I am currently dealing with an operator-valued function $f(\hat{T})$ of the following kind: $$f(\hat{T}) =\sqrt{1 + b\hat{T}^2} $$ where $b$ $\in \mathbb{R}$ and $\hat{T}$ is a linear operator acting on the usual Hilbert space of quantum mechanics, i.e. $L^2(\mathbb{R})$. What I am interested in is to define its action on a generic physical state $|\Psi\rangle$ in the representation of its own eigenbasis (of course I am supposing that this exists).

If I am correct, I could proceed with a series expansion of the previous operator. This leads me to write: $$\sqrt{1 + b\hat{T}^2}=\sum_{n=0}^{\infty} c_n (b\hat{T}^2)^n $$ where $c_n$ is the proper generalized binomial coefficent and the series should converge for $\left\|b\hat{T}^2\right\|\leq 1$.

This should not be a problem in principle, nevertheless, it turns out that, in my framework, the operator $\hat{T}$ is the usual momentum operator of quantum mechanics $\hat{p}$, which - to my knowledge - is unbounded with respect its domain of definition (which should be a dense subset of the Hilbert space $L^2(\mathbb{R})$). From this, I deduce that the previous procedure is not suitable for the present case. Hence what can be done? My question can be exposed from two different points of view:

1)How can I find the action of this operator if the series expansion fails to be a proper instrument?

2)Is it possible to "impose" that the momentum operator $\hat{p}$ be bounded - like some kind of constraint, which I imagine could lead to a redefinition of the operator itself? From a mathematical and physical point of view is this even possible?

Thank you all in advance.


1 Answer 1


I'm not sure what you are after, but, normally, for finite, as for infinite-dimensional operators, one uses some version of Sylvester's formula, and one investigates convergence properties afterwards.

If, by "its own eigenbasis", you mean the eigenbasis of momentum, $$ |\Psi\rangle= \int\!\!dp~~|p\rangle \langle p| \Psi\rangle, $$ then you evidently have $$\sqrt{I+b\hat p^2}|\Psi\rangle= \int\!\!dp~~|p\rangle ~\sqrt{1+b p^2} ~\langle p| \Psi\rangle,$$ as you appear to anticipate. (In sadistic undergraduate QM, in momentum space, this presents as $f(p)\Psi(p)$.)

You may be familiar with the operator based on the Casimir operator of angular momentum, $$ \sqrt{\hat L^2+1/4} -1/2 , $$ whose eigenvalues are $\ell$ when acting on irreducible spin multiplets.

  • $\begingroup$ Thank you for your answer. What you have written is clear for me and it allows me to clarify my final goal. I am sorry if I was somehow 'obscure' in the question. First of all, I would like to ask you one thing: in what you have written is implicit that $f(\hat{p})|\Psi\rangle=f(p) |\Psi\rangle$; this does sound obvious and intuitive and maybe it is but how can you demonstrate this? For sure I am missing something elementary here. $\endgroup$
    – RH_ss
    Feb 7, 2022 at 19:52
  • $\begingroup$ Secondly, and this is the point where I am clarifying myself, I was insisting on the 'series expansion' road because I would like to relate the mean value of the square root operator with the mean value of $\hat{p}$. This is something that apparently you cannot do with Sylvester's formula, since the mean value of course depends on the function $\Psi(p)$. $\endgroup$
    – RH_ss
    Feb 7, 2022 at 19:53
  • $\begingroup$ On first comment. ? NO! p is strictly an eigenvalue of $\hat p$ for its eigenstate $|p\rangle$. So p is always a dummy variable present inside a p-integral. In momentum-space wavefucntions, of course it is f(p), by the definition of Dirac bra-ket mechanics! $\endgroup$ Feb 7, 2022 at 20:03
  • $\begingroup$ On second comment. By mean value you mean expectation value? That always depends on the state. Unless you have a spectrum/ distribution of eigenvalues given. Define your terms. $\endgroup$ Feb 7, 2022 at 20:05
  • 1
    $\begingroup$ Yes, of course, a function of a diagonal operator is the diagonal operator with entries being functions of the respective eigenvalues! If you thought about it, it is the heart of Sylvester's formula. $\endgroup$ Feb 7, 2022 at 21:20

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