While trying to calculate the angular momentum states for the first non trivial even and odd states ($N=2$ and $N=3$). When $N=n_x + n_y + n_z$

By solving the radial problem one can see that there 6 states for $N=2$ and 10 states for $N=3$, it stems in the degeneracy of states: $(N+1)(N+2)/2$

I want to work out the angular momentum states using Schwinger's annihilation and creation operators:

$$a_+ = \frac{-1}{\sqrt 2} (a_x - ia_y) \qquad a_- = \frac{1}{\sqrt 2}(a_x + ia_y)$$

For $N=3$ we have 7 states: $\{|n=3;\ l=3;\ m \rangle\}_{m=-3}^3$ We can get the highest using: $$|n=3;\ l=3;\ m=3 \rangle = \frac{1}{\sqrt{3!}}(a^\dagger_+)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$

And then to apply $L_-$ until we get to $|n=3;\ l=3;\ m=-3 \rangle$ which also equals to: $$ \frac{1}{\sqrt{3!}}(a^\dagger_-)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$

I don't fully understand the posts about spherical tensors $T_q^{(k)}$ and how to actually translate it into practical actions.

Is there a better way to get the smaller subspace then the following:

As I said, we know that the possible $l$ states are $l=3$ and $l=1$. They are orthogonal, therefore we can build:

$$| l=3;m=3\rangle = (a^\dagger_+)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$

Then we can apply the operator: $$L_-= L_x - iL_y = \hbar \left((a_x^\dagger - ia_y)a_z - a_z^\dagger (a_x +i a_y)\right) = \hbar \sqrt{2}(a_-^\dagger a_z + a_z^\dagger a_+)$$

$2\cdot3+1 = 7$ times to find all the states in the irreducible representation of $l=3$. Then we use the fact that $| l=3;m=1\rangle$ is orthogonal to $| l=1;m=1\rangle$ and jump to the other subspace and apply the operator: $L_-$ $2\cdot 1 +1 = 3$ times on $| l=1;m=1\rangle$

My concerns: There should be more general way to jump between subspaces. Note that I didn't use the spherical tensor properties nor Wigner–Eckart theorem and Spherical harmonics $Y_l^m$.

I would expect to have an expression for each angular momentum state as a sequence of operators $a_+, a_+, a_+^\dagger, a_-^\dagger, a_z$ that builds an $|nlm\rangle$ state namely finding $\{n_i\}_{i=1}^6$ such that:

$$| l;m\rangle \propto (a_+)^{n_1} (a_+)^{n_2} (a_+^\dagger)^{n_3} (a_-^\dagger)^{n_4} (a_z)^{n_5} (a_z^\dagger)^{n_6} | n_x=0;n_y=0;n_z=0 \rangle$$

Where $\{n_i\}_{i=1}^6$ are integers between $0$ and $l$.

I am mostly missing the approach to build the state of $l=1$ in $N=3$ or $l=0$ for $N=2$, which is more complicated than simply applying $(L_-)$ $2l$ times. I can use the orthogonality, but I prefer to understand a more systematic approach to jump between subspaces.

Also after reading this wonderful document I couldn't get an answer of how to jump between subspaces of angular momentum. I picked $N=2$ and $N=3$ because those are the first states that involves 2 subspaces, but any generalization of how to jump between 2 subspaces is most welcome.

One of my motivation of understanding this problem is to be familiar more spherical tensors, lie algebra, metamorphism between the SHO, Angular momentum and bosons.

Moreover I am thinking to implement the algorithm in Mathematica, using this code to calculate all of angular momentum in the Cartesian basis.

The question boils up to the following:

Let be $N$ be the number states of a 3D isotropic SHO. Is it possible to find the expression that will give the $|n=N; l,m \rangle$ in the base of $|n_x;n_y;n_z \rangle$. Or how to write the spherical tensor operator $T_m^q$ as boson operators. e.g $T_0^1=a_z^\dagger$


How to jump between 2 subspaces. For example from $|l=3;m=1\rangle$ to $|l=1;m=1\rangle$

  • $\begingroup$ Possible duplicate of Spherical Tensor Operators and the isotropic harmonic oscillator $\endgroup$ Mar 12, 2017 at 13:06
  • $\begingroup$ Thanks @ZeroTheHero, I am also referring to this wonderful question. But it's not a duplicate for me as I don't really know how to translate its content for what I am trying to do. Can you please add an example for N=2 or N=3. Although the great posts and after long reading I do "understand" the theory, but not sure how to translate it into practice. $\endgroup$
    – 0x90
    Mar 12, 2017 at 13:07

2 Answers 2


As you have probably already worked out, the two eigenspaces that you're interested in, $N=2$ and $N=3$, are formed by a direct sum of subspaces of different angular momentum; thus, the $N=2$ eigenspace has one $l=2$ and one $l=0$ subspace, of dimensions $5+1=6$, and the $N=3$ eigenspace has one $l=3$ and one $l=1$ subspace, of dimensions $7+3=10$. You therefore have two distinct tasks: moving around within each subspace, and jumping to a different representation.

The first task is relatively easy, and in fact you do not need much knowledge of the internal structure of the hamiltonian to do this. You already know that $H$ commutes with $\mathbf L$, which guarantees you the shared eigenbasis $|nlm⟩$, but more than that you know that acting on the $|nlm⟩$ with components in the angular momentum, and particularly the ladder operators $L_\pm = L_x+iL_y$, will keep you inside that subspace. In this vein, for example, you can take the $|3,3,3⟩$ state you've already found to get $$L_-|3,3,3⟩\propto |3,3,2⟩,$$ and so on.

What you really need to implement this in your language, though, is to bring the $L_\pm$ into the language of your creation and annihilation operators. We've already done most of the heavy lifting in the previous question, which gave us the identity $$ L_i=-i\hbar\varepsilon_{ijk}a_j^\dagger a_k^\phantom{\dagger}, $$ and now we just need to apply this to the ladder operators: \begin{align} L_\pm & = L_x \pm i L_y \\ & = -i\hbar\varepsilon_{1jk}a_j^\dagger a_k^\phantom{\dagger} \pm \hbar\varepsilon_{2jk}a_j^\dagger a_k^\phantom{\dagger} \\ & = -i\hbar(a_2^\dagger a_3^\phantom{\dagger} -a_3^\dagger a_2^\phantom{\dagger} )\pm \hbar(a_3^\dagger a_1^\phantom{\dagger}-a_1^\dagger a_3^\phantom{\dagger}) \\ & = \hbar(\mp a_1^\dagger-ia_2^\dagger)a_3^\phantom{\dagger} + \hbar a_3^\dagger(\pm a_1^\phantom{\dagger}+ia_2^\phantom{\dagger}) \\ & = \sqrt{2}\hbar(a_\pm^\dagger a_0^\phantom{\dagger} + a_0^\dagger a_\mp^\phantom{\dagger}), \end{align} where I've used the forms $a_\pm = \frac{1}{\sqrt{2}}(\mp a_1 +ia_2)$ and $a_\pm^\dagger =\frac{1}{\sqrt{2}}(\mp a_1^\dagger -ia_2^\dagger) $. This result for $L_\pm$ can be understood rather easily: you destroy one quantum with $a_0$ and then recreate it with $\pm$more angular momentum, or you destroy one quantum with $a_\mp$ and then recreate it with $\pm1$ more unit of angular momentum. Easy!

As applied to your $N=3$ state, $|3,3,3⟩ = \frac{1}{\sqrt{3!}}(a_+^\dagger)^3|0⟩$, you can then proceed to get \begin{align} |3,3,2⟩ &= \frac{1}{\sqrt{6}}L_- |3,3,3⟩ \\ & = \frac{1}{\sqrt{6}} \, \sqrt{2} \hbar(a_-^\dagger a_0^\phantom{\dagger} + a_0^\dagger a_+^\phantom{\dagger}) \ \frac{1}{\sqrt{3!}}(a_+^\dagger)^3|0⟩ \\ & = \frac{1}{3\sqrt{2}}\hbar(0 + a_0^\dagger a_+^\phantom{\dagger}) \cdot (a_+^\dagger)^3|0⟩ \\ & = \frac{1}{3\sqrt{2}}\hbar\, a_0^\dagger \left((a_+^\dagger)^3 a_+^\phantom{\dagger}+3(a_+^\dagger)^2\right) |0⟩ \\ & = \frac{1}{\sqrt{2!}} \hbar \, a_0^\dagger (a_+^\dagger)^2 |0⟩, \end{align} and so on, until you reach $|3,3,-3⟩ = \frac{1}{\sqrt{3!}}(a_-^\dagger)^3|0⟩$.

The other subspace, with $l<N$, is a bit more tricky, because you cannot get it from the naive argument of just doing $(a_+^\dagger)^{N} |0⟩$ and then using the angular momentum algebra to cover you.

I always find the emergence of these shells to be easier to understand by looking at the way the cartesian components combine, so let me start by building the $N=2$, $l=0$ state directly on the cartesian basis $|n_x,n_y,n_z⟩_\mathrm{C}$. In particular, consider the state $$ |\psi⟩ = \frac{1}{\sqrt{3}} \left( |2,0,0⟩_\mathrm{C} + |0,2,0⟩_\mathrm{C} + |0,0,2⟩_\mathrm{C}\right), $$ which has a position-space representation \begin{align} ⟨x,y,z|\psi⟩ & = \frac{1}{\sqrt{3}} \left( ⟨x,y,z|2,0,0⟩_\mathrm{C} + ⟨x,y,z|0,2,0⟩_\mathrm{C} + ⟨x,y,z|0,0,2⟩_\mathrm{C}\right) \\ & = \frac{1}{\sqrt{3}} \bigg[ H_2(x)+H_2(y)+H_2(z)\bigg]e^{-r^2/2} \\ & \propto \frac{1}{\sqrt{3}} \bigg[ (4x^2-2)+(4y^2-2)+(4z^2-2)\bigg]e^{-r^2/2} \\ & = \frac{1}{\sqrt{3}} \left(4r^2-6\right)e^{-r^2/2} , \end{align} so it depends only on $r$ and therefore belongs to the $l=0$ representation. This is all well and good, but how do we represent this using the bosonic operators? This can be read directly from the state above, by rephrasing it as \begin{align} |2,0,0⟩ & = \frac{1}{\sqrt{3!}}\left((a_1^\dagger)^2+(a_2^\dagger)^2+(a_3^\dagger)^2\right)|0⟩ \\ & = \frac{1}{\sqrt{3!}}\left(\frac12(-a_+^\dagger +a_-^\dagger)^2-\frac12(a_+^\dagger +a_-^\dagger)^2+(a_0^\dagger)^2\right)|0⟩ \\ & = \frac{1}{\sqrt{3!}}\left((a_0^\dagger)^2-2a_+^\dagger a_-^\dagger\right)|0⟩ \end{align} where I've used $a_1^\dagger = \frac{1}{\sqrt{2}}\left(-a_+^\dagger +a_-^\dagger\right)$ and $a_2^\dagger = \frac{i}{\sqrt{2}}\left(a_+^\dagger +a_-^\dagger\right)$. This makes a lot of sense: $|2,0,0⟩$ is generated up from the vacuum state through two pathways that increase $N$ by two but leave $L_0$ unchanged, $(a_0^\dagger)^2$ and $a_+^\dagger a_-^\dagger$, and then those two are superposed in a way that will make $L^2$ vanish. There is an obvious question, though - what is that operator, exactly, and how do we generalize it?

The answer here is that the above example is illustrative, but it is not quite the right approach, yet. We don't really want operators that will take us from one energy eigenspace to another, because those will not commute with the hamiltonian and will therefore have some relatively complicated algebraic properties. Instead, what we really want is a way to generate the $|2,0,0⟩$ state by jumping down the $l$ ladder from the $N=2$, $l=2$ space: if we do this, then our new class of ladder operators can commute with the hamiltonian (but not with $\mathbf L$).

This gives you a couple of clear candidates, because the isotropic oscillator has very few independent symmetries: the angular momentum, which we've already used, the Laplace-Runge-Lenz vector, and something called the Fradkin tensor, which is defined as $$ F_{ij} = p_ip_j +x_ix_j $$ in the cartesian frame, and which turns out to commute with the hamiltonian (which you should check explicitly). The Fradkin tensor tends to play nicer with the harmonic oscillator than the Runge-Lenz vector, and I've tried to make it mesh below but to be honest it's not quite there so you're going to need to fill in some blanks.

To bring this over to our old language, let's start by changing the quadratures over into bosonic operators, giving us \begin{align} F_{ij} & = p_ip_j +x_ix_j \\ & = \frac{a_i^\phantom{\dagger}-a_i^\dagger}{\sqrt{2}\,i} \frac{a_j^\phantom{\dagger}-a_j^\dagger}{\sqrt{2}\, i} + \frac{a_i^\phantom{\dagger}+a_i^\dagger}{\sqrt{2}} \frac{a_j^\phantom{\dagger}+a_j^\dagger}{\sqrt{2}} \\ & = \frac12 \left[-(a_i^\phantom{\dagger}-a_i^\dagger)(a_j^\phantom{\dagger}-a_j^\dagger)+(a_i^\phantom{\dagger}+a_i^\dagger)(a_j^\phantom{\dagger}+a_j^\dagger)\right] \\ & = a_i^\phantom{\dagger}a_j^\dagger+a_i^\dagger a_j^\phantom{\dagger} . \end{align} (Exercise: is this explicitly hermitian? If not, why is it hermitian?) This is, of course, on a cartesian basis for the tensor part, but of we want to be talking about spherical tensors here, so we really want to be using its spherical components, $F_{0,0}$ and $F_{2,m}$; here the scalar component is easy, since $$ F_{0,0} = \sum_{i=1}^3 F_{ii} = 2H $$ is just the hamiltonian, and to get the quadrupole components the easiest route is by analogy with the electric quadrupole $Q_{ij}=x_ix_j$, so you get \begin{align} F_{2,2} &= F_{11}+2iF_{12}-F_{22} \\ &= (a_1^\phantom{\dagger}a_1^\dagger+a_1^\dagger a_1^\phantom{\dagger}) +2i (a_1^\phantom{\dagger}a_2^\dagger+a_1^\dagger a_2^\phantom{\dagger}) - (a_2^\phantom{\dagger}a_2^\dagger+a_2^\dagger a_2^\phantom{\dagger}) \\ & = (a_1^\phantom{\dagger} + i a_2^\phantom{\dagger})(a_1^\dagger+ia_2^\dagger) +(a_1^\dagger+ia_2^\dagger)(a_1^\phantom{\dagger} + i a_2^\phantom{\dagger}) \\ & = -4\,a_+^\dagger a_-^\phantom{\dagger}, \\[2mm] F_{2,1} &= F_{13}+iF_{23} \\ & = (a_1^\phantom{\dagger}a_3^\dagger+a_1^\dagger a_3^\phantom{\dagger}) +i(a_2^\phantom{\dagger}a_3^\dagger+a_2^\dagger a_3^\phantom{\dagger}) \\ & = (a_1^\phantom{\dagger}+ia_2^\phantom{\dagger})a_3^\dagger+(a_1^\dagger +ia_2^\dagger )a_3^\phantom{\dagger} \\ & = \sqrt{2}\left(a_-^\phantom{\dagger}a_0^\dagger-a_+^\dagger a_0^\phantom{\dagger} \right), \\[2mm] F_{2,0} & = 2F_{33}-F_{11}-F_{22} \\ & = 2(a_3^\phantom{\dagger}a_3^\dagger+a_3^\dagger a_3^\phantom{\dagger}) -(a_1^\phantom{\dagger}a_1^\dagger+a_1^\dagger a_1^\phantom{\dagger}) -(a_2^\phantom{\dagger}a_2^\dagger+a_2^\dagger a_2^\phantom{\dagger}) \\ & = 2(a_0^\phantom{\dagger}a_0^\dagger+a_0^\dagger a_0^\phantom{\dagger}) -(a_+^\phantom{\dagger}a_+^\dagger+a_+^\dagger a_+^\phantom{\dagger}) - (a_-^\phantom{\dagger}a_-^\dagger+a_-^\dagger a_-^\phantom{\dagger}) \\ & = 2a_0^\dagger a_0^\phantom{\dagger} -a_+^\dagger a_+^\phantom{\dagger} - a_-^\dagger a_-^\phantom{\dagger} , \\[2mm] \text{and similarly} \\[2mm] F_{2,-1} &= F_{13}-iF_{23} \\ & = (a_1^\phantom{\dagger}a_3^\dagger+a_1^\dagger a_3^\phantom{\dagger}) -i(a_2^\phantom{\dagger}a_3^\dagger+a_2^\dagger a_3^\phantom{\dagger}) \\ & = (a_1^\phantom{\dagger}-ia_2^\phantom{\dagger})a_3^\dagger+(a_1^\dagger -ia_2^\dagger )a_3^\phantom{\dagger} \\ & = \sqrt{2}\left(a_-^\dagger a_0^\phantom{\dagger}-a_+^\phantom{\dagger}a_0^\dagger\right) \quad\text{and} \\[2mm] F_{2,-2} &= F_{11}-2iF_{12}-F_{22} \\ &= (a_1^\phantom{\dagger}a_1^\dagger+a_1^\dagger a_1^\phantom{\dagger}) -2i (a_1^\phantom{\dagger}a_2^\dagger+a_1^\dagger a_2^\phantom{\dagger}) - (a_2^\phantom{\dagger}a_2^\dagger+a_2^\dagger a_2^\phantom{\dagger}) \\ & = (a_1^\phantom{\dagger} - i a_2^\phantom{\dagger})(a_1^\dagger-ia_2^\dagger) +(a_1^\dagger-ia_2^\dagger)(a_1^\phantom{\dagger} - i a_2^\phantom{\dagger}) \\ & = -4\,a_-^\dagger a_+^\phantom{\dagger}. \end{align} These operators obviously commute with the hamiltonian, and while not hermitian they obey $F_{2,m}^\dagger = F_{2,-m}$.

This is as far as I'm going to take it for now, as I'm out of time. However, this should give you a good flavourful of where to take this to complete all the relationships in the relevant algebras; the correct ladder operators are there somewhere and it should just be a matter of whittling things down to get the correct relationships.

  • $\begingroup$ Thank you, I was editing the question again while you posted it. Few questions: 1. What's $a_0$ in your notation, why do we need 4 ladder operators? ($a_0, a_1, a_2, a_3$)? 2. When building the states for $N=3$ Is it correct assumption that I can use the orthogonality between $|n=3;l=3;m=1\rangle$ and $|n=3;l=1;m=1\rangle$ to jump to the second irrep? 3. How are spherical tensors related? 4. I didn't understand how you jumped for the $N=3$ to the subspace of $l=1$ Thank you so much again $\endgroup$
    – 0x90
    Mar 12, 2017 at 22:31
  • 1
    $\begingroup$ 1. $a_0$ is identical to $a_3$; it's just spherical notation of the form $a_m$ for $m=-1,0,1$. 2. yes, anything with definite $l=3$ is orthogonal to anything with definite $l=1$. 3. Question makes no sense. 4. I didn't. $\endgroup$ Mar 12, 2017 at 22:52
  • $\begingroup$ I have looked into Fradkin papar. First it considers classical potential. How can I show that the Tensor $\mathbf{F}$ commutes with $H$, I mean why $F_{ij}$ for $i \ne j$ will commute with $H$ $\endgroup$
    – 0x90
    Mar 13, 2017 at 13:23
  • $\begingroup$ 1. How do you identifies $F_{i,j}$ e.g. why $F_{2,2} = F_{11}+2iF_{12}-F_{22}$? 2. why $F_{2,0}=2a_0^\dagger a_0^\phantom{\dagger} -a_+^\dagger a_+^\phantom{\dagger} - a_-^\dagger a_-^\phantom{\dagger}$ isn't the same as the one you are describing $|2,0,0⟩ = \frac{1}{\sqrt{3!}}\left((a_0^\dagger)^2-2a_+^\dagger a_-^\dagger\right)|0⟩$ ? $\endgroup$
    – 0x90
    Mar 13, 2017 at 13:35
  • 1
    $\begingroup$ @0x90 The expressions for the $F_{2m}$ are obtained by analogy from the expressions of $Y_{2m}$ in terms of $Q_{ij}=x_ix_j$ (so for instance $Y_{22} = (x+iy)^2=Q_{11}+2iQ_{12}-Q_{22}$); ultimately you just use the analogy as a suggestion in order to find the linear combinations of the $F_{ij}$ that will play nice with the spherical components. Double-check all the algebra in this answer; I won't be able to look further into this for the foreseeable future but it can hopefully point you in fruitful directions. $\endgroup$ Mar 13, 2017 at 13:40

As mentioned in previous posts, given $N=n_x+n_y+n_z$, it's straightforward to calculate the $|n=N;l=N;m=N\rangle$ state if we use: $$\frac{1}{\sqrt{N!}}(a_+^\dagger)^N |0\rangle$$

Then we can apply $L_-$ until we get to $|n=N;l=N;m=-N\rangle$

In order to jump to the second subspace in which $|n=N;l=N-2;m=0\rangle$, we can use the irreducible spherical tensors theorem from [Sakurai 3.10.27]:

$$T_{q}^{(k)} = \sum_{q_1 q_2} \langle k_1, k_2 ; q_1 , q_2 |k_1,k_2; k, q \rangle T_{q_1}^{(k_1)} T_{q_2}^{(k_2)}$$

Where the followings hold:

  1. $k=N-2$
  2. $k_1+k_2 = N$
  3. $q_1 = -k_1, \dots , k_1$ and $q_2 = -k_2, \dots , k_2$
  4. We are using the $a_+, a_-, a_+^\dagger a_-^\dagger$ which are spherical tensor.

For example for $N=2, l=0$:

We choose define 2 positive integers $k_1=1$, $k_2=1$ s.t. $k_1 + k_2 =N=2$.

$$T_0^{(0)} = \sum_{q_1=-1 q_2=-1}^{1,1} \langle 1, 1 ; q_1 , q_2 |1,1; k, q \rangle T_{q_1}^{(1)} T_{q_2}^{(1)}$$

And $T_0^{(0)}|0,0,0\rangle_C$ will give us the $|n=2;l=0;m=0\rangle$ states, hence we jumped into the second subspace and now we can apply $L_{\pm}$ to list all the angular momentum states within that subspace.

  • $\begingroup$ This was there all along here: physics.stackexchange.com/q/307552 $\endgroup$ Mar 18, 2017 at 14:49
  • $\begingroup$ @ZeroTheHero, it took me time to grasp it. Though I still don't understand why naming the orthogonal state of $n=3;l=3;m=1$ as $n=3;l=1;m=1$ yielded a non normalize $n=3;l=1;m=0$ (I have checked and I had none calculation errors) $\endgroup$
    – 0x90
    Mar 18, 2017 at 18:35
  • $\begingroup$ the tensors as constructed are not normalized. The normalization is quite complicated. $\endgroup$ Mar 18, 2017 at 18:40
  • $\begingroup$ @ZeroTheHero isn't it simply normalizing the result? Secondly I meant that jumping from $l=3$ to $l=1$ and using ladder operator yields non normalized state for $l=1 m=0$ $\endgroup$
    – 0x90
    Mar 19, 2017 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.