Question

Show that there is no acceptable solution to the Schrodinger equation for the infinite square well with $\mathbf{E}\mathbf{=}\mathbf{0}$or$\mathit{E}\mathbf{<}\mathbf{0}$(This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schrodinger equation, and showing that you cannot meet the boundary conditions.)

Short answer

$E$cannot be less than zero because $n$is squared, also, it cannot be zero because if $E=0$then this will violate the uncertainty principle.

Step-by-step solution

Significance of the Schrodinger equation

The position of a particular electron is described in space and time using Schrodinger equation in the form of a mathematical expression. It considered the wave nature of the electron.

Identification of solution of the Schrodinger equation for infinite square wall

Write equation (2.27).

${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{ma}}^2}$

$E_n$cannot be negative because all the constants are positive or squared (that is., if we take $n=\pm 1,\pm 2,\cdots$ the square will kill the negative signs, so ${\mathrm{E}}_{\mathrm{n}}$cannot be less than zero).

The other case in this problem when $E=0$(i.e.,$n=0$ ) is not accepted because $E$is the eigenvalue of the total energy operator (we can think about it as the total energy of the wave function) which cannot be zero because this will violate the uncertainty principle.

Since if $E=0$this corresponds to zero kinetic energy (because $E$ in the potential energy in the infinite potential well is zero), and zero kinetic energy corresponds to zero momentum, and the particle will rest inside the potential.

The calculation of the minimum kinetic energy by restricting the particle to move in a region of width $\Delta x\approx a$(i.e., the length of the well) can be done, then using the uncertainty principle.

The calculation of the minimum uncertainty in the momentum can be done, so the result can be$\Delta p\approx$$\hslash /\left(2a\right)$ , so the minimum kinetic energy is${\hslash }^2/\left(8m{a}^2\right)$where this result qualitatively agrees with the exact value of the ground state.

Thus, $E$cannot be less than zero because $n$is squared, also, it cannot be zero because if $E=0$then this will violate the uncertainty principle.