particles, we would be able to make one more step and the commutation relations would take form of something like ${\hat{a}^\dagger_k\cong\operatorname{diag} \{1,\,1,\,?\}}$ in the ${\{|0\rangle,\,|\underset{k}{1}\rangle,\,|\underset{k}{2}\rangle\}}$ basis.

By the end of the day, we can say that:

$\quad 1.\quad$ We managed to construct the operators defined through \begin{equation} \begin{alignedat}{6} \hat{a}_k\, |n_1,\,n_2,\,... ,\,n_k,\,... \rangle &= \sqrt{n_k}\,&&|n_1,\,n_2,\,... ,\,n_k-1,\,... \rangle \quad,\\ \hat{a}_k^\dagger\, |n_1,\,n_2,\,... ,\,n_k,\,... \rangle &= \sqrt{n_k+1}\,&&|n_1,\,n_2,\,... ,\,n_k+1,\,... \rangle \quad. \end{alignedat} \end{equation}

$\quad 2.\quad$ Those operators only obeyed the Heisenberg commutation relations when sandwiched by the vacuum states.

QUESTION 1. Does everything look correct so far?

Now, let's try to make a step further and extend the construction to a system of two identical particles. First of all, at this point one should stop thinking of one-particle states as of eigenstates of the Hamiltonian. Such basis would be inconvenient due to the issues with the degeneracy. It's better to think of the $|k\rangle$ states as of states with definite momentum or position. Then, the one-particle Hamiltonian writes as: \begin{equation} \hat{H}_{1} = \int |k\rangle H_{k\,m} \langle m|\, \operatorname{d}k \operatorname{d}m \quad. \end{equation}

The Fock space of the two-particle system is the symmetrised tensor product of two one-particle spaces. The basis vectors are now defined as: \begin{equation}\begin{alignedat}{4} |0,\,0,\,0,\,...\rangle &\equiv |0\rangle\otimes|0\rangle \equiv |0\rangle \quad&&,\\ |\underset{k}{1} \rangle &\equiv \dfrac{1}{\sqrt{2}} (|0\rangle\otimes|k\rangle + |k\rangle\otimes|0\rangle) \quad&&,\\ |\underset{k}{1},\,\underset{m}{1} \rangle &\equiv \dfrac{1}{\sqrt{2}} (|k\rangle\otimes|m\rangle + |m\rangle\otimes|k\rangle) \quad&&,\\ |\underset{k}{2} \rangle &\equiv |k\rangle\otimes|k\rangle \quad&&. \end{alignedat}\end{equation}

From now, we will denote the one-particle operators by ${{}^{(j)}\hat{a}_k^\dagger}$: \begin{equation}\begin{alignedat}{4} &{{}^{(1)}\hat{a}_k^\dagger}\,|0\rangle &&\equiv ({{}^{(1)}\hat{a}_k^\dagger} \otimes \hat{1}) \,|0\rangle &&\equiv |k\rangle \otimes |0\rangle \quad&&,\\ &{{}^{(2)}\hat{a}_k^\dagger}\,|0\rangle &&\equiv (\hat{1} \otimes {{}^{(2)}\hat{a}_k^\dagger}) \,|0\rangle &&= |0\rangle \otimes |k\rangle \quad. \end{alignedat}\end{equation}

We define the operators $\hat{a}_k^\dagger$ and $\hat{a}_k$ by their action on the basis vectors in the following way: \begin{equation}\begin{alignedat}{4} \hat{a}^\dagger_k\,|0\rangle &= &&|\underset{k}{1}\rangle \quad, \qquad \hat{a}_k \, |...,\,\underset{k}{0},...\rangle = 0 \quad, \\ \hat{a}^\dagger_k\,|1\rangle &= 2\,&&|\underset{k}{2}\rangle \quad. \end{alignedat}\end{equation}

QUESTION 2. How to express these operators in terms of one-particle operators? I.e. in the form: \begin{equation} \sum \limits_k \alpha_k\,{}^{(1)}\hat{A}_k\otimes {}^{(1)}\hat{B}_k \quad. \end{equation}

I think, in principle this should be possible, since the tensor product of bases of the spaces of operators should form a basis in the space of operators acting in the tensor product vector space. More precisely, if any two linear operators ${}^{(1)}\hat{A}$ and ${}^{(1)}\hat{B}$ can be written in the form \begin{equation} {}^{(1)}\hat{A} = \sum \limits_{i,j} A^{ij}\,{}^{(1)}\hat{e}_{ij} \quad,\qquad {}^{(2)}\hat{B} = \sum \limits_{k,l} B^{kl}\,{}^{(2)}\hat{e}_{kl} \quad, \end{equation} then any linear operator acting in the tensor product vector space can be written in the form \begin{equation} \hat{C} = \sum \limits_{i,j,k,l} C^{ijkl}\,{}^{(1)}\hat{e}_{ij}\otimes{}^{(2)}\hat{e}_{kl} \quad. \end{equation}

It seems natural to suggest something like \begin{equation} \tilde{a}^\dagger_k = \alpha\,({{}^{(1)}\hat{a}_k^\dagger} \otimes \hat{1} + \hat{1} \otimes {{}^{(2)}\hat{a}_k^\dagger}) \quad. \end{equation} However, not only this option fails to produce the correct numerical coefficients in the definition of the operator $\hat{a}^\dagger_k$, but it also provides us with unsatisfactory commutation relations. For $\alpha = 1/\sqrt{2}$ one gets: \begin{equation}\begin{alignedat}{4} \langle 0| \, [\tilde{a}_k,\,\tilde{a}^\dagger_k] \,|0\rangle &= &&1 \quad&&, \\ \langle \underset{k}{1}| \, [\tilde{a}_k,\,\tilde{a}^\dagger_k] \,|\underset{k}{1}\rangle &= &&2\quad&&,\\ \langle \underset{k}{2}| \, [\tilde{a}_k,\,\tilde{a}^\dagger_k] \,|\underset{k}{2}\rangle &= -&&1\quad&&. \end{alignedat}\end{equation}

Good news is that if we simply follow the definitions of $\hat{a}_k^\dagger$ and $\hat{a}_k$ above (the ones through the action on the basis vectors), the commutation relations are, of course, satisfied: \begin{equation}\begin{alignedat}{4} \langle 0| \, [\hat{a}_k,\,\hat{a}^\dagger_k] \,|0\rangle &= 1 \quad&&, \\ \langle \underset{k}{1}| \, [\hat{a}_k,\,\hat{a}^\dagger_k] \,|\underset{k}{1}\rangle &= 1\quad&&,\\ \langle \underset{k}{2}| \, [\hat{a}_k,\,\hat{a}^\dagger_k] \,|\underset{k}{2}\rangle &= \hspace{.35em}?\quad&&. \end{alignedat}\end{equation} I am not sure about the last equation, since, as it was already mentioned, in the system of $n$ particles, it does not make much sense to act with the operator $\hat{a}^\dagger_k$ on the state $|\underset{k}{n}\rangle$, and so, I guess, we should not really worry about it.

Now let me explain why all these questions arise. Basically, I'm not satisfied with the process of canonical quantisation of the fields. When quantising the field mode by mode, people formally perform same operation as in the case of harmonic oscillator. However, the meaning we give for the results is quite different.

Let's first ask ourselves why we call the $5^{\text{th}}$ energy state of harmonic oscillator ''the particle with energy $5\hbar\omega/2$'', not ''the five-particle state''. Is it just a matter of convention? What stops us from using one-particle QM for describing the motion of few particles?

We all we told that to some extent it makes sense to treat the particle as a Gaussian wave packet. Let \begin{equation} |\psi(x,\,t)\rangle = |g\,(x-x_0-v_{\text{group}}t)\rangle \end{equation} be the time-dependent state, a particle localised around $(x_0+v_{\text{group}}t)$ at the time $t$. Let's now consider the superposition of two such states: \begin{equation} |\widetilde{\psi}(x,\,t)\rangle = |g\,(x-x_0-v_{\text{group}}t)\rangle + |g\,(x-x_1+v_{\text{group}}t)\rangle\rangle \end{equation}

Why do we say that such wave vector corresponds to a single particle? From the theoretical point of view it's because we have started with the classical system of just a single particle. The more interesting is experimental approach. After the measurement, the wave function collapses and forgets about both 'separate' Gaussians. Had we set two detectors in the different spacial points, only one of them would observe the particle (this is basically the double slit experiment). That's why we treat the excited states of harmonic of harmonic oscillator (the potential actually does not play any role) as different energy levels of a single particle, but not as multi-particle state.

Now, let's perform first few steps of the field quantisation. For simplicity, let's use Klein-Gordon field. Even though it can be thoght of as a quantum-mechanical Schrödinger field in one-particle formalism (despite some problems with the negative energies, e.g. see Davydov), the trick is to consider it first as a classical wave equation. \begin{equation} (\partial^\mu\partial_\mu+m^2)\,\phi(x)=0\quad. \end{equation} After a (non-gauge-invariant) Fourier transform \begin{equation} \phi(\vec{x},t) = \int \dfrac{d^3 p}{(2\pi)^3} \operatorname{e}^{i\,\vec{p}\cdot\vec{x}} \phi(\vec{p},t) \quad, \end{equation} we arrive to \begin{equation} \left(\partial^2_t+(\vec{p}^{\,2}+m^2)\right)\,\phi(\vec{p},t)=0\quad, \end{equation} which looks like an equation for the harmonic oscillator with frequency $\omega_{\vec{p}} = \sqrt{\vec{p}^{\,2}+m^2}$. In what follows we assume that $\vec{p}$ remains unchanged. Here it's just an index enumerating the independent oscillation modes. Let's rewrite the equation and compare it to the regular harmonic oscillator: \begin{equation}\begin{alignedat}{4} &(\partial^2_t+\omega_{\vec{p}}^2)\,\phi_{\vec{p}}(t)&&=0\quad&&,\\ &(\partial^2_t+\omega^2)\,q(t)&&=0\quad&&. \end{alignedat}\end{equation} The quantisation turns $q(t)$ into the operator $\hat{q}(t)$ which obeys same differential equation (in the Heisenberg formalism). \begin{equation} (\partial^2_t+\omega^2)\,\hat{q}(t)=0\quad. \end{equation} In the Schrödinger picture it can be written as \begin{equation} \hat{q} = \hat{q}^\dagger = \dfrac{1}{\sqrt{2\omega}}(\hat{a} + \hat{a}^\dagger) \quad. \end{equation} As well as $\hat{a}^\dagger$ (up to a multiplier), $\hat{q}$ can be used for creating the first excited state from the vacuum: \begin{equation} \sqrt{2\omega}\,\hat{q} |0\rangle = \hat{a}^\dagger |0\rangle = |1\rangle. \end{equation} If we want to climb the ladder further, it's only $\hat{a}^\dagger$ who does the job properly.

We have already discussed that, in case of the harmonic oscillator, the states $|n\rangle$ cannot be treated as multi-particle states; one necessarily should call them the excited states of a single particle.

Nevertheless, in QFT we treat the state ${\dfrac{1}{2}(\hat{a}_{\vec{p}}^\dagger)^2\,|0\rangle}$ as a two-particle state! \begin{equation} \dfrac{1}{2}(\hat{a}_{\vec{p}}^\dagger)^2\,|0\rangle = | \underset{\vec{p}}{2} \rangle \quad, \end{equation} which looks meaningless when considered as a state of a single harmonic oscillator. The full Fock space of the system is then, of course, just a tensor product of those.

QUESTION 3. In QFT, why do we treat those excitations as multi-particle states?

Again, let me try to find an answer myself. In reality, the procedure of the field quantisation is a kind of a formal trick. A more rigorous approach is to consider the large number $N$ of particles in the Schrödinger formalism (they same way we did with two particles) and then take the limit~$N \to \infty$.

This approach is often considered old-school. It's poorly explained in Landau-Lifshitz, but also in great detail in 'Quantum Mechanics' by Blokhintsev. It requires few steps:

$\quad 1. \quad$ Determine the action of the multi-particle Hamiltonitan \begin{equation} \hat{H} = \sum \limits_{m=1}^N \hat{H}_m \end{equation} on the symmetrised states $| ...,\,\underset{k-1}{N_{k-1}},\,\underset{k}{N_k},\,\underset{k+1}{N_{k+1}},\,... \rangle$. I.e. calculate the matrix elements \begin{equation} \langle...,\,\underset{k-1}{M_{k-1}},\,\underset{k}{M_k},\,\underset{k+1}{M_{k+1}},\,... | \, \hat{H}\, | ...,\,\underset{k-1}{N_{k-1}},\,\underset{k}{N_k},\,\underset{k+1}{N_{k+1}},\,... \rangle\quad. \end{equation}

$\quad 2. \quad$ Define the the creation and annihilation operators by their action on those states: \begin{equation} \begin{alignedat}{6} \hat{a}_k\, |n_1,\,n_2,\,... ,\,n_k,\,... \rangle &= \sqrt{n_k}\,&&|n_1,\,n_2,\,... ,\,n_k-1,\,... \rangle \quad,\\ \hat{a}_k^\dagger\, |n_1,\,n_2,\,... ,\,n_k,\,... \rangle &= \sqrt{n_k+1}\,&&|n_1,\,n_2,\,... ,\,n_k+1,\,... \rangle \quad. \end{alignedat} \end{equation} Of course, such definition implies that the commutation relations are satisfied (unless we try to create more particles it's allowed by the formalism).

$\quad 3. \quad$ Show that the Hamiltonian can be nicely written in terms of creation and annihilation operators acting on the multi-particle states in the usual manner. \begin{equation} \hat{H} = \sum \limits_{m,n} H_{m\,n} \hat{a}^\dagger_m\,\hat{a}_n + ...\quad, \end{equation} where $H_{m\,n}$ are just the one-particle matrix elements, while the dots stand for the interaction terms. The resulting statement is not as trivial as it seems and requires some work.

The final expression for the Hamiltonian is same as in the case of replacing the classical fields by their 'second-quantised' versions. The particular reason why I like this approach is because it allows us to avoid using two most questionable assumptions of the canonical field quantisation:

$\quad\bullet\quad$''Let's treat the Schrödinger / Klein-Gordon / ... ~ field as a classical field and the quantise its Fourier modes...''

$\quad\bullet\quad$ ''Let's treat the excitations of quantised Fourier modes as multi-particle states...''

Instead, we define the creation operators by their action on vacuum and postulate that they create a symmetrised state in the multi-particle system.

One may wonder how this approach should work in EM, since in that case we have a field already at the classical level. Here is my guess formulated as a question.

QUESTION 4. Can we 'second-quantise' the EM field by treating the Maxwell equations as a Schrödinger equation (e.g. see books by Fushchich and Nikitin) and then considering the multi-particle states of those?

I guess, the typical conclusion people make after thinking on such topics is: ''Well, both approaches (symmetrising single-particle states and quantising the Fourier modes) should work; both have various pros and cons; the field quantisation if more useful because it leads us to the correct answer faster (so that we could finally dive into calculating the amplitudes and cross-sections).''

Finally, I would like to ask my most serious question, the one which bothered me for years, and which is perfectly illustrated by the Coleman's 'missing box method' illustration (from his Physics 253a lectures). One way to go from classical mechanics to field theory, through the 'QM' box, is: \begin{equation} \begin{gathered} \begin{gathered} \text{Quantise the}\\ \text{classical system}\\ \text{in a usual way} \end{gathered} \quad\to\quad \begin{gathered} \text{Take many copies}\\ \text{of such systems}\\ \text{and construct}\\ \text{symmetrised states} \end{gathered} \quad\to\\ \to\quad \begin{gathered} \text{Define creation}\\ \text{and annihilation}\\ \text{operators by}\\ \text{action on those states} \end{gathered} \quad\to\quad \begin{gathered} \text{Express the}\\ \text{Hamiltonian in}\\ \text{in terms of}\\ \text{such operators} \end{gathered} \end{gathered} \end{equation} Another way , through the 'Classical field theory' box, is: \begin{equation} \begin{gathered} \text{Make a Fourier}\\ \text{transform of the}\\ \text{classical field}\\ \text{equation} \end{gathered} \quad\to\quad \begin{gathered} \text{Quantise each}\\ \text{independent mode as a}\\ \text{harmonic oscillator}\\ \end{gathered} \quad\to\quad \begin{gathered} \text{Postulate that the}\\ \text{climbing the ladder}\\ \text{is but the creation}\\ \text{of new particles} \end{gathered} \end{equation} Interestingly, in the second approach we never explicitly discuss the symmetrisation. We just say that the excitations of a harmonic oscillator labeled by certain quantum number are identical particles (at this point, one may think about the spin-statistics theorem).

QUESTION 5. Why are the two procedures equivalent and lead to the same result?

More precisely, why do we end up with same Fock spaces, same creation and annihilation operators?

I know, all ingredients are right here, in front of me... but still cannot collect the puzzle. And I am not satisfied with a simple comparison of the outputs of two black boxes.

Any ideas or references to various sources will be greatly appreciated.

Many thanks!

Answer 1

@Andrew's answer provides the big picture, but I'd like to give a few more specific pointers that hopefully may help.

Questions 1-2: Does everything look correct so far? How to express these operators in terms of one-particle operators?

So you want to set up single-particle analogues of the ladder operators using eigenstates of the 1st quantized Hamiltonian, then use those to construct the second quantization ladder operators in the symmetrized multi-particle space. The problem is that such a procedure may not be possible. Here is why:

The entire second quantization framework rests on an isomorphism between the (anti)symmetric subspace of the N-particle Hilbert space and an abstract direct product of "mode" Hilbert spaces, each constructed around its own ladder operator algebra. The ladder operators ${\hat a}_n$, ${\hat a}^\dagger_n$ in the abstract/"mode" Hilbert space obviously do have equivalents on the original N-particle (anti)symmetric subspace. But cannot be expressed as symmetrized sums of similar single-particle operators. To see why, let us suppose the ${\hat a}_n$, ${\hat a}^\dagger_n$ can indeed be expressed as such symmetrized sums, reading $$ {\hat a}_n \sim \sum_{k=1}^N{ {\hat \alpha}^{(k)}_n},\;\;\; {\hat a}^\dagger_n \sim \sum_{k=1}^N{ \left( {\hat \alpha}^{(k)}_n\right)^\dagger} $$ up to some suitable normalization factor, where ${\hat \alpha}^{(k)}_n$, $\left( {\hat \alpha}^{(k)}_n\right)^\dagger$ are the desired "single-particle ladder operators" for particle $k$ and "mode"/eigenstate $n$, and each term is to be understood in the sense of ${\hat \alpha}^{(k)}_n\otimes \left[\bigotimes_{j\neq k}{\hat I}^{(j)}\right]$. The latter can be naturally assumed to (anti)commute for different particles and eigenstates (pre-symmetrization), so $\left[{\hat \alpha}^{(k)}_m, {\hat \alpha}^{(j)}_n\right]_\pm = \left[{\hat \alpha}^{(k)}_m, \left( {\hat \alpha}^{(j)}_n\right)^\dagger\right]_\pm = 0$ for any $j\neq k$ and for $n\neq m$ when $j = k$. Then the (anti)commutation relations for the ${\hat a}_n$-s, $$ \left[{\hat a}_m, {\hat a}_n\right]_\pm = 0,\;\;\; \left[ {\hat a}_m, {\hat a}^\dagger_n\right]_\pm = \delta_{mn}{\hat I} $$ require $$ 0 = \left[{\hat a}_m, {\hat a}_n\right]_\pm \sim \sum_{k=0}^N{\left[{\hat \alpha}^{(k)}_m, {\hat \alpha}^{(k)}_n\right]_\pm }, \;\;\;\; \delta_{mn}{\hat I} = \left[{\hat a}_m, {\hat a}^\dagger_n\right]_\pm \sim \sum_{k=0}^N{\left[{\hat \alpha}^{(k)}_m,\left( {\hat \alpha}^{(k)}_n\right)^\dagger\right]_\pm } $$ The important point here is that in either case the lhs does not depend on $N$. Then the first eq. above implies that each term on the rhs must vanish identically, which is great. But things are no longer clear cut for the 2nd eq. Whatever prescription we might propose for $\left[{\hat \alpha}^{(k)}_m,\left( {\hat \alpha}^{(k)}_n\right)^\dagger\right]_\pm $, there is no way to normalize the sum such that the result is non-zero yet independent of $N$ for any $N$.

Bottom line: however intuitive it may seem at first sight, this is not the way to go.

Question 3: In QFT why do we treat those excitations as multi-particle states?

Short answer: Due to the isomorphism with the many-particle framework. Sometimes, as in solid state physics, the excitations are referred to as quasi-particles for this exact reason. Think phonons and excitons. Same goes for photons and any other field quanta, but for historical reasons they are referred to as "particles".

Question 4: Can we 'second-quantise' the EM field by treating the Maxwell equations as a Schrödinger equation (e.g. see books by Fushchich and Nikitin) and then considering the multi-particle states of those?

Unfortunately I'm not familiar with the book you mention, but you may want to google "Maxwell equations in Dirac form", for instance this paper and this Wikipedia page (especially refs. within). Never saw this used as a starting point for second quantization though, but why not? Perhaps an interesting idea?

Question 5: Why are the two procedures equivalent and lead to the same result?

Another short answer, same as for Question 3: Because both procedures rely on an isomorphism with the same type of abstract Hilbert space and the associated operator algebra. As mentioned before, the "Classical Mechanics"/"particle" procedure observes an isomorphism between the finite (anti)symmetric subspace of the N-particle Hilbert space and a "fixed particle number" subspace of the abstract second quantized space, while the "Field Theory" procedure yields a true isomorphism (the "postulate" you mention).

Answer 2

Your question amounts to "explain slowly the first month of a quantum field theory course," so I am not going to be able to address everything you've brought up in your question. But, I will try to discuss some of the key points.

For a free theory there is an equivalence (at least at physicist level of rigor) between the quantum field picture and the Fock space/particle picture. (There is a relationship between particles and fields in interacting theories too but it becomes much more subtle and requires a fair amount of group theory to describe, so let's just focus on free theories).

(Of course, free theories are nice mathematical tools and the basis of perturbation theory, but are not themselves very physical. Indeed there are some very important facts about free theories that are simply not true for realistic physical theories. For example in a free theory particle number is conserved, which is very much an artifact of working with an overly simplified model and not indicative of real physics. In some sense we really wouldn't need quantum fields if particle number were conserved (although it can still be convenient, and in some condensed matter applications of field theory particle number is conserved but the formalism of field theory is still useful).)

Anyway, in terms of free theories, I think a lot of your questions can be answered by working with a finite lattice in 1+1 dimensions (1 space and 1 time dimension, and we will discretize the spatial direction). Say there are $N$ points on the lattice, representing the spatial locations. Then the free field theory Lagrangian can be written as something like \begin{equation} \mathcal{L} = \sum_{j=1}^N \left[ \frac{1}{2} \dot{\phi}_j^2 - \frac{1}{2 a^2}\left(\phi_{j+1} - \phi_j\right)^2 - \frac{m^2}{2} \phi_j^2 \right] \end{equation} where $a$ is the lattice spacing and $m$ is the mass of the scalar field. I will suppose for simplicity that there are periodic boundary conditions so that $\phi_{N+j}=\phi_j$ for any integer $j$.

In fact this is the lagrangian for $N$ harmonic oscillators. It behooves us to work with the normal modes of the system (in which these harmonic oscillators decouple) by using the (discrete) Fourier transform \begin{equation} \tilde{\phi}_k = \frac{1}{N} \sum_{j=1}^N \phi_i e^{2 \pi i k j / N} \end{equation} Then you end up with (possibly up to factors of 2 or $\pi$) \begin{equation} \mathcal{L} = \sum_{k=0}^{N} \left( |\dot{\tilde{\phi}}_k|^2 - \omega_k^2|\tilde{\phi}_k|^2 \right) \end{equation} where $\tilde{\phi}_k^\star = \tilde{\phi}_{-k}$, and where the spectrum is given by \begin{equation} \omega_k^2 = m^2 + \frac{4}{\pi^2 a^2} \sin^2\left(\frac{\pi k}{2 N}\right) \end{equation} For $k/N \ll 1$, the spectrum is \begin{equation} \omega_k^2 = m^2 + \left(\frac{k}{Na}\right)^2 \end{equation} which is of course no accident, this is a discrete version of the relativistic dispersion relationship $\omega^2 = k^2 + m^2$.

Each of these harmonic oscillators, labeled by $k$, gets a creation and annhilation operator, so there are $N$ total creation operators and $N$ total annhilation operators. Each creation and annhilation operator is labelled by $k$.

The state describing one particle with momentum $k / N a$ is then defined to be the creation operator for $k$ acting on the vacuum. The state describing $n$ particles with momentum $k / N a$ is similarly defined by be (up to a normalization) the creation operator for $k$ acting on the vacuum $n$ times.

Why do we call these particle states? You are right that we don't define something to be a 'particle' just because the quantum harmonic oscillator shows up. At this level (for the free theory), we identify the energy levels of the harmonic oscillators (labeled by $k$) with particles because these states represent a discrete packet of energy, with the same relationship between momentum and energy as a classical relativistic particle.

To give a better answer, you want to ask "what appears in a detector when I do scattering experiments?" Or "what shows up in a photon counter?" To answer that kind of question you really want to do a more sophisticated calculation: for example you want to form wave packets and ask how they propagate, and maybe look at how these wave packets interact with your detector. The net result is that the above identification matches onto what you get by looking at these more careful calculations.

Lastly two quick other points:

We can also go the other way--by starting with the diagonalized Lagrangian in Fourier space (the particle picture) we can go back to real space (the field picture) by using the inverse Fourier transform.

Finally to connect with field theory we can imagine taking a continuum limit $a\rightarrow 0$ as well as an infinite volume limit $N\rightarrow \infty$ to recover field theory. It's also very easy to generalize the above to lattices with more than one spatial direction.

Answer 3

There are two nice ways to understand "the Origins of the Second Quantization" - one is to try to arrive at it from standard 'first quantized' quantum mechanics without forcing it, the second is promoting Poisson brackets to commutators arriving at it directly (though interpreting things is less clear, and usually involves harmonic oscillator analogies). It's good to see how the second is arrived at from the first - note Heisenberg called this duality between the first and second quantization description of a theory the mathematical counterpart to wave-particle duality, the mathematical equivalence between particles and waves [4].

Maybe the most straight-forward way to do this is explained in ([1], Chapter IX). A similar but stream-lined approach is given in ([2], Chapter VII). This thinking is then directly applied to QFT in [3].

Thus, to address Q1: a lot of your post is assuming things like creation operators let alone their normalizations and matrix elements out of thin air with no derivation which is not useful from a first principles perspective, which I'll try go give now for bosons (which will cover Q2 as well), there is a similar discussion for fermions in [1] and again streamlined in [2].

It begins by considering multi-particle quantum mechanics for the special case of identical multiple particles. The Heisenberg uncertainty principle implies we can't follow the path of any of the individual particles thus we can never determine which of the identical particles was measured at a given point at an instant. In other words - we can't 'paint' identical particles.

This leads one to wave functions with Bose or Fermi symmetry on interchanging particles, and forces us to work with symmetrized or anti-symmetrized wave functions, and for a system with stationary states we can construct the total wave function from appropriately symmetrised sums of products of the stationary states which I wont review.

Thus a Bose wave function for a system of $N$ particles with energies $E_{l_1},..,E_{l_N}$ (where $l_1,..,l_N$ are a set of $N$ integers some or all of which may repeat), and where there are $n_k$ particles in the $k$'th stationary state, and is $$\psi_{n_1,n_2,...}(x_1,..,x_N) = \sqrt{\frac{N!}{n_1! n_2! ..}} \sum \psi_{l_1}(x_1) .. \psi_{l_i}(x_i) ... \psi_{l_N}(x_N)$$ with the sum going over all distinct permutations of the numbers $l_1,l_2,..$.

We can re-write this as $$\langle x_1,..,x_N|n_1,n_2,.. \rangle = \langle x_1,..,x_N| \sqrt{\frac{N!}{n_1! n_2! ..}} \sum |\psi_{l_1} .. \psi_{l_i} ... \psi_{l_N} \rangle $$ $$= \langle x_1,..,x_N| \sqrt{\frac{N!}{n_1! n_2! ..}} \sum |\psi_{l_1} \rangle .. | \psi_{l_i} \rangle ... | \psi_{l_N} \rangle $$ and ignore $\langle x_1,..,x_N|$. It's also useful to introduce a finite ket notation $$|n_1,n_2,.. \rangle = \sqrt{\frac{N!}{n_1! n_2! ..}} |l_1, .. , l_i, .. l_N \rangle . $$

We now apply a one-particle operator $$\hat{H} = \sum_{i=1}^N \hat{h}(x_i) = \sum_i \hat{h}_i,$$ for example $$\hat{H} = \sum_i \hat{h}_i = \sum_i [-\frac{1}{2m}\nabla_i^2 + \hat{V}(x_i)],$$ and note that any given state $|l \rangle$ will repeat $n_l$ times so we can then re-write the action of $\hat{H}$ on the state as a sum of $\hat{h}(x) = \hat{h}$'s acting on the $|l \rangle$'s $$ \hat{H} |n_1,n_2,.. \rangle = \sum_i \sqrt{\frac{N!}{n_1! n_2! ..}} \sum |l_1 \rangle | l_2 \rangle .. \hat{h}_i |l_i \rangle .. | l_N \rangle $$ $$ = \sum_l n_l \sqrt{\frac{N!}{n_1! n_2! .. }} \sum |l_1 \rangle | l_2 \rangle .. \hat{h} |l \rangle .. | l_N \rangle $$ we can now introduce a matrix element by introducing a resolution of the identity $I = \sum_k |k \rangle \langle k |$ $$\hat{H} |n_1,n_2,.. \rangle = \sum_{k,l} n_l \sqrt{\frac{N!}{n_1! n_2! .. }} \langle k | \hat{h} |l \rangle \sum |l_1 \rangle | l_2 \rangle .. |k \rangle .. | l_N \rangle $$ In finite ket notation this reads as $$\hat{H} |n_1,n_2,.. \rangle = \sum_{k,l} n_l \sqrt{\frac{N!}{n_1! .. n_k! .. n_l! .. }} h_{k,l} |l_1,..,k+1,..,l-1,..,l_N \rangle $$ since a $|l \rangle$ was taken out of the product and put in the matrix element, at the expense of adding a $|k \rangle$. We now fix the normalization to re-write the RHS in occupation number notation via $$\sqrt{\frac{N!}{n_1! .. n_k! .. n_l! .. }} = \sqrt{\frac{n_k+1}{n_l}} \sqrt{\frac{N!}{n_1! .. (n_k + 1)! .. (n_l - 1)! ..}} $$ so that we get $$\hat{H} |n_1,n_2,.. \rangle = \sum_{k,l} \sqrt{n_k+1} h_{k,l} \sqrt{n_l} |n_1,n_2,..,n_k+1,..,n_l-1,.. \rangle $$ Thus deriving the matrix element $$\langle n_1,n_2,..,n_k+1,..,n_l-1,.. | \hat{H} |n_1,n_2,..,n_k,..,n_l,.. \rangle = \sum_{k,l} \sqrt{n_k+1} h_{k,l} \sqrt{n_l} $$ is the main goal here, which gets cluttered when one plows into it with integrals.

Now it becomes obvious that we find a new expression for the operator $\hat{H} = \sum_i \hat{h}_i$ if we define operators $\hat{a}_i, \hat{a}_i^{\dagger}$ satisfying $$\hat{a}_i^{\dagger} |n_1,n_2,..,n_i,.. \rangle = \sqrt{n_i+1} |n_1,n_2,..,n_i+1,..\rangle $$ $$\hat{a}_i |n_1,n_2,..,n_i,.. \rangle = \sqrt{n_i} |n_1,n_2,..,n_i-1,..\rangle $$ which (from these properties) satisfy $$ [\hat{a}_i,\hat{a}_j^{\dagger}] = \delta_{ij} \ \ , \ \ [\hat{a}_i,\hat{a}_j] = [\hat{a}_i^{\dagger},\hat{a}_j^{\dagger}] = 0 \ \ , $$ so that $$\hat{H} |n_1,n_2,.. \rangle = \sum_{k,l} \hat{a}_k^{\dagger} h_{k,l} \hat{a}_l |n_1,n_2,..,n_k,..,n_l,.. \rangle $$ We thus have $$\hat{H} = \sum_{k,l} \hat{a}_k^{\dagger} h_{k,l} \hat{a}_l = (\sum_{k} \langle k | \hat{a}_k^{\dagger}) \hat{h} (\sum_l \hat{a}_l |l \rangle )$$ which is the usual second quantized one-particle Hamiltonian $$\hat{H}= \int dx \hat{\psi}^{\dagger}(x) \hat{h}(x) \hat{\psi}(x)$$ in terms of quantum field operators $$\hat{\psi}(x) = \sum_i \hat{a}_i \psi_i(x) \ \ \ , \ \ \ \hat{\psi}^{\dagger}(x) = \sum_i \hat{a}_i^{\dagger} \psi_i^*(x)$$ which then satisfy the usual CCR's $$[\psi(x),\psi^{\dagger}(x')] = \delta(x - x'),...$$ Since $\psi^{\dagger}(x')$ is the conjugate momentum in the Lagrangian associated to this Hamiltonian, this reproduces the 'P.B. to Commutator' approach to quantization (justified in [1]), but now the meaning of a quantum field operator and it's relevance to multi-particle QM even though it involves a single-particle quantum field operator is unmistakable. Notice also it is all in the Schrodinger picture, but we can easily bring it to the Heisenberg picture.

This answers Q5 on why the two approaches give the same result. It also answers Q3 in the sense that it shows starting from a classical field theory and then quantizing the P.B. relations to end up with steps like the above has to describe a multi-particle theory of identical particles in the same spectrum (one should be aware of this also applying to particles and their anti-particles in this context, see below).

For the above-stated non-relativistic Hamiltonian it reads as $$\hat{H}= \int dx \hat{\psi}^{\dagger}(x) [-\frac{1}{2m}\nabla^2 + \hat{V}(x)] \hat{\psi}(x)$$ The above also directly applies to free QFT examples which have analogous steps at each stage of the derivation. For example, Klein-Gordon, Dirac etc... do because the only difference in these steps is that the sum over stationary states now includes negative energy stationary states, which one can then deal with as explained in the second half of my answer here, even the Hamiltonian's have the above bilinear form, noting that e.g. the canonical momentum above (or Lagrangian it comes from) is in no way unique (since the form of $\hat{h}$ is not fixed above) and can be different in the different theories. Thus in free QFT we have the big difference that the quantum field operators have particle and anti-particles made explicit and treated as different spectra of particles, though they are in the same quantum field operator, and the $\hat{h}$ operator is more complicated than above (and leads to $N$ not being conserved, which is another big issue in interpreting things).

This hopefully also answers Q4 in the sense that the quantized theory has to have a multi-particle interpretation, see [3] ch. I around (2.25) where this point is explicitly made.

Regarding interacting QFT, one can now try to trace the origin of any given statement made in terms of quantum field operators back to the usual time-dependent perturbation theory as in [1], just expanded in a basis of multi-particle stationary states.

References:

1. Landau and Lifshitz, "Quantum Mechanics", 3rd Ed.
2. Vliet, "Equilibrium and Non-Equilibrium Statistical Mechanics", 1st Ed.
3. Landau and Lifshitz, "Quantum Electrodynamics", 2nd Ed.
4. Heisenberg, "The Physical Principles of Quantum Theory", 1st Ed.