Question

We learned that to be normalizable, a wave function must not itself diverge and must fall to 0 faster than ${\left|x\right|}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}$as x get large. Nevertheless. We find two functions that slightly violate these requirements very useful. Consider the quantum mechanical plane wave Ae^i(kx-ax)and the weird function${\mathbf{\Psi }}_{{\mathbf{x}}_{\mathbf{0}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$pictured which we here call by its proper name. the Dirac delta function.

(a) Which of the two normalizability requirements is violated by the plane wave, and which by Dirac delta function?

(b) Normalization of the plane wave could be accomplished if it were simply truncated, restricted to the region -b<x<+bbeing identically 0 outside. What would then be the relationship between b and A, and what would happen to A as b approaches infinity?

(c) Rather than an infinitely tall and narrow spike like the Dirac delta function. Consider a function that is 0 everywhere except the narrow region$\mathbf{-}\mathbf{E}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{+}\mathbf{\partial }$ where its value is a constant B. This too could be normalized, What would be the relationship between s and B, and what would happen to B as s approaches 0? (What we get is not exactly the Dirac delta function, but the distinction involves comparing infinities, a dangerous business that we will avoid.)

(d) As we see, the two "exceptional" functions may be viewed as limits of normalizable ones. In those limits, they are also complementary to each other in terms of their position and momentum uncertainties. Without getting into calculations, describe how they are complementary.

Short answer

(a) The function of plane wave is never fall off but Dirac delta function is diverged.

(b) b Approaches to infinity, a goes to zero.

(c) As the $\epsilon$goes to 0 B approaches to infinity.

(d) The nature of the two functions is complementary.

Step-by-step solution

Explanation

It’s given that a wave function must not be diverge and must not fall to 0 or faster than$\left|{\mathrm{x}}^{-1/2}\right|$ .

A plane wave is actually a monochromatic radiation in a field of infinite extent within space travelling in a specified direction. The wave function for the plane wave is always sinusoidal and depends on time.

For the limit of zero width Dirac delta function is used. Thisfunction alwaysdiverges but the wave function for the planewave neverfallsoff.

Thus, the function of plane wave never fall off but Dirac delta function is diverged.

To determine

It’s given that: It’s given that a wave function must not be diverge and must not fall to 0 or faster than $\left|{\mathrm{x}}^{-1/2}\right|$.

Formula used:

Write the expression for the wave function of plane wave.

$\mathrm{\Psi }\left(\mathrm{x}\right)={\mathrm{Ae}}^{\mathrm{i}(\mathrm{kx}-\mathrm{\omega t})}$……………………………….(1)

Here, $\mathrm{\Psi }\left(\mathrm{x}\right)$is the wave function for the plane wave, A is the amplitude of the wave, K is the propagation constant, x is the position of the plane wave, $\omega$is the angular frequency and t is the time.

Write the expression for the complex conjugate of the wave function $\mathrm{\Psi }\left(\mathrm{x}\right)$.

${\mathrm{\Psi }}^{*}\left(\mathrm{x}\right)={\mathrm{Ae}}^{-\mathrm{i}(\mathrm{kx}-\mathrm{\omega t})}$…………………………………..(2)

Now, multiply equation (1) and (2).

$\mathrm{\Psi }\left(\mathrm{x}\right){\mathrm{\Psi }}^{*}\left(\mathrm{x}\right)=\left({\mathrm{Ae}}^{\mathrm{i}(\mathrm{kx}-\mathrm{\omega t})}\right)\left({\mathrm{Ae}}^{-\mathrm{i}(\mathrm{kx}-\mathrm{\omega t})}\right)$

${\left|\mathrm{\Psi }\left(\mathrm{x}\right)\right|}^2={\mathrm{A}}^2$

Integrate both sides of the above equation.

$\int {\left|\mathrm{\Psi }\left(\mathrm{x}\right)\right|}^2\mathrm{dx}={\mathrm{A}}^2\underset{-\mathrm{b}}{\overset{\mathrm{b}}{\int }}\mathrm{dx}$

Substitute 1 for$\int {\left|\mathrm{\Psi }\left(\mathrm{x}\right)\right|}^2\mathrm{dx}$ due to normalization and solve the above integration.

$\begin{array}{rcl}1& =& {\mathrm{A}}^2(\mathrm{b}-(-\mathrm{b}\left)\right)\\ {\mathrm{A}}^2& =& \frac{1}{2\mathrm{b}}\\ \mathrm{A}& =& \frac{1}{\sqrt{2\mathrm{b}}}\end{array}$

Here, b is some position of the plane wave.

Thus as b approaches to infinity, A goes to zero.

relationship between s and B.

A function is infinitely tall and has narrow spike like a delta function.

Write the expression for the amplitude of the function.

$\mathrm{B}=\frac{1}{\sqrt{2\mathrm{\epsilon }}}$……………………………………….. (3)

Here, B is the amplitude of the wave andis some postion of the wave.

Substitute 0 for $\epsilon$in equation (3),

$\mathrm{B}=\infty$

Thus, as the $\epsilon$goes to 0 B approaches to infinity.

Define the complementary nature of the plane wave function and delta function.

A plane wave is actually a monochromatic radiation in a field of infinite extent within space and travelling in a specified direction. The wave function for the plane wave is always sinusoidal and depends on time.

For the limit of zero width Dirac Delta function is used. This function is always diverges but the wave function for plane wave is never fall off.

The uncertainty of the momentum of plane wave is well defined $(\mathrm{\Delta p}=0)$so the uncertainty in position is closed to infinity $(\mathrm{\Delta x}=\infty )$according to Heisenberg principle.

The uncertainty of the position of delta function is well defined $(\mathrm{\Delta x}=0)$, so the uncertainty in momentum is closed to infinity$(\mathrm{\Delta x}=\infty )$according to Heisenberg principle.

Thus, the nature of the two function is complementary.