# Single sequence of angular momentum ladder in quantum mechanics? -- Why there is only a

How do you prove that there is only one sequence of angular momentum eigenstates connected by the ladder operator, within the subspace where the squared modulus of the angular momentum has a given eigenvalue?

To be more precise, let $$\hat{\mathbf{J}}$$ be a (generalized) angular momentum operator defined as a Hermitian operator whose Cartesian components satisfy the commutation relations, \begin{align} [\hat{J}_x, \hat{J}_y] =& i\hbar \hat{J}_z, & [\hat{J}_y, \hat{J}_z] =& i\hbar \hat{J}_x, & [\hat{J}_z, \hat{J}_x] =& i\hbar \hat{J}_y. \end{align} We define the raising ($$\hat{J}_+$$) and lowering ($$\hat{J}_-$$) operators as \begin{align} \hat{J}_+ =& \hat{J}_x +i \hat{J}_y, & \hat{J}_- =& \hat{J}_x -i \hat{J}_y. \end{align} Let us give a generic name, ladder operator, to $$\hat{J}_+$$ and $$\hat{J}_-$$. Let us call the following operator as the squared modulus of the angular momentum, $$$$\hat{J}^2 = \hat{J}_x^2 + \hat{J}_y^2+\hat{J}_z^2.$$$$

From the commutation relation, $$[\hat{J}^2, \hat{J}_x] =0$$, these two operators has simultaneous eigenstates, which we write as $$|X_{\alpha}, Y_{\beta}\rangle$$, meaning that \begin{align} \hat{J}^2 |X_{\alpha}, Y_{\beta}\rangle =& X_{\alpha}|X_{\alpha}, Y_{\beta}\rangle, \\ \hat{J}_z |X_{\alpha}, Y_{\beta}\rangle =& Y_{\beta}|X_{\alpha}, Y_{\beta}\rangle . \end{align} From the commutation relations, $$[\hat{J}^2, \hat{J}_{\pm}]=0$$ and $$[\hat{J}_z, \hat{J}_{\pm}] =\pm \hbar \hat{J}_{\pm}$$, it follows that $$$$\hat{J}_{\pm}|X_{\alpha}, Y_{\beta}> = |X_{\alpha}, Y_{\beta}\pm \hbar> C_{\pm}(X_{\alpha},Y_{\beta}),$$$$ where $$C_{\pm}(X_{\alpha},Y_{\beta})$$ is a complex number. From the fact that $$$$\langle X_{\alpha}, Y_{\beta}| (\hat{J}^2 -\hat{J}_z^2) |X_{\alpha}, Y_{\beta}\rangle = \langle X_{\alpha}, Y_{\beta}| (\hat{J}_x^2 +\hat{J}_y^2 ) |X_{\alpha}, Y_{\beta}\rangle \geq 0,$$$$ it follows that $$$$X_{\alpha} - Y_{\beta}^2 \geq 0.$$$$ Hence, the eigenvalue of $$\hat{J}_z$$ has both the lower and upper bounds determined by the eigenvalue of $$\hat{J}^2$$. So far so good to me.

All of a few textbooks I have looked into follow essentially one type of argument to determine the values of $$X_{\alpha}$$ and $$Y_{\beta}$$ [1]. This argument is based on the claim that

• successive multiplication of $$\hat{J}_+$$ on $$|X_{\alpha}, Y_{\text{min}}\rangle$$ must result in $$|X_{\alpha}, Y_{\text{max}}\rangle D$$, ..... (A1)

where $$Y_{\text{min}}$$ and $$Y_{\text{max}}$$ are the minimum and maximum value taken by $$Y_{\beta}$$, and $$D$$ is a complex number.

Then, it is derived that $$X_j = j(j+1)\hbar^2$$ with $$j=0,1/2, 1, 3/2, \dots$$, and $$Y_m = m\hbar$$ with $$m=-j,-j+1,\dots,j-1,j$$, where I changed the indexes $$\alpha$$ and $$\beta$$ to $$j$$ and $$m$$ to follow the conventional notation. However, I do not see any justification of the claim (A1) cited above.

Suppose that $$0 and that $$Y_{\beta}= Y_{\text{min}}+a$$ is an eigenvalue of $$\hat{J}_z$$. What contradiction results? Application of $$\hat{J}_+$$ on $$|X_{\alpha}, Y_{\text{min}}+a\rangle$$ would just reveal another sequence of eigenstates of $$\hat{J}_z$$ with eigenvalues $$Y_{\text{min}}+a, Y_{\text{min}}+a+\hbar, \dots, Y_{\text{min}}+a+M\hbar$$, where $$M$$ is the integer such that $$Y_{\text{max}}-\hbar< Y_{\text{min}}+a+M\hbar \leq Y_{\text{max}}$$. Especially, what contradiction results if the equality holds in the last expression, i.e., if there exists $$a$$ such that $$(Y_{\text{max}} - Y_{\text{min}} -a)/\hbar$$ is an integer, and hence claim (A1) is false?

This question,

Angular momentum - proof for integer or half-integer eigenvalues

might probably intended to ask the same question when it was posed in that (unknown) textbook studied by the OP of that question, but both of the two answers therein assumes the claim (A1) concerned here, and the OP accepted one of the aswers. Therefore, I am asking this new question here, making the point of concern precise.

This question,

Why do $sl(2,\mathbb{C})$ raising and lowering operators $J_{\pm}$ guarantee quantized eigenvalues?

asks about a slightly different point, i.e., existence of ladder operators with fractional step in $$\hbar$$.

[1] The argument in textbooks is essentially the same as in this answer post:

https://physics.stackexchange.com/a/128918/6399

I took the notation partly from this post.

• In order to preserve invariance of wavefunction under a complete rotation (of $2\pi$ or $4\pi$), we get a restriction on the possible angular momentum states as integers (or half integers). Aug 26, 2019 at 10:58

Angular momentum operator $$J$$ is the generator for rotations on a wavefunction. That is if you have a state $$|\psi\rangle$$ and you want to express it in terms of coordinates that are rotated about an axis $$\hat n$$ by an angle $$\theta$$, this is given by $$\exp\left(-i\frac{J\cdot \hat n \theta}{\hbar}\right)|\psi\rangle$$ Now invariance of $$|\psi\rangle$$ under rotation of $$2n\pi$$ restricts the eigenvalues of $$J$$ to be integral multiples of $$\hbar$$. Thus we exhaust all possible states while counting via ladder operators.

• Let's consider $|\psi\rangle$ such that $\hat{J}_z |\psi\rangle = (\hbar/2)|\psi \rangle$. Upon $2\pi$ rotation around the $z$ axis, $\exp(-i 2\pi \hat{J}_z/\hbar)|\psi \rangle = |\psi\rangle e^{-i\pi}$, and it acquires the phase factor. This $|\psi\rangle$ is invariant only upon $4\pi$ rotation. Now suppose $|\phi\rangle$ is such that $\hat{J}_z |\phi\rangle = (\hbar/3)|\phi\rangle$. It acquires the phase $e^{-i2\pi/3}$ upon $2\pi$ rotation, and it is invariant under $6\pi$ rotation. Why such $|\phi\rangle$ does not exist, while $|\psi\rangle$ is a valid eigenstate of $\hat{J}_z$? Aug 26, 2019 at 11:24
• That is because $J_z$ is bound by $J$. And the allowed values for $J$ are integral multiples of $\frac{\hbar}{2}$ and the reason for which can be seen in physics.stackexchange.com/questions/174018/… Aug 26, 2019 at 11:46
• Thank you for your answer and comment. However, I do not think my question is answered. What does $J$ in your answer and comment mean? Is it a vector of operator, or is it a scaler? Let me interpret that you want to say that the eigenvalue of $\hat{J}_z$ is bounded by an integral multiples of $\hbar/2$. I may take this upper bound to be $\hbar/2$. Clearly, $\hbar/3$ is below this bound. (You may want to say the lower bound is $-\hbar/2$, and $\hbar/3$ is above this bound.) Besides, I do not understand where specifically in the link you are referring to. Aug 26, 2019 at 12:22

In short, any ladder of eigenvalues of $$\hat{J}_z$$, other than those $$-j\hbar, (-j+1)\hbar, \dots, j\hbar$$ where $$j$$ is integer or half-integer, would not terminate at least in one direction and hence would contradict with the finite bounds shown in the question text.

To be more precise, pick any eigenstate $$|X_{\alpha}, Y_{\beta}\rangle$$, where $$Y_{\beta}$$ is not assumed to be integer or half-integer multiple of $$\hbar$$. By applying the raising operator repetitively on it, we climb up the ladder as $$$$\hat{J}_+^k |X_{\alpha},Y_{\beta}\rangle = |X_{\alpha},Y_{\beta}+k\hbar\rangle C_+(X_{\alpha}, Y_{\beta}+(k-1)\hbar) \cdots C_+(X_{\alpha}, Y_{\beta})$$$$ for $$k=1,2,\dots$$. Since there is the upper bound for the eigenvalue of $$\hat{J}_z$$, there must be integer $$K\geq 0$$ such that \begin{align} C_+(X_{\alpha}, Y_{\beta}+(k-1)\hbar)\not=&0, & k=&1,\dots,K, \\ C_+(X_{\alpha}, Y_{\beta}+K\hbar)=&0. \end{align} Let us write the largest eigenvalue on this ladder as $$$$\mu := Y_{\beta} + (K-1)\hbar,$$$$ and the corresponding eigenstate as $$$$|v_0\rangle := |X_{\alpha}, \mu\rangle.$$$$ From here, we climb down the ladder and will see a condition that the ladder truncates at finite steps.

We write $$$$|v_k \rangle := \hat{J}_-^k |v_0 \rangle = |X_{\alpha}, \mu-k\hbar\rangle C_-(X_{\alpha},\mu-(k-1)\hbar) \cdots C_-(X_{\alpha},\mu), \tag{1}\label{eq:1}$$$$ for $$k=1,2,\dots$$. Then, from the commutation relation, $$[\hat{J}_+, \hat{J}_-] =2\hbar \hat{J}_z$$, i.e., $$$$\hat{J}_+ \hat{J}_- = \hat{J}_- \hat{J}_+ +2\hbar \hat{J}_z, \tag{2}\label{eq:2}$$$$ it follows that, for $$k=1,2,\dots$$, $$$$\hat{J}_+ |v_k\rangle = |v_{k-1}\rangle k\hbar [2\mu-(k-1)\hbar]. \tag{3}\label{eq:3}$$$$ To verify this formula, we can check the $$k=1$$ case by applying both hand sides of eq. (\ref{eq:2}) on $$|v_1\rangle$$, and then we can use mathematical induction for $$k>1$$.

For the ladder of $$\mu-k\hbar$$ ($$k=1,2,\dots$$), appearing in eq. (\ref{eq:1}), to truncate at a finite value, there must be integer $$K'\geq 0$$ such that \begin{align} C_-(X_{\alpha}, Y_{\beta}-(k-1)\hbar)\not=&0, & k=&1,\dots,K', \\ C_-(X_{\alpha}, Y_{\beta}-K'\hbar)=&0. \end{align} This means that \begin{align} |v_{K'}\rangle \not=&0, \\ |v_{K'+1}\rangle =&0. \end{align} By applying the left hand side of eq. (\ref{eq:2}) on $$|v_{K'}\rangle$$, $$$$\hat{J}_+ \hat{J}_- |v_{K'}\rangle = \hat{J}_+ |v_{K'+1}\rangle =0.$$$$ By applying the right hand side of eq. (\ref{eq:2}) on $$|v_{K'}\rangle$$ and using eq. (\ref{eq:3}), $$$$\begin{split} [\hat{J}_- \hat{J}_+ +2\hbar J_z] |v_{K'}\rangle =&\hat{J}_- |v_{K'-1}\rangle K'\hbar [2\mu-(K'-1)\hbar] \\& +|v_{K'}\rangle 2\hbar [\mu-K'\hbar] \\ =&|v_{K'}\rangle (K'+1)\hbar [2\mu-K'\hbar]. \end{split}$$$$ The results of the above two equations must be equal, and since $$K'+1>0$$, it follows that $$$$\mu = \frac{K'\hbar}{2}.$$$$ Therefore, we see that the largest eigenvalue $$\mu$$ on the ladder must have been an integer or half-integer multiple of $$\hbar$$, or otherwise that the ladder would continue downward infinitely and contradict with existence of the lower bound.

As a result, the eigenvalues of $$\hat{J}_z$$ must be $$$$\frac{K'\hbar}{2} -K'\hbar, \frac{K'\hbar}{2} -(K'-1)\hbar, \dots, \frac{K'\hbar}{2},$$$$ for some integer $$K'\geq 0$$.

I took this argument from the book,

• B.C. Hall, Quantum Theory for Mathematicians (Springer, 2013).

See Proof of Theorem 17.4 therein. Eq. (17.12) of the book corresponds to eq. (\ref{eq:3}) here. [$$v_j$$ on the right hand side of Eq. (17.12) should read $$v_{j-1}$$.]