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Robertson–Schrödinger uncertainty relations
The most common general form of the uncertainty principle is the Robertson uncertainty relation.

For an arbitrary Hermitian operator $$\hat{\mathcal{O}}$$ we can associate a standard deviation


 * $$\sigma_{\mathcal{O}} = \sqrt{\langle \hat{\mathcal{O}}^2 \rangle-\langle \hat{\mathcal{O}}\rangle^2},$$

where the brackets $$\langle\mathcal{O}\rangle$$ indicate an expectation value. For a pair of operators $$\hat{A}$$ and $$\hat{B}$$, we may define their commutator as


 * $$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A},$$

In this notation, the Robertson uncertainty relation is given by


 * $$\sigma_A \sigma_B \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|,$$

The Robertson uncertainty relation immediately follows from a slightly stronger inequality, the Schrödinger uncertainty relation,

where we have introduced the anticommutator,
 * $$\{\hat{A},\hat{B}\}=\hat{A}\hat{B}+\hat{B}\hat{A}.$$

Uncertainty by the Pauli matrices
In 1976, an inequality that refines the Robertson relation by applying high-order commutators was found in  Our approach is based on the Pauli matrices. Later V.V. Dodonov used the method to derive relations for a several observables by using Clifford algebra.

According to Jackiw, the Robertson uncertainty is valid only when the commutator is C-number. Found here method is effective for variables that have commutators of high-order - for example for the kinetic energy operator and for coordinate one. Consider two operators $$ \hat A $$ and $$ \hat B $$ that have commutator $$ \hat C $$:


 * $$ \left [ \hat A,\hat B\right ]=\hat C .$$

To shorten formulas we use the operator deviations:


 * $$ \delta \hat A = \hat A -\left\langle \hat A \right\rangle $$,

when new operators have the zero mean deviation. To use the Pauli matrices we can consider the operator:
 * $$ \hat F=\gamma_1 \, \delta \hat A\, \sigma_1 + \gamma_2 \,\delta \hat B \, \sigma_2 + \gamma_3 \, \delta \hat C \, \sigma_3  ,$$

where 2×2 spin matrices $$ \sigma_i $$ have commutators:


 * $$ \left [ \sigma_i ,\sigma_k \right ]= \, i\, e_{ikl} \sigma_l ,$$

where $$ e_{ikl} $$ antisymmetric symbol. They act in the spin space independently from $$ \delta \hat A {,}\,\delta \hat B {,}\,\delta \hat C $$. Pauli matrices define the Clifford algebra. We take arbitrary numbers $$ \gamma_i $$ in operator $$ \hat F$$ to be real. Physical square of the operator is equal to:


 * $$ \hat F \hat F^+ = \gamma_1^2 ( \, \delta \hat A \delta \hat A^+)+ \gamma_2^2 ( \,\delta \hat B \delta \hat B^+) + \gamma_3^2 (\, \delta \hat C \delta \hat C^+)+ \gamma_1 \gamma_2 \,  \hat C\,  \sigma_3 + \gamma_2 \gamma_3\, \hat C_2 \, \sigma_1 - \gamma_1 \gamma_3 \,  \hat C_3 \, \sigma_2,$$

where $$ \hat F^+$$ is adjoint operator and commutators $$\hat C_2 $$ and  $$ \hat C_3$$ are following:


 * $$\hat C_2 = i\left [ \delta\hat B,\hat C \right ],\qquad \hat C_3 = i\left [ \delta\hat A,\hat C \right ] . $$

Operator $$ \hat F \hat F^+$$ is positive-definite, what is essential to get an inequality below. Taking average value of it over state $$ \left | \psi \right \rangle $$, we get positive-definite matrix 2×2:


 * $$\begin {align}

\left \langle\psi\right | \hat F \hat F^+\left |\psi \right \rangle &= \gamma_1^2 \, \left \langle (\delta \hat A )^2 \right \rangle +  \gamma_2^2 \,\left \langle ( \delta \hat B )^2 \right \rangle + \gamma_3^2 \, \left \langle ( \delta \hat C )^2 \right \rangle +\\ & +\gamma_1 \gamma_2 \, \left \langle \hat C \right \rangle \, \sigma_3 + \gamma_2 \gamma_3\, \left \langle\hat C_2\right \rangle \, \sigma_1 - \gamma_1 \gamma_3 \, \left \langle \hat C_3 \right \rangle \, \sigma_2 , \end {align} $$ where used the notion:


 * $$ \left \langle (\delta \hat A )^2 \right \rangle = \left \langle (\delta \hat A \delta A^+) \right \rangle $$

and analogous one for operators $$ B,\,C $$. Regarding that coefficients $$ \gamma_i $$ are arbitrary in the equation, we get the positive-definite matrix 6×6. Its leading principal minors  are non-negative. The Robertson uncertainty follows from minor of forth degree. To strengthen result we calculate determinant of sixth order:

The equality is observed only when the state is an eigenstate for the operator $$\hat F$$ and likewise for the spin variables:


 * $$\hat F\, {\left | \psi \right \rangle} {\left | \hat s \right \rangle} $$ = 0.

Found relation we may apply to the kinetic energy operator $$ \hat E_{kin} =\frac { {\mathbf \hat p^2}}{2}$$ and for operator of the coordinate $$\mathbf \hat x $$:

In particular, equality in the formula is observed for the ground state of the oscillator, whereas the right item of the Robertson uncertainty vanishes:
 * $$ \left \langle \, \mathbf \hat p \,\right \rangle = 0 $$.

Physical sence of the relation is more clear if to divide it by the squared nonzero average impulse what yields:

where $$ (\delta t)^2=\left \langle (\delta \mathbf \hat x)^2 \right \rangle \left \langle \mathbf \,\hat p \, \right \rangle^{-2}   $$ is squared effective time within which a particle moves near the mean trajectory.

The method can be applied for three noncomuting operators of angular momentum $$ \mathbf \hat L $$. We compile the operator:


 * $$ \hat F=\gamma_1 \, \delta \hat L_x\, \sigma_1 + \gamma_2 \,\delta \hat L_y \, \sigma_2 + \gamma_3 \, \delta \hat L_z \, \sigma_3  .$$

We recall that the operators $$ \sigma_i $$ are auxiliary and there is no relation between the spin variables of the particle. In such way, their commutative properties are of importance only. Squared and averaged operator $$ \hat F $$ gives positive-definite matrix where we get following inequality from:

To develop method for a group of operators one may use the Clifford algebra instead of the Pauli matrices.

Essence of the problem
Let us recall the background proceeding Fock’s accomplishment. Two classical vector integrals, the angular momentum and the Laplace-Runge-Lenz vector, in quantum mechanics correspond to vector operators that commute with the energy operator, i.e.,with the Hamiltonian. An analysis of their commutators carried out in   shows that they generate  a Lie algebra (a linear space  with a commutation operation) coinciding with a Lie algebra of operators of small (infinitesimal) rotations in 4D space.

For physics, the correspondence means that some transformation of variables and operators maps the original quantum Coulomb problem  into the problem of  free motion  of a particle over a 3D sphere embedded  in a 4D space. The energy operator is then invariant under rotations of the 3-D sphere. This is reminiscent of the remarkable effect of Lewis  Carroll’s  soaring grin  of the  Cheshire cat. .

Fock's approach struck contemporaries.The starting point in his theory is integral Schrödinger’s equation (SE) in momentum space. The space can be considered as 3D plane in a 4-D space. Fock then wraps 3D plane  into  a 3D sphere using  stereographic  projection, known since  antiquity   as convenient transformation of a globe into a flat  map. (Fock’s globe is three-dimensional as is the map.) At the same time, additionally, Fock surmises the factor for wave functions  such that  original integral  equation  turns into an equation  for spherical functions on 3D sphere (to be distinguished from  functions  on two-dimensional  sphere.)

This equation, rarely used in physics but well-known in the theory of special functions,  , is invariant under rotations in 4D space. Fock does not explain the physical meaning of transformation he found. As a result, the fundamental questions remain: why is the SO(4) symmetry realized in the wrapped momentum space rather than in the position space, and how has the electron 'learned about the stereographic projection?'

Recently,  Fock's theory is further developed  with the help of transformation of eigen-functions from the momentum space into 4-D position one. It was found that final transition of 4D spherical harmonics into physical space is algebraic and does not need an integration at all

Fock's theory
The Schrödinger equation for eigenfunctions (SE), using atomic units (the  unit of energy beeing $$ \frac{Z^2me^4 } {\hbar^2 } $$  and the unit of length being Bohr's radius $$ \frac{\hbar^2 }{Zme^2 }  $$ ), has the form
 * $$ \left( -\frac{1 }{2 } \Delta-\frac{ 1 }{ r   }  \right)\Psi_{nlm  }=-\frac{1 }{2n^2 }\Psi_{ nlm }$$.

In what follows, it is convenient to reduce the orbits of all radii $$na_B  $$  to a single radius,  i. e., change the radius vector for each eigen function as $$ \mathbf{  r' }= \frac{ \mathbf{r}}{n} $$. The equation then takes a deceptively simple form.

where $$ r $$ is the modulus of the vector $$\mathbf{r} $$. The eigenfunctions in the momentum representation then have scaled argument $$\mathbf{p'}=n\mathbf{p}$$.

The Schrödinger equation (with $$(\hbar=1)$$)when moving to the momentum representation:
 * $$ \Psi_{nlm}(\mathbf{r})=\frac{1}{ (2\pi)^3 }\int \ a_{nlm}(\mathbf {p} ) e^{(i\mathbf{pr})}

d^3\mathbf{ p} $$, contains a convolution with respect to momenta. Because the potential $$ \frac{1}{ r} $$ goes to   $$ \frac{4\pi }{ \left|\mathbf{ p^2  } \right|  }  $$, the SE in the momentum space is nonlocal:

Fock's sphere
The first step of the theory is to multiply the function $$a_{nlm}( \mathbf{ p }) $$ (done without an explanation)      by factor $$ ( 1+\mathbf{p^2 }  )^2    $$. The second step is to wrap the 3D plane into 3-D sphere (with four  coordinates $$( \boldsymbol{\xi},\mathit{\xi_0    } ) $$;      see Fig.1). Figure 1 shows that the tangent of the slope of the projecting (red) straight line is $$ \tan \varphi =\frac{1 } {\left| \mathbf{p} \right|}. $$

Hence follow the formulas
 * $$\left|\boldsymbol \xi    \right| =\sin 2\varphi=\frac{2\left|\mathbf {p}  \right| } {(1+\mathbf{p^2})}, $$
 * $$\boldsymbol \xi =\frac{2\mathbf {p} } {(1+\mathbf{p^2})}, $$
 * $$\mathit{ \xi_0} =\cos 2\varphi  =\frac{(\mathbf {p^2}-1) } {(\mathbf{p^2}+1)}, $$
 * $$ \boldsymbol\xi^2+\xi_0^2=1. $$

The stereographic projection doubles the tilt angle $$\varphi $$ and this is the effect it produces. The flat drawing correctly reflects the 4-D transformation.

In the new variables, with Fock’s factor taken into account, the eigen-function becomes
 * $$ b_{nlm }(\boldsymbol{ \xi}, \xi_0 ) =( \mathbf {p^2}+1)^2 a_{ nlm }(\mathbf { p } ). $$

It is essential that the projection be given by a conformal transformation. In this case, the angles between intersecting curves are preserved. The metric on the sphere in the momentum space coordinates ( of the p-plane) is expressed as
 * $$\frac{4 }{\mathbf{ (p^2+1)^2 }}(d\mathbf p )^2.  $$

Hence, the contraction coefficient for elements of the p-space is $$\frac{ (\mathbf{ p^2}+1   ) }{ 2   }   $$. The volume element in Fock's equation can be  expressed in terms of  the  3D surface element of sphere:


 * $$ d^3\mathbf{p}=\frac{1 }{8} (\mathbf{p^2}+1)^3 dS_\xi  $$

Additionally, the kernel of the integral  can be (very fortunately but not obviously)  transformed as


 * $$\frac{ 1 }{ \mathbf{(p-p')^2 } }= \frac{ 2 }{ \mathbf{(p^2+1) } } \frac{1 }{[\boldsymbol{(\xi-\xi')^2}+(\xi_0-\xi'_0   )^2 ]}  \frac{ 2 }{ \mathbf{(p'^2+1) } },  $$

which doesn’t follow from the conformal property. This relations allow Fock to obtain the integral equation

where as can be seen from the figure, the surface element of the unit 4D sphere with the volume $$2\pi^2  $$ is
 * $$ dS_\xi=\frac{d{\xi_1} d\xi_2 d{\xi_3}  }{\xi_0 }=\frac{ dV_\xi    }{ \xi_0 }   $$.

Next, V. Fock refers to the theory of spherical functions in 4-D space. These functions contain product of 3D spherical functions by   the Gegenbauer polynomials of  the argument  $$ \xi_0  $$.

Found equation is rather complicated while Gegenbauer polynomials are very simple and useful for physicists.

Anomalous magnetic moment of the electron
Here, it is worth pointing out, the empiric formula is valid for the anomalous magnetic moment        $$\delta\mu_e $$
 * $$\delta\mu_e =\frac{e\hbar}{2 c    }\left[\frac{ 3 }{ \sqrt{ 2 }\left(m_p-\exp\left(1\right )\delta m \right) }\right]$$,

where $$m_p $$ is the proton mass and $$\delta m $$ is difference between neutron and proton masses $$\left( m_p-m_n \right )$$. In terms of the Bohr's magneton $$\mu_B=\frac{e\hbar}{2m_e c    }  $$, the accuracy of the formula as follows
 * $$\delta\mu_e = 0.0011596523\times \mu_B  $$.

General properties
Four properties of symmetric tensor  $$\mathbf{M}_{i...k} $$   lead to the use of it in physics.

A. Tensor is homogeneous polynomial:
 * $$\mathbf{M}_{i...k}(k\mathbf{x})=k^l\mathbf{M}_{i...k}(\mathbf{x}) $$,

where l is the number of indices,i.e., tensor rank ;

B. Tensor is symmetric with respect to indices;

C. Tensor is harmonic, i.e., it is a solution of the Laplace equation:
 * $$\Delta \mathbf{M}_{i...k}(\mathbf{ x })=0 $$;

D. Trace over any two indices vanishes:
 * $$\mathbf{M}_{ii[...]}( \mathbf{x }  )=0 $$,

where symbol $$[...] $$ denotes remaining $$(l-2) $$ indices after equating $$i=i $$.

Components of tensor are solid spherical functions. Tensor can be divided by factor $$r^l $$ to acquire components in the form of spherical functions.

Multipole tensors in electrostatics
The multipole potentials arise when the potential of a point charge is expanded in powers of coordinates $$x_{oi } $$  of the radius vector  $$\mathbf{r }_{o} $$  ('Maxwell poles') . For potential
 * $$\frac{1 }{\left|\mathbf{r}-\mathbf{r}_{o }\right| }$$,

there is well known formula:
 * $$\frac{1 }{\left|\mathbf{r}-\mathbf{r}_{o }\right| }=

\sum_l(-1)^l\frac{(\mathbf{r}_0\nabla )^l  }{ l!}\frac{ 1}{ r }= \sum_l\frac { x_{0i }...x_{0k}    }{l!r^{ 2l+1 }     } \mathbf{M}_{i...k }^{(l)} (\mathbf{r}  ) = \sum_l\frac{ \mathbf{r}_0^{\otimes l} \mathbf{M}^{(l)}_{[i]}}{l!r^{2l+1  }}  $$, where the following notation is used. For the $$ l$$th tensor power of the radius vector
 * $$x_{0i}...x_ {0k}=\mathbf{r}_0^{\otimes l} $$,

and for a symmetric harmonic tensor of rank $$ l$$,
 * $$ \mathbf{M}^{( l ) }_{i...k }(\mathbf{r }  )= \mathbf{M}^{( l ) }_{[i] }(\mathbf{r }    )$$.

The tensor is a homogeneous harmonic polynomial with described the general properties. Contraction over any two indices (when the two gradients become the $$\Delta $$ operator) is null. If tensor is divided by $$r^{2l+1  }  $$, then a multipole harmonic tensor arises


 * $$ \frac{\mathbf{M}^{( l ) }_{i...k }(\mathbf{r })}{r^{2l+1 } } =\frac{ \mathbf{M}^{( l ) }_{[i] }(\mathbf{r }) }{r^{2l+1  }   }$$,

which is also a homogeneous harmonic function with homogeneity degree -(l+1).

From the formula for potential follows that
 * $$ \frac{\mathbf{M}^{( l+1 ) }_{im...k }(\mathbf{r })}{r^{2l+3 } } = -\nabla _i \frac{\mathbf{M}^ {(l)} _{m...k }(\mathbf{r })}{r^{2l+1}} $$,

which allows to construct a ladder operator.

Theorem on power-law equivalent moments in electrostatics
There is an obvious property of contraction
 * $$ (2l-1)!! ( \mathbf{ r}_0^{\otimes l } ,\mathbf{M }^{( l )}_{[i] }(\mathbf{r })) =

( \mathbf{M }^{( l )}_{[i] }(\mathbf{r }_0), \mathbf{M }^{( l )}_{[i] }(\mathbf{r } )      )   $$, that give rise to a theorem simplifying essentially the calculation of moments in theoretical physics.
 * Theorem

Let $$ \rho (x) $$ be a distribution of charge. When calculating a multipole potential, power-law moments can be used instead of  harmonic tensors (or instead of spherical  functions ):
 * $$ (\int\rho(\mathbf{r}_0 )  \mathbf{ r}_0^{\otimes l }dV_{\mathbf{r}_0  } ,\frac{\mathbf{M }^{( l )}_{[i] }(\mathbf{r })}{l!r^{ 2l+1} }) =

( \int \rho (\mathbf{r}_0 )\mathbf{M }^{( l )}_{[i] }(\mathbf{r }_0)dV_{\mathbf{r}_0  },\frac{ \mathbf{M }^{( l )}_{[i] }(\mathbf{r }  )}{(2l-1)!!l!r^{2l+1}   } )  $$.

It is an advantage in comparing with using of spherical functions.

Example 1.

For the quadrupole moment, instead of the integral

one can use 'short' integral

Moments are different but potentials are equal each other.

Formula for a harmonic tensor
Formula for the tensor was considered in using a ladder operator. It can be derived using the Laplace operator. Similar approach is known in the theory of special functions. The first term in the formula, as is easy to see from expansion of a point charge potential, is equal to
 * $$\mathbf{M}^ {(l ) }_{[i] } ( \mathbf{ r } ) = (2l-1)!!x_{i_1}...x_{i_l }+...=(2l-1)!!\mathbf{r}^{\otimes l   }+...   $$.

The remaining terms can be obtained by repeatedly applying the Laplace operator and multiplying by an even power of the modulus $$r $$. The coefficients are easy to determine by substituting expansion in the Laplace equation. As a result, formula is following:

.

This form is useful for applying differential operators of quantum mechanics and electrostatics to it. The differentiation generates product of the Kronecker symbols.

Example 2



The last quality can be verified using the contraction with  is a Legendre polynomial.

Special contractions
In perturbation theory, it is necessary to expand the source in terms of spherical functions. If the source is a polynomial, for example, when calculating the Stark effect, then the integrals are standard, but cumbersome. When calculating with the help of invariant tensors, the expansion coefficients are simplified, and there is then no need to integrals. It suffices, as shown in, to calculate contractions that lower the rank of the tensors  under consideration. Instead of integrals, the operation of calculating the trace \hat{ T }r of a tensor over two indices is used. The following rank reduction formula is useful:

where symbol [m] denotes all left (l-2) indices.

If the brackets contain several factors with the Kronecker delta, the following relation formula holds:

Calculating the trace reduces the number of the Kronecker symbols by one, and the rank of the harmonic tensor on the right-hand side of the equation decreases by two. Repeating the calculation of the trace k times eliminates all the Kronecker symbols:



Harmonic 4D tensors
The Laplace equation in four-dimensional 4D space has its own specifics. The potential of a point charge in 4D space is equal to  . From the expansion of the point-charge potential  with respect to powers   the multipole 4D potential arises:



The harmonic tensor in the numinator has a structure similar to  3D harmonic tensor. Its contraction with respect to any two indices must vanish. The dipole and quadruple 4-D tensors, as follows from here, are expressed as

The leading term of the expansion, as can be seen, is equal to

The method described for 3D tensor, gives relations . Four-dimensional tensors are structurally simpler than 3D tensors.

Decomposition of polynomials in terms of harmonic functions
Applying the contraction rules allows decomposing the tensor with respect to the harmonic ones. In the perturbation theory, even the third approximation often considered good. Here, the decomposition of the tensor power up to the rank  l=6 is presented:







To derive the formulas, it is useful to calculate the contraction with respect two indices,i.e., the trace. The formula for l=6 then implies the formula for l=4. Applying the trace, there is convenient to use  rules of previous section. Particular, the last term of the relations for even values of l has the form

Also useful is the frequently occuring contraction over all indices,

which arises when normalizing the states.

Decomposition of polynomials in 4D space
The decomposition of tensor powers of a vector is also compact in four dimensions:









When using the tensor notation with indices suppressed, the last equality becomes

Decomposition of higher powers is not more difficult using contractions over two indices.

Ladder operator
Ladder operators  are useful for representing eigen functions in a compact form. They are a basis for constructing coherent states . Operators considered here, in mani respects close to the 'creation' and 'annihilation' operators of an oscillator.

Efimov's operator that increases the value of rank by one was introduced in  . It can be obtained from expansion of point-charge potential:

Straightforward differentiation on the left-hand side of the equation yields a vector operator acting on a harmonic tensor:

where operator

multiplies homogeneous polynomial by degree of homogeneity l. In particular,

As a result of an l- fold application to unity, the harmonic tensor arises: written here in different forms.

The relation of this tensor to the angular momentum operator (\hbar=1 ) is as follows:

Some useful properties of the operator in vector form given below. Scalar product



yields a vanishing trace over any two indices. The scalar product of vectors and is

and, hence, the contraction of the tensor with the vector \mathbf{ x} can be expressed as

where l is a number.

The commutator in the scalar product on the sphere is equal to unity:
 * \mathbf {x} \mathbf{D} - \mathbf{D}\mathbf {x} =r^2 $$.

To calculate the divergence of a tensor, a useful formula is
 * $$ \mathbf{\nabla}\hat \mathbf{D} = (\hat l+1 )(2 \hat l+3  )$$,

whence
 * $$\nabla_i \hat M^{(l)}_{ik...m }=l (2l+1 )M^{(l-1)}_{(k...m )} $$

($$l $$ on the right-hand side is a number).

Four-dimensional ladder operator $$ \mathfrak{ D } $$
The raising operator in 4D space
 * $$ \hat \mathfrak{D}_i \mathfrak {M}^{(n)}_{k...m }(\mathbf{ r },\tau)= \mathfrak {M}^{(n+1)}_{ik...m }(\mathbf{ r },\tau) =\mathfrak {M}^{(n+1)}_{ik...m }(\mathbf{y })$$

has largely similar properties. The main formula for it is

, where $$ y_i $$ is a 4D vector, $$ i=1,2,3,4 $$,
 * $$ \mathbf{y}=(\mathbf{r },\tau ),\quad \left|\mathbf{ y }\right|^2=\rho^2_{\mathbf{y}}=\mathbf{x}^2+\tau^2 $$,

and the $$ \hat n$$ operator multiplies a homogeneous polynomial by its degree. Separating the $$ \tau$$ variable is convenient for physical problems:
 * $$\hat n=(\mathbf{x}\mathbf{\nabla}_\mathbf{ x } +\tau\frac{\partial  } {\partial \tau})=\mathbf{ y  }\mathbf{\nabla}_{\mathbf{y}} $$.

In particular,
 * $$\hat \mathfrak{D}_i\mathbf 1=2y_i,\quad\hat \mathfrak{D}_iy_k=2(4y_iy_k -\rho_{\mathbf{y}}^2 \delta_{ ik } )$$,
 * $$ \hat \mathfrak{D}_i \hat \mathfrak{D}_k\hat \mathfrak{D}_m=4!![6x_ix_kx_m-\rho^2_{\mathbf{y}} (\delta_{ik}x_m+ \delta_{km}x_i+ \delta_{im}x_k) ] $$.

The scalar product of the ladder operator $$\hat \mathfrak{D} $$ and $$y$$ is as simple as in 3D space:
 * $$\mathbf{y} \hat \mathfrak{D}=\hat n_{\mathbf{y}}\rho^2_{\mathbf{y}},\quad \hat \mathfrak{D}\mathbf{y}=(\hat n-2)\rho^2_{\mathbf{y}} $$.

The scalar product of $$\hat \mathfrak{D} $$ and $$ \mathbf{ \nabla } $$ is
 * $$\mathbf{\nabla} \hat \mathfrak{D}=2(\hat n+2)^2$$.

The ladder operator  is now associated  with the angular momentum operator  and additional operator of rotations in 4D space $$ \hat \mathbf{A } $$. They perform Lie algebra as the angular momentum and the  Laplace-Runge-Lenz operators. Operator $$ \hat \mathbf{A } $$ has the simple form
 * $$  \mathbf{\hat A } = i(\tau \mathbf{\nabla}-\mathbf{ x } \frac{\partial}{\partial\tau } )         $$.

Separately for the 3D $$\mathbf{ r } $$ -component and the forth coordinate $$ \tau $$ of the raising operator, formulas are
 * $$\hat\mathfrak D_{\mathbf{r}} = (\hat n+1 )\mathbf{r}+i[\mathbf{r}\times\hat\mathbf{L}]+i\tau \hat \mathbf{A }  $$,


 * $$\hat\mathfrak D_{\tau} = (\hat n+1 )\tau+i(\mathbf{r}\hat\mathbf{A}) $$.