# Quantum Harmonic Oscillator Raising and Lowering operators

The commutator of the operators, $$[a,a^\dagger] = 1$$ is useful in rewriting the Hamiltonian in a neat way in terms of the creation and annihilation operators.

So my question is, Is there a physical meaning to $$[a,a^\dagger] = 1$$? for except containing the canonical relationship of $$x$$ and $$p$$ and helping in a neat description of the Hamiltonian?

Like how a commutator of two operators being "$$0$$" says about common eigenfunctions , is there a meaning to it being $$1$$? Or is it just to keep further calculations and numbers simple?

• I don't think there is one correct answer to your question about the physical meaning of $[a,a^\dagger] = 1$. Moreover, $a$ and $a^\dagger$, unlike $x$ and $p$, are not physical quantities. Commented Aug 3, 2019 at 9:18

The real meaning of $$[x, p] = i\hbar$$ can be made clearer when you use it to further derive the Heisenberg's uncertainty principle for these quantities: $$\Delta x \Delta p \ge \frac{\hbar}{2}$$, which tells us that there's a fundamental limit to the precision with which this pair of physical quantities can be known/measured.
Unfortunately, $$a$$ and $$a^\dagger$$ are not physical properties, and thus, not measurable (even though you could derive similar Heisenberg principle of them). That said I don't think the CCR of $$a$$ and $$a^\dagger$$ bears any physical meaning.
• So what I should be taking from this is that $a$ and $a\dagger$ are a tool of math with no direct physical interpretation even tho they are composed of actual physical quantities $x$ and $p$ and have a physical effect of particle creation/annihilation or energy raising/lowering in the system? Commented Aug 4, 2019 at 12:45
• OK, this may not be obvious at first, but if you go further into quantum field theory, you'll realize that $a$ and $a^\dagger$ are basically the representation of a wavefunction (if you're confused about this statement, I would explain it further). So what I want to emphasize here is that, just like a wavefunction doesn't have a physical meaning, so are $a$ & $a^\dagger$. Remember what actually has a physical meaning is the squared norm of the wavefunction. Commented Aug 5, 2019 at 5:38
2. If we define the number operator $$N~:=~a^{\dagger}a,$$ the CCR $$[a,a^{\dagger}]~=~\mathbb{1}$$ leads to $$[N,a^{\dagger}]~=~a^{\dagger},$$ which states that the creation operator $$a^{\dagger}$$ raises the number (that $$N$$ is counting) by $$1$$ unit, cf. a Fock space. See also this related Phys.SE post.