Source of Ladder Operators

Question

This might be a silly question but I am curious as to where the ladder operators in quantum mechanics come from. For example, in introductory texts on quantum mechanics, they try to solve the eigenvalue/eigenstate problem for angular momentum. In the process, you find the commutation relations $$ [S_i, S_j] = i\hbar \epsilon_{ijk}S_k,$$ and so on. Then, we are introduced to the so called 'ladder operators' $$ S_{\pm} = S_x \pm i S_y$$ and use that to compute a bunch more commutators and get our result in the end. However, these ladder operators just came out of nowhere, they were handed to us. Is there are deeper meaning to them in a sense, or rather where do they come from?

Answer 1

The origin of the ladder operators is the representation theory of Lie groups and Lie algebras.

Lie groups are sets of continuous transformations with a group structure. Each continuous transformation of a given group is identified as a group element. In physics these elements or transformations are important because they are mapped into symmetry transformations, such as the rotation symmetry which gives rise to the rotation group. The latter being relevant to the angular momentum theory.

Due to the smooth character of these sets (in fact Lie groups are manifolds) one can study a small region nearby the identity of the group and see how the manifold is spanned. This is done through the so called generators of the group. These generators satisfy certain conditions, in particular, commutation relations, and they form a Lie algebra. It happens that one can obtain much of the information about the group (although not all information) just by studying the algebra (that is, the local structure of the group).

Many of the most relevant Lie algebras in physics are the so-called compact semi-simple Lie algebras. For these algebras it is possible to decompose its generators into two subsets, the Cartan algebra (the maximal set of linearly independent generators $H_i$ that commute among themselves) and the step or ladder operators $E_\alpha$. In general they satisfy $$[H_i,E_\alpha]=\alpha_iE_\alpha,$$ and the properties of these generators can be used to classify all possible compact and simple Lie algebras.

The generators as well as any (group or algebra) element are in principle abstract and defined just by their actions (such as a given rotation) or by their commutation relations. In order to make things more explicit or (even more important) in order to "fit" these groups and algebra with physical theories one has to adopt a particular representation.

Given a linear representation, the groups or algebra elements are now mapped into linear operators which act on linear spaces. The vectors of these linear space are normally identified with physical states so it is important to find the linearly independent vectors. This is achieved by the representation theory in pretty much the same way for every compact semi-simple Lie algebra, with the help of the ladder operators. By successively acting with them on a given state we can obtain all the states of the theory (or all the states of the representation).

Answer 2

Basically it arises when we have a commutation relation of the spin operator and ladder operator (we arrived at the result) $$[S_z, S_+]=S_+$$ (I had only used $S_z$ and $S_+$ in conventional means) Now we needed an operator which satisfies this relation. In other hand we have $$J^2-J_z=J_x^2+J_y^2$$ Factoring RHS we get $$J_x^2 +J_y^2 =(J_x+iJ_y) (J_x-iJ_y) -\hbar J_z$$ (Here we had an extra term $\hbar J_z$) but by the trick $J_x+iJ_y$ and $J_x-iJ_y$ satisfy the commutation relation. From which we got to know that these are ladder operator for spin and where each is hermitian conjugate of other.

Answer 3

Imagine you have an electron sitting at rest. We know that the electron has two possible values for measurements of its spin in the z direction, +1/2 and -1/2. The Hilbert space to describe the electron then has two basis states, $|+\rangle$ and $|-\rangle$ that satisfy $S_z |\pm \rangle = \pm \frac{\hbar}{2}|\pm\rangle$.

Now suppose we know the electron system radiates a photon, and in particular it radiates a circularly polarized photon in the z direction. This photon then carries away one unit of angular momentum in the z direction. What is the final state of the electron spin? It must be one unit less than before.

So the question becomes, what operator is associated with this photon process? Which operator takes us from the initial electron state to the final electron state? We are looking for $O$ such that $|e'\rangle = O|e\rangle$. Whatever the operator is, it has to lower the $S_z$ value by one unit. That is, $S_z|e'\rangle = (s_{z,i}-1)|e'\rangle$. To get arrange this, we need $S_z O|e\rangle = OS_z |e\rangle - O|e\rangle$, for every possible initial $|e\rangle$. This implies that $[S_z,O]=-O$.

Can we construct an operator that has this property? The answer is yes: $O = S_x - i S_y$ does the job. Similarly, we can find an operator that increases the spin by one unit: $[S_z,S_+] = +S_+$. Photons carrying angular momentum away in the +z direction must couple in the Hamiltonian to $S_-$ and photons carrying angular momentum in the -z direction couple to $S_+$.

This is the physics motivation for raising and lowering operators. There are also very important math motivations. It turns out that many useful operator algebras can be understood entirely in terms of diagonal operators (like $S_z$) and raising and lowering operators that move between the different eigenstates.

Answer 4

Note that $$ [S_{\pm},S_{z}] = \pm 2S_{z} $$ This means that generators $S_{\pm}$ have well-defined "isospin" (i.e., the quantum number associated with the generator $S_{z}$). This is important when we work with eigenstates of $S_{z}$.