I know that the energy eigenstates of the 3D quantum harmonic oscillator can be characterized by three quantum numbers: $$ | n_1,n_2,n_3\rangle$$ or, if solved in the spherical coordinate system: $$|N,l,m\rangle$$

The relationship between capital $N$ and the little $n_i$'s is straightforward: $N=n_1+n_2+n_3$, but this can't be said for the other quantum numbers. I want to find a way of relating the two representations, but I'm not sure how to go about doing this (my linear algebra background is pretty weak).

Let's say I fix the energy to be $\frac{5\hbar \omega}{2}$, which is equivalent to saying $N=1$. There are three states corresponding to this situation in the first representation: $| 1,0,0\rangle, |0,1,0\rangle,| 0,0,1\rangle$. But what are the corresponding states in the second representation? If $N$ is fixed to be $1$, what are the allowed values of $l$ and $m$? I remember that $l=0,...,N-1$, and $-l\leq m\leq l$, but this makes no sense, since it would mean both $m$ and $l$ would have to be $0$...

  • $\begingroup$ You probably want to read up in Messiah QM vI, Ch XII,  § 15, p 456. $\endgroup$ Commented Dec 27, 2018 at 16:04
  • $\begingroup$ The allowed vaues of $l$ are $n, n-2, n-4,\dots [0,1]$ and $-l\le m\le l$ as you stated. $\endgroup$
    – drer
    Commented Jul 23, 2022 at 23:56

2 Answers 2


We introduce the ladder operators $a^\dagger_i, a_i$ such that \begin{align}x_i & = \sqrt{\frac{\hbar}{2m\omega}} (a^\dagger_i + a_i) \\ p_i & = i\sqrt{\frac{\hbar m\omega}{2}} (a^\dagger_i - a_i) \end{align} where $i = 1,2,3$. The commutators are of course $$[a_i, a^\dagger_j] = \delta_{ij}.$$

Then the angular momentum operator is $$L_i = \epsilon_{ijk}x_j p_k$$ with $\epsilon_{ijk}$ the Levi-Civita symbol and sums over $j, k$ implied. On expanding $x_j p_k$ only $-a^\dagger_j a_k$ and $a_j a^\dagger_k$ contribute, since $a_k a_j$ is symmetric in $k,j$. These two terms give equal contributions since their commutator is symmetric in $k,j$. It follows that $$L_i = -i\hbar\epsilon_{ijk} a^\dagger_j a_k.$$

Now defining $$a_+ = \frac{-1}{\sqrt 2} (a_x - ia_y) \qquad a_- = \frac{1}{\sqrt 2}(a_x + ia_y)$$ we have $[a_\pm, a_\pm^\dagger]= 1$ and $$L_z = \hbar(a^\dagger_+ a_+ - a^\dagger_- a_-).$$ It is rather clear that $a^\dagger_\pm$ raise $N$ by $1$, and $a_\pm$ adds an excitation with $L_z = \pm\hbar$: $L_z$ is the difference between the number operators corresponding to $a_\pm$.

Using these operators you can in principle work out the matrix for $L_z$ (and also $L_x$ and $L_y$) and $L^2$. Since the $L_i$ operators contain only products one creation and one annihilation operator they do not connect states with different $N$. It follows that neither does $L^2$, so you can consider each $N$ separately. Once you have these matrices, diagonalizing them will tell you how to express the $|N, l, m\rangle$ in terms of the $|n_x,n_y, n_z\rangle$.

Note that for each $N$ there are $(N+2)(N+1)/2$ states, so you probably don't want to do this by hand except maybe for $N = 2$ (the $N = 0,1$ cases are trivial). Maybe you can get Mathematica or Maple to do it for you for some larger $N$.

For $N = 1$ we can calculate as follows. $$a_+ |0,0,1\rangle =\frac{-1}{\sqrt 2}(a_x - ia_y)|0,0,1\rangle = 0.$$ What we have used here is that $$a_x |n_x, n_y, n_z\rangle = \sqrt{n_x}|n_x-1,n_y,n_z\rangle$$ and similarly for $y,z$. We obtain the same with $a_-$, so this means that $L_z |0,0,1\rangle = 0$, so the state $|0,0,1\rangle$ has $m = 0$. For $|1,0,0\rangle$, we have $$a_+|1,0,0\rangle = -a_- |0,1,0\rangle = -\frac{1}{\sqrt 2} |0,0,0\rangle$$ from which we get $$a^\dagger_+ a_+ |1,0,0\rangle = \frac{1}{2} \big( |1,0,0\rangle + i |0,1,0\rangle\big)$$ $$a^\dagger_- a_- |1,0,0\rangle = \frac{1}{2} \big(|1,0,0\rangle - i|0,1,0\rangle\big).$$ Thus $$L_z |1,0,0\rangle = i\hbar |0,1,0\rangle.$$ Now you can probably work out on your own that $L_z |0,1,0\rangle = -i\hbar |1,0,0\rangle.$ This gives the matrix for $L_z$ on $N = 1$ states as $$L_z = \begin{pmatrix}0 & i & 0 \\ -i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ The eigenvalues of $L_z$ are given by the solutions to $$m(m^2-1) = 0$$ which are $m = 0, \pm 1$. We already know that $|0,0,1\rangle$ is the eigenvector corresponding to $m = 0$. To find the eigenvector corresponding to $m = \pm 1$, we have to solve the system of linear equations \begin{align}iy & = \pm x \\ -ix & = \pm y\end{align} which just says $x = \pm i y$ (the two equations are equivalent). Thus $$L_z \big( |1,0,0\rangle \pm i |0,1,0\rangle\big) = \pm\hbar \big( |1,0,0\rangle \pm i |0,1,0\rangle\big).$$

Since we found three states, with $m = -1,0,1$, we must have $l = 1$.

Of course in this simple case we could also have reasoned like this: $a^\dagger_\pm$ adds an excitation with angular momentum $\pm \hbar$. Knowing that $L_z|0,0,0\rangle = 0$, we get states with $m = \pm 1$ just by acting with $a^\dagger_\pm$ on $|0,0,0\rangle$. Indeed up to a normalization this is just what we found.

  • $\begingroup$ Thank you for your answer. As I said, my linear algebra is very weak, I've never had a proper course in it and so I've just been kind of picking up bits of it as I studied QM. Could you explain in some more detail (given, say, $N=1$) how to obtain for example the matrix for $L_x$ and how diagonalization allows one to find the expressions for $N,l,m$ in terms of $n_1,n_2,n_3$? Or maybe you know where I can find some worked examples? I've been unsuccessful in finding any excercises online that dealt with this sort of problem. $\endgroup$ Commented Jun 23, 2014 at 19:28

Those three states you have listed are all equivalent. Think of those 3 state, and then realize that your choice of which is n1, which is n2, and which is n3 is totally arbitrary, Thus those states are entirely equivalent. Thus, if N=1, l and m are both zero. That is the only allowable state.

  • $\begingroup$ But then what is meant by a matrix representation of an operator? If we focus for the moment on the first basis, I remember there being a 3x3 matrix representation of, each of the $L_{x,y,z,+,-}$ and $L^2$ operators... but if there is only one state, what is there to represent? $\endgroup$ Commented Jun 23, 2014 at 17:29
  • $\begingroup$ Yes, but that selection of a "special" z-axis is entirely that--it's a selection. You pick one axis to to be the special one, to be the one around which you can find out information and have smaller uncertainty. Those representations allow you to manipulate your eigenstates once you have chosen your preferred axis. $\endgroup$
    – jhobbie
    Commented Jun 23, 2014 at 17:37

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