# Does the shift operator on $\ell^2(\mathbb{Z})$ have a logarithm?

Consider the Hilbert space $\ell^2(\mathbb{Z})$, i.e., the space of all sequences $\ldots,a_{-2},a_{-1},a_0,a_1,a_2,\ldots$ of complex numbers such that $\sum_n |a_n|^2 < \infty$ with the usual inner product. Let $S$ be the shift operator: $$(S a)_n = a_{n-1}.$$ If there a linear operator $A:\ell^2\to\ell^2$ such that $S=e^A$? I really doubt there is, but I'm not sure.

Context: I'm thinking about the momentum operator as a generator of the translation operator in quantum mechanics. One could think of $\ell^2(\mathbb{Z})$ as a discrete version of a particle on the line.

$l^{2}(\mathbb{Z})$ becomes $L^{2}[-\pi,\pi]$ under the unitary Fourier map $U : l^{2}(\mathbb{Z})\rightarrow L^{2}[-\pi,\pi]$ given by \begin{align} U\{ a_{n}\}_{n=-\infty}^{\infty} & = \sum_{n=-\infty}^{\infty} a_{n}e^{in\theta},\\ U^{-1}f & = \{ (f,e^{in\theta})_{L^{2}[-\pi,\pi]}\}_{n=-\infty}^{\infty}. \end{align} (The the inner-product $(\cdot,\cdot)_{L^{2}[-\pi,\pi]}$ on $L^{2}[0,2\pi]$ is normalized so that $\|1\|_{L^{2}[-\pi,\pi]}=1$.) The shift $S$ on $l^{2}(\mathbb{Z})$ becomes multiplication by $e^{i\theta}$ on $L^{2}[-\pi,\pi]$. That is, $S=U^{-1}EU$, where $$(Ef)(\theta)=e^{i\theta}f(\theta).$$

The 'log' operator $L : L^{2}[-\pi,\pi]\rightarrow L^{2}[-\pi,\pi]$ defined by $(Lf)(\theta)=i\theta f(\theta)$ is a bounded normal linear operator such that $e^{L}=E=USU^{-1}$. So $$S=U^{-1}e^{L}U=e^{U^{-1}LU}=e^{A},\;\;\; A = U^{-1}LU.$$ Sanity check: The spectrum of $L$ is $\{ i\theta : -\pi \le \theta \le \pi \}$ so that the spectrum of $e^{L}$ is the entire unit circle, as expected. It is possible to determine the explicit form for $A$ on $L^{2}(\mathbb{Z})$ from $$\{ a_{n}\}_{n=-\infty}^{\infty} \mapsto \sum_{n}a_{n}e^{in\theta} \mapsto \left\{ \frac{1}{2\pi}\int_{-\pi}^{\pi} i\theta\sum_{n}a_{n}e^{in\theta}e^{-im\theta}\,d\theta\right\}_{m=-\infty}^{\infty}$$

• This looks great. As you say, operator $A$ should then be such that $(A a)_m = \sum_n A_{mn} a_n$ where $A_{mn} = \frac{i}{2\pi}\int_{-\pi}^{\pi} \theta e^{i(n-m)\theta}\;d\theta$. This works out to $A_{mm}=0$ and $A_{mn} = \frac{(-1)^{n-m}}{n-m}$. The only problem now is that this $A$ doesn't satisfy the fundamental commutator relationship $[x,A]=I$. – Will Nelson Jun 15 '14 at 8:18
• In this context you have $[\frac{d}{d\theta},\theta]=I$. There's an issue when dealing with the full domain. – COVID-20 Jun 15 '14 at 9:28

I think the answer is yes, by the spectral theorem, but $A$ does not admit an easy description. Since $S$ is a normal operator it admits a representation as $S = \int_{S^1} \lambda\,d\pi(\lambda)$ for some projection-valued measure $\pi$. One can now just take $L:S^1\to i\mathbf{R}$ to be any bounded Borel branch of the logarithm and put $A = \int_{S^1} L(\lambda)\,d\pi(\lambda).$

The above is a bit abstract and technical, and it's helpful at least for me to think about a finite-dimensional analogue. Let $S:\ell^2(\mathbf{Z}/N\mathbf{Z})\to\ell^2(\mathbf{Z}/N\mathbf{Z})$ be the operator defined by $(Sa)_n = a_{n+1\bmod{N}}$. Then $S$ has eigenvector $v_\zeta = (1,\zeta,\dots,\zeta^{N-1})$ with eigenvalue $\zeta$ for each $N$th root of unity $\zeta$. Since the vectors $v_\zeta$ are a basis for $\ell^2(\mathbf{Z}/N\mathbf{Z})$ there is an operator $A$ mapping each $v_\zeta$ to $L(\zeta)v_\zeta$, and this $A$ satisfies $e^A = S$.

• This looks like the right answer, but I'm not sure about the conditions under which spectral theorem will hold in this case. (I don't have a good reference on the subject.) But I took your helpful suggestion to look at $\ell^2(\mathbb{Z}/N\mathbb{Z})$ to heart. It's quite easy to compute $\log S$ in that case, and it turns out that the kernel (matrix coefficients) for $\log S$ converges nicely as $N\to\infty$. – Will Nelson Jun 15 '14 at 1:04
• (cont.) The coefficients are $\log S_{ij} = i\pi$ if $i=j$ and $\log S_{ij} = 1/(j-i)$ for $i\ne j$. That should correspond to $i$ times the momentum operator (or $\hbar \frac{\partial}{\partial x}$) in quantum mechanics. I find this really remarkable! – Will Nelson Jun 15 '14 at 1:04