This is only true for a parabola (harmonic oscilator in QM). Its called ground state energy.

In that case 1/2 wave length can fit in the imagrey of old QM. There are many ways to get a ground state energy.

This arises in time independent QM.

It's not a paradox. It is a standard exercise for the particle in a quadratic potential. A similar result can be found for a particle in a square potential and the same holds for a particle in any potential, that is bounded from below and confines at infinity-the lowest eigenstate has an energy that's not zero. So what?

In Niels Bohr original model which was entirely based on Newtons laws of motion supplimented  by the Bohr quantization hypothesis,By classical quantum mechanics we mean N. Bohr's original QM of the hydrogen atom before the Schrödinger equation and the Bohr/Copenhagen superposition interpretation.We assume that classical quantum mechanics is somehow the basis or foundations of modern quantum mechanics and, therefore, a professional mathematician/physicist would have to master classical quantum mechanics before they can start learning modern quantum mechanics, sometimes called complete quantum mechanics.We therefore recommend a good level of understanding of what is called classical quantum mechanics.To be more specific, we provide the following two illustrative examples:Niels Bohr introduced his atomic Hydrogen model in 1913 where he described it as, electrons orbiting  the nucleus in circular orbits similar to solar system.I-Bohr model for Hydrogen atom

mv 2Pi r(n) = n . h ,n=1,2,3,. . .etc.

he did not say a word about electron cloud in a quadratic potential or any of the quantization numbers (n,l,m,s) other than the principal quantum number n.

We present Fig.1 to explain briefly the precedent model.

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Fig1. Bohr original description of quantized energy levels in Hydrogen atom.

II- Classical quantum mechanics model for a particle in 1D infinite potential well

It describes a subatomic particle free to move in a small 1D space surrounded by impenetrable barriers.

Here we do not have a precise classical solution but only a Schrodinger equation based solution.

Figure 2 presents a schematic diagram for the time-independent solution of the Schrödinger equation for a quantum particle in a 1D infinite potential well.

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Fig.2. A schematic diagram for the time-independent solution of the Schrödinger equation for a quantum particle

in a 1D infinite potential well.

And yet, to our knowledge fixing the minimum energy of a quantum particle in a box (Emin=hf/2) has no solid physical or mathematical basis.

This remains a paradox.

However, the statistical chains of the transition matrices B predict an explanation of the previous problem.

To be continued.

II-Answer II continued.

In fact, the statistical chains of the transition matrices B predict a numerical solution of time-dependent partial differential equations.

It was shown previously that the time-dependent solution of the energy density [U(x,t)] follows the following recurrence relation[1,2,3],

U(x,t+dt)= U(x,t) + B(x,t) . (b+S). . .(1)

where b is the vector of boundary conditions arranged in the correct order and S is the source/sink term at point (x, t).

Equation (1) also applies to the solution of the heat diffusion equation and to the solution of the Schrodiger equation.

However, there are differences in procedural techniques due to the differences in nature between the microscopic quantum world and the world of macroscopic classical physics:

I-The source/sink term in heat diffusion is external and mainly due to the boundary conditions b while the internal source S is neglected or assumed to be zero.

In the Schrödinger equation, the source/sink term is internal and is equal to a constant * V(x,t) at the point in space x while the contribution of the boundary conditions is set to zero since the space extends to infinity.

Note that the latter is an eigenvalue problem and that the constant of proportionality could be imaginary.

II- In heat diffusion, the time-independent steady state is reached when B^N tends to zero while the steady state condition in the Schrödinger equation is reached by convergence of the value of the energy density towards the eigenvalues ​​of equation 1.

From the previous analysis, it is simple to show that the transfer matrix (M) required for the Schrödinger equation is given by:

M=

{1+ ( md^2 V(X1)/h^2 } -1/2 0 0 0 0 0

-1/2 {1+ (md^2 V(X1)/h^2 } -1/2 0 0 0 0........................ .. ...........0 0

0 -1/2 {1+ (md^2 V(X1)/h^2 } -1/2 0 0 0........................ .. ........0 0

.........................

.......................................

0 0 0 0..................................0 0 0 -1 {2+ (2 md^2 V( )/h^2 } -1

0 0 0 0............................................. ..- 1 {2+ (2 md^2 V(X1)/h^2 }

It is worth mentioning that the matrix M is precisely adequate to the matrix obtained for the numerical solution of the time-independent Schrödinger equation and that the matrix M is obtained via statistical methods bypassing the Schrödinger equation itself.

To be continued.

Answer III - continued.

Here is a brief response aiming on the one hand to thank our fellow contributors for their useful answers and on the other hand to shed light on the so-called quantum particle in a box paradox, namely:

Emin=hf/2.

Some physicists and mathematicians claim that the solution to the time-dependent Schrödinger equation, in which the applied potential is a function of time, can prove this to be true.

Yet this does not exist in the current literature on modern quantum mechanics.

we therefore assume that:

I- Emin=hf/2 is not true and therefore there is no physical or mathematical proof.

II-Emin=hf/2 is true but the mathematical/physical proof is inaccessible.

What the Schrödinger equation and its statistical equivalence on the B matrix clearly show is that a quantum particle of zero energy cannot exist inside a closed potential box.

Obviously, Emin =0 is not the same as Emin=hf/2.

We assume that currently neither the Schrödinger equation nor its statistical equivalence on matrix B are capable of resolving the paradox.

To be continued.

That is correct. In the solution of the time-dependent Schrödinger partial differential equation, it is assumed that the minimum energy of a quantum particle is not zero, but rather is equal to hν/2, where h is Planck's constant and ν is the frequency of the particle. This minimum energy is known as the zero-point energy and arises from the Heisenberg uncertainty principle, which states that the position and momentum of a particle cannot both be precisely known at the same time. As a result, even when a particle is at rest, it still possesses a minimum energy due to this uncertainty in its position and momentum. This energy is important to take into account in quantum mechanics, as it can affect the behavior and properties of quantum systems.

Another way to describe quantum particle dynamics is to use statistical transition matrices that completely ignore the Schrödinger equation as if it never existed in the same way that one solves the heat diffusion equation without going through thermal PDE itself.

today we only know one physical transition matrix which is the transition matrix B resulting from the so-called Cairo technique.

Step 1

Construct the 2D statistical matrix B corresponding to Figure 1 which represents a quantum particle in a 2D infinite potential well.

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Fig 1. A quantum particle in a 2D infinite potential well

The basis for generating an eigen or proper matrix is the 2D matrix B with 9 equidistant free nodes as shown in Figure 2, nodes 1-9.

Note that RO=0 because PE is zero.

It is expressed by the same matrix M1 explained previously in the example of 2D thermal conduction.

step 2

Compose the proper or eigen matrix M2 as given by,

M2=M1+S(x,y)

Where S(x,y ) is a diagonal matrix and S=C1*V(x,y)

The resulting eigenmatrix M2 will be given by,

M2=

1/14 1/4 0 1/4 0 0 0 0 0

1/4 4/14 1/4 0 1/4 0 0 0 0

0 1/4 1/14 0 0 1/4 0 0 0

1/4 0 0 4/14 1/4 0 1/4 0 0

0 1/4 0 1/4 9/14 1/4 0 1/4 0

0 0 1/4 0 1/4 4/14 0 0 1/4

0 0 0 1/4 0 0 1/14 1/4 0

0 0 0 0 1/4 0 1/4 4/14 1/4

0 0 0 0 0 1/4 0 1/4 1/14

Where C1 is substituted for by the factor 1/14.

step 3

The energy eigenvector E(x,y) is equal to the principal diagonal of the matrix A which gives the following eigenvector equation,

2/14 1/4 0 1/4 0 0 0 0 0

1/4 4/14 1/4 0 1/4 0 0 0 0

0 1/4 2/14 0 0 1/4 0 0 0

1/4 0 0 4/14 1/4 0 1/4 0 0

0 1/4 0 1/4 9/14 1/4 0 1/4 0

0 0 1/4 0 1/4 4/14 0 0 1/4

0 0 0 1/4 0 0 2/14 1/4 0

0 0 0 0 1/4 0 1/4 4/14 1/4

0 0 0 0 0 1/4 0 1/4 2/14

*

[2/14  4/14  2/14  4/14  9/14  4/14  2/14  4/14  2/14]T

is equal to,

[8/49 123/392 8/49 123/392 137/196 123/392 8/49 123/392 8/49] T

Showing that the energy eigenvector is=

[2/14 4/14 2/14 4/14 9/14 4/14 2/14 4/14 2/14 ] T

with a dominant eigenvalue almost equal to 1.

The reason why we multiply the nodes 1,3,6 and 9 by the factor 2 is that these nodes are located at the four intersections of the two axes x and y where the rule E=Ex+Ey applies.

The x-oriented eigenvectors and the y-oriented eigenvectors are shown in Figure 2 in black and red lines.