Relation between exponential behavior and energy denominators -- Weak Coupling Limit

TL;DR

This paper explores the relationship between exponential functions and energy denominators in weak coupling limits, revealing how matrix properties influence wave functions and transition rates in quantum systems.

Contribution

It demonstrates a novel connection between exponential behavior and energy denominators, and explains unique features of transition rates in pentadiagonal matrices.

Findings

Ground state wave function amplitudes match Taylor coefficients of e^{(-v/E)}

A dip in transition rates is explained for pentadiagonal matrices

An explicit link between matrix structure and exponential behavior is established

Abstract

We show some interesting properties of tridiagonal and pentadiagonal matrices in the weak coupling limits. In the former case of this limit the ground state wave function amplitudes are identical to the Taylor expansion coefficients of the exponential function e$^{(-v/E)}$. With regards to transition rates a dip in the pentadiagonal case which is not present in the tridiagonal case is explained. An intimate connection between energy denominators and exponential behavior is demonstrated.

Tables

Table 1. Table 4: Wave functions for tridiagonal case in the weak coupling limit. Note that u = v E 𝑣 𝐸 \frac{v}{E}

a0	a1	a2	a3	a4	a5	a6	a7	a8	a9	a10
1	-u	u2/2!	-u${}_{}{}^3/3!$	u4/4!	-u5/5!	u6/6!	-u7/7!	u8/8!	-u9/9!	u10/10!
u	1	-u	u2/2!	-u${}_{}{}^3/3!$	0	0	0	0	0	0
u2/2!	u	1	-u	u2/2!	- u3/3!	0	0	0	0	0
u3/3!	u2/2!	u	1	-u	u2/2!	- u3/3!	0	0	0	0
u4/4!	u3/3!	u2/2!	u	1	-u	u2/2!	- u3/3!	0	0	0
0	u4/4!	u3/3!	u2/2!	u	1	-u	u2/2!	- u3/3!	0	0
0	0	u4/4!	u3/3!	u2/2!	u	1	-u	u2/2!	- u3/3!	0
0	0	0	u4/4!	u3/3!	u2/2!	u	1	-u	u2/2!	- u3/3!
0	0	0	0	u4/4!	u3/3!	u2/2!	u	1	-u	u2/2!
0	0	0	0	0	u4/4!	u3/3!	u2/2!	u	1	-u
u10/10!	u9/9!	u8/8!	u7/7!	u6/6!	u5/5!	u4/4!	u3/3	u2/2	u	1

Equations

$$\Psi =\Phi +\frac{1}{E_0-H_0}QV\Psi =(1+\frac{1}{E_0-H_0}QV+\frac{1}{E_0-H_0}QV\frac{1}{E_0-H_0}QV+.....)\Phi$$

$$O=(a_0{b}_1+,..+a_9{b}_{10})+(b_0{a}_1+....+b_9{a}_{10})\text{textless}n{T}_1(n+1)\text{textgreater}$$

Full text

*Keywords: Distribution ,

PACs Number: 21.60.Cs*

1 Introduction

This work is a continuation of work done before on matrix models of strength distributions[1,2].Matrix mechanics was of course introduced into quantum mechanics by Heisenberg[2] and Born and Jordan[3]. In the previous work we had a matrix in which the diagonal elements were En= nE with E=1 MeV. We introduced a constant coupling v which for a level En occurs only with the nearest neighbors E*(n-1)* and E${}_{(n+1)}.$ The matrix is shown in Table 1. Note that the only relevant parameter is $\frac{v}{E}$. Here we extend the work to pentadiagonal matrices, as shown in Table 2. In both cases we discuss for the first time the weak coupling limits for the ground state wave functions.

Table 1: The Matrix Hamiltonian-Tridiagonal

[TABLE]

Table 2: The Matrix Hamiltonian-Pentadiagonal

[TABLE]

.

2 The calculation

The eigenfunctions resulting from the diagonalizations of the above matrices are of the form

{a0,a1,a2,, …..a10}. We are interested in the values of these an in the limit where $\frac{v}{E}$ is very small (NB In previous publications we used the notation a1 to a11).

We ran Mathematica programs for small v to get an idea of the coefficients of the ground state eigenfunctions. For the tridiagonal case we used $\frac{v}{E}$=0.01 . The results were as follows:

{0.99995, –0.009999, 0.0000499933, -1.666410-7*, 4.16592* 10*-10*, -8.33171* 10*-13*, 1.3886* 10*-15*, -1.98269* 10*-18*, 2.47958* 10${}^{-21},-2.75506$* 10*-24*, 2.75504* 10*-27*}

These numbers can be put in a more suggestive way with a bit of rounding up.

1, -$\frac{v}{E}$, ($\frac{v^{2}}{E^{2}2}$), -($\frac{v^{3}}{E^{3}6}$) ……… (-1)n $\frac{v^{n}}{E^{n}n!}$. We recognize these ai’s as coefficients in the Taylor series of e*-v**/E*.

To derive this result we should realize that to get an we have to go the nth order in perturbation theory. Let H=H0+V with H0 the diagonal part of the matrix and V the off diagonal.

[TABLE]

where Q prevents the unperturbed ground state from being an intermediate state.

To get an to the lowest power in $\frac{v}{E}$ we have to go in n steps 0 to 1, 1 to 2,…n-1 to n. In the numerator all the matrix elements < (n+1) QVn> are the same, namely v. In the denominator we get (E0-E${}_{1})$…..(E0 -En). Since we have En = nE the denominator is $(n!)$En. This proves the result and shows an intimate connection between exponential behavior and equally spaced levels.

We next consider the pentadiagonal case. This time we chose $\frac{v}{E}$ to be 10*-10*. The results are are as follows:

{ {1, - 1.0000610-10*, - 5.010-11*, 5.0009110-21*, 1.2510-21*, - 1.2507110-31*, - 2.0833310-32*, 2.0895110-42*, 2.6041710-43*, - 2.6337610-53*, - 2.6041710-54*} },

These numbers can be put in a more suggestive way with a bit of rounding up.

{1,- $\frac{v}{E}$, -$\frac{v}{2E}$, ($\frac{v^{2}}{E^{2}2}$), ($\frac{v^{2}}{E^{2}8}$), -( $\frac{v^{3}}{E^{3}8}$) …..

We now have 2 types of non-vanishing matrix elements -one from m to(m+1) and another from m to (m+2). Both have a value v so it is still the energy denominators which come into play. We now consider selected an:

a1: We get $\frac{v}{E_{0}-E_{1}}$ =- $\frac{v}{E}$.

a2: In order to get a result linear in v we have only the direct connection from 0 to 2 which yields a value $-\frac{v}{2}$

a3: there are 2 paths:

A: 0 to 2 and 2 to 3

B: 0 to 1 and 1 to 3.

In the former case the value is $\frac{1}{2*3}$ and in the latter $\frac{1}{3}$. So we get $\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$–answer $-\frac{v}{2E}$.

a4: We look for($\frac{v}{E}$)2 terms. There is actually only one: 0 to 2 and 2 to 4. This gives a value $\frac{1}{2*4}$=$\frac{1}{8}$-answer $\frac{1}{8}$ ($\frac{v}{E}$)2

a5: There are 3 paths:

C: 0 to 1, 1 to 3, 3 to five: $-\frac{1}{1*3*5}$

D: 0 to 2, 2 to 3, 3 to 5: $-\frac{1}{2*3*5}$

E: 0 to 2, 2 to 4, 4 to 5: $-\frac{1}{2*4*5}$

sum = $-\frac{1}{8}$ - answer $-\frac{1}{8}$ ($\frac{v}{E}$)3

And so it goes.

3 Results for <n T1(n+1)> and <n

T2 (n+1)>

In previous works we introduced 2 possible transition operators:

<n T1(n+1)> is a constant. We choose this constant to be one.

<(n+1)T2n>is equal to $\sqrt{n+1}$

In both cases all other transition elements are taken to be zero. Since this is a model we do not have to assign units to T1 and T2, but it should be pointed out that if we were to say double T1 we would quadruple the transition rate. This would affect the scale but not the overall shape of the graph.

The strength matrix element between a state {a} and a state {b} for T1 is simply

[TABLE]

with the last factor taken to be unity (1). For T2 we take the previous expression and multiply the an b*(n+1)* by $\sqrt{n+1}$ -likewise bn a*(n+1)*. The strength is O2 and we will be plotting ln(O2) versus excitation energy E∗.

In Fig.1 and 2 we show results for T1, v=0.1 MeV for the tridiagonal and pentadiagonal cases. In Fig 3 and 4 we show corresponding results for T2. We see a near exponential decrease of the strength in the the tridiagonal case with T1. This shows up on our log plot as a near straight line with a negative slope, however the last point (the tenth excited state) is much lower than the exponential projection. This has been discussed in [1,2] where it was shown that for $T_{1}$ the value is actually minus infinity. Also it should be pointed out that although for v=0.1 MeV the tridiagonal curve for T1 looks exponential and this persists until v=1 and even beyond, this is not the case for extremely small v or extremely large v. This can be seen in the figures of ref[2].

In Table 3 we show what happens for very small v. In the weak coupling limit the T1 matrix element O can be shown to be of the form O(1$\rightarrow$n)= $(\frac{v}{E})^{m}$A(1$\rightarrow$n) where A is independent of $\frac{v}{E}$. We define gk=$\frac{1}{k!}$ and give the values of A in the third column of the table. In the fourth column of Table 3 we give the value of ln(O2) using the expressions in the second and third columns i.e. keeping only the lowest powers of $\frac{v}{E}$. We do this for $\frac{v}{E}$= 0.0001. The behavior is somewhat complicated.The results are not monotonic in n. For example although m increases from n=1 to 5, there is a decease in going from n=5 (m=6) to n=6 (m=5). Thus we get a non-exponential behavior with the value of ln(O2 ) being larger for n=6 than for n=5. However when we go to larger $\frac{v}{E}$ e.g. v= 0.1, the behavior looks quite exponential. For T2, on the other hand ,we do get an exponential behavior in the extreme weak coupling limit i.e. m=n for all n.

Table 3: Weak coupling expression for The T1 amplitude and ln (O2) .

[TABLE]

For very large v there is an even-odd effect resulting in 2 exponential behaviors-one for even n and one or odd n. In Fig. 2 we show corresponding results for the pentadiagonal case. There is one significant difference between tri and penta for T1. As seen in Fig 2 in the transition from the ground state to the second excited state there is a big dip for the pentadiagonal case. This is not the case for the tridiagonal case. After the dip in the pentadiagonal case one gets a near exponential behavior.

To explain the dip we examine the second excited state for the pentadiagonal case in the weak coupling limit. Recall that for the ground state the values of { a0, a1, a2} are respectively {1,-$\frac{v}{E}$,$-\frac{v}{2E}$}. For the second excited state the values of {b0,b1,b2} can be shown to be {$\frac{v}{2E}$, v, 1}. For the T1 case note that a0**b1*=$\frac{v}{E}$ while a1**b2* = -$\frac{v}{E}$ so we have complete cancellation of the leading terms.

To get {b0,b1,b2} we note that clearly b2=1 in the weak coupling limit. For b0 we go from state 2 to state 0 so that we get a contribution (from the energy denominator) $\frac{v}{2E}$ i.e. ( E2-E0)=2. For b1 we go from state 2 to state 1 and so we get $\frac{v}{E}$ i.e. (E2-E1)=1.

For the T${}_{{}_{2}}$ case , figs 3 and 4, the dip is not as pronounced. Because of the factor $\sqrt{(n+1)}$ there is only a partial cancellation.

We had previously discussed the strong coupling limit for T1 [1,2]. In that case all transitions vanished. This was explained by the fact that in the strong coupling limit the transition matrix becomes identical to the Hamiltonian.

4 All v’s.

We briefly consider the case where on the diagonal we still have nE but all other matrix elements are v (there are no zeros). One can easily show that the ground state column vector {a0, a1,….a10)} in the weak coupling limit is {1,-$\frac{v}{E}$,$-\frac{v}{2E}$,$-\frac{v}{3E}$ ,…$-\frac{v}{10E}$ }.

There are other matrix models which address problems related to, but different, from what we have here considered. As previously mentioned, [1] Bohr and Mottelson [5] used matrix models to derive the Breit-Wigner formula for a resonance. Brown and Bolsterli described the giant dipole resonances in nuclei in a schematic model using a delta interaction[6]. In that work they made the approximation that certain radial integrals were constant. Abbas and Zamick [7] removed this restriction. Generally speaking matrix models are very useful for casting insights into the physics of given problems where the more accurate but involved calculations fail.

Although the models here are not geared to fit specific experiments we do keep our eyes on the empirical data. In refs [8-11] we cite works in which exponential behavior is seen and explained. Here we show that with unperturbed equally spaced energy levels we generate factorials which are needed to get exponential behavior. These come from the energy denominators. Also, our perturbations are simple enough that we can easily perform nth order perturbation theory for any n and get analytic results in the weak coupling limit.

5 Appendix

Levering Wolfe was supported by a Rutgers Aresty Research Assistant award and a Richard J. Plano award.