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