# Relation between destruction/creation operators of harmonic oscillator in QM and Second Quantization

It is well known that in elementary QM the so-called destruction/creation operators

$$a_i = \frac{Q_j + i P_j }{\sqrt{2}}, \quad a_i^* = \frac{Q_j - i P_j }{\sqrt{2}},$$

are introduced when studying the $$n$$-dimensional harmonic oscillator, where operators $$Q_j$$ and $$P_j$$ are defined by

$$(Q_j \psi) (x) = x_j \psi(x), \quad (P_j \psi) (x) = -i\hbar\frac{\partial\psi}{\partial x_j}(x),$$

on suitable domains in $$L^2(\mathbb{R}^n)$$, say the Schwartz space $$\mathcal{S}(\mathbb{R}^n)$$ of smooth "rapidly decreasing" functions. So they are operators $$L^2(\mathbb{R}^n) \supset \mathcal{S}(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$$.

However, in Second Quantization formalism we have destruction/creation operators acting on the Fock space $$\mathcal{F}(\mathfrak{h}) = \bigoplus_{k=0}^\infty \mathfrak{h}^{\otimes^k}$$, where $$\mathfrak{h}$$ is some one-particle Hilbert space, and they are defined as operator-valued distributions $$\psi \to a(\psi), \psi \to a(\psi)^*$$, where $$\psi \in \mathfrak{h}$$.

Is there is any connection between the two notion, e.g. by taking as one-particle Hilbert space exatcly $$\mathfrak{h} = L^2({\mathbb{R}^n})$$, or the fact that they share the same name is just an "accident"? In the former case, I tried to exploit the very definition of destruction/creation operators in second quantization for the harmonic oscillator eigenfunctions (or Hermite functions) $$\varphi_k$$ by defining $$a_k \equiv a(\varphi_k)$$, but I think this approach fails miserably if I'm looking for any link between the two definitions...

They are indeed part of the same general construction, called "Fock representation of the canonical commutation relations".

The case of quantum mechanical creation and annihilation operators is recovered from the general case by setting $$\mathfrak{h}=\mathbb{C}^n$$ in OP's definition of Fock space. In fact, mathematically it is not difficult to see that the resulting Fock space is isomorphic (in the sense of representations of canonical commutation relations) to $$L^2(\mathbb{R}^n)$$ together with the standard creation and annihilation operators. (The Fock vacuum vector being mapped to the ground state vector of the harmonic oscillator)

In the general case of an arbitrary $$\mathfrak{h}$$, a basis of eigenvectors of the number operator (the generalized harmonic oscillator) is obtained starting from the vacuum by successive action of one of the creation operators $$a^*(e_n)$$, where $$\{e_n\}_{n\in\mathbb{N}}$$ is an orthonormal basis of $$\mathfrak{h}$$. (and if $$\mathfrak{h}=\mathbb{C}$$ this gives exactly the usual hermite polynomials)

• This was exactly what I was looking for! Thank you very much. – ErroDSF Dec 20 '18 at 8:49

My answer might not be sufficient, but I am hoping to cast some light on it.

I tried to exploit the very definition of destruction/creation operators in second quantization for the harmonic oscillator eigenfunctions

When we are talking about destruction/creation operators, we are not talking about actual particles, but a quasiparticle - phonons.

I tried to exploit the very definition of destruction/creation operators in second quantization for the harmonic oscillator eigenfunctions (or Hermite functions) φk by defining ak≡a(φk),

I think it is possible to write this expression, but not in the form of Hermite functions, since the ladder operator are entirely symbolic.

... defined as operator-valued distributions ψ→a(ψ),ψ→a(ψ)∗, where ψ∈h.

Well this is not a definition. You can actually arrive at a conslusion that ψ doesn't commute. This means that they are NOT ordinary functions or state vectors, they MUST be operators. This realization is called the second quantization.

• I gave a very poor definition of destruction/creation operators in Second Quantization on purpose because I’m actually (more or less) familiar with them, and I didn’t mean to be too explicit. What I was asking is whether one can recover the ladder operators of harmonic oscillator by a certain choice of $\mathfrak{h}$ (see yuggib’s response below). By the way, thanks for your answer! – ErroDSF Dec 20 '18 at 8:47