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EDITED POST

Suppose we have a classical problem where the Hamiltonian is defined as: $$H = c\frac{p^2}{x}$$ This Hamiltonian emerges in the context of Hamiltonian 1D cosmology, where we define $x= a(t)x_0$ as the position of some quantum particle in a flat isotropic homogeneous universe. In this case we take $x\geq 0$ since we are considering it to be a sort of radial coordinate.

For some reasons, one wishes to quantize this system in the momentum space, so that: \begin{equation} \begin{cases} \hat{p}^2 \phi(p) = p^2\phi(p)\\ \hat{x}\phi(p) = i\frac{d}{dp}\phi(p) \end{cases} \end{equation} On some dense domain D. To solve the Eigenvalue equation for $\phi$ one poses: $$\hat{H}\phi(p)=E\phi(p)$$

The operator order taken in consideration, for easier calculations, is: $$\frac{p^2}{x}\rightarrow \left(\frac{1}{\hat{x}}\right)(\hat{p}^2)$$

The problem thus reads:

$$c\left(\frac{1}{\hat{x}}\right)(\hat{p}^2)\phi(p)=E\phi(p)$$

At first sight i thought to solve the problem acting on the left with $\hat{x}$ obtaining: $$\left(c\hat{p}^2 - E\hat{x}\right)\phi(p)=0$$ But im not convinced this is well posed.

My question is:

Is this a correct way to solve the eigenvalue problem when dealing with the classical term $1/x$?

I cannot find another way to implent such an operator with different results.

The momentum Hilbert space should be $\mathcal{H}=L^2(\mathbb{R})$ but i think that those wave functions $\phi(p)$ are not normalizable, just like the plane waves

LolloBoldo
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    You would need to find a way to make sense of the operator ${1\over \hat x}$. One way to do this would be to interpret such an expression as the inverse of $\hat x$. Further, does one mean:$(\hat p)({1\over \hat x})$ or $({1\over \hat x})(\hat p)$? So decide what you want for ${1\over \hat x}$, write $\hat H$ in terms of products not ratios and apply it to the Hilbert space element. – Albertus Magnus Dec 26 '23 at 17:56
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    Note that, besides your question, you have an ordering ambiguity here, of a similar nature as the one in this question. In other words, there is no unique quantum mechanical operator for you classical $H$. – ZeroTheHero Dec 26 '23 at 17:56
  • If you go with the inverse operator definition, you would get an anti-derivative operator in momemtum space. Via analytic continuation you could define negative power and even fractional power derivatives using the Cauchy integral formula. – Albertus Magnus Dec 26 '23 at 18:00
  • This comment will become an answer (hopefully) but my condensed matter theory prof. said these operators (e.g. inverse $x$) can be understood under a taylor expansion. – JohnA. Dec 26 '23 at 18:01
  • Thats the problem with operators like these, there is no one unique way in which to understand them, however, they can still make for interesting mathematical study. If there is a solid state application then that at least fixes some motivation for assuming a particular definition. It seems that if you use a taylor expansion, one would have to terminate the series somewhere? – Albertus Magnus Dec 26 '23 at 18:07
  • Related: https://physics.stackexchange.com/questions/198190/can-the-inverse-operator-be-expressed-as-a-series?rq=1 – JohnA. Dec 26 '23 at 18:18
  • I am aware of my operator ordering ambiguity, so before fixing the behaviour related to the operators order i wanted to make sense of this inverse operator i found – LolloBoldo Dec 26 '23 at 21:39
  • It’s a shame this is still without an answer the community likes. @LolloBoldo would you clarify the operator order to maybe entice more people to answer. Also, add context (as someone suggested, is this a solid state physics problem?) – JohnA. Dec 27 '23 at 00:35
  • @JohnA. I can do that – LolloBoldo Dec 27 '23 at 10:23
  • @LolloBoldo. If a particle has coordinate $x_0$ at some time $t_0$, then it remains at that coordinate for all $t$ unless disturbed. The only thing that changes is the particle's displacement relative to other points in space. – Albertus Magnus Dec 27 '23 at 13:31
  • I know, the model defined in this way gives an equation to model the "motion", really a displacement, from the origin at $x=0$ given by the universe expansion rate – LolloBoldo Dec 27 '23 at 13:46
  • To Weyl-attched readers of the question, 1/x behaves abominably in the Weyl correspondence, so it is advisable to use a different ordering, even implicitly. – Cosmas Zachos Dec 29 '23 at 17:17

2 Answers2

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Quantization is an art, not a functor, as comments reminded you.

Here is a path to start thinking about your problem. You may arbitrarily choose the quantization $$ \hat H=c \hat p {1\over \hat x} \hat p, $$ which has your classical Hamiltonian as its classical limit, and is manifestly Hermitean, among 37897 similar hamiltonians with this property. (The other answer chooses $H={c\over 2}\left ( \hat p ^2 {1\over \hat x} + {1\over \hat x} \hat p ^2 \right )$, instead, complicating the footnote ODE.)

Your proposal is not Hermitean.

Working in the coordinate representation, equivalent to the momentum one by Fourier transformation (the representation has little to do with quantization!), you have $$ -c\hbar^2 \left({1\over x} \psi’\right )’=E\psi, $$ which you might solve$^\natural$, and Fourier transform, as you have learned in elementary QM; any system may be Fourier transformed, including those with a brick-wall step function potential. You may impose the restriction of variables by a boundary condition, just as you handle the solutions of the infinite square well free hamiltonian.

$^\natural$ Try $$ x y''-y'+b x^2 y =0 $$ in Wolfram α.

Cosmas Zachos
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  • The reason why i am forced to work in momentum space is that my x coordinate lives in $x>0$, so is not a good space for a lot of reasons. And i cannot go from x to p in my case – LolloBoldo Dec 30 '23 at 17:17
  • You should have specified the brick wall potential right from the start. In position space, you impose the suitable boundary conditions, as in infinite wall and infinite well potentials. You can always switch representation, even if technically challenging. – Cosmas Zachos Dec 30 '23 at 17:38
  • Yes thats true, i assumed it was clear by the definition since the scale factor is never negative in cosmology. Should i edit my original post to make this detail more clear? – LolloBoldo Dec 30 '23 at 18:02
  • Absolutely! I dont believe the majority of the readers care about the cosmology context; they’d care about quantization, negative powers, and hermiticity…. – Cosmas Zachos Dec 30 '23 at 18:43
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It's not a fine way to present an operator product of two noncommuting operators in a seemingly symmetric way, as $$x^{-1} p^2 \ne p^2 x^{-1}.$$ Your operator is not symmetric, let alone self-adjoint.

Take the first one and try to find a measure space such that $$ \int\ \rho(x) \ f^*(x) x^{-1} g''(x) dx = -\int \ \rho(x) \ f''^*(x) x^{-1} g(x) dx .$$

Background: In order to define an eigenvalue problem, $A f - \lambda f= 0$, in an infinite-dimensional space of functions, unbounded operators like $x, \partial_x$ have to be restricted to a domain in Hilbert space that yields finite scalar products $$\left< f, A g \right> = \int f(x) A g(x) dx \lt \infty.$$ In oder to get real eigenvalues, $A$ has to be symmetric: $$\left< f, A g \right> = \int f(x)^* A g(x) dx = \int (A f(x))^* g(x) dx .$$ If an operator of derivatives has left coefficients depending on x, the symmetry, by integration by parts, shifts the coefficients from left to right, $$\int f(x) a(x) \partial_{xx} g(x) dx = -\int \partial_x (f(x) a(x)) \partial_x g(x) dx = \int \partial_{xx}\left(( a(x) f(x)\right) g(x) dx.$$

There is only one way to get rid of the unsymmetric piece product: to consider another Hilbert space with integration measure $a(x)^{-1} dx$ cancelling that coefficient.

This is a common problem, if the Laplacian is expressed in a curvilinear coordinates. E.g., polar $$\partial_{xx}+\partial_{yy} = \frac{1}{r}\partial_r r \partial_r + \frac{1}{r^2} \partial_{\phi\phi}.$$

The radial factor of the Hilbert space transforms to $$ \left< f(r) , g(r) \right> = \int_0^\infty \rho(r)\ f(r) \ g(r) dr$$ such that, in the scalar product of derivatives, the factor r disapears $$\int_0^\infty f(r) (r^{-1}\partial_r (r g'(r))) r dr = -\int_0^\infty f'(r) \ r \ g'(r) dr = \int_0^\infty g(r) (r^{-1}\partial_r (r f'(r))) r dr.$$ The expression in the middle has the required symmetry for the linear space of of functions with integrable derivatives diverging not faster than $1/\sqrt r$ at $r=0$.

Cosmas Zachos
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Roland F
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  • I am not working in the position space so this is not a relevant example, also not working with 2D laplacians – LolloBoldo Dec 27 '23 at 10:22
  • @LolloBoldo The Laplacian is merely an illustration. Any and all work in the position representation is always equivalent to such in the momentum representation, through Fourier transformation: the representation is irrelevant to the quantization ordering scheme, which is what you are asking about. – Cosmas Zachos Dec 29 '23 at 17:29