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Anonymous Mon 31 May 16:10:04 2021 No.13217184 View Reply Original Report

Quoted By: >>13217564 >>13217970 >>13218080 >>13218125 >>13218134

What is an excitation of a quantum field in mathematical terms?

If I would take the analogy of a quantum field being an infinite collection of coupled harmonic oscillators I could see the excitation as the presence of a quanta at a position in space, or per extension I could understand it through the occupation number representation. Both feel like wrapping a concept that can be defined more rigorously, however.

If I define the quantum field $\Phi: M \rightarrow B$ as an operator-valued distribution on a spacetime manifold $M$ , could an excitation at $m \in M$ be defined by $\langle\Omega\mid\Phi(m)\mid\Omega\rangle$ being trivial? If yes, what would constitute being trivial in this case?

Anonymous

Anonymous Mon 31 May 2021 16:13:33 No.13217206 Report

Quoted By:

I meant to say the expectation would have to be non-trivial at $m$ ofcourse

Anonymous

Anonymous Mon 31 May 2021 16:28:28 No.13217277 Report

Quoted By: >>13217518

position states generally aren't eigenstates of the hamiltonian. this is why you ordinarily see things in momentum space representation. if you act on the vacuum state with the creation operator then you get an occupation number of 1, i.e. you have one "excitation" and it looks like a plane wave. a spatially localized excitation is a superposition of a bunch of such plane waves and it is not an eigenstate of the hamiltonian and things are messy and nontrivial

Anonymous

Anonymous Mon 31 May 2021 17:22:08 No.13217518 Report

Quoted By:

>>13217277
So it's best for my sanity if I stick to the occupation number representation's explanation?

Anonymous

Anonymous Mon 31 May 2021 17:32:44 No.13217564 Report

Quoted By: >>13217587

>>13217184
What would Phi(m) mean if Phi is a distribution?

Anonymous

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Anonymous Mon 31 May 2021 17:41:47 No.13217587 Report

Quoted By: >>13217591

>>13217564
nLab proposes this
https://ncatlab.org/nlab/show/distribution#TraditionalDefinition

Anonymous

Anonymous Mon 31 May 2021 17:42:49 No.13217591 Report

Quoted By: >>13217618

>>13217587
Wrong link https://ncatlab.org/nlab/show/operator-valued+distribution#idea

Anonymous

Anonymous Mon 31 May 2021 17:52:21 No.13217618 Report

Quoted By:

>>13217591
Ah okay, I was a bit confused since you referred to a particular point on M.

Anonymous

Anonymous Mon 31 May 2021 19:23:13 No.13217970 Report

Quoted By: >>13218026 >>13220479

>>13217184
The dirty little secret of "particle physics" is that particles/excitations are basis-dependent descriptions/interpretations of QFT.
When constructing a Fock space on which your QFT lives, one typically takes a "single-particle" Hilbert space $\mathcal{H}$ and builds up the $k$ -particle sector $\mathcal{H}_k := \underbrace{\mathcal{H} \otimes_{S/A} \cdots \otimes_{S/A} \mathcal{H}}_{k \text{ times}},$ where $\otimes_{S/A}$ is the symmetrized/antisymmetrized tensor product (depending on whether you're working with bosons or fermions). The Fock space is then the direct sum of all these sectors, $\mathcal{F} := \bigoplus_{k=0}^\infty \mathcal{H}_k.$ A state with $k$ excitations lives exclusively in $\mathcal{H}_k$ , and excitation/deexcitation operators move states in between these different sectors.

A question one might have now is: what the hell is the "single-particle" Hilbert space, $\mathcal{H} \equiv \mathcal{H}_1$ ? Well it's pretty much any Hilbert space you like. If we take it to be finite dimension $\dim \mathcal{H} = d$ it's trivially isomorphic to $\mathbb{C}^d$ , and if we take it to be infinite-dimensional, then there's only one separable $\infty$ -dimensional Hilbert space up to isomorphisms. Pick your favorite, say $L^2(\mathbb{R}^{1,3})$ . I might prefer $\ell^2(\mathbb{N})$ personally, however. So how do we discern what *the* single-particle space is? And since Fock space is just another separable Hilbert space, couldn't one say something absurd like $\mathcal{F} \cong \mathcal{H}$ ?

Anonymous

Anonymous Mon 31 May 2021 19:37:14 No.13218026 Report

Quoted By: >>13218080

>>13217970
Mathematically, perhaps, but physically we know what we're constructing. We have a class of functions in mind when we talk about the wavefunction of a single particle (photon plane waves, electron orbitals, etc.), and this gives us a concrete representation of $\mathcal{H}$ to discern it from other isomorphic spaces. (Or perhaps I should say: we know that we can always choose a basis in which to represent $\mathcal{H}$ - we don't literally need a concrete description at hand.)

The upshot of all of this is that what it means to be an excitation really does depend on one's choice of how to represent a Hilbert space and build up a Fock space from that choice. At the end of the day you have to decide what the basis $\{ | n_1 \cdots n_k \cdots \rangle \mid n_k \in \mathbb{N} \} \subset \mathcal{F}$ is referring to. As an extreme converse, you might consider starting with some Hilbert space and asking whether one can partition it into orthogonal subspaces such that it "looks like" a Fock space. Assuming appropriate dimension, I believe this is always possible, but it be a potentially unphysical partitioning. For instance, it may require the use of highly nonlocal unitary equivalences.

Anonymous

Anonymous Mon 31 May 2021 19:54:46 No.13218080 Report

Quoted By: >>13218085 >>13220479

>>13217184
>>13218026
As for your example in the OP, I'm not quite sure what you mean by saying that $\langle \Omega | \Phi(m) | \Omega \rangle$ be nontrivial. Excitations are a property of states, not the operators. (Well you can go into the Heisenberg picture if you really want, but then you'll need to start talking about solving for dynamics to see anything interesting.)

Perhaps what you mean is, take the occupation number operator $a_m^\dagger a_m$ at the "mode corresponding to $m \in M$ " (actually this is a little wonky since this means our modes are Dirac deltas; we should really be working in the momentum representation to define the modes which form the operator basis, but I'll use this language to keep inline with the definition in the OP). Then one might say that a state $| \psi \rangle \in \mathcal{F}$ has an excitation at mode $m$ iff $\langle \psi | a_m^\dagger a_m | \psi \rangle \neq 0$ . We can then "probe" $| \psi \rangle$ at all points $m \in M$ and see which modes are excited in $| \psi \rangle$ . We can actually be a bit more nuanced, since the size of $\langle \psi | a_m^\dagger a_m | \psi \rangle > 0$ tells us by how much that mode $m$ is excited.

Anonymous

Anonymous Mon 31 May 2021 19:55:55 No.13218085 Report

Quoted By: >>13220479

>>13218080
*Dumb typo - obviously, the final line should read $\langle \psi | a_m^\dagger a_m | \psi \rangle \geq 0.$

Anonymous

Anonymous Mon 31 May 2021 20:03:39 No.13218125 Report

Quoted By: >>13220479

>>13217184
>What is an excitation of a quantum field in mathematical terms?
Which of those terms is non-mathematical and needs explanation? By quantum, just think of discretization in one of the variables or functions on that variable. Excitation being something that isn't in the lowest energy configuration

Anonymous

Anonymous Mon 31 May 2021 20:06:46 No.13218134 Report

Quoted By:

>>13217184
Depends what you mean by trivial. By most definitions that would probably be a ground state and not an excitation

Anonymous

Anonymous Tue 01 Jun 2021 08:55:42 No.13220411 Report

Quoted By:

Bump so I can read and reply before 404

Anonymous

Anonymous Tue 01 Jun 2021 09:15:15 No.13220479 Report

Quoted By: >>13223691

First of all thanks for the responses

>>13217970
I read about the Fock space construction, it also relates directly to the occupation number representation, right? Would it be correct to define an excited state as being a state that is contained exclusively in an $\mathcal{H}_{k}$ for $k > 0$ ? From my understanding this is ofcourse saying the same thing in essence, but I feel like referring to the occupation number isn't satisfying. Somehow it feels more like saying "it is so because we defined it so".

>>13218080
>>13218085
>Excitations are a property of states, not the operators.
I see, I thought the operator expectation in the vacuum state might betray excitations. Directly using the creation operator sounds like a good idea, but what is to say the excitations are not annihilated?

>>13218125
>Excitation being something that isn't in the lowest energy configuration
Sure, but then I would like to tie this to a statement in the form of "something is not in it's lowest energy configuration if and only if the set $X$ of predicates on the operator or state are not satisfied." for some $X$ - that is what I mean by a rigorous statement. The previous comments I replied to follow the line of reasoning I am looking for.

Anonymous

Anonymous Tue 01 Jun 2021 16:02:13 No.13221697 Report

Quoted By:

Bump

Anonymous

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Anonymous Tue 01 Jun 2021 16:04:07 No.13221706 Report

Quoted By: >>13223442

Checking for responses.

Anonymous

Anonymous Tue 01 Jun 2021 22:05:39 No.13223060 Report

Quoted By:

Last bump

Anonymous

Anonymous Wed 02 Jun 2021 00:07:47 No.13223442 Report

Quoted By:

>>13221706
ywnbaw

Anonymous

Anonymous Wed 02 Jun 2021 01:40:57 No.13223691 Report

Quoted By: >>13223704 >>13225055

>>13220479
>Somehow it feels more like saying "it is so because we defined it so"
I mean, essentially, yes. That was kinda the point of my post. There isn't enough structure on Hilbert space alone to be able to make delineations like "what is an excited state" and "what is the vacuum". You have to pick out subspaces and define excitations, occupancy numbers, etc. from there.
The reason this all works and is consistent is because we can pick out ladder operators $a, a^\dagger$ and show that they form a complete operator basis (equivalently, they generate a full set of basis states). But while this always works, the fact that they look like creation/annihilation operators in one basis may not be so simple in another.
A simple example for demonstrating this is to consider a Bogoliubov transformation. That is, a unitary matrix $U : \mathcal{F} \to \mathcal{F}$ which has the adjoint action $U a_i^\dagger U^\dagger = \sum_{j=1}^N W_{ij} a_j^\dagger + \sum_{j=1+N}^{2N} W_{ij} a_j.$ For simplicity, suppose we are working with a finite number of $N$ fermionic modes, so that $\dim \mathcal{F} = 2^N$ , hence $U \in \mathbb{C}^{2^N \times 2^N}$ and $W \in \mathbb{C}^{2N \times 2N}$ . This shows that there exists a unitary transformation on $\mathcal{F}$ which mixes annihilation and creation operators. The resulting operators $b_i^\dagger := U a_i^\dagger U^\dagger$ themselves continue to obey the canonical anticommutation relations $\{b_i, b_j\} = 0, \quad \{b_j, b_j^\dagger\} = \delta_{ij},$ and so they are perfectly good ladder operators for the fermionic algebra. The fact that they seem to mix excitations and deexcitations is just an artifact of how we defined the original basis. Furthermore, $U$ is perfectly physical (Bogoliubov transformations are generated from local one-body Hamiltonians).

Anonymous

Anonymous Wed 02 Jun 2021 01:46:42 No.13223704 Report

Quoted By:

>>13223691
nice post. if anyone else appreciates this i would suggest reading about Olive-Montonen duality, a different but conceptually related idea

Anonymous

Anonymous Wed 02 Jun 2021 10:11:00 No.13225055 Report

Quoted By:

>>13223691
I understand, thank you!

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