What is the mathematically precise definition of raising and lowering operators?

Question

As a finite dimensional example, we have spin raising and lowering operators with the defining property $$[S_z, S_+] = \omega S_+,$$ $$\quad \quad \quad \iff [S_z, S_-] = -\omega S_-$$ for some constant $\omega$ (the subscript $z$ was arbitrarily chosen).

As an infinite dimensional example, we have the harmonic oscillator mode raising and lowering operators with the defining property $$[H, a^\dagger] = \omega a^\dagger,$$ $$\quad \quad \quad \iff [H, a] = -\omega a$$ for some constant $\omega$.

Is there an "axiomatic" definition of raising and lowering operators of a given operator (e.g. spin in the $z$-direction or quantum harmonic oscillator Hamiltonian)? I am particularly interested in understanding when raising and lowering operators can be defined for a given operator.

Answer 1

In general, a ladder operator $O_{\pm}$ is an operator defined for some states $|n\rangle$ as

$$O_{\pm}|n\rangle = C_{\pm}(n)|n\pm1\rangle \ \ \mathrm{and} \ \ O_+^\dagger =O_- .$$

The term $C_\pm$ is a constant coefficient that generally depends on the state $|n\rangle$. The other condition states that ladder operators form a hermitian conjugate pair. That's an important condition because, usually, you can define another operator $\Omega$ as

$$\Omega = O_+O_-, $$

and the hermitian conjugate pair condition imposes that $\Omega$ be hermitian

$$\Omega^\dagger = (O_+O_-)^\dagger = O_-^\dagger O_+^\dagger = O_+O_- =\Omega,$$

in other words, $\Omega$ is an observable. In the well known case of the quantum harmonic oscillator, $\Omega$ corresponds to the number operator.

Answer 2

There are two main different ways in which one can generalize the notion of ladder operators, but neither is particularly useful in practice:

1. Naive ladder operators. The nice thing about linear algebra is that sometimes you can just make a very straightforward and naive definition and it works. Consider any self-adjoint operator $A$ with eigenvalues $a_i$ and eigenvectors $v_i$ and assume the $v_i$ form a basis, i.e. the spectrum of $A$ is purely discrete. Then, by virtue of how bases work, \begin{align} \bar{b}v_i & = \bar{b}_i v_{i+1} \tag{1a}\\ bv_i & = b_i v_{i-1} \tag{1b} \end{align} are definitions of operators $\bar{b}$ and $b$ that "raise" and "lower" the eigenvalues of $A$. This is a perfectly workable definition, but it doesn't give us any of the nice things that we usually also get from ladder operators, in particular this does not necessarily lead to nice commutation relations: $[A,\bar{b}] v_i = (a_{i+1} - a_i) \bar{b}_i v_{i+1}$, so unless $a_{i+1} - a_i$ is a constant for all $i$, $[A,\bar{b}]$ is not equal to a multiple of $\bar{b}$. Note also that $\bar{b}^\ast_{i-1} = b_i$ would be required so that $\bar{b} = b^\dagger$. Also, while this definition certainly works always, the resulting operators $b,b^\dagger$ may be arbitrarily ugly and hard to deal with in terms of the "natural" physical operators like $x$ and $p$.

2. Adjoint eigenvectors. The question attempts to do this the other way around: Start with the commutation relation $[A,b] = cb, c\in\mathbb{R}$, i.e. $b$ being an eigenvector under the adjoint action of $A$. Then $[A,b^\dagger] = -cb^\dagger$ immediately follows. For any eigenvector (under the normal action) $v_i$ of $A$ with $Av_i = a_i v_i$, we then have that $b^\dagger v_i$ and $b v_i$ are also eigenvectors of $A$: \begin{align} Ab^\dagger v_i = b^\dagger A v_i - [b^\dagger, A] v_i = (a_i + c) b^\dagger v_i \end{align} But unless $a_i + c$ was an eigenvalue of $A$ to begin with, this yields only rather useless operators. In general, the adjoint action $[A,-]$ of $A$ has eigenvalues $\alpha_{ij} = a_i - a_j$ (see also this math.SE post), so in general this simply yields operators $b_{ij}$ that act in a non-zero fashion exactly on one of the $v_i$, i.e. $b_{ij} v_i = v_j$ and $b_{ij} v_k = 0$ for any $k\neq i$. Note that this means that the matrix representation of the $b_{ij}$ in the basis of the $v_i$ is simply a matrix with one non-zero entry (which is equal to one). So if you want useful ladder operators that fulfill the commutation relation $[A,b] = cb$, $A$ has to have evenly spaced spectrum, which is an extremely restrictive requirement - essentially this is only fulfilled by the harmonic oscillator, angular momentum operator-likes and by number operators on Fock spaces (i.e. any kind of occupation representation) in practice.

From the above, you might get the idea that one could rescale the operator $A$ to get evenly spaced eigenvalues so that ladder operators arise. Indeed, this is possible - in the extreme case we could simply define an operator $A'$ by $A' v_i = i v_i$ to do this - but again in general this yields intractable expressions for $A'$. However, for the particular case of differential equations in $L^2(\mathbb{R})$ with Hamiltonians of the spherically symmetric form $\nabla^2 + V(r)$ like the one for the hydrogen atom, the "method of factorizations" has enjoyed some limited popularity, where such a rescaling of the variables of the equation allows us to write $A' = b^\dagger b$ in certain subspaces. See this answer by DanielC for the hydrogen atom and "The Factorization Method" by Infeld and Hull for the general case.

Answer 3

There is a simple way in which I understand this concept of raising and lowering operators. I'll try to explain it in two stages.

First, given the assumed relation $$ [\hat{a},\hat{a}^{\dagger}] = \mathbf{1} , $$ where $\hat{a}$ and $\hat{a}^{\dagger}$ are the ladder operators, and $\mathbf{1}$ is the identity, we can construct a number operator $$ \hat{n} = \hat{a}^{\dagger}\hat{a} . $$ It then follows that $$ [\hat{n},\hat{a}] = -\hat{a} ~~~ \text{and} ~~~ [\hat{n},\hat{a}^{\dagger}] = \hat{a}^{\dagger} . $$ So, it produces mechanism for the raising and lowering action. When we now have another operator constructed from the same product of ladder operators as found in the number operators, we will get a similar action. For example, Hamiltonian $$ \hat{H} = \frac{1}{2}\hbar\omega(\hat{a}^{\dagger}\hat{a} +\hat{a}^{\dagger}\hat{a}) , $$ leads to $$ [\hat{H},\hat{a}] = -\hbar\omega\hat{a} ~~~ \text{and} ~~~ [\hat{H},\hat{a}^{\dagger}] = \hbar\omega\hat{a}^{\dagger} . $$ which we can interpret as the increase or decrease in the number of quanta in the eigenstates of the Hamiltonian.

With this basic structure in place, one can ask how to apply this structure to specific scenarios which may have different numbers of dimensions. That leads to the second phase of the discussion, which is to give a kind of back story.

The question is, how would one construct these ladder operators in the first place? We start with any Hermitian operator $\hat{q}$ that can be express in terms of its eigenstates and eigenvectors as (the summations can be replaced by integrals for the continuous case) $$ \hat{q} = \sum_q |q\rangle q \langle q| . $$ The eigenstates $|q\rangle$ is a complete orthogonal basis for the space we consider here.

Next, construct another basis which is mutually unbiased with respect to this basis. It is easily done with $$ |p\rangle=\sum_q |q\rangle \exp(ipq) . $$ Then we can define another Hermitian operator $$ \hat{p} = \sum_p |p\rangle p \langle p| . $$

Without too much effort, one can now show that $$ [\hat{q},\hat{p}] = i\mathbf{1} . $$ The final step is the define the ladder operators in terms of these Hermitian operators by $$ \hat{a} = \frac{1}{\sqrt{2}} (\hat{q}+i\hat{p}) ~~~ \text{and} ~~~ \hat{a}^{\dagger} = \frac{1}{\sqrt{2}} (\hat{q}-i\hat{p}) . $$ Their commutation relation will then reproduce that which I initially assumed above.

The above analysis readily works for the infinite dimensional case. In the finite dimensional case, the operators are often given in terms of matrices, which are already known. These matrices will have mutually unbiased eigenvectors if they do not commute. As an instructive exercise, you can consider the two-dimensional case in terms of the Pauli matrices, noting that their eigenvectors are mutually unbiased.