Modification of Constants in the Schrödinger Equation Giving Access to Neutrino Behavior

The paper deals with a study of the Schrödinger equation with an original approach. Recalling the well-known relation: $p\to i\hslash {\text{Δ}}_r$ . It considers this equation for which the kinetic factor is $E_{Kin}=\frac{p^2}{2M}=-\frac{{\hslash }^2}{2M}{\text{Δ}}_r^2$ . Making the kinetic factor $E_{Kin}=-{\text{Δ}}_r^2$ can be obtained if one defines a mass $M=\frac{{\hslash }^2}{2}$ , very small and close to the accepted mass of a neutrino ${\nu }_e$ . The Schrödinger equation reduces to: $-{\text{Δ}}_r^2\varphi \left(r\right)=E\varphi \left(r\right)$ . The energy E is that given by Dirac (1927), (c being the light velocity), with his remark that two solutions exist $E=\pm \sqrt{p^2{c}^2+M^2{c}^4}$ . The body of this paper shows all solutions obtained when solving the simplified Schrödinger equation.

• Schrödinger Equation
• Planck Constant
• Particle Physics
• Neutrino Physics

1. Introduction

This paper studies all combinations and solutions for which the kinetic operator $E_{Kin}=-\frac{{\hslash }^2}{2M}{\text{Δ}}_r^2=-{\text{Δ}}_r^2=-1$ that is: $M=\frac{{\hslash }^2}{2}$. Thus:

$E_{Kin}=-{\text{Δ}}_r^2$(1)

Applying this transformation to the Schrödinger equation: [1] For free particles as photons or neutrinos, it is correct to put V = 0, and separate variables r and time t.

$i\hslash \frac{\partial }{\partial t}\Psi \left(r,t\right)=-\frac{{\hslash }^2}{2M}{\Delta }_r^2\Psi \left(r,t\right)=\left(E-V\right)\Psi \left(r,t\right)$(2)

$i\hslash \frac{\partial }{\partial t}\Psi \left(r,t\right)=-\frac{{\hslash }^2}{2M}{\Delta }_r^2\Psi \left(r,t\right)=E\Psi \left(r,t\right)$(3)

$\Psi \left(r,t\right)=\varphi \left(r\right)\times {\text{e}}^{-i\frac{E\times t}{\hslash }}$(4)

$-{\text{Δ}}_r^2\varphi \left(r\right)=\frac{2M}{{\hslash }^2}E\varphi \left(r\right)$(5)

The phase factor ${\text{e}}^{-i\frac{E\times t}{\hslash }}=1$ for time $t=0$ and $\Psi \left(r,t\right)$ reduces to $\varphi \left(r\right)$.

It is important to note that the kinetic factor $-{\text{Δ}}_r^2$ makes this equation a wave equation that propagate.

Defining the Planck constant as: $\hslash =1.054571\times {10}^{-34}\text{J}\cdot \text{s}$ and $h=2\pi \times \hslash$, a neutrino has a velocity close to the light velocity c, thus it is possible and reasonable to write $V=0$.

If one considers a photon, the energy is $E=pc=h\nu =\hslash \omega$. $\nu$ is the frequency of the photon.

In a 1D approach, it is possible to write:

$-\frac{{\hslash }^2}{2M}{\varphi }^{″}\left(x\right)=E\varphi \left(x\right)$(6)

$-\frac{{\hslash }^2}{2M}{\varphi }^{″}\left(x\right)=M{c}^2\varphi \left(x\right)$(7)

${\varphi }^{″}\left(x\right)=-\frac{2M}{{\hslash }^2}E\varphi \left(x\right)$(8)

Equation (8) is the expression of a wave equation, with very simple trigonometric solutions. E is positive and restricted to the kinetic energy of the neutrino.

If the neutrino has a small mass, inferior to $0.8\frac{e^2}{c^2}$, this mass is accepted for the ${\nu }_e$, [2] the electronic neutrino: mass of the neutrino $m_{{\nu }_e}=1.425976\times {10}^{-36}\text{kg}$.

Dealing with relativity, the energy has to be changed according to the famous Albert Einstein formula:

$E=M{c}^2$ or $\gamma \times M{c}^2$, where $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$ is the Lorentz factor and $\beta =\frac{v}{c}$.

This relation requires $\gamma =1$ that is $v=0$, thus $M=M_0$ is the rest mass. The impulse p is: $p=\gamma M\times v$ combining these relations. The relativistic energy follows from: [3]

The theory shows that in a frame where the velocity of the particle is v, the energy E and the impulse $p=M\times v$ are given by: $E=\gamma M{c}^2\equiv M{c}^2/\sqrt{1-\left(v^2/c^2\right)}$, $p=\gamma Mv\equiv Mv/\sqrt{1-\left(v^2/c^2\right)}$.

Finally:

$E^2-p^2{c}^2=M^2{c}^4.$

Assuming a small but real mass M for the neutrino, it is correct to write:

$E^2=p^2{c}^2+M^2{c}^4$(9)

$E=\pm \sqrt{p^2{c}^2+M^2{c}^4}$(10)

Retaining the positive root that is: $E=\sqrt{p^2{c}^2+M^2{c}^4}$.

2. Obtaining the Kinetic Term E

Setting $E=M\times c^2$ the solution of Equation (5) is: with a 1D approach Equation (8) admits the simple solution:

${\varphi }^{″}\left(x\right)=-\frac{2M}{{\hslash }^2}E\varphi \left(x\right)$(11)

$\varphi \left(x\right)=c_2\sin \left(\sqrt{2}\sqrt{\frac{M^2{c}^2}{{\hslash }^2}}x\right)$(12)

Numerically:

$\varphi \left(x\right)=c_2\sin \left(2.84278\times {10}^{42}Mx\right)$(13)

$c_2$ is an integration constant taken as $c_2=1$. It is possible to define a mass M so that this mass M in this equation is written as: $M=\frac{\hslash }{c\times x}$.

It has a correct dimension of a mass x being a length, taking $x=\sqrt{2}\frac{\pi }{2}$, quantity that is a length; thus, the dimension of M is correct. Numerically $M=3.518\times {10}^{-42}\text{kg}$ gives a simple argument in the $\varphi \left(x\right)$ and this is compatible with an accepted neutrino mass, leaving aside the three different neutrino masses. Because of the smallness of the neutrino mass, [2] it is possible to write as a first-order development:

$E=pc\times \sqrt{1+\frac{M^2{c}^4}{p^2{c}^2}}\approx pc\times \left(1+\frac{M^2{c}^4}{2p^2{c}^2}\right)$(14)

The quantity $pc=h\nu =\frac{hc}{\lambda }$ is the energy of a photon, or a neutrino with no mass.

Setting $E=M\times c^2$ the mass M in this equation to $M=\frac{\hslash }{c\times x}$. It has a correct dimension of a mass x being a length, taking $x=\sqrt{2}\frac{\pi }{2}$, quantity that is a length; thus, the dimension of M is correct. Numerically $M=3.518\times {10}^{-42}\text{kg}$ compatible with an accepted neutrino mass, leaving aside the three different neutrino masses. [4] [5] The neutrino estimated mass is less than $m_{\nu }=\frac{0.8\text{eV}}{c^2}$ giving a numerical estimate: $m_{\nu }=1.42685\times {10}^{-36}\text{kg}$. It appears that the sin function is oscillatory as a photon wave function (not normalizable).

Because the neutrino estimated mass is very small, $m_{\nu }\le \frac{0.8\text{eV}}{c^2}$ choosing the development at first order in Equation (14).

That can be written $\text{Δ}E=\frac{M^2{c}^4}{2h\nu }$.

Replacing $M^2$ by the value $M^2=\frac{{\hslash }^2}{c^2\times x^2}$.

$E=pc\times \sqrt{1+\frac{M^2{c}^4}{p^2{c}^2}}\approx pc\times \left(1+\frac{M^2{c}^4}{2p^2{c}^2}\right)$(15)

$E=pc+\frac{M^2{c}^4}{2pc}$(16)

$\text{Δ}E=\frac{M^2{c}^4}{2pc}$(17)

$M=\sqrt{\frac{{\hslash }^2}{c^2{x}^2}}$(18)

$E=pc+\frac{{\hslash }^2{c}^4}{2c^{2}pc\times x^2}$(19)

$E=pc+\frac{{\hslash }^{2}c}{2p\times x^2}$(20)

For a photon (massless particle) $E=pc=h\nu$ thus $p=\frac{h\nu }{c}$. Thus, it proves that the neutrino (when considering its mass) differs from the photon by an amount of energy:

$\text{Δ}E=+\frac{{\hslash }^{2}c}{2p\times x^2}$(21)

Let’ s solve the correct equation easily obtained with Mathematica: $E>0$ is the photon energy: $pc=h\nu =h\frac{c}{\lambda }$.

$E=pc+\frac{M^2{c}^4}{2pc}$(22)

The mass M is defined: $M=\frac{\hslash }{c}$.

Inserting the mass M in the energy equation $E=pc+\frac{M^2{c}^4}{2pc}$ gives numerically:

$E=\frac{1.66702\times {10}^{-60}}{p\times x^2}+2.99792\times {10}^{8}p$(23)

The solution to find with Mathematica is a little more complex:

$M=\frac{\hslash }{c\times x}$(24)

$E=pc+\frac{{\hslash }^{2}c}{2p\times x^2}$(25)

$-{\varphi }^{″}\left(x\right)=E\times \varphi \left(x\right)$(26)

$-{\varphi }^{″}\left(x\right)=\left(pc+\frac{{\hslash }^{2}c}{2p\times x^2}\right)\times \varphi \left(x\right)$(27)

¹Using the relation $E=\frac{1.34719\times {10}^{25}{M}^2}{p}+2.99792\times {10}^{8}p$, the Mathematica solution is:

$\varphi \left(x\right)=(\cos \left(\frac{2\sqrt{2.99792\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2}x}{\sqrt{Px}}\right)$(28)

$+i\sin \left(\frac{2\sqrt{2.99792\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2}x}{\sqrt{Px}}\right)-1)$(29)

$(\cos \left(\frac{\sqrt{1.34719\times {10}^{25}{M}^2+2.99792\times {10}^{8}P{x}^2}x}{\sqrt{Px}}\right)$(30)

$-i\sin \left(\frac{\sqrt{1.34719\times {10}^{25}{M}^2+2.99792\times {10}^{8}P{x}^2}x}{\sqrt{Px}}\right))$(31)

Finally, taking the real part of $\varphi \left(x\right)$ the expression reduces to:

$\begin{array}{l}\varphi \left(x\right)=\cos \left(\frac{2\sqrt{2.99792\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2}x}{\sqrt{Px}}-1\right)\\ \left(\cos \left(\frac{\sqrt{1.34719\times {10}^{25}{M}^2+2.99792\times {10}^{8}P{x}^2}x}{\sqrt{Px}}\right)\right)\end{array}$(32)

or else:

$\begin{array}{l}\varphi \left(x\right)=\cos \left(\frac{x\sqrt{1.34719\times {10}^{25}{M}^2+2.99792\times {10}^{8}P{x}^2}}{\sqrt{Px}}\right)\\ \left(\cos \left(\frac{2x\sqrt{\left|2.99792\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2\right|}}{\sqrt{Px}}\right)-1\right)\end{array}$(33)

To be closer to a full solution, it is possible to obtain the squared module of these functions:

$\begin{array}{c}{\left|\varphi \left(x\right)\right|}^2={\cos }^2\left(\frac{\sqrt{1.34719\times {10}^{25}{M}^2+2.99792\times {10}^{8}P{x}^2}x}{\sqrt{Px}}\right)\\ {\left(\cos \left(\frac{2x\sqrt{\left|2.99792\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2\right|}}{\sqrt{Px}}\right)-1\right)}^2\\ +(\sin \left(\frac{2x\sqrt{\left|2.9979\times {10}^{8}P{x}^2-1.34719\times {10}^{25}{M}^2\right|}}{\sqrt{Px}}\right)\\ -{\sin \left(\frac{\sqrt{1.34719\times {10}^{25}{M}^2+2.9979\times {10}^{8}P{x}^2}x}{\sqrt{Px}}\right))}^2\end{array}$(34)

Using the relation $E=pc+\frac{{\hslash }^2\times c}{2p\times x^2}$,

$E=\frac{1.34719\times {10}^{25}{M}^2}{p}+2.99792\times {10}^{8}p$(35)

leads to another solution is:

$\begin{array}{c}\varphi (x)=\sqrt{x}(BesselJ[\frac{6.6681038\times {10}^{-60}}{2\sqrt{Px}}(17314.5\sqrt{Px}x]\\ +BesselY[\frac{6.6681038\times {10}^{-60}}{2\sqrt{Px}}(17314.5\sqrt{Px}x]\end{array}$(36)

The relation $E=-\sqrt{p^2{c}^2+M^2{c}^4}$ holds with the negative sign and gives another solution for $\varphi \left(x\right)$ expressed as an exponential. Changing the sign of the energy, that is:

$-{\varphi }^{″}\left(x\right)=-E\times \varphi \left(x\right)$(37)

${\varphi }^{″}\left(x\right)-\frac{2M^2{c}^2}{{\hslash }^2}\varphi \left(x\right)=0$(38)

leads to other solutions: That is with M, Px, x variables:

$\varphi \left(x\right)={\text{e}}^{\frac{x\sqrt{{1.3471910}^{25}{M}^2+{2.9979210}^{8}P{x}^2}}{\sqrt{Px}}}-{\text{e}}^{-\frac{x\sqrt{{1.3471910}^{25}{M}^2+{2.997910}^{8}P{x}^2}}{\sqrt{Px}}}$(39)

$\varphi \left(x\right)=2\sinh \left(\frac{x\sqrt{1.34719\times {10}^{25}{M}^2+2.9979\times {10}^{8}P{x}^2}}{\sqrt{Px}}\right)$(40)

That can be shown with several Plot3D functions with Mathematica:²

3. Conclusions

The main idea in this article is to change the kinetic term that makes the Schrödinger equation:

$-\frac{{\hslash }^2}{2M}{\text{Δ}}_r^2\varphi \left(r\right)=E\varphi \left(r\right)$(41)

It is possible to change the factor $-\frac{{\hslash }^2}{2M}{\text{Δ}}_r^2$, that is $M=\frac{{\hslash }^2}{2}$, then the wave equation is simpler:

$-{\text{Δ}}_r^2\varphi \left(r\right)=E\varphi \left(r\right)$(42)

In a 1D approach, it reduces to:

$-\frac{{\text{d}}^2}{\text{d}{x}^2}\varphi \left(x\right)=E\varphi \left(x\right)$(43)

$-{\varphi }^{″}\left(x\right)=E\varphi \left(x\right)$(44)

The wave solution is simpler. With such small mass $M=\frac{{\hslash }^2}{2}=0.55\times {10}^{-68}\text{kg}$, it can concern neutrinos with their maximal estimated mass: $M_{\nu }=0.8\frac{\text{eV}}{c^2}=1.42598\times {10}^{-36}\text{kg}$.

Figure 1

Figure 1. E = simple oscillating 3D plot (Mass, x) with a 0 ≤ M ≤ 4.48 × 10 − 26 kg and 0 ≤ x ≤ 4 π as a variable.

Figure 2

Figure 2. E = p c + 0.5 c 4 M 2 2 p c ( 0 ≤ M ≤ 4.48 × 10 − 23 kg ) pc is the photon energy.

Figure 3

Figure 3. Simple oscillating solution free wave function for photon (M = 0) and neutrinos with a small neutrino fixed mass M = 2.84 × 10 − 26 kg with x as a variable.

Figure 4

Figure 4. Exponential solution ϕ ( x ) = 2 sinh ( 2.9979 × 10 8 P x 2 + 1.6670 × 10 − 60 x P x ) the variable 0 ≤ P x ≤ 1.42598 × 10 − 36 kg Px maximum value for the neutrino with 0 ≤ x ≤ 4 π as a variable.

Figure 5

Figure 5. Solution | ϕ ( x ) | 2 the mass M is fixed M = 8.4 × 10 − 20 kg and 0 ≤ x ≤ 4 π with Px (impulse) ≤ 1.42598 × 10 − 36 kg, for the neutrino with 0 ≤ x ≤ 4 π as a variable.

Owing to the sign of the Einstein formula $E=\pm \sqrt{p^2{c}^2+M^2{c}^4}$ two solutions exist, one is a wave equation that is the usual wave function of a photon or a neutrino propagating in space-time.

The second solution gives a new insight into neutrino theory. The complex wave function appears, a new phenomenon shown in this paper.

That can be shown with several Plot3D functions with Mathematica. (See Figures 1-5 )

NOTES

¹Here, the light velocity is fixed $c=2.99792\times {10}^8\text{m}\cdot {\text{s}}^{-1}$.

²The integration constant is fixed to $c_1=1$.

References

1. Schrödinger, E. (1926) An Undulatory Theory of the Mechanics of Atoms and Molecules. Physical Review, 28, 1049-1070. >https://doi.org/10.1103/PhysRev.28.1049

2. Lipari, P. (2003) Introduction to Neutrino Physics. In: 2001 CERN—CLAF School of High-Energy Physics, CERN, 115-200.

3. Einstein, A. (1905) Does the Inertia of a Body Depend Upon Its Energy-Content? Annalen der Physik, 323, 639-641. (In German)>https://doi.org/10.1002/andp.19053231314

4. Dirac, P.A.M. (1930) Principles of the Quantum Mechanics. Oxford University Press.

5. Giunti, C. and Kim, C.W. (2007) Fundamentals of Neutrino Physics and Astrophysics. OUP Oxford. >https://books.google.fr/books?id=2faTXKIDnfgC