(The idea to this question came from my answer to Uniqueness of quantum ladder for the harmonic oscillator)

The Hamiltonian $H$ for quantum harmonic oscillator can be written in terms of the ladder operators $a_+$ and $a_-$ as $$ H=\hbar\omega(a_+ a_-+1/2)=\hbar\omega(N+1/2), $$ where $N$ is the number operator. Then $$ [N,a_+]=a_+ \qquad \text{and} \qquad [N,a_-]=-a_-, $$ and if $|\psi\rangle$ is an eigenstate for $N$ with eigenvalue $c$ then $$ Na_+|\psi\rangle=(c+1)a_+|\psi\rangle\qquad\text{and}\qquad Na_-|\psi\rangle=(c-1)a_-|\psi\rangle. $$

Any operator $M$ such that $[N,M]=\lambda M$, where $\lambda$ is a number, produces the same effect, getting new eigenvalues and eigenstates: $$ NM|\psi\rangle=(c+\lambda)M|\psi\rangle. $$ In fact, defining the grade of a product of ladder operators as $$ \text{grade}(a_+^n a_-^m)=n-m, $$ where $n$ and $m$ are positive integers, any sum of same grade operators satisfies the same relation as $M$ with $\lambda=n-m$. In particular, any grade zero operator commutes with the Hamiltonian.

The question

Can operators with non-integer grade be defined?

For example, if the operator $\sqrt{a_+}$ can be defined, then $$ [a_+a_-,\sqrt{a_+}]=a_+[a_-,\sqrt{a_+}]=\frac{1}{2}\sqrt{a_+}, $$ where the formal rule $[a_-,f(a_-,a_+)]=\frac{∂f(a_-,a_+)}{\partial a_+}$ was used, $f$ being an arbitrary function of $a_+$ and $a_-$. But this implies a difference of one half between eigenvalues associated to different eigenstates: $$ N\sqrt{a_+}|\psi\rangle=(c+\frac{1}{2})\sqrt{a_+}|\psi\rangle. $$

An operator such as the above one would produce a different spectrum and it is very well established that this is impossible in the following questions:

How do we know that we have captured the entire spectrum of the Harmonic Oscillator by using ladder operators?

Proof that energy states of a harmonic oscillator given by ladder operator include all states

How do we know that we have captured the entire spectrum of the Harmonic Oscillator by using ladder operators?

So the answer to the above question is negative, but all the answers cited above resort to the actual spectrum to get a proof and my real question is:

Is it possible to prove that non-integer powers of the operators $a_+$ and $a_-$ do not exist without resort to the spectrum?

I mean a proof like that ladder operators do not have inverse for finite-dimensional vector spaces: if the ladder operator $M$ has inverse then $N-MNM^{-1}=\lambda 1$, but the trace of the left-hand side is zero, while the trace of right-hand side is not, a contradiction.

Position representation

In position representation, the question is if differential operators such as $\sqrt{x-\frac{d}{dx}}$ exist. I search a lot for fractional differential operators, but I didn't find anything that could provide any help. I thought in expressing the operator as $\sqrt{x}\sqrt{1-\frac{d/dx}{x}}$ and developing the second square root as a power series, but there is some ambiguity as $x$ and $d/dx$ do not commute.


2 Answers 2


The answer is negative. Suppose your operators could be defined on a domain including the natural domain of $a_+a_-$ (indifferently made of rapidly vanishing smooth functions or all possible finite linear combinations of vectors $a_+^n|0\rangle$). And suppose that they satisfy there the "anomalous" commutation relation you point out.

As a consequence, as you notice, they would produce a different spectrum for $a_+a_-$ on the said domain. Consequently also any self adjoint extension of $a_+a_-$ would gain a different spectrum.

Since $a_+a_-$ is essentially self-adjoint on its natural domain, there is only one self-adjoint extension of $a_+a_-$ and the spectrum of that unique extension is the known one. The spectrum is therefore rigidly fixed and your operators cannot exist: Every attempt to define them would encounter some obstruction at the level of domains.

  • $\begingroup$ Thank for your answer. I've never heard of self-adjoint extensions and I am going to do some research on it to better appreciate your answer. I want to say I already knew that such operators could not exist: that fact is very well established in former answers by you and others. What I wanted is an argument for nonexistence without explicitly resort to the spectrum. My point is that such an argument could be useful in other problems with different spectra. It is my fault not making this clear in the question. $\endgroup$ Commented Jan 25, 2018 at 11:12
  • $\begingroup$ @ValterMoretti Could you also please consider addressing this question: physics.stackexchange.com/q/380655 (it is relevant because it was the root behind this post, as "jobe" has written on top of page here.) Your insight into this would be much appreciated. $\endgroup$
    – user135626
    Commented Jan 29, 2018 at 14:42
  • $\begingroup$ Sorry I am too busy now, I will try to have a look next week. $\endgroup$ Commented Jan 29, 2018 at 18:47
  1. Let us for simplicity just consider the square root as an example of a non-integer power. Square roots of operators are usually only defined for semipositive operators, but $a_{\pm}=a_{\mp}^{\dagger}$ are not even normal operators, cf. the CCR $$ [a_-,a_+]~=~\hbar {\bf 1} .\tag{1}$$

  2. Nevertheless, if we ignore this fact, then we must demand for consistency $$ [\sqrt{a_-},a_+]~=~\frac{\hbar}{2\sqrt{a_-}}, \qquad [a_-,\sqrt{a_+}]~=~\frac{\hbar}{2\sqrt{a_+}},\tag{2}$$ as OP essentially already deduced. Eq. (2) clashes with the fact that $a_{\pm}$ are usually taken to be not invertible.

  3. Nevertheless, if we are willing to ignore this as well, then we should next find a consistent formula for $$ [\sqrt{a_-},\sqrt{a_+}]~=~?\tag{3} $$ This turns out to be harder than it looks.

  4. We conjecture that the appropriate formula (3) is an infinite series $$ [\sqrt{a_-},\sqrt{a_+}]~=~\sum_{k=1}^{\infty} \frac{((2k-1)!!)^2\hbar^k}{2^{2k} k!}a_+^{1/2-k}a_-^{1/2-k},\tag{3} $$ and more generally $$[a_-^r,a_+^s]~=~\sum_{k=1}^{\infty} \frac{r!s!\hbar^k}{(r-k)!(s-k)! k!}a_+^{s-k}a_-^{r-k}, \qquad r,s~\in~ \mathbb{C},\tag{4}$$ where $r!:=\Gamma(r+1)$. The conjecture (4) is mainly based on the fact that eq. (4) is correct for non-negative integers $r,s\in \mathbb{N}_0$, cf. e.g. this Phys.SE post.

  5. A non-trivial consistency check of eq. (4) (which we haven't performed) is whether operator composition remains associative with the rule (4).

  • $\begingroup$ Thank you for your answer. I was looking for a argumentation for the nonexistence of such operators like yours: without resorting to the actual spectrum. I didn't understand what is "OP" in point 2. In point 3, I think the commutator is proportional to $1/\sqrt{a_-a_+}$, as is the case for classical phase-space. A last point: do you know any proof that $a_+$ and $a_-$ have no inverse without make explicit use of the spectrum, like the one I exemplified for finite-dimensional vector spaces? $\endgroup$ Commented Jan 25, 2018 at 10:44
  • $\begingroup$ OP means 'Original Poster'. In this case: You. $\endgroup$
    – Qmechanic
    Commented Jan 25, 2018 at 10:48

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