0

If we apply the raising (creation) operator to $Ψ_n(x)$ and the apply to it the lowering (annihilation) operator, we get $Ψ_n(x)$ times a constant. Does it physically say something? Can we get any intuition out of it?
Now, i know that the wavefunction times a constant does not mean anything physically, but we we did the same thing at a wavefunction that is in a superposition of two number eigenstates, then the constant factor in front of each eigenstate will change, so we will get a different wavefunction since it will now "consist of different amounts of each eigenstate" than before. So, i think that the non-commutativity has a physical significance.

Note: This question refers to he cases in which $Ψ_n(x)$ is a number eigenstate.

TheQuantumMan
  • 7,664
  • 2
    You should specify the context of this. It is not always true that $a a^\dagger \lvert \Psi \rangle$ equals a constant. It is true when $\lvert \Psi \rangle$ is a number eigenstate, like $\lvert n \rangle$. In that case the constant you get is $n+1$. The question in the title is different from the one in the text, though, and has already been addressed on the site, see e.g. http://physics.stackexchange.com/q/9194/58382, http://physics.stackexchange.com/q/130800/58382 – glS Mar 11 '16 at 18:08
  • @glS Hello, i edited the question! Thanks. I checked out the link, but it refers to the commutativity of two operators. So, this means that we can't simultaneously measure the two ladder operators? If so, does it tell us anything about the energy and the way that a particle jumps from one energy level to another? – TheQuantumMan Mar 11 '16 at 18:13
  • 1
    The ladder operators are not Hermitian, so they do not correspond to observables. Consequently we cannot talk about measuring ladder operators at all, let alone simultaneously. – By Symmetry Mar 11 '16 at 18:52
  • @BySymmetry Yes, that what i was thinking. So, what does their non commutativity means? It must be something! I just think that it has something to do with the energy when the particle jumps from one energy level to another. – TheQuantumMan Mar 11 '16 at 18:53
  • the non commutivity makes sense cause otherwise any state would be equivalent to both the presence and absence of particles ie. the vaccum state would then be treated as both having 1 and zero particles. Non-commutivity ensures that no such paradox arises. I hope my argument was clear. – Abhishek Pal Mar 11 '16 at 18:55
  • @AbhishekPal Why would the vacuum state be treated as both having 1 and 0 particles? In honesty, i do not understand your argument. – TheQuantumMan Mar 11 '16 at 18:57
  • commuting operators imply existence of a set of eigen states that simultaneously diagonalize the operators. Now if the creation and annhilation operators commute then vaccum would be an eigen state of both a and "a dagger" but application of 'a' on vaccum gives zero i.e no particles are present in the vaccum. a dagger on the other hand acting on vaccum creates a particle and then 'a' acting on the single particle state annhilates it but if operators commute then a a-dagger on vaccum should also be zero...i.e a-dagger should not produce a particle. This contradicts the definition of a-dagger. – Abhishek Pal Mar 11 '16 at 19:06
  • Related: http://physics.stackexchange.com/q/82746/2451 and links therein. – Qmechanic Mar 11 '16 at 20:38
  • It's not really clear to me what you are trying to ask here. A constant factor is physically irrelevant. What has this to do with commutativity? – ACuriousMind Mar 11 '16 at 23:58
  • @ACuriousMind The constant factor has to do with the commutativity because it makes a difference if we act on the function with the ladder operators with different order. I am just asking if this gives us any insight to the problem of the harmonic oscillator(if any). – TheQuantumMan Mar 12 '16 at 19:23

0 Answers0