# 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

## 3 Answers

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).

Since that answer seems controversial...
Here is a peer-reviewed paper reference on the subject for the angular momentum ladder operators:

D.Bhaumik et al 1975 J. Phys. A: Math. Gen. 8 1868

• OP is asking about the angular momentum ladder operators, not about the bosonic creation and annihilation operators. – Emilio Pisanty Oct 8 '18 at 9:36
• @EmilioPisanty I'm not answering specifically about the bosonic creation and annihilation operators. I'm talking about ladder operators in general: en.wikipedia.org/wiki/Ladder_operator – Kostya Oct 8 '18 at 9:43
• 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
• This is not correct. Coherent states are only eigenvectors of the lowering operators in the harmonic oscillator case. Otherwise one must use a different definition of coherent state : the most common definition is by Perelomov and gives the coherent state as a lowest weight state translated by a group operation. This coindices with the definition of CS for the harmonic oscillator and works for angular momentum and other cases. – ZeroTheHero Oct 8 '18 at 12:51