# Annihilation, creation, and ladder operators

These are some notes, mostly for my own benefit, on annihilation, creation, and ladder operators in quantum mechanics, with a few remarks towards the end on angular momentum, spin and Clebsch–Gordan coefficients.

First, the abstract definition: if T, LV → V are linear operators on a vector space V over a field K, then L is said to be a ladder operator for T if there is a scalar cK such that the commutator of T and L satisfies

$\displaystyle [T, L] := TL - LT = cL.$

The operator L is called a raising operator for T if c is real and positive, and a lowering operator for T if c is real and negative.

The motivation behind this definition is that if (λv) ∈ K × V is an eigenpair for T (i.e. Tv = λv), then a quick calculation reveals that (λ + cLv) is an eigenpair for T:

$\displaystyle T(Lv) = (TL)v = (LT - [T,L])v = LTv + cLv = (\lambda + c) (L v).$

Ladder operators come up in quantum mechanics because many of the elementary operations on quantum systems act as ladder operators and increase or decrease the eigenvalues of other operators. Those eigenvalues often encode important information about the system, and the increments and decrements provided by the ladder operators often come in discrete, rather than continuous, values. Annihilation and creation operators are a prime example of this phenomenon.

Annihilation and Creation Operators

It took me a little while to marry the mathematicians’ and physicists’ notations in my head, so I’ll make the connection explicit here, and illustrate it with a running toy example. The starting point is to take the above vector space V to be a Hilbert space ℋ over the field ℂ of complex numbers; the unit norm elements of ℋ represent the possible quantum states of a single particle. The corresponding Hilbert space for some whole number n ≥ 0 of such particles is the n-fold tensor power ℋn. (For the moment, I’m gliding over issues of symmetry and anti-symmetry for bosons and fermions respectively.) We have

$\displaystyle \mathcal{H}^{\otimes 0} = \mathbb{C},$
$\displaystyle \mathcal{H}^{\otimes 1} = \mathcal{H},$
$\displaystyle \mathcal{H}^{\otimes 2} = \mathcal{H} \otimes \mathcal{H},$
$\displaystyle \vdots$

and the Fock space is

$\displaystyle \mathcal{F} := \bigoplus_{n \in \mathbb{N}_{0}}\mathcal{H}^{\otimes n} .$

Now suppose that ℋ has an orthonormal basis (ei)iI. A useful way of thinking of ℋn is as the vector space with orthonormal basis given by formal elements of the form

$\displaystyle \underbrace{e_{i_{1}} \otimes e_{i_{2}} \otimes \dots \otimes e_{i_{n}}}_{\mathrm{exactly\,\,} n \mathrm{\,\,factors}}$

subject to some fairly obvious algebraic requirements: for example, in ℋ ⊗ ℋ, the equalities

$\displaystyle (u + v) \otimes w = u \otimes w + v \otimes w ,$
$\displaystyle u \otimes (v + w) = u \otimes v + u \otimes w ,$
$\displaystyle (c u) \otimes v = u \otimes (c v) = c(u \otimes v) .$

hold for every u, v, w ∈ ℋ and every scalar c ∈ ℂ, and similarly in higher tensor powers of ℋ. More details on tensor products of Hilbert spaces can be found on Wikipedia.

Toy Model. A useful finite-dimensional toy model is to take ℋ = ℂm as a model for a quantum system in which the particle (or particles) that we consider may live in one of m “boxes”. The standard orthonormal basis of ℋ is {e1, …, em}, where ei is the vector with 1 in the ith component and 0 elsewhere. A unit-norm wave-function (vector) ψ ∈ ℋ describes the probability distribution of the particle in the sense that the particle has probability |ψ(i)|2 to be in “box” i. Similarly, the probability distribution of two particles is described by a unit-norm wave-function (m-by-m matrix) ψ ∈ ℋ ⊗ ℋ; |ψ(i, j)|2 is the probability that the first particle is in “box” i and that the second particle is in “box” j.

In quantum mechanics, it helps to consider not just tensor powers of ℋ, but symmetric and anti-symmetric tensor powers of ℋ, which represent bosonic and fermionic systems respectively. Define the symmetrization operator S+: ℋn → ℋn on basis elements by

$\displaystyle S^{+} \bigl( e_{i_{1}} \otimes \dots \otimes e_{i_{n}} \bigr) := \frac{1}{n!} \sum_{\pi \in \mathrm{Sym}(n)} e_{i_{\pi(1)}} \otimes \dots \otimes e_{i_{\pi(n)}} ,$

and extend to the rest of ℋn by linearity; similarly, define the anti-symmetrization operator S: ℋn → ℋn on basis elements by

$\displaystyle S^{-} \bigl( e_{i_{1}} \otimes \dots \otimes e_{i_{n}} \bigr) := \frac{1}{n!} \sum_{\pi \in \mathrm{Sym}(n)} \mathrm{sgn}(\pi) e_{i_{\pi(1)}} \otimes \dots \otimes e_{i_{\pi(n)}} ,$

and extend to the rest of ℋn by linearity. Physically, the difference between bosons and fermions is that bosons can be in the same quantum state, and indeed two bosons in the same state are indistinguishable, whereas fermions obey the Pauli exclusion principle and no two of them can be in the same state. Translated into the terms used above, this means that the wave-functions for bosonic systems must be symmetric, whereas those for fermionic systems must be anti-symmetric. (Of course, mixed boson-fermion systems need have no symmetry at all.)

Toy Model (cont.). When ℋ = ℂm, elements of ℋ ⊗ ℋ are basically m-by-m complex matrices. The tensor symmetrization operator S+ corresponds to the matrix conjugate-symmetrization operator, which adds a matrix to its conjugate-transpose and divides the sum by 2; S+(ℋ ⊗ ℋ) is the space of conjugate-symmetric m-by-m complex matrices, such as

$\displaystyle \begin{bmatrix} 1/\sqrt{2} & 1/2 \\ 1/2 & 0 \end{bmatrix} ,$

which represents the quantum state in which two indistinguishable bosonic particles a and b reside together in box 1 with probability ½; reside together in box 2 with probability 0; and have a in box 1 and b in box 2 (or vice versa, and by indistinguishability we can’t tell which way around the two particles are) with total probability ½. Similarly, the tensor anti-symmetrization operator S corresponds to the matrix skew-conjugate-symmetrization operator, which takes the difference of a matrix and its conjugate-transpose and divides the sum by 2; S(ℋ ⊗ ℋ) is the space of skew-conjugate-symmetric m-by-m complex matrices, such as

$\displaystyle \begin{bmatrix} 0 & 1/\sqrt{2} \\ -1/\sqrt{2} & 0 \end{bmatrix} ,$

which represents the quantum state in which two distinguishable fermionic particles a and b reside together in box 1 with probability 0; reside together in box 2 with probability 0; have a in box 1 and b in box 2 with probability ½; and have a in box 2 and b in box 1 with probability ½.

An alternative notation for the elements of ℋn is Dirac notation. As usual, this is easiest to explain in terms of basis elements. What we seek is a compact notation for describing the state of a system of precisely n particles. Let n = (ni)iI be a sequence of non-negative integers with sum n. Then we define the ket |n to be

$\displaystyle |\boldsymbol{n}\rangle_{\pm} := S^{\pm} \bigotimes_{i \in I} \underbrace{e_{i} \otimes \dots \otimes e_{i}}_{\mathrm{exactly\,\,} n_{i} \mathrm{\,\,factors}} .$

This should be read as saying that of the total n particles, ni are in the state described by the basis vector ei. This notation elegantly glosses over all issues of normalization — we just assume that an appropriate normalization factor has been applied to make sure that |n has norm 1 in the n-particle space ℋn. In general, a system’s state may be described by a linear combination of such basic kets, each of which may lie in the same or different tensor powers S±n, with the sum appropriately normalized to have norm 1 in the Fock space ℱ± (see the Wikipedia article for details of how the inner products and norms on ℋn and ℱ are built out of those of ℋ).

Toy Model (cont.). When ℋ = ℂm with m = 2, the above-mentioned skew-symmetric matrix

$\displaystyle \begin{bmatrix} 0 & 1/\sqrt{2} \\ -1/\sqrt{2} & 0 \end{bmatrix}$

(which represented two fermions equiprobably located in the two “boxes” of the system) corresponds to the anti-symmetrized ket

$\displaystyle |(1,1)\rangle_{-} := \frac{e_{1} \otimes e_{2} - e_{2} \otimes e_{1}}{2} = \frac{1}{2} \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \right)$,

appropriately normalized to have norm 1.

From now on, the particles will be bosons; to adapt the discussion to fermions, one needs to replace the commutator [ · , · ] by the anti-commutator

$\displaystyle \{ A, B \} := AB + BA .$

For each iI, the annihilation operator ai (which can be thought of a removing one particle whose state is the basis vector ei) is defined by

$\displaystyle a_{i} |\boldsymbol{n}\rangle_{+} := \sqrt{n_{i}} \, | \boldsymbol{n} - \delta_{i}\rangle_{+} .$

In the other direction, the creation operator ai (which can be thought of a removing one particle whose state is the basis vector ei) is defined by

$\displaystyle a_{i}^{\dagger} |\boldsymbol{n}\rangle_{+} := \sqrt{n_{i} + 1} \, | \boldsymbol{n} + \delta_{i}\rangle_{+} .$

The dagger notation reinforces the fact that the creation and annihilation operators are, in fact, each other’s Hermitian adjoints. The number operator Ni is defined by

$\displaystyle N_{i} := a_{i}^{\dagger} a_{i} .$

The number operator gets its name from the fact that

$\displaystyle N_{i} |\boldsymbol{n}\rangle_{+} = n_{i} |\boldsymbol{n}\rangle_{+} ,$

which a mathematician would read as saying that ni is an eigenvalue for Ni with eigenvector |n+, but a physicist would interpret as saying that Ni reveals how many particles of the particles in the state |n+ are in state i.

Toy Model (cont.). When ℋ = ℂm, the annihilation operator ai is a map from ℋn down to ℋ⊗(n−1) (e.g. from matrices to vectors) that removes a particle from “box” i and then renormalizes and symmetrizes appropriately. Similarly, the creation operator ai is a map from ℋn up to ℋ⊗(n+1) (e.g. from vectors to matrices) that adds a particle to “box” i and then renormalizes and symmetrizes appropriately. For example, with m = 3, starting with a particle that is certainly in box 1, a2 adds a particle to box 2 thus:

$\displaystyle a_{2}^{\dagger} |(1,0,0)\rangle_{+} = |(1,1,0)\rangle_{+} .$

The creation and annihilation operators satisfy the commutation relation

$\displaystyle [a_{i}, a_{j}^{\dagger}] = \delta_{ij}.$

In particular,

$\displaystyle a_{i} a_{i}^{\dagger} = a_{i}^{\dagger} a_{i} + 1 = N_{i} + 1.$

A quick calculation using this commutation relation reveals that the annihilation operator ai is a lowering operator for the number operator Ni:

$\displaystyle [N_{i}, a_{i}] = a_{i}^{\dagger} a_{i} a_{i} - a_{i} a_{i}^{\dagger} a_{i} = [a_{i}^{\dagger}, a_{i}] a_{i} = - a_{i}.$

Similarly, the creation operator ai is a raising operator for Ni:

$\displaystyle [N_{i}, a_{i}^{\dagger}] = a_{i}.$

These simple results underwrite the initial heuristic that ai removes a particle from state i, since it decrements the eigenvalue of the state i number operator Ni by 1, and that eigenvalue is interpreted as the number of particles in state i; similarly, ai increments the eigenvalue of Ni by 1.

Expected Values. As a final point, the fact that most of the material above has talked in terms of basic states may obscure the fact that Ni is not “integer-valued”: it is an operator whose eigenvalues on basic states happen to be integers. For non-basic states, the situation is mildly more complicated. For example, consider the following normalized wave-function for a single particle, which is equiprobably in “boxes” 1 and 2:

$\displaystyle \psi = S^{+} \left( \frac{1}{\sqrt{2}} e_{1} + \frac{1}{\sqrt{2}} e_{2} \right) ,$

or, in Dirac notation,

$\displaystyle |\psi\rangle_{+} = \frac{1}{\sqrt{2}} |(1,0)\rangle_{+} + \frac{1}{\sqrt{2}} |(0,1)\rangle_{+} .$

Then

$\displaystyle N_{1} |\psi\rangle_{+} = \frac{1}{\sqrt{2}} |\psi\rangle_{+} ,$

but this should not be read as saying that the number of particles in “box” 1 is 1⁄√2. Instead, one should take the inner product of ψ with N1ψ to get the expected value of the number of particles in “box” 1. Doing that yields

$\displaystyle (\psi, N_{1} \psi)_{\mathcal{H}} \equiv |\psi\rangle^{\dagger} N_{1} |\psi\rangle \equiv \langle\psi| N_{1} |\psi\rangle = \frac{1}{2},$

which should be read as saying that the expected number of particles in “box” 1 is ½.

Angular Momentum and Spin

One can make very similar remarks about angular momentum. In quantum mechanics, the angular momentum of a particle is quantized, i.e. it can take only certain discrete values, rather than the continuous spectrum of values that is possible in the classical (Newtonian) situation. In fact, there are two kinds of angular momentum, orbital angular momentum and an intrinsic measure of rotation called spin. Spin functions very much like angular momentum, and indeed the operators for each will have the same algebraic rules. The discussion below is phrased in terms of spin, but everything applies to other angular momentum operators.

For the treatment of spin, we posit three self-adjoint operators Sx, Sy and Sz, which measure the particle’s spin about the x, y, and z axes. These three operators satisfy the commutator relations

$\displaystyle [S_{j}, S_{k}] = i \hbar \epsilon^{jk\ell} S_{\ell} ,$

where j, k, range over x, y, z and εjkℓ is the Levi–Civita symbol

$\displaystyle \epsilon^{jk\ell} := \begin{cases} +1, & \mbox{if } (j, k, \ell) \mbox{ is an even permutation of } (x, y, z) \\ -1, & \mbox{if } (j, k, \ell) \mbox{ is an even permutation of } (x, y, z) \\ 0, & \mbox{if } j = k, k = \ell \mbox{ , or } \ell = i. \\ \end{cases},$

We also define

$\displaystyle S^{2} := S_{x}^{2} + S_{y}^{2} + S_{z}^{2},$

and it is easy to verify that S2 commutes with each of Sx, Sy and Sz:

$\displaystyle [S^{2}, S_{x}] = [S^{2}, S_{y}] = [S^{2}, S_{z}] = 0.$

This is a rather interesting situation: the three spin operators Sx, Sy and Sz do not commute, but S2 does commute with any single one of them (though not more than one). This fact has two important consequences:

1. By the uncertainty principle, arbitrarily accurate simultaneous measurements of two of Sx, Sy and Sz cannot be achieved.
2. We can work in a basis of states that are eigenvectors for both S2 and, say, Sz. In what follows, these eigenvectors will be indexed by two (discrete) parameters s and m; the eigenvalues for S2 will be indexed by s, and the eigenvalues for Sz will be indexed by m, so that, in Dirac notation, |s, m denotes a typical eigenstate.

Mathematically, of course, one could work with the eigenvalues themselves; from a physical point of view, the parametric point of view is much cleaner. It is conventional to work in units so that the eigenvalues of S2 and Sz are given by

$\displaystyle S^{2} | s, m \rangle = \hbar^{2} s (s + 1) | s, m \rangle,$
$\displaystyle S_{z} | s, m \rangle = \hbar m | s, m \rangle.$

It is a fact of nature that s ≥ 0 is always an integer (for a boson) or a half-integer (for a fermion). It follows after a couple of pages of calculation that m ranges between −s and s in integer steps. So, for example, if s = 32 (the particle has “spin 32”), then m = −32, −12, 12 or 32. A common very simple setting is spin 12, in which case there are two possible states:

$\displaystyle | \uparrow \, \rangle := | \tfrac{1}{2}, \tfrac{1}{2} \rangle \quad \mbox{spin up'',}$
$\displaystyle | \downarrow \, \rangle := | \tfrac{1}{2}, -\tfrac{1}{2} \rangle \quad \mbox{spin down''.}$

Now define raising (+) and lowering (−) operators by

$\displaystyle S_{\pm} := S_{x} \pm i S_{y}.$

A quick calculation confirms that S± are ladder operators for Sz:

$\displaystyle [S_{z}, S_{\pm}] = \pm \hbar S_{\pm}.$

So, for example, application of S+ increments the eigenvalue ℏm of the Sz operator by one unit of ℏ, i.e. adds one ℏ’s worth of spin about the z axis. In this situation, another useful calculation to perform is to work out the normalizing constants C±(s, m) — called Clebsch–Gordan coefficients — that will make

$\displaystyle S_{\pm} |s,m\rangle = C_{\pm}(s,m) | s, m \pm 1 \rangle.$

an equality of unit-norm states. To do this, note that S is the Hermitian adjoint of S+ and so

$\displaystyle \| S_{+} |s,m\rangle \|^{2} = \langle s,m| S_{-} S_{+} |s,m\rangle .$

Expanding the SS+ using the definition of the ladder operators and the commutation relations of the Sx, Sy and Sz yields that

$\displaystyle S_{-} S_{+} = S^{2} - S_{z}^{2} - \hbar S_{z}$

and so

$\displaystyle \| S_{+} |s,m\rangle \|^{2} = \hbar^{2} \bigl( s(s+1) - m(m+1) \bigr) \| |s,m\rangle \|^{2} .$

From this, and the corresponding calculation with S+ and S interchanged, it follows that

$\displaystyle C_{\pm}(s,m) = \hbar \bigl( s(s+1) - m(m\pm1) \bigr)^{1/2} .$