# Why does the lowering operator applied to lowest state have to be 0? [closed]

When solving a problem in QM with raising and lowering operators ($$\hat{a},\hat{a}^\dagger,L_{\pm},..$$) it is often assumed that:

$$L_-|\Omega>=0$$

Why is this assumed?

Couldn't the result of applying the lowering operator to the lowest eigenstate give a mathematical function without physical meaning?

Or the same for applying the raising operator to the eigenstate with the highest eigenvalue (say energy, angular momentum, etc) ?

• What do you mean by it resulting in 'a mathematical function without physical meaning'? The result is not a function but a state vector. Aug 14, 2019 at 16:50
• Possible duplicate: physics.stackexchange.com/q/112807 Aug 14, 2019 at 16:53
• Clearly if $\vert x\rangle$ is the lowest and $L_-\vert \ell,m\rangle$ lowers to $\sim \vert \ell,m-1\rangle$, then it must be that $L_-\vert x\rangle=0$ else one would have $L_-\vert x\rangle\sim \vert x-1\rangle$, which would mean $\vert x\rangle$ was NOT the lowest state. Aug 14, 2019 at 17:02
• I don't think it is correct to say it is so by "assumption". It is so by construction; you first identify eigenstates of spin-z operator and then you can define the ladder operators that make the eigenstates travel up and down the scale and annihilate the edge states (it is very easy to explicitly construct such operators).
– oleg
Aug 14, 2019 at 17:17
• Possible duplicate of After using annihilation operator on vacuum state, why it is $0$ instead of vacuum? Aug 15, 2019 at 12:35

## Harmonic oscillator

For the harmonic oscillator ladder operators the reason is the following:

A physical system where there is no lower bound on the energy is unstable.$${}^{1}$$ This is why in QM we always assume that there is one state which attains the smallest values of the energy: $$|\Omega\rangle$$.

As a consequence of the commutation relations, the energy of $$\hat{a}|\Omega\rangle$$ is one unit of $$\omega \hbar$$ less than that of $$|\Omega\rangle$$. $$H \,\hat{a}|\Omega\rangle = (E_{\mathrm{min}} -\hbar\omega)\,\hat{a}|\Omega\rangle\,.$$ So the only way out is that $$\hat{a}|\Omega\rangle$$ is not an eigenstate, therefore it has to be the zero state (not $$0$$ as a number but the null element of the Hilbert space).

## Angular momentum

For the angular momentum the reason is the following:

We want to build finite dimensional representations. For the same reasoning as above the eigenvalue under $$\hat{L}_z$$ of $$\hat{L}_-|\Omega\rangle$$ is $$\hbar$$ less than that of $$|\Omega\rangle$$. Since we are diagonalizing an Hermitian operator ($$\hat{L}_z$$), then $$\hat{L}_-|\Omega\rangle$$ is orthogonal to $$|\Omega\rangle$$, so it's either a new state or the zero state.

If it's not the zero state then this procedure never ends and the representation becomes infinite dimensional. There is nothing wrong with infinite dimensional representations (they do exist), but particles typically transform under finite dimensional ones.

Note: the two arguments are really the same argument, but I wanted to emphasize the fact that the crucial point (boundedness of energies vs. finite dimension) is different.

$$\quad{}^1$$ Having negative energies is not problematic per se. The theory is fine as long as there is a lower bound. Naturally we can always change the offset so that this lower bound is zero

Couldn't the result of applying the lowering operator to the lowest eigenstate give a mathematical function without physical meaning?

No. Why would someone be interested in mathematics that is physically meaningless when doing physics?

Though you may already know, the reasoning is that we cannot repeatedly apply the lowering operator because at some point we will reach a state with, say, negative energy -- a state which does not physically exist. And thus $$a_- \psi_0 = 0$$ is required.