# Am I correct to say that ladder operators have complex eigenvalues?

From the definition: $$\left. \begin{array} { l } { \hat { L } _ { + } = \hat { L } _ { x } + i \hat { L } _ { y } } \\ { \hat { L } _ { - } = \hat { L } _ { x } - i \hat { L } _ { y } } \end{array} \right.$$ We know that $$\hat { L } _ { x }$$ and $$\hat { L } _ { y }$$ have real eigenvalues, thus $$\hat { L } _ { + }$$ and $$\hat { L } _ { - }$$ should have complex eigenvalues.

Am I correct to say so? Is there any physical meaning in this?

I do know that $$\hat { L } _ { + }$$ and $$\hat { L } _ { - }$$ do produce real coefficients, but only for a different state (the "next/previous" states of $$\hat { L } _ { z }$$. What I wonder is what happen to the same state :D

• Try playing with Pauli matrices. The eigenvalues of $L_{+}$ and $L_{-}$ are real but no independent eigenvectors. – K_inverse Oct 8 '18 at 8:15

Since $$L_+$$ and $$L_-$$ are not hermitian it is perfectly reasonable, as a possibility, that they might have complex eigenvalues.

However, for these specific operators, this is not the case. You can check this explicitly by taking the well-known relation $$L_+|l,m\rangle = \sqrt{l(l+1)-m(m+1)}|l,m+1\rangle,$$ expressing it as an explicit matrix, and taking the eigenvalues. The structure of the matrix is of the form $$L_+ = \begin{pmatrix} 0 & & \\ \sqrt{2l} & 0 & \\ & \sqrt{4l-2} & 0\\ & & \ddots & \ddots & & \\ & & &\sqrt{4l-2} & 0 \\ & & & &\sqrt{2l} & 0\\ \end{pmatrix}$$ where all the empty entries are zero, and that means that the characteristic polynomial can be calculated fairly simply, using row-reduction techniques, to the bare expression \begin{align} \det(L_+-\lambda) & = \det\begin{pmatrix} -\lambda & & \\ \sqrt{2l} & -\lambda & \\ & \sqrt{4l-2} & -\lambda\\ & & \ddots & \ddots & & \\ & & &\sqrt{4l-2} & -\lambda \\ & & & &\sqrt{2l} & -\lambda\\ \end{pmatrix} \\ & = (-1)^{2l+1}\lambda^{2l+1}. \end{align} In other words: the only eigenvalue of $$L_+$$ is zero, with multiplicity $$2l+1$$.

As for the eigenvectors of that eigenvalue, there is only one: $$L_+|l,l\rangle = 0. \tag{*}$$ The rest of the matrix is one big Jordan block, for which there are provably no more eigenvectors than the base case in $$(*)$$ above. In fact, $$L_z$$ is almost already in Jordan-Block form in the $$|l,m\rangle$$ basis, and all you need to do is to take a non-unit-normalized multiple of the $$|l,m\rangle$$ basis to bring $$L_z$$ into explicit Jordan-block form, $$L_+ = \begin{pmatrix} 0 & & & & & \\ 1 & 0 & \\ & 1 & 0\\ & & \ddots & \ddots & & \\ & & & 1 & 0 \\ & & & & 1 & 0\\ \end{pmatrix} .$$

The eigenvalues are $$0$$. The simplest way to see this is to suppose you work in a finite dimensional space of size $$2\ell+1$$. Then for any state $$\vert \ell m\rangle$$ you have $$L_+^k\vert\psi\rangle=0$$ for $$k\ge 2\ell+1$$ since $$L_+^{2\ell+1}\vert \ell,-\ell\rangle=0\, ,\qquad L_+^{2\ell+1}\vert \ell, -\ell+1\rangle=0\, \ldots$$ i.e. you can raise a state at most $$2\ell$$ times before you kill it. Now suppose $$\vert \psi\rangle=\sum_m c_m\vert\ell m\rangle$$ is such that $$L_+\vert\psi\rangle=\lambda\vert\psi\rangle$$. Apply $$L_+$$ again, and then again and then apply it $$2\ell+1$$ times to find $$L_+^{2\ell+1}\vert\psi\rangle= \lambda^{2\ell+1}\vert\psi\rangle =\sum_m c_m L_+^{2\ell+1}\vert\ell m\rangle=0$$ from which one must conclude $$\lambda=0$$. The same argument can be make to show that eigenvalues of $$\hat L_-$$ are $$0$$. Given that the eigenvalues are $$0$$ one must then find states $$\vert \psi\rangle$$ such that $$L_+\vert\psi\rangle=0$$. The only state that satisfies this is (up to normalization) $$\vert \ell,\ell\rangle$$.

This is unlike the situation for harmonic oscillator, where states are never killed by the raising operator $$\hat a^\dagger$$ because the space contains states $$\vert n\rangle$$ for any $$n\ge 0$$, i.e. the space is infinite dimensional. It is moreover possible to find some states which are eigenstates of $$\hat a$$: these are the famous coherent states and they are a sum containing all $$\vert n\rangle$$ state.

Eigenvectors of ladder operators are called "coherent states". Quoting the Wikipedia article:

Since $$\hat{a}$$ is not hermitian, $$\alpha$$ is, in general, a complex number.

So, yeah, you are correct to say that in general the eigenvalues are complex (yet it is possible to have some coherent states with real eigenvalues).

• The Bhaumik et al. paper that you reference provides coherent states with complex eigenvalues for the $I_-$ and $K_-$ operators which act as ladder operators on the total angular momentum $l$, and which have very different properties to the $L_\pm$ operators that OP is asking about. You can go hunting for references that construct complex-eigenvalue eigenvectors for $L_\pm$ all you want and come back when you find such a reference, of course. (Hint: you'd be wasting your time. $L_\pm$ can easily be shown to have zero as their one and only eigenvalue.) – Emilio Pisanty Oct 8 '18 at 10:17