This is just a curiosity I had recently. I am going to let the reader interpret the context of the question, in their own way!

  • $\begingroup$ Very fun question, but probably more math SE-dwelling. Although I think asking for physical interpretations (although I can't think of anything) would make it a physics question too. $\endgroup$ Sep 14 '15 at 1:30

There are two main problems in defining the logarithm, and are related to the fact that the creation/annihilation operators are unbounded and not self-adjoint, neither normal.

The first problem is that there is not a functional calculus for non-normal operators; in addition an usual power series expansion is difficult to manage for unbounded operators.

It is true however that there is a nice common domain for these operators and their powers, namely the finite particle vectors. So you may hope to define the logarithm by means of its power series expansion. However, here it is another problem: the complex logarithm is multivalued, and you change branch when you approach the negative real axis (that, alas, is in the numerical range of both $a$ and $a^*$).

I do not know of any method to define such a multi-valued function of unbounded operators, either "physical" (not perfectly rigorous) or mathematical.

Different is the case of the field operator $\varphi=a+a^*$. Since it is self-adjoint, and the logarithm should be measurable wrt its spectral family (not sure, but I think it is), then it is possible to define at least $\log\lvert\varphi\rvert$ by means of the functional calculus (spectral theorem) for self-adjoint operators. You cannot get rid of the absolute value because $\varphi$ is not a positive operator in general.


This can be most easily done by looking at the eigenstates of the annihilation operator. They are the coherent states \begin{align} \left| \alpha \right\rangle \propto e^{\alpha a^\dagger}\left|0\right\rangle \end{align} such that $a \left| \alpha \right\rangle = \alpha \left| \alpha \right\rangle$.

This now implies that the operator $\text{log }a$ has the following eigenstates \begin{align} \text{log }a\left| \alpha \right\rangle &= \text{log }\alpha \left| \alpha \right\rangle\\ \Rightarrow \text{log }a\left| e^\alpha \right\rangle &= \alpha \left| e^\alpha \right\rangle \end{align} so that the eigenstates of $\text{log }a^\dagger$ with eigenvalue $\alpha$ are \begin{equation} \text{exp}\left(e^\alpha a^\dagger\right)\left|0 \right\rangle \end{equation}

and as you know, finding all the eigenvectors and eigenvalues of an operator completely defines it...

$\textbf{However}$ upon reconsideration this analysis fails because $a$ is singular as demonstrated by $a\left|0\right\rangle = 0$, and equivalently non invertible. This leads to the elementary result that its logarithm is not defined (in linear algebra one says that a singular matrix does not have a well defined logarithm)

This problem appears when you want to find how $\text{log }a$ acts on the vacuum \begin{align*} \text{log }a \left| 0 \right\rangle &= \text{log }a\text{ } e^{-\alpha a^\dagger}e^{+\alpha a^\dagger}\left| 0 \right\rangle \\ &= \alpha \left| 0 \right\rangle +\left[ \text{log }a,\text{ } e^{-\alpha a^\dagger}\right]e^{+\alpha a^\dagger}\left| 0 \right\rangle \end{align*} but then you realize that \begin{align*} \left[ \text{log }a,\text{ } e^{-\alpha a^\dagger}\right] &= \text{exp}\left( -\alpha \frac{\partial}{\partial a}\right) \text{log }a \\&= \text{not defined because of }a^{-1} \end{align*}


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