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Let´s consider a system, that consists out of $N$ bosonic particles, that are not interacting with each other. The Hamiltonian of this system would be given as

$$H = \sum_{i=1}^N \frac{\hbar^2}{2m}\hat{p}^2_i \quad .$$

This Hamiltonian acts on the Hilbert space of $N$ particles $\mathcal{H_N}$. We can rewrite the Hamiltonian using ladder operators

$$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m}c^\dagger_{\vec{k}} c_{\vec{k}} \quad . $$

What bothers me is the following: the Hamiltonian is an operator, that acts on the Hilbert space $\mathcal{H_N}$. By taking a look at the Hamiltonian using second quantization, it looks like the ladder operators "connect" different Hilbert spaces:

$$c_{\vec{k}}: \mathcal{H_N} \rightarrow \mathcal{H_{N-1}}$$

and

$$c^{\dagger}_{\vec{k}}: \mathcal{H_{N-1}} \rightarrow \mathcal{H_{N}}$$

However using the commutation relations, we can re-express the given Hamiltonian:

$$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} (c_{\vec{k}}c^\dagger_{\vec{k}}+\mathbb{1})\quad $$

In this case it looks like the operators switched the spaces, they act on:

$$c^{\dagger}_{\vec{k}}: \mathcal{H_{N}} \rightarrow \mathcal{H_{N+1}}$$

and

$$c_{\vec{k}}: \mathcal{H_{N+1}} \rightarrow \mathcal{H_{N}} \quad.$$

It appears to me very wrong, that we change the nature of the ladder operators simply by taking advantage of commutation relation. Therefore I hope for some clarification about this problem.

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    $\begingroup$ Do you know what a Fock space is? $\endgroup$
    – NDewolf
    Mar 11, 2022 at 11:24
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    $\begingroup$ I don't understand what the problem is - your second set of maps is the same as with the first, you just replaced the label $\mathcal{N}$ by $\mathcal{N} + 1$ (note that $\mathcal{N}$ is a free variable). To make an analogy: There is no difference between saying a sequence is given by $a_i = i$ starting at $i=0$ or saying it is given by $a_i = i - 1$ starting at $i=1$. Can you elaborate what sort of clarification you're looking for? $\endgroup$
    – ACuriousMind
    Mar 11, 2022 at 12:10

3 Answers 3

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I think the problem lies in the fact that although one can define the creation and annihilation operators for a fixed $N$, their well-known commutation relations only make sense if the operators are extended to the Fock space: $$\mathcal F = \bigoplus\limits_{N=0}^\infty \mathcal H_N \quad . $$

Indeed, writing

$$[c,c^\dagger] = c c^\dagger - c^\dagger c \tag{1} $$

is ill-defined if both $c$ and $c^\dagger$ are the creation and annihilation operators defined for an arbitrary but fixed $N$, e.g. as maps \begin{align} c^\dagger&: \mathcal H_N \longrightarrow \mathcal H_{N+1} \\ c&:\mathcal H_{N+1} \longrightarrow \mathcal H_N \quad . \end{align} This is rather easy to see, since first term on the RHS in equation $(1)$ is a map from $\mathcal H_N$ to $\mathcal H_N$, but the second term cannot act on any state $|\psi\rangle \in\mathcal H_N$, as $c$ is only defined for states in $\mathcal H_{N+1}$. So the commutation relation you've used is actually not valid.

However, if the operators are extended to $\mathcal F$ (cf. this), then we can define their corresponding commutator and obtain the usual commutation relations. So we have $c,c^\dagger :\mathcal F \longrightarrow\mathcal F$ and we can identify a $N$-particle state $|\psi\rangle \in \mathcal H_N$ with $ \mathcal F \ni \psi = (0,\ldots,|\psi\rangle,0,\ldots)$. Note that it still holds that e.g. $c^\dagger$ maps a state with $N$ particles to a state with $N+1$ particles. But we only have one $c^\dagger$ (per mode) on $\mathcal F$ and not one operator (per mode) for each $N$.

Consequently, your Hamiltonian is an operator $H:\mathcal F\longrightarrow \mathcal F$, with the property that if $\psi \in\mathcal F$ is a $N$-particle state, then $H\psi \in F$ is a $N$-particle state too, i.e. $H$ is number conserving: $[H,N]=0$, where $N = \displaystyle \sum\limits_k c^\dagger_k c_k$ is the number operator on $\mathcal F$.

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One of the problems is that your two Hamiltonians aren't actually equal.

The second one, $$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} c^†_{k}c_k$$ is the Hamiltonian for a single boson, not for multiple bosons.

The creation and annihilation operators here create and annihilate particular frequency modes for the single particle. They are not creating and annihilating particles. To really make the two Hamiltonians match your second version should be $$H = \sum_{i}\sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} c^†_{(i),k}c_{(i),k}$$ Now each $c_{(i),k}$ is an operator that removes the mode $\vec{k}$ from the state of the $i$th particle.

The other part of your question is how changing the order of the $c$ operators appears to change which Hilbert space they act on. Your mistake is the part where you write $$c_k : \mathcal{H}_N \rightarrow \mathcal{H}_{N-1}$$

This notation implies that $c_k$ is an operator that only acts on $\mathcal{H}_N$. But that's definitely not true. The $c$ operators act on the entire Hilbert space, which is made up of subspaces with different occupation numbers. Put in math notation: $$\mathcal{H} = \mathcal{H}_1\oplus\mathcal{H}_2\oplus \dots$$ $$c_{k} : \mathcal{H}\rightarrow \mathcal{H}$$ $$c_k(\mathcal{H}_N) \subseteq \mathcal{H}_{N-1}\text{ for any $N$}$$.

Put this way, $c^†_kc_k$ and $c_k c^†_k$ clearly act on the same Hilbert space, and there is no problem with transposing them.

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    $\begingroup$ I think your argument that the "two Hamiltonians aren't actually equal" is incorrect. The operators $c_k^\dagger$ and $c_k$ are creating and annihilating particles. There is a single type of boson, with multiple particles of this type, and because they are indistinguishable there is no meaning to "the $i$th particle". So $c_k$ does not "remove the mode $k$ from the state" of a specific particle, but rather destroys a particle in mode $k$ (or returns the null vector when there is no such particle). ... $\endgroup$
    – nanoman
    Mar 12, 2022 at 5:18
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    $\begingroup$ ... Put another way: $c_k^\dagger + c_l^\dagger$ creates one particle in a superposition of modes $k$ and $l$, whereas $c_k^\dagger c_l^\dagger$ creates two particles, one in $k$ and one in $l$. There is no need for an index $(i)$ here. $\endgroup$
    – nanoman
    Mar 12, 2022 at 5:18
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I believe it is probably incorrect to be thinking about creation/destruction of particles for a seemingly non-relativistic system or even more generally a system without interactions. Second quantization in this context still means a fixed number of particles while the interpretation of the ladder operators is to excite or relax the modes of a given particle.

Having said that, you must understand that the operator $c_i$ always acts on the portion (factor) of the full Hilbert space (Fock Space) ${\cal F} = {\cal H}_1 \otimes \cdots\otimes {\cal H}_i \otimes \cdots\otimes {\cal H}_N$ that corresponds to the $i$-th particle, namely only on ${\cal H}_i$. So within ${\cal H}_i$ you have all the excitations. And the operator $c_i$ is always going from ${\cal H}_i$ to itself. As long as there are no interaction terms between the particles the ladder operators by themselves do not connect different one-particle Hilbert spaces.

PS: For relativistic systems particles can actually decay into other particles or radiation or vice versa be created in certain processes. This only happens when the energy is high enough that relativistic speeds are involved.

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  • $\begingroup$ Can you elaborate on your first sentence. Why is it important in this case whether it is relativistic or not ? $\endgroup$
    – Martin
    Mar 11, 2022 at 14:02
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    $\begingroup$ The creation/annihilation formalism is used in non-relativistic many body systems all the time, this answer is very odd. The second paragraph is even worse since it's flat out wrong, the creation/annihilation operators explicitly connect the sectors of the Hilbert space with different particle numbers. $\endgroup$
    – AfterShave
    Mar 11, 2022 at 15:26
  • $\begingroup$ First paragraph states I am considering a fixed number of particles, in the Hamiltonian given without any sort of interactions this is the case. Second paragraph states that there are well defined ladder operators for each particle that act in the particles Hilbert space. I will be even more explicit to avoid confusions, let me know if something else can be improved $\endgroup$
    – ohneVal
    Mar 11, 2022 at 15:36
  • $\begingroup$ I agree with @AfterShave that you cannot define a ladder operator purely within a single-particle (or any fixed-particle) Hilbert space. Perhaps you're thinking of the ladder operators of ordinary harmonic oscillators (non-free particles in an $x^2$ potential)? In second quantization, different levels of the free-particle "oscillators" represent different numbers of particles, not different excitations of the same particles. $\endgroup$
    – nanoman
    Mar 12, 2022 at 5:29
  • $\begingroup$ I am indeed thinking in terms of a Harmonic oscillator and defining the ladder operators in the exact same way. So they can be understood as raising the energy level. But there are no changes in the number of particles in what the OP writes. $\endgroup$
    – ohneVal
    Mar 12, 2022 at 14:17

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