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Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

Under the finite dilation, the volume element transforms as $d^4{x}^{\prime }=e^{4\alpha }{d}^{4}x$, whereas ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{-2\alpha }{\partial }_{\mu }\varphi (x)$. Thus the kinetic term and quartic interaction both acquire the factor $e^{-4\alpha }$. Their factors cancel the Jacobian determinant, provingA mass term would instead acquire a net factor $e^{2\alpha }$ in the action and break this scale invariance.

For the Noether current, use the variation at the same coordinate, rather than comparing values at different coordinates:At $m=0$, the corresponding Lagrangian density variation is ${\delta }_0\mathcal{L}=-\alpha (4+x\cdot \partial )\mathcal{L}=-\alpha {\partial }_{\mu }(x^{\mu }\mathcal{L})$. Thus $J_0^{\mu }=-x^{\mu }\mathcal{L}$, and the stated Noether current convention givesThe frequently used dilatation current $D^{\mu }=x_{\nu }{T}^{\mu \nu }+\varphi {\partial }^{\mu }\varphi$ is $-j^{\mu }$; this is solely an overall parameter convention. In the convention $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$, the virial current is $V^{\mu }=-\varphi {\partial }^{\mu }\varphi$.

Using solution and the trace computed in solution gives the useful identityIt follows that $\boxed{{\partial }_{\mu }{j}^{\mu }=0}$ in the massless theory on shell. Equivalently, $x_{\nu }{\Omega }^{\mu \nu }$ is a conserved current, since its divergence is the trace of the improved stress-energy tensor on shell. These are classical scale invariance statements; the nonzero beta function in solution explains why the interacting quantum theory need not retain them.

The Fock vacuum condition and the creation operator commutator giveEquality of the square-root product and $2E_{\mathbf{p}}$ holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measureHere both $p^2$ and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is $J=E_{\mathbf{\Lambda }p}/E_{\mathbf{p}}$. Transforming the Dirac delta function yieldsso its factor exactly cancels that of $2E_{\mathbf{\Lambda }p}$. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is $d^{3}p/((2\pi {)}^{3}2E_{\mathbf{p}})$.^[ref]

Write $b=(\ell +1)a$, $R(r)=r^{\ell }{e}^{-r/b}$ and $Y(\theta ,\varphi )={\sin }^{\ell }\theta e^{i\ell \varphi }$, for nonnegative integer $\ell$. The angular Laplacian satisfies ${\Delta }_{S^2}Y=-\ell (\ell +1)Y$. Direct radial differentiation givesBecause $(\ell +1)/b=1/a$, the $1/r$ contribution cancels the Coulomb potential in the Hamiltonian. ThereforeThe volume element contributes $r^2\sin \theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular normalization cancels in every radial expectation. With $I_j={\int }_0^{\infty }{r}^j{e}^{-2r/b}dr$,The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)b{I}_j/2$. HenceIn particular $\mathrm{Var}(r)/{\mu }^2=(2\ell +3{)}^{-1}$. Put $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality gives

For the polar variable set $J_j={\int }_0^{\pi }{\sin }^j\theta d\theta$. The polar density is proportional to ${\sin }^{2\ell +1}\theta$, soAn exact expression for the requested tail, as well as its useful estimate, isThe inequality is Markov's inequality applied to ${\cos }^2\theta$. If a finite expression is desired, expand $(1-z^2{)}^{\ell }$: the numerator is ${\sum }_{j=0}^{\ell }(-1{)}^j(\genfrac{}{}{0ex}{}{\ell }{j})(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.

By the union bound, with probability at least $1-2h$ both $|r/\mu -1|\le h$ and $|\cos \theta |\le h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with probability tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The probability concentration of a circular Coulomb eigenstate is a precise version of the large-angular-momentum orbit picture.

For a nonnegative integer $\ell$, this Coulomb potential energy eigenstate has principal quantum number $\ell +1$ and maximal magnetic quantum number. Put $b=(\ell +1)a$. Including the spherical volume element, its radial probability density is a gamma distribution proportional to $r^{2\ell +2}{e}^{-2r/b}$; its polar density is proportional to ${\sin }^{2\ell +1}\theta$, while its azimuth is uniform. Radial and polar variables are independent. Their moments satisfyThe radial identities follow by successive integration by parts; the angular identity follows from ${\int }_0^{\pi }{\sin }^{j+2}\theta d\theta =(j+1){\int }_0^{\pi }{\sin }^j\theta d\theta /(j+2)$.

Set $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality and Markov's inequality give $\mathbb{P}(|r/\mu -1|>h)\le h$ and $\mathbb{P}(|\cos \theta |>h)\le h$. By the union bound, outside an event of probability at most $2h$, the point lies at distance $O(\mu h)$ from the equatorial circle of radius $\mu$. The quantum state therefore concentrates near a classical circular orbit in relative units as $\ell$ grows. The absolute orbit radius and absolute transverse width both grow; this statement does not assert localization in a fixed narrow tube.

With polar angle $\theta$ measured from positive $z$ and azimuth $\varphi$, the Cartesian formulas $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$ have Jacobian determinant $r^2\sin \theta$. For $r\ge 0$ and $0\le \theta \le \pi$ this gives the unsigned volume element shown. The inverse determinant is $1/(r^2\sin \theta )$ off the origin and polar axis.

Under a smooth locally invertible coordinate map $\Phi$ into Euclidean space, the positive volume element in a multiple integral is multiplied by the absolute Jacobian determinant. In three-dimensional spherical coordinates, $dV=r^2\sin \theta drd\theta d\phi$; in cylindrical coordinates, $dV=rdrd\theta dz$. The absolute value distinguishes positive integration measure from a signed orientation convention.