p>

${E}_{n}=\frac{{m}_{e}}{2}{v}_{n}^{2}$ (6)

where the absolute value of the electron velocity ${v}_{n}$ is

${v}_{n}={\left(\frac{2{E}_{n}}{{m}_{e}}\right)}^{1/2}={\left(\frac{{n}^{2}{h}^{2}}{4{m}_{e}^{2}{L}^{2}}\right)}^{1/2}=\frac{nh}{2{m}_{e}L}.$ (7)

This value of the velocity is expected to be dominating in course of the electron oscillation within the interval

$0 (8)

Evidently the electron motion is going from $x=0$ to $x=L$ and vice versa.

The time period ${T}_{n}$ of the oscillation satisfies the formula

$\frac{2L}{{T}_{n}}={v}_{n},$ (9)

so

${T}_{n}=\frac{2L}{{v}_{n}}=2L×\frac{2{m}_{e}L}{nh}=\frac{4{m}_{e}{L}^{2}}{nh}.$ (10)

The motion frequency $\nu$ due to ${T}_{n}$ is

${\nu }_{n}=\frac{1}{{T}_{n}}=\frac{nh}{4{m}_{e}{L}^{2}}.$ (11)

In effect the energy provided by ${\nu }_{n}$ becomes

${E}_{n}^{osc}=h{\nu }_{n}=\frac{n{h}^{2}}{4{m}_{e}{L}^{2}}=\hslash {\omega }_{n}$ (12)

so

${\omega }_{n}=\frac{n{h}^{2}}{4{m}_{e}{L}^{2}}\frac{1}{\hslash }=\frac{\pi nh}{2{m}_{e}{L}^{2}}$ (13)

and its reversal is

$\frac{1}{{\omega }_{n}}=\frac{2{m}_{e}{L}^{2}}{\pi nh}$ (14)

Now our idea is to match results of Sec. 2 with those of Sec. 3.

Before we do such comparison let us note that parameters Tn and vn satisfy the original Heisenberg relation :

$\frac{d}{dn}\left(nh\right)=h=\frac{d}{dn}{\oint {m}_{e}\stackrel{˙}{x}}^{2}dt$ (14a)

For, from (10) and because of $\stackrel{˙}{x}={v}_{n}$ given in (9), we obtain for (14a):

$\frac{d}{dn}{\oint {m}_{e}\stackrel{˙}{x}}^{2}dt=\frac{d}{dn}\left({m}_{e}{v}_{n}^{2}{T}_{n}\right)=\frac{d}{dn}\left({m}_{e}\frac{4{L}^{2}}{{T}_{n}}\right)=\frac{d}{dn}\left({m}_{e}\frac{nh}{{m}_{e}}\right)=h$ (14b)

which is identical to the result in  and (14a).

In the same way we have

$\oint pdq=\oint {m}_{e}\stackrel{˙}{x}dx={m}_{e}{v}_{n}2L={m}_{e}\frac{{\left(2L\right)}^{2}}{{T}_{n}}={m}_{e}4{L}^{2}\frac{nh}{4{m}_{e}{L}^{2}}=nh.$ (14c)

If the result in (14c) is considered as the action J, it becomes evident that the derivative of energy in (5) done with respect to J provides us with the electron oscillation frequency (11).

4. Heisenberg Approach Applied to the Electron Oscillators

In the first step of the Heisenberg approach to the electron gas enclosed in a potential box we consider the Hamiltonian of an oscillator moving in direction of the axis x :

$\stackrel{^}{H}=\frac{{\stackrel{^}{p}}_{x}^{2}}{2{m}_{e}}+\frac{{m}_{e}{\omega }^{2}}{2}{x}^{2}={m}_{e}{\omega }^{2}‖\begin{array}{cccc}{x}_{01}{x}_{10}& 0& 0& \cdots \\ 0& {x}_{01}{x}_{10}+{x}_{12}{x}_{21}& 0& \cdots \\ 0& 0& {x}_{12}{x}_{21}+{x}_{23}{x}_{32}& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}‖$ (15)

where

${x}_{01}{x}_{10}=\frac{\hslash }{2{m}_{e}\omega },$ (16)

${x}_{12}{x}_{21}=2\frac{\hslash }{2{m}_{e}\omega },$ (16a)

${x}_{23}{x}_{32}=3\frac{\hslash }{2{m}_{e}\omega },$ (16b)

${x}_{34}{x}_{43}=4\frac{\hslash }{2{m}_{e}\omega },$ (16c)

$⋮$

The ${x}_{mn}$ are the matrix elements of x calculated between the oscillator states m and n, the frequencies $\omega$ are those calculated in (13) taken for $n=1$.

When the results in the (16) formulae are substituted to (15) they give the following diagonal elements for the matrix presented in (15):

${x}_{01}{x}_{10}=\frac{{L}^{2}}{2{\pi }^{2}},$ (17)

${x}_{01}{x}_{10}+{x}_{12}{x}_{21}=\left(1+2\right)\frac{{L}^{2}}{2{\pi }^{2}}=\frac{3{L}^{2}}{2{\pi }^{2}},$ (17a)

${x}_{12}{x}_{21}+{x}_{23}{x}_{32}=\left(2+3\right)\frac{{L}^{2}}{2{\pi }^{2}}=\frac{5{L}^{2}}{2{\pi }^{2}},$ (17b)

${x}_{23}{x}_{32}+{x}_{34}{x}_{43}=\left(3+4\right)\frac{{L}^{2}}{2{\pi }^{2}}=\frac{7{L}^{2}}{2{\pi }^{2}},$ (17c)

$⋮$

which give the following result for (15):

$\stackrel{^}{H}={m}_{e}{\omega }^{2}‖\begin{array}{ccccc}\frac{{L}^{2}}{2{\pi }^{2}}& 0& 0& 0& \cdots \\ 0& \frac{3{L}^{2}}{2{\pi }^{2}}& 0& 0& \cdots \\ 0& 0& \frac{5{L}^{2}}{2{\pi }^{2}}& 0& \cdots \\ 0& 0& 0& \frac{7{L}^{2}}{2{\pi }^{2}}& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}‖$ (18)

Because of $\omega ={\omega }_{1}$ calculated in (13), the diagonal matrix elements in (18) give the oscillator energies

${E}_{m}^{osc}={m}_{e}{\omega }^{2}\frac{{L}^{2}}{2{\pi }^{2}}\left(2m+1\right)={m}_{e}{\left(\frac{\pi h}{2{m}_{e}{L}^{2}}\right)}^{2}\frac{{L}^{2}}{2{\pi }^{2}}\left(2m+1\right)=\frac{{h}^{2}}{8{m}_{e}{L}^{2}}\left(2m+1\right)$ (19)

where

$m=0,1,2,3,\cdots$ (19a)

Let us note that the oscillator frequency $\omega$ in  is considered as a known parameter. In our calculations this frequency is deduced from the electron motion in the potential box; see Sec. 3.

This feature enables us to present the matrix elements entering the oscillator energy in terms of the matrix elements dependent on the properties characteris tic for the motion in the potential box. Typically for the Heisenberg’s treatment of an oscillator we choose only a single oscillation frequency $\omega$ for calculating all quantum states. In the Schrödinger picture this frequency is associated with the lowest quantum state $n=1$ ; see (13).

A passage to the particle energy in the box is very simple. We note that the sums entering the partial traces of the diagonal matrix elements given in (19), viz.

$\underset{m=0}{\overset{m=n-1}{\sum }}\left(2m+1\right),$ (20)

give respectively

$\begin{array}{l}{n}^{2}={1}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}n=1,\\ {n}^{2}=1+3={2}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}n=2,\\ {n}^{2}=1+3+5={3}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}n=3,\\ {n}^{2}=1+3+5+7={4}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}n=4,\\ {n}^{2}=1+3+5+7+9={5}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}n=5,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⋮\end{array}$ (21)

In effect the (21)—combined with the factor entering the last term in (19)—provide us with the Schrödinger energy results presented in (5).

5. An alternative Heisenberg Treatment of Free Electrons in the Potential Box

In this case we apply the Hamiltonian different than in the oscillator case examined in Sec. 4. This kind of treatment takes into account only the kinetic part of the Hamiltonian and neglects the whole of the x-dependent (potential) part in the first row of (15):

$\stackrel{^}{H}=\frac{{\stackrel{^}{p}}_{x}^{2}}{2{m}_{e}}.$ (22)

Certainly the constant parameters become different than those applied in the Hamiltonian case of Sec. 4.

The first of the diagonal terms of ${p}_{x}^{2}$ belonging to the energy matrix representing the Hamiltonian in (22) is:

${p}_{01}{p}_{10}={x}_{01}{x}_{10}2{m}_{e}^{2}{\omega }^{2}=\frac{\hslash }{2{m}_{e}\omega }2{m}_{e}^{2}{\omega }^{2}=\frac{\hslash }{2}2{m}_{e}\omega =\frac{h}{2\pi }{m}_{e}\frac{\pi h}{2{m}_{e}{L}^{2}}=\frac{{h}^{2}}{4{L}^{2}}.$ (23)

In fact

${p}_{01}={m}_{e}{v}_{1}={m}_{e}\frac{h}{2{m}_{e}L}=\frac{h}{2L}$ (24)

if we note that ${v}_{1}$ entering (24) and

${E}_{1}=\frac{1}{2{m}_{e}}{p}_{01}{p}_{10}=\frac{{h}^{2}}{8{m}_{e}{L}^{2}}$ (25)

calculated from (23) are equal respectively to ${v}_{1}$ and ${E}_{1}$ obtained in the Schrödinger theory; see Sec. 3.

The next diagonal element of the matrix ${\stackrel{^}{p}}_{x}^{2}$ is

$\begin{array}{c}{p}_{02}{p}_{20}={x}_{12}{x}_{21}2{m}_{e}^{2}{\omega }^{2}=\left(\frac{\hslash }{2{m}_{e}\omega }+\frac{2\hslash }{2{m}_{e}\omega }\right)2{m}_{e}^{2}{\omega }^{2}\\ =3\hslash 2{m}_{e}\omega =3\frac{h}{2\pi }{m}_{e}\frac{\pi h}{2{m}_{e}{L}^{2}}=\frac{3}{4}\frac{{h}^{2}}{{L}^{2}}.\end{array}$ (26)

When divided by $2{m}_{e}$ the expression (26) becomes

$\frac{1}{2{m}_{e}}{p}_{02}{p}_{20}=\frac{3{h}^{2}}{8{m}_{e}{L}^{2}}.$ (27)

This result added to that obtained in (25) gives

$\frac{1}{2{m}_{e}}{p}_{01}{p}_{10}+\frac{1}{2{m}_{e}}{p}_{02}{p}_{20}=\frac{4{h}^{2}}{8{m}_{e}{L}^{2}}=\frac{{2}^{2}{h}^{2}}{8{m}_{e}{L}^{2}}$ (28)

which is precisely the next Schrödinger value for the free-electron energy, i.e. it is corresponding to the quantum level $n=2$.

If we take the next diagonal term for the matrix of ${\stackrel{^}{p}}_{x}^{2}$ which is

$\begin{array}{c}{p}_{03}{p}_{30}=\left(\frac{\hslash }{{m}_{e}\omega }+\frac{3}{2}\text{ }\frac{\hslash }{{m}_{e}\omega }\right)2{\left({m}_{e}\omega \right)}^{2}=\left(1+\frac{3}{2}\right)\hslash 2{m}_{e}\omega \\ =\frac{5\hslash }{2}2{m}_{e}\omega =5\frac{h}{2\pi }{m}_{e}\frac{\pi h}{2{m}_{e}{L}^{2}}=\frac{5}{4}\frac{{h}^{2}}{{L}^{2}},\end{array}$ (29)

it gives, when multiplied by $\frac{1}{2{m}_{e}}$, the result

$\frac{1}{2{m}_{e}}{p}_{03}{p}_{30}=\frac{5}{8}\frac{{h}^{2}}{{m}_{e}{L}^{2}}.$ (30)

The sum of terms (25), (27) and (30) becomes

$\frac{1}{2{m}_{e}}\left({p}_{01}{p}_{10}+{p}_{02}{p}_{20}+{p}_{03}{p}_{30}\right)=\frac{1+3+5}{8{m}_{e}{L}^{2}}{h}^{2}=\frac{{3}^{2}{h}^{2}}{8{m}_{e}{L}^{2}}$ (31)

which is the Schrödinger energy of a free electron on the level $n=3$.

The procedure can be readily extended to an arbitrary quantum level n.

6. Some Special Statistical Behaviour of the Electron Energy Quanta Present in a One-Dimensional Potential Box

Till the present point of our considerations we neglected the properties of the electron statistics applied to the electron gas. In fact the problem of the electron spin, and the Pauli exclusion principle connected with it, were not developed enough at the time of an early competition of the Heisenberg and Schrödinger theories. In principle both the boson and fermion statistics seem to be here applicable, first because of the electron oscillation waves considered in the Heisenberg picture, next because of the double spin-dependent occupation of the energy levels connected with the Schrödinger electron gas state. The fermion-like behaviour of electrons seems however to predominate and our task is to make only a supplementary insight into the Fermi statistical distribution considered before.

Our point concerns the question whether the highest occupied Fermi level ${n}_{max}$ in the one-dimensional gas should be considered as identical with the Fermi energy, or it does represent a distinct energy value. By assuming the second point of view, the Fermi energy—in accordance with former investigations   —should be considered as an inflexion point on the Fermi distribution function $F\left(E\right)$ plotted as a function of the electron energy E.

The function $F\left(E\right)$ —as it is well known—depends also on the absolute temperature parameter T:

$F\left(E\right)=\frac{1}{{\text{e}}^{-E/kT}+1}.$ (32)

Let us assume E to be an abbreviation of a small value of the difference between the Fermi energy ${E}_{F}$ and the electron energy on the highest occupied level ${n}_{max}$ in the gas which is

$E\left({n}_{max}\right)=\frac{{h}^{2}{n}_{max}^{2}}{8{m}_{e}{L}^{2}}$ (33)

so

$E=\Delta {E}^{\left(1\right)}={E}_{F}^{\left(1\right)}-{E}^{\left(1\right)}\left({n}_{\mathrm{max}}\right)>0.$ (34)

The superscript (1) indicates that—for simplicity—only the gas having a single electron on each of its energy levels is considered. In principle we assume that E in (34) is a small number.

The first derivative of $F\left(E\right)$ in (32) calculated with respect to the energy E gives

$\frac{\text{d}F}{\text{d}E}=\frac{-1}{{\left({\text{e}}^{-E/kT}+1\right)}^{2}}\left(-\frac{1}{kT}\right){\text{e}}^{-E/kT}=\frac{{\text{e}}^{-E/kT}}{{\left({e}^{-E/kT}+1\right)}^{2}}\frac{1}{kT},$ (35)

whereas the second derivative of $F\left(E\right)$ is represented by the derivative of the result obtained in (35):

$\begin{array}{c}\frac{{\text{d}}^{2}F}{\text{d}{E}^{2}}=\frac{1}{kT}\frac{\text{d}}{\text{d}E}\left[\frac{{\text{e}}^{-E/kT}}{{\left({\text{e}}^{-E/kT}+1\right)}^{2}}\right]=\frac{1}{{\left(kT\right)}^{2}}\left[\frac{2{\text{e}}^{-2E/kT}}{{\left({\text{e}}^{-E/kT}+1\right)}^{3}}-\frac{{\text{e}}^{-E/kT}}{{\left({\text{e}}^{-E/kT}+1\right)}^{2}}\right]\\ =\frac{1}{{\left(kT\right)}^{2}}\text{ }\frac{1}{{\left({\text{e}}^{-E/kT}+1\right)}^{3}}\left[2{\text{e}}^{-E/kT}-{\text{e}}^{-E/kT}\left({\text{e}}^{-E/kT}+1\right)\right]\\ =\frac{1}{{\left(kT\right)}^{2}}\text{ }\frac{{\text{e}}^{-E/kT}}{{\left({\text{e}}^{-E/kT}+1\right)}^{3}}\left[2-{\text{e}}^{-E/kT}-1\right].\end{array}$ (36)

The E in the inflexion point should make (36) equal to zero. To attain that it is enough to require the square-bracket to be vanishing

${\text{e}}^{-E/kT}-1=0$ (37)

which for small $E/kT$ gives the equation

$1-1+\frac{E}{kT}-\frac{1}{2!}{\left(\frac{E}{kT}\right)}^{2}\cong \left(1-\frac{1}{2!}\frac{E}{kT}\right)\frac{E}{kT}=0.$ (38)

The last equation is satisfied when

$E=\Delta {E}^{\left(1\right)}=2!kT,$ (39)

so in this case [see (32)]

$F\left(E=\Delta {E}^{\left(1\right)}=2kT\right)\cong \frac{1}{\frac{1}{{\text{e}}^{2}}+1}$ (40)

is obtained at the inflexion point.

In the next step we consider a double occupation of the energy levels in the gas by the electrons having an opposite spin. By assuming that

${E}_{F}^{\left(2\right)}=2{E}_{F}^{\left(1\right)}$ (41)

and putting for E in (32) the expression

$E=\Delta {E}^{\left(2\right)}=2\Delta {E}^{\left(1\right)},$ (42)

we obtain the energy E twice as large as $E=\Delta {E}^{\left(1\right)}$.

A substitution of $E=2\text{ }\Delta {E}^{\left(1\right)}$ instead of $E=\Delta {E}^{\left(1\right)}$ into Equation (38) gives:

$\left(1-\frac{1}{2!}\frac{2\Delta {E}^{\left(1\right)}}{kT}\right)\frac{2\Delta {E}^{\left(1\right)}}{kT}=0$ (43)

so

$\Delta {E}^{\left(2\right)}=2\Delta {E}^{\left(1\right)}=2kT.$ (44)

Therefore the result (44) obtained for a double occupation of the quantum states implies a reduction of $\Delta {E}^{\left(1\right)}$ defined in (39) to a single $kT$.

Both results for $\Delta {E}^{\left(1\right)}$ and $\Delta {E}^{\left(2\right)}$ vanish at $T=0$ giving respectively

${E}_{F}^{\left(1\right)}=E\left({n}_{max}\right)$ (45)

and

${E}_{F}^{\left(2\right)}=2E\left({n}_{max}\right).$ (45a)

7. Possible Duality of Statistics Applied to the Electron Quantum Levels

A duality of the boson and fermion statistics which can be applied to the electron levels can be detected by examining the energy of the level ensembles obtained in two different ways. For the electron-gas case a better insight seems to be provided by the Schrödinger’s method because of its simplicity.

The eigenvalues of the free-electron Hamiltonian considered by Schrödinger (see Sec. 2) are:

${E}_{n}=\frac{{n}^{2}{h}^{2}}{8{m}_{e}{L}^{2}}$ (46)

where

$n=1,2,3,\cdots$ (46a)

giving the results identical to those obtained in (5). These results can be successfully considered with aid of the Fermi-Dirac statistics.

But another approach than that in (46) and (46a) can be obtained for the electrons in a one-dimensional box when instead of the stationary states of the energy Hamiltonian in (1) a spectrum of energies due to the electron oscillations within the potential box is considered.

Because of a free-electron character of the particles we obtain from (11) and (12):

${E}_{n}^{\text{osc}}=\frac{n{h}^{2}}{4{m}_{e}{L}^{2}}.$ (47)

Clasically the electron having the velocity ${v}_{n}$ undergoes the way 2L along the box in course of the time period ${T}_{n}$ ; see (9) and (10).

Evidently the energy in (47) is by a factor of

$n/2$ (48)

smaller than the energy obtained in (5) and (46). This means that $n/2$ oscillators are required to provide the energy equal to a single eigenenergy state labeled by a given n indicating a large degeneracy of the oscillator energies necessary for any large n.

In effect the energy of a one-dimensional electron gas can be considered also as a superposition of a large number of the boson energy quanta due solely to the electron oscillations.

8. Summary

The paper compares two approaches to the energy levels of a free-electron one-dimensional gas done respectively from the point of view of the Heisenberg and Schrödinger quantum theory. This comparison seems to be absent in the literature.

As a starting point we take into account the Schrödinger wave-mechanical calculation which is very simple. In the next step the electrons are considered as oscillators and the Heisenberg matrices are applied. In fact two different kinds of the Heisenberg’s Hamiltonian can be examined for free electrons on condition the constant parameters entering the matrices are suitably modified.

A short calculation concerning the position of the Fermi level in the gas as a function of the absolute temperature has been added. It should be noted that the statistics of quantum energy levels presented in both Heisenberg and Schrödinger theories can be different from that valid for the fermions alone.

Cite this paper
Olszewski, S. (2020) Dualism of the Heisenberg and Schr&ouml;dinger Approaches to the Quantum States Entering a One-Dimensional Electron Gas. Journal of Modern Physics, 11, 475-485. doi: 10.4236/jmp.2020.113030.
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