We know that the eigenvalues for $\hat{J}^2$ is $2j+1$ - fold degenerate. However, most textbooks I have read, when explaining the ladder operator, shows the action of the ladder operator to be like this: $$ \hat{J}_z \bigg(\hat{J}_{-} |j, m\rangle \bigg) = \hbar(m-1) \ \hat{J}_{-} |j, m\rangle $$

they then proceed to conclude that $\hat{J}_{-} |j, m\rangle$ is an eigenket of $\hat{J}_z $ with eigenvalue $(m-1)$ and hence: $$ \hat{J}_{-} |j, m\rangle \propto |j, m - 1\rangle \Longrightarrow \hat{J}_{-} |j, m\rangle = C |j, m - 1\rangle $$ where $C$ is a constant.

However, the above implies that the eigenkets corresponding to $m$ are non-degenerate, that is there do not exist $\geq 2$ eigenkets corresponding to the eigenvalue of $m-1$. Why is this so?

I have read Angular Momentum Operators Non-Degenerate in which the answers say that this is an assumption. Why can we make this assumption?

  • $\begingroup$ The non-degeneracy assumption is not always satisfied. Possible duplicate: physics.stackexchange.com/q/46848/2451 $\endgroup$
    – Qmechanic
    May 30, 2020 at 6:37
  • $\begingroup$ @Qmechanic Then why do textbooks always assume non-degeneracy? $\endgroup$
    – D. Soul
    May 30, 2020 at 7:47

2 Answers 2


In fact, we do not know!

The point is that in almost all textbooks two facts are confused: one is a physical fact the other is a mathematical fact.

Let us start from physics, by assuming that a physical system $S$, which is described in the Hilbert space $H_S$, admits a (strongly continuous) unitary representation of $SU(2)$ (or projective of $SO(3)$) representing the action of the physical rotations with respect to an inertial reference frame and an origin in the rest space of it: $$SU(2) \ni r \mapsto U(r) = e^{-\frac{i}{\hbar} \sum_{k=1}^3 \theta_k J_k}\: \quad \left(\mbox{where}\quad r = e^{-\frac{i}{2} \sum_{k=1}^3 \theta_k \sigma_k}\right)$$ By definition the three selfajoint generators $J_1,J_2,J_3$ of that representation are the observables total angular momentum of the physical system.

Now we pass to mathematics. The famous Peter-Weyl theorem establishes that every strongly continuous unitary representation of a compact group $G$ (as $SU(2)$) can be decomposed as a direct orthogonal sum of finite-dimensional irreducible representations of the said topological group. Hence, up to unitary isomorphisms, $$H = H^{(j_1)} \oplus H^{(j_2)} \oplus \cdots\tag{1}$$ where the set of parameters $j_k$ labels the irreducible representation $$G \ni g \mapsto U^{(j)}(g) : H^{(j)} \to H^{(j)}$$ and it can happen that $j_k=j_h$ (giving rise to degeneration as we discuss shortly).

In the case of $G=SU(2)$, the irreducible representations are labelled by $j =0, 1/2, 1, \ldots$. Each space $H^{(j)}$ is spanned by the familar basis $$|j, m\rangle$$ where $$m= -j, -j+1, \ldots, j-1, j$$ so that $\dim H^{(j)} = 2j+1$.

Here, by construction, there is no degeneration in each space $H^{(j)}$: once you fix $m$ therein you have completely fixed the vector $|j, m\rangle$.

The popular procedure that uses $J_\pm$ is actually exploited to construct the irreducible representations $H^{(j)}$ of $SU(2)$ and irreducibility is, in this case, equivalent to non-degenerateness. But this is a pure mathematical fact, it does not mean that the angular momentum is non-degenerate, since it is defined in $H$ which contains many copies of the spaces $H^{(j)}$. $$J_k = \oplus J_k^{(j_m)}\quad k=1,2,3\:.$$

The fact that a vector is completely fixed when choosing the eigenvalue of $J^2$ and $J_z$ is decided by physics analysing the above decomposition (1) in every concrete case as I am going to show.

Let us come back to physics to see how the generation may take place. In view of the said theorem, our Hilbert space $H_S$ will be decomposed as $$H_S = \oplus_{k \in K} H^{(j_k)}$$ where $j_k \in \{0, 1/2, 1, \ldots\}$ and, depending on the nature of the physical system, some values can be repeated. Suppose that, in a certain case, we find $$H_S = H^{(j_1)} \oplus H^{(j_2)}$$ and $j_1=j_2 =1$. In this case $H^{(1)}$ occurs twice and we can describe $H^{(1)}\oplus H^{(1)}$ using a basis $|1,m,l\rangle$ where $m=-1,0,1$ and $l=1,2$ with $$|1, m, 1> = |1,m> \oplus \: 0\quad \mbox{and}\quad \quad |1, m, 2> = 0 \oplus |1,m>\:.$$ Since $J_k = J^{(1)}_k\oplus J^{(1)}_k$, here your type of degenearation pops out: both $|1, m, 1>$ and $|1, m, 2>$ are eigenstates of $J^2$ and $J_z$ with the same eigenvalues $j(j+1)=2$ and $m$.

Mathematically speaking, degeneration is always due to the occurrence of many irreducible representations of the same type and this is decided by physics in the concrete case. It strictly depends on the nature of the studied physical system.

Let us consider a popular example of physical relevance. For a (spin $0$) particle in $\mathbb{R}^3$ the previous theory applies. In this case the Hilbert space is the tensor product $$H_S = L^2(\mathbb{R}^3) \simeq L^2(S^2)\otimes L^2([0,+\infty), r^2dr)$$ The first factor in the right most side describes the angular degrees of freedom, whereas the second one describes the radial degree of freedom. The first factor can be decomposed as $$L^2(S^2) = \oplus_{j=0,1,2, \ldots}H^{(j)}$$ where in terms of spherical harmonics $$H^{(j)} \ni |j,m\rangle = Y^j_m$$ Here no repetitions arise since the values of $j$ are all different. However repetitions show up when taking the second factor $L^2([0,+\infty), r^2dr)$ into account. Fix a Hilbert basis $\{y_n\}_{n=0,1,\ldots} \subset L^2([0,+\infty), dr)$ with some physical meaning (it does not matter now). We have $$L^2([0,+\infty),r^2 dr) = \oplus_{k=0,1,\ldots} K_n\:,$$ where $$K_n := span(y_n)\:.$$ In summary, $$H_S = (\oplus_{j=0,1,2, \ldots}H^{(j)}) \otimes (\oplus_{k=0,1,\ldots} K_n)$$ $$=\oplus_{j, n}(H^{(j)}\otimes K_n) $$ every space $H^{(j)}\otimes K_n$ is isomorphic to $H^{(j)}$ since $K_n$ is one dimensional.

The decomposition $$H_S =\oplus_{j, n}(H^{(j)}\otimes K_n) $$ is therefore another way to write (1) specialised to this case.

In this case, we have infinite degeneration: for every fixed $j,m$ we have an infinite number of vectors $$|j,m, n\rangle := |j, m\rangle |y_n\rangle$$ (one for each representation $H^{(j)}\otimes K_n \simeq H^{(j)}$). Every vector $|j,m, n\rangle$ is eigenvector of $J^2$ and $J_z$ with the same eigenvalues $j(j+1)$ and $m$, but we can also fix $n$ arbitrarily in $\mathbb{N}$.

  • $\begingroup$ Really appreciate your answer. Apologies but the math is a bit confusing for me. Actually, you mentioned that because every physical system can be broke into a direct sum of irreducible representations. In our case, this would imply that: $$ H = H^{-j} \oplus \cdots \oplus H^{j} $$ However, you also mentioned that it is possible for the case where $j_k = j_h$. If that is the case, then it would be possible for: $$ H = \cdots \oplus H^{k} \oplus \cdots \oplus H^{k} \oplus \cdots \oplus H^{j} $$ Wouldn't this contradict the statement "By construction there is no degeneration"? $\endgroup$
    – D. Soul
    May 31, 2020 at 12:20
  • $\begingroup$ Sorry but could you also elaborate more on what you mean by "this is decided by physics in the concrete case"? Does this means that if we had solely taken into account the "math" part, which is purely the angular momentum, we would have encountered no degeneracy. But adding the "physics" part, the radial portion, would result in degeneracy? If this is the case, then because ladder operations were derived solely from the "math" part. How can we be assured it will be valid after including the radial portions? $\endgroup$
    – D. Soul
    May 31, 2020 at 12:26
  • 3
    $\begingroup$ Absence of degeneration concerns each space $H^{(j)}$ separately and not the total space $H_S$, OK? The physical space is however, in general, a direct sum of various such subspaces. This fact produces the degeneration. $\endgroup$ May 31, 2020 at 14:59
  • 2
    $\begingroup$ To see how the physics decides, look at the last example. There it is evident that for a particle (without spin) in 3D the states at fixed $j$ and $m$ must be always infinitely degenerated. $\endgroup$ May 31, 2020 at 15:07
  • 2
    $\begingroup$ Here the physical input is the Hilbert space is $L^2(R^3)$ because we are studying a spinless particle, and that the angular momentum has the standard form $J_k = \epsilon_{kjr} X_l P_r$. $\endgroup$ May 31, 2020 at 15:13

Well, just like the principal ($n$) and azimuthal quantum numbers ($l$), the magnetic quantum number ($m$) is also an index used for referencing the wave functions. The degeneracy of fermionic states depends on several factors, which is out of scope for this discussion.

But eigenkets of $m$ are $\textbf{indeed degenerate}$. For each permissible value of $m$, we have $2S+1$ quantum states, indexed by their spin quantum number $S$. For electrons, we have ($s=\pm \frac{1}{2}$) and the states are given by $\lvert \frac{1}{2} \rangle$ and $\lvert -\frac{1}{2} \rangle$.

The degree of degeneracy of a quantum particle depends on the internal degrees of freedom of the particle. Till now, experimental conditions known have exacted only $4$ quantum numbers, which is a correct estimation of the quantum numbers, as proven by particles following Pauli's Exclusion Principle.

Hope this helps.

  • $\begingroup$ So the reason why we say that the eigenvalue $m_j$ is non - degenerate, is because we have verified in experiments that it is so? Is there some kind of mathematical proof for it? $\endgroup$
    – D. Soul
    May 30, 2020 at 7:48
  • 1
    $\begingroup$ @D.Soul First of all, it is not non-degenerate. The spin quantum number exists. And each quantum number corresponds to some degree of freedom, or better some sort of symmetry. When solving the Schrodinger equation with the corresponding symmetry broken, you can observe how the quantum number comes into existence and it can be mathematically shown. What I meant is that the existence of all the quantum numbers was theoretically estimated, and subsequently experimentally verified. $\endgroup$ May 30, 2020 at 7:58
  • $\begingroup$ Then why do all textbooks always assume this degeneracy in order to prove the action of the ladder operations? I have read 3 textbooks so far and they seem to assume that. Also here: physics.stackexchange.com/questions/46848/… you can see on the 2nd answer, that an assumption needs to be made to prove ladder operations $\endgroup$
    – D. Soul
    May 30, 2020 at 8:13
  • 1
    $\begingroup$ @D.Soul I checked the linked question and the accepted answer has the right facts. Please check that. True, the $m_j$ eigenvalues are non-degenerate but remember that to lift the degeneracy of a system and make the states non-degenerate, you need to break the corresponding symmetry. This may sound like a difficult concept, but that relates to the corresponding Hamiltonian. So if the Hamiltonian involves very strong potentials, or if it is affected by a strong perturbation, the symmetry will be broken and the spin degeneracy lifted. But if the perturbation is weak, then the... $\endgroup$ May 30, 2020 at 14:46
  • 1
    $\begingroup$ (contd.) .. spin degeneracy is not lifted. And the magnetic states act as degenerate states. See if you can wrap your head around this. $\endgroup$ May 30, 2020 at 14:47

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