Ladder operators are found in various contexts (such as calculating the spectra of the harmonic oscillator and angular momentum) in almost all introductory Quantum Mechanics textbooks. And every book I have consulted starts by defining the ladder operators. This makes me wonder why do these operators have their respective forms? I.e. why is the ladder operator for the harmonic oscillator

$$\hat{a}=\sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right) $$

and not something else?

On a similar note, does anyone know the physicist/paper who/which proposed this method? Wikipedia mentions Dirac, but I have been unable to find any leads.

  • $\begingroup$ When you say "why do these operators have their respective forms?" do you mean "why is it useful to consider this particular combination of operators?" or perhaps "are the ladder operators of quantum mechanics a special case of a construction that has wider applicability?" or perhaps both? $\endgroup$ Commented May 24, 2013 at 22:57
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    $\begingroup$ Perhaps something along the lines of the creation and annihilation operators arising when you try to factorize the Hamiltonian? This link has the details for the harmonic oscillator. $\endgroup$
    – zkf
    Commented May 25, 2013 at 7:57
  • $\begingroup$ Have a look to the Quantum harmonic oscillator $\endgroup$
    – Trimok
    Commented May 25, 2013 at 9:38
  • $\begingroup$ Joshphysics, my question is more along the lines of the first. Zkf, the link is pretty insightful, thanks. $\endgroup$ Commented May 25, 2013 at 10:52
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    $\begingroup$ @zkf Hate to comment on a 5-year old question, but any chance there's another link or title of what you've linked to. That one seems to be dead. $\endgroup$ Commented May 23, 2018 at 1:40

3 Answers 3


Ladder operators are usually constructed to form a Lie algebra (we want them to have specific conmutation relations). The mathematical basis is weight theory.

The important thing of Lie algebras is that they are a vector space and their elements, which are called generators obbey this conmutation rule: $$[X_i,X_j]=f_{ijk}X_k$$ Where we have used the summation convention. $f_{ijk}$ are just constants, so we call them structure constants.

In our case generators will be just matrices.

In general, we will have n number of generators, which will form an algebra. There will be m simultaneously diaganolizable generators (i.e. they conmute with each other). These generators are called Cartan generators and they form the Cartan subalgebra. We will denote them by $H^i$ and the non Cartan generators by $E^i$.

Each eigenvector asociated to the Cartan generators is called a weight vector, $|t_i\rangle$. Their components $t_i$ are called weights. Weight vectors will correspond to physical states.

A Cartan generator will act on a weight vector as: $$H^i|t_j\rangle =t^i_j|t_j\rangle$$

At this point I should explain roots, but we shall just skip them.

Now, here is when ladder operators come into play. When a non Cartan generator acts on a state (weight vector) the new eigenvalue will be shifted by $\pm e_j^i+t_k^i$ . When the value is raised we denote the generator by $E^j$ and when its lowered $E^{-j}$. We take that they are the hermitian conjugates of each other.

Then, it possible to prove that $[H^i, E^j]=e^i_jE^j$ and $[E^j,E^{-j}]=e^k_jH^k$. These conmutation relations are very important and they will be used in the angular momentum and harmonic oscillator case.

So we are done, we just need to identify our Cartan and non Cartan generators. Then, the non Cartan generators will move us around the possible states.

Angular momentum

We have that $J^1,J^2,J^3$ are the generators of SU(2). We choose one of this generators to be diagonal one, typically it's $J_3$ (this is the Cartan generator). Then, each state $|j,m\rangle$ is labeled by the eigenvalues of $J_3$, which we'll identify as the angular momentum $m$ and the maximum angular momentum is $j$.

Since $J^1,J^2$ don't satisfy $[J^3,J^i]=\alpha J^i$ nor $[J^i,J^{-i}]=\alpha J^3$, we have to take linear combinations of them. We could show, solving a linear system, that this combination is: $$N^\pm=\frac{1}{\sqrt{2}}(J_1\mp J_2)$$

These operators will change the value of the angular momentum. We can check that they satisfy the conmutation rules. $$[J^3,J^\pm]=\pm J^\pm$$ $$[J^+,J^-]=J^3$$

Harmonic oscillator

(I'm a bit confused with SU(1,1) algebras and that stuff, so someone else should explain it)

In this case the Cartan generators are two, the identity $\mathbb{I}$ and the Hamiltonian $H$ (I think that the Hamiltonian could be interchanged by the number operator $N=a^\dagger a$). We also know from QM that $[x,p]=i$ ($h=1$). As in the previous case, we take linear combinations to form the ladder operators. We obtain: $$[H,\hat{a}]=-\hat{a}$$ $$[H,\hat{a}^\dagger]=\hat{a}^\dagger$$ $$[\hat{a},\hat{a}^\dagger]=\mathbb{I}$$ $$[\hat{a},\hat{a}]=0$$ $$[\hat{a}^\dagger,\hat{a}^\dagger]=0$$

The harmonic oscillator can be extended in QFT to study bosons and fermions.

If you want more information about the math of ladder operators in angular momentum you should have a look at Georgi's book. For the harmonic oscillator there is not so much information, I like this notes: https://web.archive.org/web/20140804024331/http://www.math.columbia.edu/%7Ewoit/QM/old-fermions-clifford.pdf .

  • $\begingroup$ Wow, this is pretty detailed, but I can't say I understand ; especially since I don't know about group theory orLie algebras! $\endgroup$ Commented May 25, 2013 at 10:53
  • $\begingroup$ @Comp_Warrior No problem, I'll try to explain it better. $\endgroup$
    – jinawee
    Commented May 25, 2013 at 11:38
  • $\begingroup$ Can you give a link to the Georgi's book you refered? $\endgroup$
    – Ziqian Xie
    Commented Jun 4, 2013 at 17:43
  • $\begingroup$ @Urukec The book is this. It explains weight representations and almost every type of algebra in Particle Physics (SU(n), SO(n), E6, etc). $\endgroup$
    – jinawee
    Commented Jun 5, 2013 at 17:55
  • $\begingroup$ @jinawee A few questions upon reading your answer: (1) You say that the action of a non-Cartan generator on a weight vector will simply produce an eigenvalue $\pm e^i_j + t^i_k$. How can you even say "eigenvalue" for a non-Cartan generator? If a non-Cartan generator acts on a weight vector, which by definition is an eigenvector of one of the Cartan generators, couldn't it in general produce a complicated linear combination of weight-vectors? $\endgroup$ Commented Oct 10, 2016 at 20:27

You may recall from high school algebra that $x^2 + y^2 = (x + iy)(x - iy)$. Because the way the adjoint operator works, you could define an operator $\hat a = x + iy$, and its adjoint becomes $\hat a^\dagger = x - iy$. The hamiltonian for the quantum oscillator is just this relation with some constants. You have to be careful because the ladder operators don't commute; that causes the constant $\frac{1}{2}\hbar\omega$ to show up. Of all the sources that I've seen discuss the oscillator with the ladder operators Griffiths (section 2.3.1) is the only one who actually explains the problem this way. The others just pull the ladder operators seemingly out of nowhere, then demonstrate that they work.

The ladder operators date at least to Dirac's Principles of Quantum Mechanics, first published in 1930. That's a really good example of Dirac just inventing the ladder operators and then showing that they solve the problem. Dirac had a tendency to bring in math that physicists at the time weren't familiar with. So it's possible that he saw the ladder operators in math, realized they could solve physics problems, and introduced them to physics. He doesn't provide a citation in Principles, so it's also possible that he invented them. The best citation for where Dirac got the ladder operators should be in one of his original papers.

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    $\begingroup$ Comment to the answer (v1): It may be worthwhile stressing that $x^2 + y^2 = (x + iy)(x - iy)$ is only true if $[x,y]=0$. $\endgroup$
    – Qmechanic
    Commented Dec 12, 2013 at 21:22

Why do they have that form and not some other? I suppose one answer is "the form of the Hamiltonian".

Because of the form of the Hamiltonian for the QHO, there is a "number" basis for the states.

Suppose you don't use the ladder operator algebra to solve for the energy eigenstates of the Hamiltonian. You still find that the energy eigenvalues are of the form $(n + \frac{1}{2})\hbar \omega, \ n = 0,1,2,...$

Thus, there is a basis, the number basis, consisting of states with eigenvalue $n$ and an associated number operator, $\hat N$.

$$\hat N | n \rangle = n | n \rangle$$

Then the Hamiltonian can be written as:

$$\hat H = (\hat N + \frac{1}{2}) \hbar \omega$$

Factor $\hat N$ into the product of an operator and its Hermitian adjoint:

$$\hat N = \hat a^\dagger \, \hat a$$


$$\hat H = ( \hat a^\dagger \, \hat a+ \frac{1}{2}) \hbar \omega$$

But, we also have:

$$\hat H = \frac{\hat P^2}{2m} + \frac{m \omega^2 \hat X}{2}$$

Equating these gives the form for $\hat a$ and $\hat a^\dagger$.

But what do these operators, $\hat a$ and $\hat a^\dagger$ do?

Using the commutation relations for $\hat X$ and $\hat P$, find that:

$$[\hat a, \hat a^\dagger] = 1$$


$$[\hat N, \hat a^\dagger] = \hat a^\dagger$$

Operating on a number eigenstate with the above, find that:

$$\hat N (\hat a^\dagger | n \rangle) = (n+1)(\hat a^\dagger | n \rangle) = \lambda |n+1\rangle$$

So, we find that $\hat a^\dagger$ is a raising operator connecting the number state $|n\rangle$ to the state $|n+1\rangle$.

By similar reasoning, we find that $\hat a$ is a lowering operator.

So, without assuming ladder operators or their form, we necessarily arrive at them.

  • $\begingroup$ Related physics.stackexchange.com/questions/54691/… $\endgroup$ Commented May 25, 2013 at 16:40
  • $\begingroup$ I wonder for your last equation how can you form |n+1> ? It should be eigenvalue not ket. $\endgroup$
    – Outrageous
    Commented Nov 28, 2013 at 15:45
  • $\begingroup$ @Outrageous, You would contract the ket with its associated bra to get just the eigenvalue. The number operator on a number eigenket is the number eigenket scaled by the eigenvalue. $$\hat N | n \rangle = n | n \rangle$$ $\endgroup$ Commented Nov 29, 2013 at 20:42

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