Comment on 'On the realisation of quantum Fisher information'

Recently, a calculation of the position I_ρ and momentum I_γ Fisher information was presented for (i) a linear harmonic oscillator, (ii) a quasi-one-dimensional hydrogen atom and (iii) an infinite potential well [1]. By general definition [2, 3], these quantum-information measures for arbitrary one-dimensional (1D) structures are defined as

$$\begin{eqnarray}&&{I}_{{\rho }_{n}}=\int {\rho }_{n}(x){\left[\displaystyle \frac{{\rm{d}}}{{\rm{d}}{x}}\mathrm{ln}{\rho }_{n}(x)\right]}^{2}{\rm{d}}{x}=\int \displaystyle \frac{{\rho }_{n}^{\prime }{(x)}^{2}}{{\rho }_{n}(x)}\,{\rm{d}}{x},\end{eqnarray} \tag{ 1a }$$

$$\begin{eqnarray}&&{I}_{{\gamma }_{n}}={\int }_{-\infty }^{\infty }{\gamma }_{n}(p){\left[\displaystyle \frac{{\rm{d}}}{{\rm{d}}{p}}\mathrm{ln}{\gamma }_{n}(p)\right]}^{2}{\rm{d}}{p}={\int }_{-\infty }^{\infty }\displaystyle \frac{{\gamma }_{n}^{\prime }{(p)}^{2}}{{\gamma }_{n}(p)}{\rm{d}}{p},\end{eqnarray} \tag{ 1b }$$

where a position integration is carried out over all available interval x and positive integer index n counts all possible quantum states. In these equations, ${\rho }_{n}(x)$ and ${\gamma }_{n}(p)$ are position and momentum probability density, respectively:

$$\begin{eqnarray}&&{\rho }_{n}(x)={{\rm{\Psi }}}_{n}^{2}(x),\end{eqnarray} \tag{ 2a }$$

$$\begin{eqnarray}&&{\gamma }_{n}(p)=| {{\rm{\Phi }}}_{n}(p){| }^{2},\end{eqnarray} \tag{ 2b }$$

and corresponding waveforms ${{\rm{\Psi }}}_{n}(x)$ and ${{\rm{\Phi }}}_{n}(p)$ are related through the Fourier transformation:

$$\begin{eqnarray}&&{{\rm{\Phi }}}_{n}(p)=\displaystyle \frac{1}{\sqrt{2\pi {\hslash }}}\int \exp \left(-\displaystyle \frac{{\rm{i}}}{{\hslash }}{px}\right){{\rm{\Psi }}}_{n}(x){\rm{d}}{x}.\end{eqnarray} \tag{ 3 }$$

Both of them satisfy orthonormality conditions:

$$\begin{eqnarray}&&\int {{\rm{\Psi }}}_{n^{\prime} }(x){{\rm{\Psi }}}_{n}(x){\rm{d}}{x}={\delta }_{{nn}^{\prime} },\end{eqnarray} \tag{ 4a }$$

$$\begin{eqnarray}&&{\int }_{-\infty }^{\infty }{{\rm{\Phi }}}_{n^{\prime} }^{* }(p){{\rm{\Phi }}}_{n}(p){\rm{d}}{p}={\delta }_{{nn}^{\prime} },\end{eqnarray} \tag{ 4b }$$

where ${\delta }_{{nn}^{\prime} }=\left\{\begin{array}{cc}1, & n=n^{\prime} \\ 0, & n\ne n^{\prime} \end{array}\right.$ is a Kronecker delta, $n,n^{\prime} =1,2,\ldots$. Real position wave function ${{\rm{\Psi }}}_{n}(x)$ and associated eigen energy E_n are found from the 1D Schrödinger equation:

$$\begin{eqnarray}&&-\displaystyle \frac{{{\hslash }}^{2}}{2{m}_{p}}\displaystyle \frac{{{\rm{d}}}^{2}{{\rm{\Psi }}}_{n}(x)}{{{\rm{d}}{x}}^{2}}+V(x){{\rm{\Psi }}}_{n}(x)={E}_{n}{{\rm{\Psi }}}_{n}(x),\end{eqnarray} \tag{ 5 }$$

with m_p being a mass of the particle and V(x) being an external potential.

First, we point out that for the infinite potential well of the width a the Fisher informations were calculated before [4] where the position component was evaluated directly from equation (1a) whereas for finding ${I}_{{\gamma }_{n}}$ an elegant and didactically instructive method was used; namely, since for this geometry both ${{\rm{\Psi }}}_{n}(x)$ and ${{\rm{\Phi }}}_{n}(p)$ are real, the integrand in equation (1b) becomes $4{[{{\rm{\Phi }}}_{n}^{\prime }(p)]}^{2}$, and using the reciprocity between position and momentum spaces, equation (3), one replaces infinite p integration by the finite x one:

$$\begin{eqnarray}&&{I}_{{\gamma }_{n}}=4{\int }_{-a/2}^{a/2}{x}^{2}{{\rm{\Psi }}}_{n}^{2}(x){\rm{d}}{x}.\end{eqnarray} \tag{ 6 }$$

Turning to the discussion of the quasi-1D hydrogen atom, it has to be noted that a problem of the quantum motion along the whole x axis, $-\infty \lt x\lt \infty$, in the potential $V(x)\sim -| x{| }^{-1}$, despite its long history, still remains (owing to the strong singularity at the origin and concomitant difficulty of matching right and left solutions at x = 0) a topic of debate and controversy, see, e.g., [5, 6] and literature cited therein. A situation is somewhat simplified when one confines the motion only to the right half space terminated at x = 0 by the infinite barrier. Accordingly, let us consider solutions of the Schrödinger equation (5) with the potential [1]

$$\begin{eqnarray}V(x)=\left\{\begin{array}{cc}-\tfrac{\alpha }{x}, & x\gt 0,\\ \infty , & x\leqslant 0,\end{array}\right.\end{eqnarray} \tag{ 7 }$$

$\alpha \gt 0$. Upon introducing Coulomb units where energies and distances are measured in terms of ${m}_{p}{\alpha }^{2}/{{\hslash }}^{2}$ and ${{\hslash }}^{2}/({m}_{p}\alpha )$, respectively [7], and momenta—in units of ${m}_{p}\alpha /{\hslash }$, one arrives at the differential equation

$$\begin{eqnarray}&&-\displaystyle \frac{1}{2}\displaystyle \frac{{{\rm{d}}}^{2}{\rm{\Psi }}(x)}{{{\rm{d}}}^{2}x}-\displaystyle \frac{1}{x}{\rm{\Psi }}(x)-E{\rm{\Psi }}(x)=0,\end{eqnarray} \tag{ 8 }$$

whose general solution for the negative energies, $E=-| E|$, corresponding to the bound states, reads:

$$\begin{eqnarray}{\rm{\Psi }}(x) & = & {{\rm{e}}}^{-{(2| E| )}^{1/2}x}{(2| E| )}^{1/2}x\left[{c}_{1}M\left(1-\displaystyle \frac{1}{{(2| E| )}^{1/2}},2,{(2| E| )}^{1/2}x\right)\right.\\ & & \left.+{c}_{2}U\left(1-\displaystyle \frac{1}{{(2| E| )}^{1/2}},2,{(2| E| )}^{1/2}x\right)\right].\end{eqnarray} \tag{ 9 }$$

Here, $M(a,b,x)$ and $U(a,b,x)$ are Kummer, or confluent hypergeometric, functions (we follow the notation adopted in [8]), and c₁ and c₂ are normalisation constants. Physically, this mathematical solution vanishes at the origin and, since the second item in the square brackets of the right-hand side of equation (9) diverges at $x\to 0$ [8], it has to be neglected, ${c}_{2}=0$. Remaining part must decay sufficiently fast at infinity. From the properties of the Kummer function $M(a,b,x)$ [8] it follows that it is possible only when its first parameter is equal to the nonpositive integer what immediately leads to the energy spectrum coinciding with the 3D hydrogen atom [7]

$$\begin{eqnarray}&&{E}_{n}=-\displaystyle \frac{1}{2{n}^{2}},\end{eqnarray} \tag{ 10 }$$

whereas the corresponding waveform simplifies to

$$\begin{eqnarray}&&{{\rm{\Psi }}}_{n}(x)=\displaystyle \frac{2x}{{n}^{5/2}}{{\rm{e}}}^{-x/n}{L}_{n-1}^{(1)}\left(\displaystyle \frac{2x}{n}\right),\end{eqnarray} \tag{ 11 }$$

with ${L}_{m}^{(\beta )}(x)$, $m=0,1,\ldots$, being a generalised Laguerre polynomial [8]. Figure 1 shows waveforms of the first four levels. Didactically, a representation of the solution in the form of the Laguerre polynomials is much more advantageous compared to that of the confluent hypergeometric functions, equation (16) in [1]; in particular, utilising properties of the Laguerre polynomials (see equation 2.19.14.18 in [9]), one instantly confirms that equation (11) does satisfy the orthonormality condition, equation (4a), for $n=n^{\prime}$.¹ Moreover, the form of solution from equation (11) allows an instructive calculation from equation (3) of the momentum waveform. For doing this, one recalls Rodrigues formula for Laguerre polynomials [8]:

$$\begin{eqnarray}&&{L}_{m}^{(\beta )}(x)=\displaystyle \frac{{x}^{-\beta }}{m!}\,{{\rm{e}}}^{x}\displaystyle \frac{{{\rm{d}}}^{m}}{{{\rm{d}}{x}}^{m}}({{\rm{e}}}^{-x}{x}^{m+\beta }).\end{eqnarray} \tag{ 12 }$$

Then, ${{\rm{\Phi }}}_{n}(p)$ becomes:

$$\begin{eqnarray}&&{{\rm{\Phi }}}_{n}(p)=\displaystyle \frac{1}{2\sqrt{2\pi }}\displaystyle \frac{{n}^{1/2}}{n!}{\int }_{0}^{\infty }{{\rm{e}}}^{(1-{\rm{i}}{n}{p})x/2}\displaystyle \frac{{{\rm{d}}}^{n-1}}{{{\rm{d}}{x}}^{n-1}}({{\rm{e}}}^{-x}{x}^{n}){\rm{d}}{x}.\end{eqnarray} \tag{ 13 }$$

Successive $(n-1)$ integrations by parts simplify this to:

$$\begin{eqnarray}&&{{\rm{\Phi }}}_{n}(p)=\displaystyle \frac{{(-1)}^{n-1}}{{2}^{n}\sqrt{2\pi }}\displaystyle \frac{{n}^{1/2}}{n!}{(1-{\rm{i}}np)}^{n-1}{\int }_{0}^{\infty }{x}^{n}{{\rm{e}}}^{-(1+{\rm{i}}np)x/2}{\rm{d}}{x}.\end{eqnarray} \tag{ 14 }$$

An elementary deformation $z=\tfrac{1}{2}(1+{\rm{i}}np)x$ of the integration contour in this equation yields ultimately:

$$\begin{eqnarray}&&{{\rm{\Phi }}}_{n}(p)={(-1)}^{n+1}\sqrt{\displaystyle \frac{2n}{\pi }}\displaystyle \frac{{(1-{\rm{i}}np)}^{n-1}}{{(1+{\rm{i}}np)}^{n+1}}.\end{eqnarray} \tag{ 15 }$$

Observe that this complex solution is completely different from the real one provided by equation (17) from [1]. With the help of residue theorem applied to calculation of the integrals with infinite limits [10], it is elementary to check that the set from equation (15) does obey equation (4b), as expected. The dependencies of the real and imaginary parts of the waveforms ${{\rm{\Phi }}}_{n}(p)$ on the momentum are shown in figure 2. It is seen that the number and amplitude of the oscillations increase for the higher quantum indices n.

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Knowledge of the functions ${{\rm{\Psi }}}_{n}(x)$ and ${{\rm{\Phi }}}_{n}(p)$ and, accordingly, of the corresponding densities

$$\begin{eqnarray}&&{\rho }_{n}(x)=\displaystyle \frac{1}{{n}^{3}}{\left(\displaystyle \frac{2x}{n}\right)}^{2}{{\rm{e}}}^{-2x/n}{L}_{n-1}^{(1)}{\left(\displaystyle \frac{2x}{n}\right)}^{2}\end{eqnarray} \tag{ 16a }$$

$$\begin{eqnarray}&&{\gamma }_{n}(p)=\displaystyle \frac{2n}{\pi }\displaystyle \frac{1}{{(1+{n}^{2}{p}^{2})}^{2}},\end{eqnarray} \tag{ 16b }$$

paves the way to calculating the associated Fisher informations. Dropping quite simple intermediate computations (which in the case of the position component ${I}_{{\rho }_{n}}$ rely on the properties of the Laguerre polynomials [8, 9] while for its momentum counterpart ${I}_{{\gamma }_{n}}$ elementary properties of the integrals of the product of the power and algebraic functions are employed), the ultimate results are given as

$$\begin{eqnarray}&&{I}_{{\rho }_{n}}=\displaystyle \frac{4}{{n}^{2}}=8| {E}_{n}| ,\end{eqnarray} \tag{ 17a }$$

$$\begin{eqnarray}&&{I}_{{\gamma }_{n}}=2{n}^{2}=| {E}_{n}{| }^{-1},\end{eqnarray} \tag{ 17b }$$

what leads to the index independent product

$$\begin{eqnarray}&&{I}_{{\rho }_{n}}{I}_{{\gamma }_{n}}=8.\end{eqnarray} \tag{ 18 }$$

Note that position Fisher information whose expression, equation (17a), does coincide with its counterpart from [1] is proportional to the absolute value of energy what is its general property while the momentum component is just the inverse of $| {E}_{n}|$. As a result, the product of the two informations stays the same for all levels. This n independence singles out the hydrogen atom from other two structures studied in [1] for which ${I}_{{\rho }_{n}}{I}_{{\gamma }_{n}}$ is a quadratic function of the quantum index.