In a book I am reading (Blank, Exner, Havlíček: Hilbert Space Operators in Quantum Mechanics), functions of operators are defined via spectral decomposition for self-adjoint operators. Spectral decomposition is also well-defined for unitary operators so I expect functions over these could also be defined analogously.

Then, however, throughout quantum mechanics, one often encounters exponentials of operators which are neither and not even normal, like the annihilation operator (with spectrum equal to $\mathbb{C}$) or the creation operator (whose spectrum is empty, AFAIK). This is usually understood in terms of the Taylor expansion,

$$\exp A = \sum_{n=0}^{+\infty} \frac1{n!} A^n$$

which in the relevant contexts turns out to be easy enough to apply term-wise when $\exp A$ acts on a vector. This is obviously a work-around with many dangerous pitfalls, keeping in mind that Taylor expansion does not work properly even in the "canonical" uses of exponential (take, for example,

$$\exp \left( a \frac{\mathrm{d}}{\mathrm{d}x} \right) f(x) \ne f(x+a)$$

for functions which are $C^\infty$ but not analytic).

My question is, is there a way of defining exponentiation rigorously for a broader class of operators? For example, if we know the spectrum and all eigenvectors of an operator, and those span the entire space (like the annihilation operator), would it be true to say its exponential is an operator with the same eigenvectors corresponding to $e$ raised to the respective eigenvalues? What would then happen in the case of the creation operator?

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    $\begingroup$ In the absence of the spectral theorem and the Borel functional calculus it provides, it would seem that you can at least get some sort of holomorphic functional calculus for so-called sectorial operators, see Section 5.2 here math.kit.edu/iana3/~schnaubelt/media/st-skript.pdf That said, given that the annihilation operator has spectrum all of $\mathbb{C}$, I'm not convinced it's sectorial, unless I'm missing something. $\endgroup$ Sep 12, 2013 at 15:14

1 Answer 1


In order for the power series definition $\exp(tA) v = \sum_{n=0}^\infty t^n A^n v/n!$ to be useful for an operator $A$ and a vector $v$ in your Hilbert space, you want $v$ to be in the domain of $A^n$ for all $n$ and $\sum_{n=0}^\infty \left\| A^n v \right\| |t|^n/n! < \infty$ for some $t > 0$. Such a vector $v$ is called an analytic vector for $A$. For a self-adjoint $A$ there's always a dense set of them. You might look at Reed and Simon, "Methods of Modern Mathematical Physics II: Fourier Analysis, Self-Adjointness", sec. X.6.

  • $\begingroup$ Thanks! My book covers analytic vectors in a later chapter, they seem to give the answer. $\endgroup$
    – The Vee
    Sep 18, 2013 at 11:00

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