The Schrödinger Equation is a Newtonian Equation


When a photon impacts a free matter particle at rest, mass and momentum are imparted to the particle, and it accelerates to velocity $v$.  The center of mass of the imparted mass is captured at such a radius that the angular momentum imparted to the system is equal to Planck’s constant $h$.  The imparted mass and the impacted particle mass remain at a fixed distance from each other, and they begin rotating about their common center of mass as the system center of mass translates in a straight line.  This motion is manifested as a matter particle undulating as it translates.  The wavelength of this undulation is $h/(mv)$, where $m$ is the matter particle mass.  The Schrödinger equation models the dynamics of the motion of the matter particle relative to a reference frame moving with the captured mass/matter particle system.  Solution to the equation gives the velocity of the matter particle as a function of the location of the particle.  We derive the Schrödinger equation by balancing the centrifugal forces against the centripetal forces.  Thus, we show that the Schrödinger equation is a Newtonian equation.

1.  Introduction

The Schrödinger equation models the dynamics of matter particles when they translate.  The equation was discovered in 1925 by the Austrian physicist Dr. Erwin Schrödinger.  Discover is used since he, nor any other scientist, has been able to derive the equation from basic principles during the 90 plus years since its discovery.  The equation, along with Einstein’s special theory of relativity, ushered in a whole new theory of physical science, namely Modern Physics, consisting of quantum theory and relativistic theory.  These theories both rejected classical, or Newtonian, theory.  The equation has been used without fail by scientists throughout the world for almost a century.  There can be no doubt about its capability for accurately predicting the behavior of nuclear particles.

The Schrödinger equation is a second order partial differential equation in two variables, space and time.  The total energy of the translating particle is needed, and physicists all over the world assume that the energy for a free translating particle is $mv^2/2$ , where m is the mass of the particle and v is its velocity.  They conclude that this is the energy since the work done on the particle by the accelerating medium is $mv^2/2$, i.e., the kinetic energy.  The difficulty with this value of energy shows up in solving the Schrödinger equation.  For the free translating particle the Schrödinger equation is solved by separation of variables giving two ordinary differential equations-a time equation and a space equation.  The wavelength for the particle is $h/(mv)$, the deBroglie wavelength.  The wavelength from the time equation, using the total energy as $mv^2/2$, is twice the deBroglie length, which is impossible.  From this we conclude that the total energy is $mv^2$.

Next, we examine the space equation.  Using the energy as $mv^2$ gives the result that the wavelength from the space equation is half the deBroglie length.  By replacing the factor 2 in the denominator of one term of the Schrödinger equation by unity results in a space equation with the same wavelength as the deBroglie wavelength.

From the above two paragraphs we have proven that the Schrödinger equation has two errors: the particle total energy is half the actual energy, and the factor 2 should not be in the original equation.  The two errors used in the space equation balance each other so that predictions of the equation are accurate.  The equation is useful, without fail.

With our model of a translating matter particle we show that the interpretation of the psi ($\psi$) function in the Schrödinger equation is not related to the probable location of the particle but is the velocity of the particle relative to a reference frame translating with the particle.  We then show that the (corrected) Schrödinger equation results from summing forces, i.e., applying Newton’s equations.

2.  Kinematics of the Motion of a Translating Particle

In the following discussion we consider the translation of an electron.  Figure 1 shows a translating electron.

Figure 1.  Geometry of the Translating Electron

The electron is labeled e in the figure.  It rotates in a circle of radius r with velocity $v_{e/g}$ relative to the moving frame about the electron mass/captured mass center of gravity, g.  The captured mass, $m_c$, rotates about g with a radius R and a velocity $v_{c/g}$ relative to the moving frame.  The angular velocity is $\omega$.  The acceleration of the electron is $a_e$, and the centripetal force is F.  The acceleration of the captured mass, $m_c$ , is $a_c$, and its centripetal force also is F.

Figure 2 shows the absolute motion of the electron and captured mass as a function of distance traveled.

Figure 2.  Paths of an Electron and Captured Mass

Figure 3 shows the path of the electron relative to the translating system.  We note that the path is a circle.

Figure 3.  Electron Path Relative to the Electron/Captured Mass System

While the electron travels one (deBroglie) wavelength the electron rotates one cycle, or a distance of $2 \pi r$.  Thus, $2 \pi r $ is its path length, and it travels this distance while the electron travels $\lambda$, the deBroglie wavelength.  Now

$2 \pi r = \lambda$  [Eq. 1]

Further, $v_{e/g}$ must be the same as v since the distances and elapsed times are the same.

The vertical and horizontal components of the velocity $v_{e/g}$ ($= v$) are harmonic functions, obviously.  Figure 4 shows the horizontal component of the electron velocity relative to the translating reference system.

Figure 4.  Horizontal Component of Electron Velocity vs. Location

This distribution is the same as obtained from the solution to the Schrödinger equation.  However, quantum mechanicists incorrectly assume $\psi$ is related to the probable dwell time.

$p_{1-2} = A \int_{x_1}^{x_2} \psi \psi^*$ [Eq. 2]

where $p_{1-2}$ is the probability that the electron is located between distances $x_1$ and $x_2$,  $A$ is a normalizing constant which makes the total probability unity, and $\psi^*$ is the complex conjugate of $\psi$.

The current quantum mechanics paradigm assumes that $\psi$ is related to the dwell time of the electron along its path.  We now determine the dwell time distribution based on our model of the translating electron.

From the equation for velocity

$v_{e/g_x} = v_x = \frac{dx}{dt}$ [Eq. 3]

we have

$dt = \frac{dx}{v_x} = \frac{dx}{v \sin \theta}$ [Eq. 4]

Then the time $\tau_{1-2}$ to move from $x_1$ to $x_2$ where the distances are measured relative to the moving frame is given by

$\tau_{1-2} = \int_{x_1}^{x_2} \frac{dx}{v \sin \theta} = \int_{x_1}^{x_2} f_t dx $ [Eq. 5]

where $f_t$ is defined as the dwell-time frequency and is given by

$ f_t = \frac{1}{v \sin (x/ƛ)} = \frac{1}{v \sin (2 \pi x / \lambda)} $ [Eq. 6]

Now we can write,

$v \tau_{1-2} = \int_{x_1}^{x_2} \frac{dx}{\sin(x/ƛ)}, \, \, v dt = \frac{dx}{\sin (x/ƛ)}$ [Eq. 7]

A plot of $f_t v$ is given in Figure 5.

Figure 5.  Electron Dwell Time Frequency Versus Distance Along Path

Comparing Figure 5 with Figure 4, the Schrödinger solution, we conclude that the psi ($\psi$) function is actually the electron velocity.  The real part and the complex part give the velocity components parallel to the $x$ and $y$ axes.

From the above we replace $\psi$ by $v$ in the Schrödinger equation.  Now $v$ is a complex number given by

$v = v \sin \theta + iv \cos \theta $ [Eq. 8]

where $v \sin \theta$ is the horizontal component of the velocity and $iv \cos \theta$ is the vertical component

3.  Correcting the Schrödinger Equation

The Schrödinger equation for the free translation of a particle of mass $m$ is

$\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + i\hbar \frac{\partial \psi}{\partial t} = 0$ [Eq. 9]

where $\psi(x,t)$ is a function associated with the probable location of the particle.  Assuming $\psi$ is a function of $x$ and $t$ separately, the equation is solved by using

$\psi(x,t) = X(x)T(t)$ [Eq. 10]

where $X$ is a function of $x$ only and $T$ is a function of $t$ only.  Substituting (10) into (9) gives the two ordinary differential equations

$\frac{i \hbar}{T} \frac{dT}{dt} = E$ [Eq. 11]


$\frac{\hbar^2}{2m} \frac{1}{X} \frac{d^2 X}{dx^2} = -E$ [Eq. 12]

In these expressions $E$ is the separation constant and is the total energy of the free particle.

Solving (11) gives

$T(t) = \cos \frac{Et}{\hbar} – i \sin \frac{Et}{\hbar}$ [Eq. 13]

Quantum electrodynamacists use the total energy as $mv^2 /2$.  Substituting the energy into the arguments of cosine and sine in (13) and using $\hbar$ given by the deBroglie wavelength of the particle, i.e.,

$\hbar = mvƛ$ [Eq. 14]


$\frac{Et}{\hbar} = \frac{(mv^2/2)t}{mvr} = \frac{v}{2r} t$ [Eq. 15]

The period of this wave, $\tau_t$, is obtained by setting $vt/(2r)$ equal to $2 \pi$.  Thus,

$\frac{v \tau_t}{2r} = 2 \pi, \,\, \tau_t = 2 \frac{2 \pi r}{v}$ [Eq. 16]

The period of the deBroglie wave, $\tau_d$, is

$\tau_d = \frac{2 \pi r}{v}$ [Eq. 17]

From (16) and (17) the ratio of the period from the time equation to the deBroglie period is

$\frac{\tau_t}{\tau_d} = \left ( 2 \frac{2 \pi r}{v} \right ) \div \left ( \frac{2 \pi r}{v} \right ) = 2 $ [Eq. 18]

which, of course, is impossible.  The ratio of the wavelengths is

$\lambda_\tau / \lambda_d = 2$ [Eq. 19]

since the time is doubled and the velocity is the same.  If in (15) we use the energy as $mv^2$ then (15) becomes

$\frac{Et}{\hbar} = \frac{(mv^2)t}{mvr} = \frac{v}{r} t$ [Eq. 20]

The period now is given by,

$\frac{v \tau_t}{r} = 2 \pi, \,\, \tau_t = \frac{2 \pi r}{v}$ [Eq. 21]


$\frac{\tau_t}{\tau_d} = \left ( \frac{2 \pi r}{v} \right ) \div \left ( \frac{2 \pi r}{v} \right ) = 1$ [Eq. 22]

From this we conclude that the particle total energy is $mv^2$.

Let us return to the space equation (12).  Letting

$e^2 = \frac{2mE}{\hbar^2}$ [Eq. 23]

We solve (12) by writing

$ \frac{d^2 X}{dx^2} + e^2 X = 0 $[Eq. 24]

A solution is

$X = A \sin (ex) + B \cos (ex)$ [Eq. 25]

which is easily seen to be a solution by using (23) and substituting (25) into (24).

$-Ae^2 \sin (ex) – B e^2 \cos (ex) + e^2 A \sin (ex) + e^2 B \cos (ex) = 0$ [Eq. 26]

The wavelength, $\lambda_x$ , for the wave represented by this equation is obtained by setting the sine and cosine arguments equal to $2 \pi$.  Thus

$e \lambda_x = 2 \pi = \sqrt{\frac{2mE}{\hbar^2}} \lambda_x$ [Eq. 27]

Using the incorrect energy  we have

$\lambda_x = \frac{2 \pi \hbar}{sqrt{2mmv^2/2}} = \frac{mv \lambda_d}{mv} = \lambda_d$ [Eq. 28]

Thus, the incorrect energy results in a space equation wavelength which agrees with the deBroglie equation.  The reason for this agreement is that the Schrödinger equation is incorrect.  Of course if in the term $\sqrt{2mmv^2/2}$ of (28) the first 2 were replaced by 1 and the second 2 also were replaced by 1, the space equation would agree with the deBroglie wavelength.  Two wrongs make a right in this case.

The Schrödinger equation, i.e., (9), is corrected by removing the 2 in the denominator of the first term.  In this case we let $e$ have the value

$e^2 = \frac{mE}{\hbar^2}$ [Eq. 29]

and $E$ has the value

$E = mv^2$ [Eq. 30]


$e \lambda_x = 2 \pi = \frac{\sqrt{mE}}{\hbar} \lambda_x = \frac{\sqrt{m^2 v^2}}{\hbar} \lambda_x = \frac{mv}{mvr} = \frac{\lambda_x}{r}$ [Eq. 31]


$2 \pi r = \lambda_x = \lambda_d$ [Eq. 32]

Using the corrected Schrödinger equation and the correct total energy results in a space equation wavelength the same as the deBroglie wavelength.  The two must be the same.

4.  Model of a Moving Matter Particle

To accelerate, a particle mass impacts the particle.  Part of the mass scatters, and part of the mass is captured by the field of the particle.  The captured mass and the particle begin rotating about the combined mass center of the two masses as they translate.  This motion is the undulation, or wave property, of translating particles.  The center of mass of the captured mass is located at a position which results in the particle having an angular momentum of $h$.  The resulting motion of the particle then has a flow component at velocity $v$ (and an energy of $(1/2) mv^2$) plus a thermal component.  The thermal component is the particle taking a circular path relative to the straight-line center of mass path.  The circumference of this circular path is the same length as the wavelength so the velocity is $v$, same as the translational velocity.  The thermal energy is, thus, also $(1/2) mv^2$.  The total energy $E$ then is

$ E=\frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 $ [Eq. 33]

We, of course, had discovered that the total energy had to be $mv^2$, from solving the Schrödinger equation.

Making the above correction we obtain the equation for a free particle as

$\frac{\hbar^2}{m} \frac{\partial^2 \psi}{\partial x^2} + i \hbar \frac{\partial \psi}{\partial t} = 0$ [Eq. 34]

and the correct particle total energy is

$ E = mv^2 $ [Eq. 35]

Now, recalling our discussion of the kinematics of the electron motion we replace $ \psi $ by the particle relative velocity, $v$, so that the corrected Schrödinger equation is

$ \frac{\hbar^2}{m} \frac{\partial^2 v}{\partial x^2} + i\hbar \frac{\partial v}{\partial t} = 0 $ [Eq. 36]

5.  Derivation of the Corrected Schrödinger Equation

An electron is accelerated by photons impacting it. The photons add momentum and mass to the particle. The mass added, $m_c$, is

$m_c = \frac{m_0}{\sqrt{1 – (v/c)^2}} – m_0$ [Eq. 37]

where $m_0$  is the matter particle mass when at rest, $v$ is the particle velocity, and $c$ is the speed of light.

The mass added, or captured mass, $m_c$, is assumed to continue moving at the velocity of the photon but in a closed loop around the electron. This is consistent with the energy of a moving matter particle having an energy of $ m_0 c^2 / \sqrt{1-\beta^2}$. The electron of mass $m_0$ and the captured mass $m_c$have a composite center of gravity $g$ which translates in a straight path, of course.

The captured mass center $c$ and the electron mass center $e$ rotate around the center of gravity as the assembly translates. They are held together by electromagnetic forces.  Figure 1 shows the two masses,  $m_0$  and $m_c$, their radii $r$ and $R$, their velocities, $v_{e/g}$  and $v_{c/g}$, the force pair $F$ and $F$ holding them together, the accelerations $a_c$ and $a_e$, the center of gravity $g$ of the electron/captured mass, and the angular velocity $\omega$. The electron (deBroglie) wavelength is $2 \pi r$. While point $g$ translates one wavelength, point $e$ travels a circle relative to point $g$, which circle has a circumference of $2 \pi r$. Thus, the velocity of $e$ relative to $g$ is $v$ – same as the absolute velocity of $g$.

The force balance for the electron, i.e., Newton’s equation for the electron is

$F = \frac{m_0 v_{e/g}^2}{r}$ [Eq. 38]

where $v_{e/g}^2 / r$ is the radial acceleration, which is inward in the radial direction. We can also write the force balance equation for the captured mass as

$F = m_c \frac{\partial v_{c/g}}{\partial t}$ [Eq. 39]

where $\partial v_{c/g} / \partial t$ is the radial acceleration of the captured mass.

We can represent the velocity $v$ using complex numbers as

$ v = v \cos \theta + iv \sin \theta $ [Eq. 40]

This has the advantage of explicitly giving vertical and horizontal components of the parameters.

In the following analyses we need some partial differential relations. For the time equation

$\frac{\partial v}{\partial t} = \frac{\partial v}{\partial \theta} \frac{\partial \theta}{\partial t} = \frac{\partial v}{\partial \theta} \omega = \frac{\partial v}{\partial \theta} \frac{v}{r}$ [Eq. 41]

For the space equation

$ \frac{\partial^2 v}{\partial \theta^2} = \frac{\partial}{\partial \theta} \left ( \frac{\partial v}{\partial \theta} \right )$ [Eq. 42]


$ \frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial \theta}$ [Eq. 43]


$\frac{\partial x}{\partial \theta} = \frac{\partial x}{\partial t} \frac{\partial t}{\partial \theta} = \frac{v}{\omega} = \frac{vr}{v} = r$ [Eq. 44]


$ \frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x} r = rf$ [Eq. 45]

where we have introduced

$f = \partial v / \partial x $ [Eq. 46]

Returning to (42) and (45)

$ \frac{\partial ^2 v}{\partial \theta^2} = \frac{\partial}{\partial \theta} \left ( rf \right ) = r \frac{\partial f}{\partial \theta}$ [Eq. 47]

where we have placed $r$ to the left of the differentiation since $r$ is constant. We can write the derivative of $f$ as

$ \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} = \frac{\partial f}{\partial x} r = \frac{\partial f}{\partial x} \frac{v}{\omega} = \frac{\partial f}{\partial x} \frac{v}{v} r = r \frac{\partial f}{\partial x}$ [Eq. 48]

Now, from (47), (48), and (46)

$\frac{\partial^2 v}{\partial \theta^2} = r \left ( r \frac{\partial f}{\partial x} \right ) = r^2 \frac{\partial^2 v}{\partial x^2} $ [Eq. 49]

We will now show that (38) can be used to give the variation of $v$ with the space coordinate $x$. We write $ m_0 v^2 /r$ as follows

$ F = \frac{m_0 v^2}{r} = \frac{m_0 v}{r} \left ( v \cos \theta + iv \sin \theta \right )$
$ = – \frac{m_0 v}{r} \frac{\partial^2}{\partial \theta^2} \left( v \cos \theta + iv \sin \theta\right )$ [Eq. 50]
$ = – \frac{m_0 v}{r} \frac{\partial^2}{ \partial x^2} \left ( v \cos \theta + iv \theta \right ) r^2$
$ = – \frac{m_0 v}{r} \frac{\partial^2 v}{\partial x^2} r^2 = \hbar \frac{\partial^2 v}{\partial x^2}$

Let us now investigate the dependence on the time coordinate, $t$.  Returning to (39) we write

$ F = m_c \frac{\partial v_{c/g}}{\partial t} = m_c \frac{\partial v_{c/g}}{\partial \theta} \frac{v}{r}$ [Eq. 51]
$ = m_c \frac{\partial}{\partial \theta} \left ( v_{c/g} \cos \theta + iv_{c/g} \sin \theta \right ) \frac{v}{r}$
$ = m_c \left (-v_{c/g} \sin \theta + iv_{c/g \cos \theta} \right ) \frac{v}{r}$
$ = im_c \left ( v_{c/g} \cos \theta + iv_{c/g} \sin \theta \right) \frac{v}{r}$
$ = i m_c v_{c/g} v/r $
$ = im_0 \frac{r}{R} v \frac{R}{r} \frac{v}{r} = im_0 v^2/r$

From the last and second terms in (51) we can express the force on the captured mass as

$\frac{m_0 v^2}{r} = \frac{1}{i} m_c \frac{\partial v_{c/g}}{\partial t} = -im_0 \frac{r}{R} \frac{\partial v}{\partial t} \frac{R}{r} = -im_0 \frac{\partial v}{\partial t}$ [Eq. 52]

where we have represented the velocity by a complex number.  The sum of the force on the electron added to the force on the captured mass must be zero. Thus from (50) and (52)

$ -\hbar \frac{\partial^2 v}{\partial x^2} – im_0 \frac{\partial v}{\partial t} = 0$ [Eq. 53]

Or, changing signs, multiplying by $\hbar$, and dividing by $m_0$ gives

$\frac{\hbar^2}{m_0} \frac{\partial^2 v}{\partial x^2} + i \frac{\hbar \partial v}{\partial t} = 0$ [Eq. 54]

Equation (54) is very similar to the Schrödinger equation for the free translation of a charged particle.  The differences are that the Schrödinger equation has a factor 2 in the denominator of the first term and $\psi$, presumably, is not a velocity.  We have shown that the factor 2 is incorrect and that $\psi$ should be replaced by the particle (complex) velocity.

6.  Analysis of Results

First, we show that there are two errors in the Schrödinger equation.  In solving the second degree partial differential equation, which is a function of space, $x$, and time, $t$, the separation of variables technique is used.  The result is two ordinary differential equations, the time equation and the space equation.  The Schrödinger equation models the dynamics of a translating matter particle.  The particle undulates, and its wavelength is given by $h/(mv)$, i.e., the deBroglie wavelength.  The time equation involves the total energy of the particle and conventional practice is to use the energy as $(1/2)mv^2$.  With this value the wavelength from the time equation is double the particle wavelength, which is impossible.  In section 3 we present a dynamic model of a translating particle.  With this model the particle total energy is $mv^2$.  The second error in the Schrödinger equation is that the factor 2 in the equation must be replaced by unity.

With the above corrections to the Schrödinger equation, along with considering the psi complex function as the complex velocity of the particle, it is shown that the (correct) Schrödinger equation is a classical mechanics equation.  The Schrödinger equation is an example of Newton’s equation $F = ma$.

In section 4, using our dynamic model of the translating particle, simply from the kinematics of the motion we show that the particle path is a periodic oscillating path same as the harmonic path resulting from the incorrect Schrödinger solution using psi as the velocity.

7.  Conclusions

We present a model of a translating matter particle consisting of the mass used to accelerate the particle and the particle itself.  The two masses rotate about each other as they translate.  This motion of the matter particle is its wave property.

We found two errors in the Schrödinger equation.  The energy of the translating particle is the translating part with an energy of $(1/2) mv^2$ and a thermal part with an energy of $(1/2)mv^2$ for a total of $mv^2$.  In solving the Schrödinger equation separation of variables is used resulting in two ordinary differential equations, a time equation and a space equation.  When the total energy is taken as $(1/2)mv^2$ for a free particle, which is almost always done, the wavelength given by the time equation is twice the particle (deBroglie) wavelength.  This is impossible, but using the correct energy, $mv^2$, the two wavelengths are the same.  Using the correct energy $mv^2$ in the space equation shows that the factor 2 in the denominator of the Schrödinger equation should be unity.

We then show that the correct Schrödinger equation is derived from Newtonian mechanics.  Furthermore, we show that our model of the translating particle gives the same result as the Schrödinger equation.