User:Alexander Adam

I am a physicist looking for a PhD position in theoretical physics in the realm of quantum information or related topics in quantum mechanics concerned with fundamental properties like entanglement or superposition and validity of theoretical assumptions and predictions.

[email protected]

To live in an asymmetric world, you should better be symmetric. However if you live in a symmetric world, asymmetry suffices.

I have also a strong interest in number theory because the universe of (natural) numbers is similar to an enigma. It is like the prototyp of all theories of maximal stability and we learn from it fundamental rules how to encode information with minimal relative overhead.

Some results of my personal research (currently most of them is a collection of identities of number theoretic functions I found by myself unless otherwise mentioned which seem important for me, if you find errors feel free to send me a hint - i will apply the correction and a note):

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When we are dealing with a natural number by the fundamental theorem of arithmetic we are dealing obviously with prime numbers. We may decompose . In the linear representation we have which means that we write down only prime numbers. However to quantify the multiplicity of each prime number we resort to natural numbers again. We may overcome this border by recursively decompose each multiplicity and switch to an exponential representation. Again we see only prime numbers, e.g. , but if we want to quantify the exponential levels where the primes appear we use again natural numbers. We might say in the last example that the multiplicity for is and the level is . We can now represent every level again in an exponential representation and also every other quantification of the number like the number of distinct prime numbers in and the number of levels of levels and so on. We say that a representation of a natural number is in prime number language if all of its quantifications are closed under prime numbers meaning that they are expressable only with prime numbers (or - including the - the set of smallest coprime numbers). The general question is now if there is a system that quantifies every natural number in prime number language. If it exists then a natural number would be like an illusion. It seems natural for this question to be answered that square-free numbers play a critical role in this game and also the number of quantifiable properties of a natural number . We now try to make the statement precise which serves us also as a check of the problem if it is well posed: Let be the set of all quantifiable properties of a natural number so that is a quantification of . It is clear that if holds we would be done but this is in general not the case. If we look at . A solution to the illusion problem would be to find such that there is always a finite chain of composition of elements from which evaluates to an element of .

Definition (Basis of a semi-group): Let be a countable semi-group and with . Then is called a basis for iff

.

Example:

Remark: Both examples are commutative and include a neutral element which makes them a commutative monoid. Please note that in the last example the square-free numbers form a semi-group if the multiplicity saturates to under multiplication. Thus we find a morphism between the lhs and the rhs.

Definition (Minimal basis):

Lemma (Semi-group): Let be a countable semi-group such that . Then .

Proof: Let . Then . But then we can decompose recursively and we get a binary tree. If the tree would have a finite level then we could take the leafes as elements of a set . So it follows that there must be at least one path of length but then by assumption.

Definition (Free abelian monoid)

Let be any set. Then with the following properties

is called the free abelian monoid with beeing the concatenation.

Corollary (Free abelian monoid)

Let be a family of monoids and . Then

• .

Let be any countable set and a bijection. Then

• where is the occurence number of in .

Proof: First statement: Let . Then . Since is from a monoid we can repeat this step a finite times for and we get a sequence with . Since this implies .

The second statement follows directly from the abelian property of the monoid.

Definition (Primorial)

Let be any countable set and a bijection. Then and .

Definition (Primorial iterates)

Let be any infinite countable set and a bijection. Then we define:

Lemma (Properties of primorial iterates)

• is an isomorphism.

Proof The first property is directly by definition since : .

Now the second one goes by induction. Because of , is a homomorphism. Since it is also invertible on , is bijective because the image is precisely . Suppose now that is an isomorphism. Now but and are isomorphisms so has to be .

The third property follows again by induction. We have . Now suppose . Then .

The last property can be proven by considering the fixpoints of . Since and we know already two fixpoints. Now consider . Since we will increase the sequence length under every iteration of .

• Let . Then .
  • Let . Then .
  • Let . Then .
  • Let . Then .
• Let and . Then

• • (The rhs follows from the special Lambert series.)

Proof:

We used for the first step Faulhaber's formula and for the last step the identity: .

• Let and . Then
  • Let . Then

• Let where . Then .

We have in general the identity: or which is invariant under reflection . First we proof the general property

Proof:

A relative tight upper bound can then be found as where the harmonic number can be coarse bounded as . Using the symmetry of the divisors we get an even better upper bound: .

• If we restrict ourself only to odd natural numbers we may get an other bound because the spacing of possible divisors is at least . We can express this as:

• I we consider instead the set we have a minimal periodic spacing of which we can express algebraically as (we have for every periodic sequence algebraically which shows that every such sequence is isomorphic to an alternating sequence) and therefore we get

We might also find always a better upper bound for any multiplicative function under the following assumption: Let and . Then .

• with

Lemma

Let the strictly increasing sequence of prime numbers and . Then

.

Proof: This follows from the fact that .

Lemma (Exchange of exponents)

Let the strictly increasing sequence of prime numbers and with and . Then

.

Proof: It is enough to consider the derivative of for . We have with because and . Therefore .

Definition (Asymptotic equivalence)

Corollary (Representation of a natural number)

Let . Then there exists with so that . In particular we make the identification .

Proof: Just consider the prime factorization of and regroup the factors according to it's prime multiples in increasing order. The co-prime condition follows immediately.

Remark: Clearly there will some so that .

Definition (Dedekind )

This representation has the advantage that we have .

Corollary (Dedekind ) Let . Then .

Lemma (Dedekind ) Let and .

Definition (Extremal primes)

Let given as above so that we have . We define:

Definition (Primorial)

Corollary (Primorial) Let . Then .

Lemma (Product of primorials)

Let be a product of primorials. Then for some sequences .

Remark: If we understand pointwise then we find a particular short notation of products of primorials: . As an example pick as the colossally abundant number. It is given in this notation as: . However we can get even more structure:

Definition (Primorial products) Let be the set of all finite primorial products. Then we define a completely multiplicative mapping with .

Lemma (Primorial isomorphism) Let be the set of all finite primorial products. Then under the multiplication of we have .

Proof This is because is a bijection and a homomorphism.

Corollary (Properties of primorial isomorphism)

• is an isomorphism.

Corollary (Product of primorials)

Let be a product of primorials. Then .

Definition (Monotonic bound)

and .

Definition (RH bound)

Lemma (Bound violation)

Let and . Then .

Theorem (Violation of montonic bound) Let and a monotonic bound. Then .

Proof: Let and . Then by our lemma we may assume that . However and . Because .

Lemma: (Violation of bound) Let and a monotonic bound. Then .

Proof: This follows directly by the Theorem and the fact that .

Corollary: (Violation of bound) Let . Then .

Lemma: (On the RH bound for even positive integers)

Let and . Then .

Proof:

We have . Now . Then .

Lemma: (On the RH bound for arbitrary powers)

Let and and such that . Then

.

Proof: Suppose that . Now . Therefore . It follows that .

Application (Square-full numbers):

All proper powers of with fulfill the RH bound. In particular for all such we have: . Because we deduce that all square-full numbers fulfill the RH bound.

Corollary: (On the RH bound for positive integer powers)

Lemma: (On the RH bound)

Let and and . Then .

Proof: . Now and therefore .

Lemma: Let with such that . Further let with . Then .

Proof:

Remark: This shows essentially that we are blind with this technique to numbers with a long square-free tail. Since for we have control over the quadratic component but not over .

Lemma: Let and such that . Then .

Remark: The second assumption is equivalent to: . But this condition is not always fulfilled. A counter example is which is a colossally abundant number. The existence of a counter example leads us to the conclusion that we have to investigate further the numbers which have .

Lemma (Bound on the largest prime factor) Let and may be eventually false. Then with .

Lemma (Necessary condition for the failure of Robin's inequality) Let , where is the exceptionel set where Robin's inequality is false. Then .

Proof: By failure of Robin's inequality we have . But and therefore .

Lemma (Sufficient condition for the truth of Robin's inequality) Let and sucht that . Then .

Proof: By assumption we have . This implies the stronger condition: .

We may assume here that we have a Hilbert space with dimension and an orthonormal base . The space of -tensors is given by and are the basis coefficients.

Now for some let be the entries of the modular multiplication table . This multiplication table has under the left and right action of the symmetric group symmetries.

We may define the generalized discrete Fourier transformation (GDFT) as an unitary matrix such that where . The connection follows then by letting . This is also the complex representation of the multiplication of elements of the finite cyclic group .

We establish here a connection to the microcanonical ensemble under the von Neumann measurement. Given some diagonal -tensor and an unitary matrix such that the new measurement base is the von Neumann measurement acts on the diagonal entries as the transition . If we demand now that we have . If we impose and this tells us also that it is always possible to push a quantum system into the state of maximal entropy. It is therefore pretty useful to investigate the space of all GDFTs.

Proof:

We set . If is given in column notation, we build a new matrix with where we identified . We may now order the according to the divisors of : every divisor of spawns an equivalence class of column vectors of . Elements of different equivalence classes are linearly independent so that our rank is determined by the size of the equivalence classes. The case where will decrease the rank by and will decrease the rank by . In the case where all these classes except have even size and we get:

In the case where also consists of only one element and we get

The result follows then together with the identity .

If we pick as a representative for the smallest natural number possible we have so that counts the number of cycles we passed already. E.g. so . Now we build the column difference matrix with which contains only zeros and ones. This is the binary representation of .

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The binary representation has many interesting properties and we will present here some of them. We will show here two derivations but first we introduce as a new -tensor vector space where and . In the first place we are interested in a special element . It is a binary rectangular matrix with all zeros except where the geometric line from to intersects one of the squares. We find now that so we can look at all the elements at once. The identification works because for some and we identify and recall that which is the triangular number and we have .

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From figure 2 we can guess that is composed as the direct sum of its irreducible element - that is we may write .

Proof: We draw a line along . To intersect the lattice nodes we demand that which is only possible if . Because this will happen exactly times.

Further from figure 3 we guess that the number of ones for two coprime natural numbers is given as .

Proof: We draw a line along . Because this adds ones. Another one is added if we transit to the next natural number in -direction meaning that we add another ones.

From that it follows that the number of ones in a generic matrix is

The last equation arises by counting the number of zeroes and removing them from the maximal amount of ones which is or by a little bit more complicated transformation of ceiling to floor. It is not new and a reference will be added. It is this trace formula which tells us that the diagonal and the spectrum of carry some important information about the natural numbers. We may also interpret as a bipartite pure quantum state and as the reduced state in subsystem and as the reduced state in subsystem . Further we have the properties:

• with
  • .
    • Notice that we can generalize it e.g. to so that we get the Euler identity for free: .
  • • The last inequality actually tells us that the most simple upper bound which we can choose is pretty worse because we have by far . The equality and upper bound can also easily extended to the slightly more general gcd-sum: . Again we get the Euler identity for free: .

Now we are turning back to the matrix and show how it is related to the binary representation . As a hint we may look at the elements and we observe that the differences of the number of white squares in every row of the half spaces created through the black line of consecutive matrices belong to the binary representation . This is because we have for the number of white squares in every row in the left half space .

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• 👁 The binary matrix under local reflection.
  
  The binary matrix under local reflection.
• 👁 Both binary matrices under logical or.
  
  Both binary matrices under logical or.
• 👁 Both binary matrices under logical and.
  
  Both binary matrices under logical and.
• 👁 Both binary matrices under logical xor.
  
  Both binary matrices under logical xor.

Let and . We note that describes the homogeneous commutative two-point nearest neighbor and self interaction of a linear chain^[1]. Then the spectrum ^[2] is given as

with the corresponding normalized eigenbasis ^[2] . Because we can calculate the entanglement information

which converges to which is less than the expected information of if we would take a pure bipartite quantum state at random under the unitarily invariant Haar measure. Anyway we conjecture here that meaning that it behaves similar like the information content of a random ensemble. Further if then .

Let such that . If we define and then . But is the squared semi-norm of the -dimensional Minkowski space where . Therefore is just another parametrization of and we have where is the Lorentz boost. It has the nice property that and forms therefore a one-parameter group. This ultimately shows that is really Lorentz scalar under bipartite factorization. Now is invariant under the dilation of with . We find that a boost of with dilates with . This makes it now easy for us to find the boost for certain lattice points.

Example (Boost of integer points): Let and . Then there exists a boost from the point to the point . This is because if we choose then .

Lemma (Boost of integer points): There exists a boost from the point to the point .

Proof: This is because if we choose then .

If we choose then . This shows that if we want to resolve a bipartite factorization of it is sufficient to know the fraction of the factors. This can be seen also by . However without a lot of work there is no chance to infer this fraction with our correspondence principle between dilation and Lorentz boost. But it allows us to state this as an inverse problem by finding so that . We find a bipartite factorization for odd numbers for example if we determine first and then which gives us then the greatest non-trivial factor of . If is not a prime we have . It is unclear how the formalism above would help us here.

Because we may think also of a different parametrization for . We might choose . Then and to make independent of we introduce so that with . Here we cannot revover from (actually we have ) but we may think of recovering from because we have the identity . From the identity it follows that there is not a such that so acts not continuous on the diagonal vector.

We did not talk about compatibility between two vectors and under a boost . We want to calculate the boost parameter such that . Under the boosts we say that is compatible with .

Corollar (Distance of compatible bipartitions): .

Proof: .

But how does the parameter looks like? We solve for in our dilation coordinates and . Remember that , and , . We find that there exists two solutions: which are only real if under the assumption that . If we exclude a complex valued we can say alternatively that lower valued factorizations are only compatible with higher valued ones. However a complex valued boost parameter shifts our points also into the complex plane which means that we can now find a boost which transforms between the complex integers to the real ones! Consider e.g. the factorization . If we take and then and . This means that . Essential we can reach any factorization of some integer under any quadratic field which means that their integer lattices are connected under hyperbolic rotation. Every bipartite factorization of a number in any quadratic field lies on the orbit . In our example we used a decomposition in the Gaussian integers which is the prototype for an imaginary quadratic field. Note that in our setting the physical interpretation of is that of time or space (dependent on the convention) so we would need here imaginary time or space to reach the Gaussian integers (or in general complex numbers).

Our considerations may seem simple but there is already a connection to a really deep problem called the Goldbach conjecture (GC). If we consider the sum of the factors then we see that this yields an even number if or . For the GC to be true we need the latter property (because we have with the only occurence of the first prime ) so we fix with such that . The GC states now that . Please note that the lower bound seems necessary only for the case . The upper and lower bound arises from the fact that the number of divisors of a number composed of two primes is in the case that with and otherwise. The transformations we need to count take to .

We can easily complicate the things further. Let us generalize our form for as where . For we get back our initial case from above. We can also write by taking elements from the Hilbert space . Here we encounter already the orthogonal symmetry because . But this is just a special kind of dilation because . So our complete symmetry is again defined over the space . We have symmetry of under the transformation with . Now we do the same again and define , and . Then where . We may infer for the finite dimensional case the symmetry group for the elements . It will turn out that they are symmetric under because we have . However if we want to establish such an isomorphism we need a permutation matrix which maps the odd components to the first half and the even components to the second half. E.g. and . Sadly this is not well defined for . Anyway we may conclude that and an additional boost between e.g and . We may write where . We may also introduce here a and a -dimensional analog of the Lorentz boost with between two neighbored components e.g. and . In the case that certain restrictions have to be taken to the symmetry group because we cannot allow transformations in the Euclidian subspaces to be taken over infinitely many dimensions at once because they are not necessary elements from . Instead we may take our space to be such that . Our symmetries tell us then which operations on the terms of the sum are allowed and which are not. Because such sums are conditionally convergent we cannot allow arbitrary manipulations of the terms. Only finite portions may be rearranged to give the same result. In general we may define such that . This implies that with and .

We consider here a free particle in a dimensional box with impenetrable walls. First we study the deterministic case where the particle follows the deterministic laws of motion of classical mechanics where we may include the relativistic effect of relative length scale if the particle speed is close to the speed of light .

Here the particle speed is unbounded and the particle is subject to no force having therefore a constant momentum between the boundaries. We will see that we get a connection under the right choice of particle speed relative to the box scale to the least common multiple (lcm). To describe the position of our particle over time we make the first observation that the components of the position are independent of each other because a perfect reflection at the boundary changes only the sign of the corresponding speed components. A perfect reflection in one dimension can then be described with the help of the triangular pulse which in turn we represent with another very important function in number theory. We have then for the position component where is the dimension of the box and its offset, the initial velocity of the particle and the offset in time which we can use also to describe an arbitrary starting position of the particle within the box. If we want an arbitrary position in the box we have to choose .

• The periodicity defines the spatial recurrence time in space component . From this we have to solve to get a non trivial solution for the spatial recurrence time. Fixing we have then which has a solution iff . Without loss of generality we can set and let with . Then which solves then our set of equations. We have made therefore a connection with the lcm and the trajectory of the particle is a periodic orbit. Because of our restriction of spatial recurrence to happen iff it is clear that if the quotient is irrational the trajectory of the particle is dense in the box over time.

• By Fourier series we have the representation:

• . If we apply here the Mellin transformation over time we get a further connection to number theoretic functions. We let be our transformation which is by our geometric interpretation the weight of a real unit ball with dimension over the density where only that in our case the density is taken over the unit frequency domain which renders a very unusual system. The weight plays therefore the role similar of position divided by times the unit of time. If is a zero where our weight vanishes this means that in this temporal dimension our particle returned to the coordinate . Interesting itself is the interpretation of the speed . Because the speed changes sign whenever the particle hits the boundary at it happens here in non uniform frequency steps which can be interpreted as a shrinking box with a shrinking factor and a length in frequency space between two consecutive frequency intervals of constant velocity while . This alone brings us to a trivial observation:

Lemma: If the speed component changes only finitely often over the unit frequency interval then the speed component changes only finitely often over time .

Proof: We have by inversion for the lowest frequency intervall .

For a particle trapped in a box with perfect reflective boundary this means that the particle sees no boundary after a finite amount of time meaning that the boundary might be already pushed to infinity or moving at least with the same speed as the particle. In our case however the box is static and the particle changes speed infinitely often.

The evaluation of the transformation leads us then to

which simplifies in the case and to

We set now , and . Then .

We have the same setting as above which means that our potential wall is infinite outside the box and zero inside and at the boundary of the -dimensional box. We note that the length of classical periodic orbits is twice the same as in the case where we have just periodic boundaries except there exists no orbit of zero length (no constant solution).

• Primitive periodic orbits:

Without loss of generality we may assume that the box has unit length in each spatial dimension. Then the set of lengths of the p.p.o. is given via

.

Now any length is of the shape where can be interpreted as the speed of the particle on the classical orbit and the number of unit time steps it takes to return to it's origin in phase space.

Taking the logarithm we see that we have trivially . In particular we have the bound .

Here we see that we have an infinity amount of p.p.o due to the fact that the minimal velocity is bounded from below by zero. This in turn implies that for fixed dimension the length of a p.p.o. is unbounded.

This interpretation needs some serious discussion since we normally take for the trace of an orbit the unit speed and it's corresponding return time. While mathematically this is just a relabeling of variables this dynamic point of view enables us to cut through the quantum numbers more easily while maintaining physical meaning which would not be obvious if we just trace the periodic orbits with .

• Momentum cut-off:

By enumerating the lengths of p.p.o. we adopted the point of view that the particle moves in subspaces with a speed of . The corresponding energy levels are for one of these mutually orthogonal subspaces which is clearly unbounded and this means also both our number of p.p.o. and it's length are unbounded.

If we want to select only a finite subset of p.p.o. we have to impose a restriction on allowed quantum numbers. From the number theoretical point of view it would be nice if the maximal length of a p.p.o. is bounded by an arithmetic function. Rejecting quantum numbers means that we restrict our classical particle to certain directions of unit velocity. However filtering directions is not so easy since they normally come in a bundle.

But what happens if impose a lower speed limit to the classical particle? Physically we do a cutoff in momentum space. If we impose for example then . Our particle on a classical p.p.o. has now a relative restriction of speed in each spatial dimension: . It follows that the maximal length of a p.p.o. behaves like: where is the second Chebyshev function.

The question which arises now is how does such a physical model look like? Ideally we want only classical particles to enter the quantum world which obey the imposed speed limit. Necessarily quantum mechanical this means that if we measured each speed component of our particle - call it - we want with probability one that .

• Primitive eigenfunctions:

To every eigenvalue of energy there is at least one corresponding eigenfunction. We say that an eigenfunction is primitive for which the quantum numbers belong to a p.p.o. and denote the set by . In our case we have . Every other imprimitive eigenfunction is just a proper rescaling of these.

Follow the link and find representations and transformations of number theoretic functions.

I am glad that you helped me to correct some of the many mistakes:

• Charles R Greathouse IV

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