Appendix A: Measure changing non-singular perturbation theory to 2nd order
Consider an operator \(\hat{{\mathcal {O}}}\) with domain D acting on the Hilbert space \({\mathcal {H}}=L^2(D\subseteq \mathrm{I\!R^3,}{\mathrm {d}}\mu )\) which is self-adjoint with respect to the measure \({\mathrm {d}}\mu .\) Then it satisfies the eigenvalue equation
$$\begin{aligned} \hat{{\mathcal {O}}}\psi _n&=\lambda _n\psi _n\quad \text {in}\ D \end{aligned}$$
(A1)
$$\begin{aligned} \psi _n&=0\quad \text {on}\ \partial D \end{aligned}$$
(A2)
with its eigenvalues \(\lambda _n\) and eigenvectors \(\psi _n\) (n may stand for multiple quantum numbers) which are assumed to be normalized such that
Further assume that the operator can be expanded perturbatively as
$$\begin{aligned} \hat{{\mathcal {O}}}&=\sum _{N=0}^\infty \epsilon ^N\hat{{\mathcal {O}}}^{(N)} \end{aligned}$$
(A4)
with \(\epsilon \ll 1\) and where all operators \(\hat{{\mathcal {O}}}^{(N)}\) are \(\hat{{\mathcal {O}}}^{(0)}\)-bounded. The latter is then called the unperturbed operator whose eigenvalues and -vectors are known. This is the starting point for perturbation theory in quantum mechanics.
A perturbative treatment of the eigenvalue problem (A1) can be complicated by the fact that the unperturbed operator might not be self-adjoint with respect to the measure \(\mu .\) In fact, it may instead be self-adjoint with respect to a measure \({\mathrm {d}}\mu _{0}.\) In usual introductions to perturbation theory \({\mathrm {d}}\mu ={\mathrm {d}}\mu _0\) is assumed.
Yet, this is not generally the case. In principle, the measures may be related as
$$\begin{aligned} {\mathrm {d}}\mu (x)={\mathrm {d}}\mu _0(x) a^2(x) \end{aligned}$$
(A5)
with a positive non-vanishing function of the coordinates \(a^2(x).\)
A simplifying assumption that has to be made here requires that the function \(a^2(x)\) may be expanded perturbatively as
$$\begin{aligned} a^2(x)=1+\sum _{N=1}^{\infty }\epsilon ^N a_{N}^{2}(x). \end{aligned}$$
(A6)
It is questionable whether this problem is solvable or even mathematically well-defined without this additional assumption. Anyway, it certainly applies to the physical example studied in Sect. 3.3.
Expanding \(a^2(x)\) as in (A6) is equivalent to introducing a perturbed scalar product as
Additionally, we assume the spectrum of \(\hat{{\mathcal {O}}}^{(0)}\) to be discrete i.e. we consider non-singular perturbation theory. Of course, this could be generalised to the singular case.
By virtue of (A6) the unperturbed operator is almost self-adjoint meaning that the non-self-adjointness is of higher order i.e.
In fact, the eigenvectors of the unperturbed operator \(\psi _n^{(0)}\) can be normalised (without loss of generality) as
Thus, they span an orthonormal basis of the Hilbert space corresponding to the unperturbed problem \({\mathcal {H}}_0=L^2(D\subseteq \mathrm{I\!R^3,}{\mathrm {d}}\mu _0)\) while the eigenvectors of the full operator \(\psi _n\) span an orthonormal basis of \({\mathcal {H}}.\) Hence, we can expand
$$\begin{aligned} \psi ^{(i)}_n&=\sum _{m}c^{(i)}_{nm}\psi _m \end{aligned}$$
(A9)
$$\begin{aligned}&=\sum _{m}c^{(1)}_{nm}\psi _m^{(0)}+{\mathcal {O}}(\epsilon ) \end{aligned}$$
(A10)
with the complex \(\text {i}\)th order coefficients \(c^{(i)}_{nm}.\) In particular, we approximate
$$\begin{aligned} \psi ^{(1)}_n&\simeq \sum _{m}c^{(1)}_{nm}\psi ^{(0)}_m. \end{aligned}$$
(A11)
Furthermore, from (A3) and the normalization of \(\psi ^{(0)}_n\), we deduce
and
The first of these relations (A12) translates to
which yields without loss of generality (discarding a global phase)
At this point we see the first influence of the non-self-adjointness of \(\hat{{\mathcal {O}}}^{(0)}\): the first order eigenvectors cannot be taken to be orthogonal to the unperturbed ones because they themselves are only orthogonal to 0th order i.e.
Perturbing (A1) yields order by order
$$\begin{aligned} {\mathcal {O}}(1)&{\left\{ \begin{array}{ll} \left( \hat{{\mathcal {O}}}^{(0)}-\lambda _n^{(0)}\right) &{}\psi _n^{(0)}=0|_{D}\\ &{}\psi _n^{(0)}=0|_{\partial D} \end{array}\right. } \end{aligned}$$
(A17)
$$\begin{aligned} {\mathcal {O}}(\epsilon )&{\left\{ \begin{array}{ll} \left( \hat{{\mathcal {O}}}^{(0)}-\lambda ^{(0)}_n\right) &{}\psi _n^{(1)}=-\left( \hat{{\mathcal {O}}}^{(1)}-\lambda _n^{(1)}\right) \psi _n^{(0)}\Big |_{D}\\ &{}\psi _n^{(1)}=0|_{\partial D} \end{array}\right. } \end{aligned}$$
(A18)
$$\begin{aligned} {\mathcal {O}}(\epsilon ^2)&{\left\{ \begin{array}{ll} \left( \hat{{\mathcal {O}}}^{(0)}-\lambda ^{(0)}_n\right) &{}\psi _n^{(2)}=-\left( \hat{{\mathcal {O}}}^{(2)}-\lambda ^{(2)}_n\right) \psi _n^{(0)}-\left( \hat{{\mathcal {O}}}^{(1)}-\lambda _n^{(1)}\right) \psi _n^{(1)}\Big |_{D} \\ &{}\psi _n^{(2)}=0|_{\partial D}. \end{array}\right. } \end{aligned}$$
(A19)
These equations will now be contracted with
for some m. The case \(m=n\) is yields the corrections to the eigenvalues while the coefficients \(c^{(1)}_{nm}\) can be determined from the opposite case Note that said contraction shifts the order of perturbation of some terms not only due to the expansion of \(\psi _m\) but also of the scalar product.
The 0th order part then really corresponds to solving the unperturbed eigenvalue problem because
takes place entirely within \({\mathcal {H}}_0.\)
Furthermore, the 1st order corrections to the eigenvalues read (applying (A17))
which with (A12) becomes
Hence, \(\hat{{\mathcal {O}}}^{(0)}\) not being self-adjoint does not significantly change the 1st order contribution to the eigenvalues. However, the expectation value is taken with respect to the measure \({\mathrm {d}}\mu _0.\)
Contracting the eigenvalue problem with
assuming \(n\ne m\) yields (after some algebra and applying (A18))
Again, this is equivalent to the case where \(\hat{{\mathcal {O}}}^{(0)}\) is self-adjoint and again the amplitude is calculated with respect to the unperturbed measure. Thus, to 1st order the only clearly visible change is in the coefficient \(c_{nn}^{(1)}\) according to (A15).
Analogously, the 2nd order correction to the eigenvalues reads after application of (A13) and (A17)
Finally, expanding \(\psi _n^{(1)}\) as in (A11) and applying (A15), (A16) and (A24), this can be expressed in terms of already known quantities as
The first two terms of this equation appear in usual perturbation theory as well (which is indicated by the fact that they are evaluated with respect to the unperturbed measure) while the last two are entirely new a fact that can be inferred from the higher order scalar products they are featuring. Thus, the change in the measure has a strong effect on the 2nd order eigenvalues and cannot be neglected.
Appendix B: Higher order corrections to the asymptotic uncertainty relation
In this appendix we expand upon the perturbative derivation of the asymptotic uncertainty relation in the main text to get to 4th order thereby explaining the origin of the mathematical constant \(\xi \) introduced in Sect. 4. The required math is in principle the same as in the text although the calculations get rather lengthy and can in their entirety only be performed numerically. The Riemann normal coordinate expansion of the metric reads to 4th order
[20]
$$\begin{aligned}&g_{ij}\simeq e^a_ie^b_j\left[ \delta _{ab}-\frac{1}{3}R_{acbd}|_{p_0}x^{c}x^{d} -\frac{1}{6}\nabla _e R_{acbd}|_{p_0}x^{c}x^{d}x^e\right. \nonumber \\&\quad \left. +\left( \frac{2}{45}R_{acdg}R^{~~~g}_{bef}|_{p_0} -\frac{1}{20}\nabla _e\nabla _f R_{acbd}|_{p_0}\right) x^{c}x^{d}x^ex^f\right] . \end{aligned}$$
(B1)
Furthermore, the root of the determinant of the metric equals to 2nd order
$$\begin{aligned} \sqrt{g}\simeq 1-\frac{1}{6}R_{ab}|_{p_0}x^ax^b \end{aligned}$$
(B2)
which induces a perturbation in the measure
$$\begin{aligned} {\mathrm {d}}\mu \simeq {\mathrm {d}}\mu _0\left( 1-\frac{1}{6}R_{ab}|_{p_0}x^ax^b\right) \end{aligned}$$
(B3)
with \({\mathrm {d}}\mu _0=\sigma ^2{\mathrm {d}}\sigma {\mathrm {d}}\Omega .\)
As in Sect. 4, the symmetries of the Riemann tensor imply that spherical coordinates constructed around \(p_0\) are automatically geodesic coordinates. In other words (B1) already satisfies (14).
Accordingly, the higher order contributions to the radial part of the Laplace–Beltrami-operator read
$$\begin{aligned} \Delta _\sigma ^{(3)}&=-\frac{\sigma ^2}{4}\nabla _c R_{ab}|_{p_0}l^al^bl^c\partial _\sigma \end{aligned}$$
(B4)
$$\begin{aligned} \Delta _\sigma ^{(4)}&=-\frac{\sigma ^3}{5}{\mathcal {R}}_{abcd}|_{p_0}l^al^bl^cl^d\partial _\sigma \end{aligned}$$
(B5)
where we defined for simplicity
$$\begin{aligned} {\mathcal {R}}_{abcd}\equiv \frac{1}{9}R^{e~f}_{~a~b}R_{ecfd} +\frac{1}{2}\nabla _c\nabla _dR_{ab}. \end{aligned}$$
(B6)
As there is no 1st order contribution to the metric. the 3rd order contribution can be treated as if it was of 1st order while the 4th order contribution can be treated as if it was of 2nd order in the sense of Appendix A.
Hence, the 3rd order contribution reads
$$\begin{aligned}&\quad =\int _0^\rho {\mathrm {d}}\sigma \frac{\sigma ^4}{4}\psi _{100}^{(0)} \partial _\sigma \psi _{100}^{(0)} \int _{S^2}{\mathrm {d}}\Omega \nabla _{c}R_{ab}|_{p_0} l^al^bl^c\end{aligned}$$
(B8)
$$\begin{aligned}&\quad =0 \end{aligned}$$
(B9)
where the last equality stems from the fact that the angular integral vanishes. Thus, there is no 3rd order contribution.
However, the 4th order eigenvalue is not as trivial. According to (A26), it reads
In order to be able to compute the values of the terms which are not summed over, we need radial integral
$$\begin{aligned} \int _0^\rho {\mathrm {d}}\sigma \sigma ^5\psi _{100}^{(0)}\partial _\sigma \psi _{100}^{(0)}&=\frac{5(3-2\pi ^2)}{48\pi ^3} \end{aligned}$$
(B11)
and the angular expression
$$\begin{aligned} \int _{S^2}{\mathrm {d}}\Omega l^al^bl^cl^d&=\frac{4\pi }{15}\left( \delta ^{ab}\delta ^{cd}+\delta ^{ac}\delta ^{bd}+\delta ^{ad}\delta ^{bc}\right) . \end{aligned}$$
(B12)
Furthermore, in three dimensions the Riemann tensor can be expressed in terms of the Ricci tensor and the metric (note that at \(p_0\) we have \(g_{ab}=\delta _{ab}\)) as
$$\begin{aligned} R_{abcd}|_{p_0}=\,&\delta _{ac}R_{bd}|_{p_0}-\delta _{ad}R_{bc}|_{p_0}-\delta _{bc}R_{ad}|_{p_0}+\delta _{bd}R_{ac}|_{p_0}\nonumber \\&-\frac{R|_{p_0}}{2}\left( \delta _{ac}\delta _{bd}-\delta _{ad}\delta _{bc}\right) \end{aligned}$$
(B13)
and by the contracted Bianchi identity
$$\begin{aligned} \nabla _c\nabla _dR_{ab}|_{p_0}\left( \delta ^{ab}\delta ^{cd}+\delta ^{ac}\delta ^{bd}+\delta ^{ad}\delta ^{bc}\right) =3\Delta R|_{p_0}. \end{aligned}$$
(B14)
Combining these, we obtain
Concerning the sums in (B10) both
and
only give non-vanishing contributions if \(l=m=0\) or \(l=2,\) \(m=-2,-1,0,1,2.\)
The exact limits of these sums are not known (the summands are complicated terms of spherical Bessel functions evaluated at special zeroes of other Bessel functions). Yet, they show fast convergence and can thus be obtained numerically. Consider the quantity
Using this notation, we can express the 4th order contribution to the eigenvalue as
Computing the remaining terms, we obtain
$$\begin{aligned} S_{00}&=\frac{3-2\pi ^2}{1944\pi ^2}R^2|_{p_0} \end{aligned}$$
(B20)
$$\begin{aligned} S_{20}&= -\frac{S_{2c}}{3} (T_{20}^{ab}R_{ab}|_{p_0})^2 \end{aligned}$$
(B21)
$$\begin{aligned} S_{21}+S_{2-1}&= -S_{2c}|T_{21}^{ab}R_{ab}|_{p_0}|^2 \end{aligned}$$
(B22)
$$\begin{aligned} S_{22}+S_{2-2}&= -S_{2c}|T_{20}^{ab}R_{ab}|_{p_0}|^2 \end{aligned}$$
(B23)
with the matrices (which are not tensors!)
$$\begin{aligned} T_{20}^{ab}&=\text {Diag}(1,1,-2) \end{aligned}$$
(B24)
$$\begin{aligned} T_{21}^{ab}&= \begin{bmatrix} 0 &{} 0 &{} 1\\ 0 &{} 0 &{} i\\ 1 &{} i &{} 0\\ \end{bmatrix} \end{aligned}$$
(B25)
$$\begin{aligned} T_{22}^{ab}&= \begin{bmatrix} 1 &{} i &{} 0\\ i &{} -1 &{} 0\\ 0 &{} 0 &{} 0\\ \end{bmatrix} \end{aligned}$$
(B26)
and the newly introduced numerical constant \(S_{2c}\simeq 6*10^{-3}.\) As the matrices (B24), (B25) and (B26) are not tensors, the contributions \(S_{20},\) \(S_{21}\) and \(S_{22}\) are not scalars. The reason for this unusual behaviour lies in the fact that \(\Delta ^{(2)}\) and
are by themselves not scalar quantities. However, summed up they yield
$$\begin{aligned} \sum _{m=-2}^2S_{2m}=S_{2c}\left( \frac{R^2}{6}-\frac{R^{ab}R_{ab}}{2}\right) \end{aligned}$$
(B27)
which is a scalar as expected.
Finally, plugging all terms into (B18), we obtain the 4th order contribution to the eigenvalue
$$\begin{aligned} \lambda _{100}^{(4)}=-\rho ^2\left[ \xi _1\left( 3R_{ab}R^{ab}-R^2\right) +\xi _2\Delta R\right] |_{p_0} \end{aligned}$$
(B28)
where we introduced the numerical constants
$$\begin{aligned} \xi _1&\simeq 3.0*10^{-3}, \end{aligned}$$
(B29)
$$\begin{aligned} \xi _2&=\frac{2\pi ^2-3}{120\pi ^2}. \end{aligned}$$
(B30)
Hence, we can approximately say that
$$\begin{aligned} \xi \equiv 72\xi _1\simeq 15\xi _2. \end{aligned}$$
(B31)
Furthermore, the result can be expressed in terms of the 0th order Cartan invariant \(\Psi _2\) which in three dimensions reads
[21]
$$\begin{aligned} \Psi _2^{~2}=\frac{3R^{ab}R_{ab}-R^2}{72} \end{aligned}$$
(B32)
thus yielding
$$\begin{aligned} \lambda _{100}^{(4)}\simeq -\rho ^2\xi \left( \Psi _2^{~2}+\frac{\Delta R}{15}\right) \Bigg |_{p_0} \end{aligned}$$
(B33)
with the numerical constant
$$\begin{aligned} \xi =\frac{2\pi ^2-3}{8\pi ^2}. \end{aligned}$$
(B34)
This concludes our treatment of the extended uncertainty relation to 4th order.