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Let's suppose I have a Hilbert space $K = L^2(X)$ equipped with a Hamiltonian $H$ such that the Schrödinger equation with respect to $H$ on $K$ describes some boson I'm interested in, and I want to create and annihilate a bunch of these bosons. So I construct the bosonic Fock space

$$S(K) = \bigoplus_{i \ge 0} S^i(K)$$

where $S^i$ denotes the $i^{th}$ symmetric power. (Is this "second quantization"?) Feel free to assume that $H$ has discrete spectrum.

What is the new Hamiltonian on $S(K)$ (assuming that the bosons don't interact)? How do observables on $K$ translate to $S(K)$?

I'm not entirely sure this is a meaningful question to ask, so feel free to tell me that it's not and that I have to postulate some mechanism by which creation and/or annihilation actually happens. In that case, I would love to be enlightened about how to do this.

Now, various sources (Wikipedia, the Feynman lectures) inform me that $S(K)$ is somehow closely related to the Hilbert space of states of a quantum harmonic oscillator. That is, the creation and annihilation operators one defines in that context are somehow the same as the creation and annihilation operators one can define on $S(K)$, and maybe the Hamiltonians even look the same somehow.

Why is this? What's going on here?

Assume that I know a teensy bit of ordinary quantum mechanics but no quantum field theory.

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Hello Qiaochu, welcome to physics.SE! Nice question and I hope we can expect many more :-) – Marek Jan 8 '11 at 3:31
What is $S^i(K)$ @Qiaochu ? – user346 Jan 8 '11 at 5:10
@space_cadet: the i^{th} symmetric power, i.e. the Hilbert space of states of i identical bosons. – Qiaochu Yuan Jan 8 '11 at 13:14
Ah ok. In the physics literature $H$, almost always, denotes the Hamiltonian and $S$ the action. – user346 Jan 8 '11 at 13:25
$H$ on $Sym^2(K)$ is really $H\otimes 1 + 1 \otimes H$ and likewise for $a$ and $a^\dagger$. So for example the energy is the sum of the (uncoupled) energies. You might have expected $H\otimes H$, for example, but $H$ generates an infinitesimal translation in time. Exponentiating gives the expected result on the propogator $U = exp(tH)$ as $U\otimes U.$ – Eric Zaslow Jan 8 '11 at 20:20

3 Answers 3

up vote 3 down vote accepted

Let's discuss the harmonic oscillator first. It is actually a very special system (one and only of its kind in whole QM), itself being already second quantized in a sense (this point will be elucidated later).

First, a general talk about HO (skip this paragraph if you already know them inside-out). It's possible to express its Hamiltonian as $H = \hbar \omega(N + 1/2)$ where $N = a^{\dagger} a$ and $a$ is a linear combination of momentum and position operator). By using the commutation relations $[a, a^{\dagger}] = 1$ one obtains basis $\{ \left| n \right >$ | $n \in {\mathbb N} \}$ with $N \left | n \right > = n$. So we obtain a convenient interpretation that this basis is in fact the number of particles in the system, each carrying energy $\hbar \omega$ and that the vacuum $\left | 0 \right >$ has energy $\hbar \omega \over 2$.

Now, the above construction was actually the same as yours for $X = \{0\}$. Fock's construction (also known as second quantization) can be understood as introducing particles, $S^i$ corresponding to $i$ particles (so HO is a second quantization of a particle with one degree of freedom). In any case, we obtain position-dependent operators $a(x), a^{\dagger}(x), N(x)$ and $H(x)$ which are for every $x \in X$ isomorphic to HO operators discussed previously and also obtain base $\left | n(x) \right >$ (though I am actually not sure this is base in the strict sense of the word; these affairs are not discussed much in field theory by physicists). The total hamiltonian $H$ will then be an integral $H = \int H(x) dx$. The generic state in this system looks like a bunch of particles scattered all over and this is in fact particle description of a free bosonic field.

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I realize I left your original Hamiltonian $H$ out of the discussion. I'll add that to the answer later. For now note that $x$ is in no way special in the above, we could have used other "basis" of $K$ like momentum and in particular energy basis of the $H$. In that case the relevant states for $S(K)$ become $\left | n_0 n_1 \cdots \right>$ with $n_i$ telling us how many particles are in the state with energy $E_i$. – Marek Jan 8 '11 at 4:07
@Marek: thanks! I would definitely appreciate some pointers about exactly what to do with the original Hamiltonian. Some follow-up questions: are the creation and annihilation operators observables? Is number going to turn out to be a conserved quantity in the general case? – Qiaochu Yuan Jan 8 '11 at 14:05
@Marek: and one more question. Given an observable A on K, what's the corresponding observable on S(K)? I can think of a few different possibilities and I'm not sure which one physicists actually use. – Qiaochu Yuan Jan 8 '11 at 14:27
@Qiaochu: true, but I thought you were asking how to promote observables from $K$ to $S(K)$. $N(\lambda)$ are completely new operators than need the structure of $S(K)$ to be defined. As for interactions: well, that is a topic for a one-semester course in quantum field theory so I recommend you ask this as a separate question. But in short: in general any $H_I$ is possible. But physical ones need to conserve energy, momentum and in fact complete Poincaré symmetry. So one uses representations of Poincaré group to restrict possible choices of $H_I$. – Marek Jan 8 '11 at 15:24
@Qiaochu: (cont.) in the end it turns out that it's really ineffective to work in this way and one is forced to pass to the language of fields. One can quantize classical fields (again enforcing Poincaré and perhaps other, gauge symmetries) by usual means (canonical quantization, path-integral, etc.) and in the end one can decompose the Hilbert space into particles (in the Fock sense) and $H_I$ falls out. In any case, there is still a lot of room for possible interactions and to get a taste, see e.g. QED Lagrangian. – Marek Jan 8 '11 at 15:30

Reference: Fetter and Walecka, Quantum Theory of Many Particle Systems, Ch. 1

The Hamiltonian for a SHO is:

$$ H = \sum_{i = 0}^{\infty}\hbar \omega ( a_i^{+} a_i + \frac{1}{2} ) $$

where $\{a^+_i, a_i\}$ are the creation and annihilation operators for the $i^\textrm{th}$ eigenstate (momentum mode). The Fock space $\mathbf{F}$ consists of states of the form:

$$ \vert n_{a_0},n_{a_1}, ...,n_{a_N} \rangle $$

which are obtained by repeatedly acting on the vacuum $\vert 0 \rangle $ by the ladder operators:

$$ \Psi = \vert n_{i_0},n_{i_1}, ...,n_{i_N} \rangle = (a_0^+)^{i_0} (a_1^+)^{i_1} \ldots (a_N^+)^{i_N} \vert 0 \rangle $$

The interpretation of $\Psi$ is as the state which contains $i_k$ quanta of the $k^\textrm{th}$ eigenstate created by application of $(a^+_k)^{i_k}$ on the vacuum.

The above state is not normalized until multiplied by factor of the form $\prod_{k=0}^N \frac{1}{\sqrt{k+1}}$. If your excitations are bosonic you are done, because the commutator of the ladder operators $[a^+_i,a_j] = \delta_{ij}$ vanishes for $i\ne j$. However if the statistics of your particles are non-bosonic (fermionic or anyonic) then the order, in which you act on the vacuum with the ladder operators, matters.

Of course, to construct a Fock space $\mathbf{F}$ you do not need to specify a Hamiltonian. Only the ladder operators with their commutation/anti-commutation relations are needed. In usual flat-space problems the ladder operators correspond to our usual fourier modes $ a^+_k \Rightarrow \exp ^{i k x} $. For curved spacetimes this can procedure can be generalized by defining our ladder operators to correspond to suitable positive (negative) frequency solutions of a laplacian on that space. For details, see Wald, QFT in Curved Spacetimes. Now, given any Hamiltonian of the form:

$$ H = \sum_{k=1}^{N} T(x_k) + \frac{1}{2} \sum_{k \ne l = 1}^N V(x_k,x_l) $$

with a kinetic term $T$ for a particle at $x_k$ and a pairwise potential term $V(x_k,x_l)$, one can write down the quantum Hamiltonian in terms of matrix elements of these operators:

$$ H = \sum_{ij} a^+_i \langle i \vert T \vert j \rangle a_i + \frac{1}{2}a^+_i a^+_j \langle ij \vert V \vert kl \rangle a_l a_k $$

where $|i\rangle$ is the state with a single excited quantum corresponding the action of $a^+_i$ on the vacuum. (For details, steps, see Fetter & Walecka, Ch. 1).

I hope this helps resolves some of your doubts. Being as you are from math, there are bound to be semantic differences between my language and yours so if you have any questions at all please don't hesitate to ask.

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Can you explain the notation in that last formula? What are the b_i? – Qiaochu Yuan Jan 8 '11 at 18:00
@qiaochu that was a typo. Its fixed now. – user346 Jan 8 '11 at 20:29
As recently as 10 years ago Welecka was still teaching as William & Mary. It's worth taking his course. Any course. Or even going to see a talk. Really. – dmckee Jan 8 '11 at 23:00

Suppose, as you do, that $K$ is the space of states of a single boson. Then the space of states of a combined system of two bosons is not $K\otimes K$ as it would be if the two bosons were distinguishable, it is the symmetric subspace which you are denoting as $S^2$. Your sum over all $i$, which you denote $S$, is then a HIlbert space (state space) of a new system whose states contain the states of one-boson system, a two-boson system, a three-boson system, etc. except not an infinite number of bosons. (that is not included in the space $S$). And your space $S$ includes superpositions, for example if $v_1$ is an element of $S$ (a state of one boson) and if $v_3 \in S^3$ (a state of a three boson system) then $0.707 v_1 - -.707 v_3$ is a state which has a fifty per cent. probability of being one boson, if the number of particles is measured, and a fifty per cent. probability of being found to be three bosons. That is the physical meaning of Fock space. It is the state space on which the operators of a quantum field act.

As already remarked by Eric Zaslow, if $H$ is the Hamiltonian of the h.o. $K$, then by definition, $H\otimes I + I \otimes H$ is the Hamiltonian on $S^2$, etc. on each $S^i$. Then one sums them all up to get a Hamiltonian on the direct sum $S$.

Unless this Hamiltonian is perturbed, the number of particles is constant, obviously, since it preserves each subspace $S^i$ of $S$. So there will be no creation or annihilation of pairs of particles. If this field comes into interaction with an extraneous particle, the Hamiltonian will be perturbed of course.

It is connected with second quantisation as follows: if you have a classical h.o. and quantise it, you get $K$. If you now second quantise $K$, you get $S$ which can be regarded as a quantum field. Sir James Jeans showed, before the quantum revolution, that the classical electromagnetic field could be obtained from the classical mechanics h.o. as a limit of more and more classical h.o.'s not interacting with each other, and this procedure of second quantisation is a quantum analogue. It is not the same procedure as if you start with a classical field and then quantise it. But it is remarkable that you can get the same anser either way, as JEans noticed in the classical case. That is, you started with a quantum one-particle system and passed to Fock space and got the quantum field theory corresponding to that system. But we could have started with a classical field and quantised it, and gotten the quantum field that way.

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