Quantum mechanics and derivative operators

Question

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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.

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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

Answer 1

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.

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

Answer 2

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 $r=0$.