Figure 2 shows that P(x) becomes sharply peaked as the variance ( σ ) goes to zero.

http://www1.uprh.edu/rbaretti/lqmch5.htm

 

The delta function potential

Some properties of the Delta function in one dimension  are

 

                             ∫ f(x) δ(x- x₀ ) dx = f (x ₀ )                    ,        -∞ < x < +∞       (5)

 

      

 

                                              ∫ δ(x- x₀ ) dx =1                    ,                              (6)

 

 

and                                          ∫ δ( a x ) dx =1/a                .                               (7)

 

The dimensions of   δ( x ) are 1/length  .

A dirac sequence is established when certain functions approach the delta function  with the help of an adjustable parameter. Two example are given. First consider the gaussian function

 

                              P(x) = (1/σ) ( 1/(2π)^1/2 ) exp[ -(x-μ)² /(2σ² )]                 ,        (8)

 

μ is the mean and σ² the variance. In other words

                                           ∫ x P(x) dx =  μ       ,      -∞ < x < +∞                      (9)

 

and                                     ∫ x² P(x) dx =  σ²  .                                             . (10)                                           

                               

 

 

 

 

Fig 2. The gaussian function P(x) will resemble Dirac delta function as σ→ 0 .

 

Figure 2 shows that P(x) becomes sharply peaked as  the variance ( σ ) goes to zero.

 

 

 

A second  sequence that converges to the Dirac delta function is 

                                                                                        (11)          

It resembles the dirac delta function as n becomes large .

 

Its integral   ∫ δ_n (x) dx = 1    ,  - ∞ < x < + ∞ .

 

 

 

Let the potential be V(x) =- η δ(x)  where η is the strength of the delta function. This peculiar potential produces a single energy eigenvalue  given by E = - {h'² /(2m)} η²  . To show this start with

 

                      d² Ψ /dx²  - (2m/h'² ) V(x) Ψ = - (2m/h'² ) E Ψ                         .   (12)

Integrate the whole expression around the origin  from  -ε to  +ε . We get 

 

 

 ∫ (d² Ψ /dx² )  dx =  ( d Ψ( +ε ) /dx   - d Ψ( -ε ) /dx ) = 2 ( d Ψ( +ε ) /dx )           ,  (13)

see next figure 3.

Fig 3. Shape of  the eigenfunction of the delta function.

 

The potential term integration gives

 

- (2m/h'² )∫ V(x) Ψ dx = - (2m/h'² )∫  -η δ(x)  Ψ dx =  (2m/h'² ) η  Ψ(0)    .      (14)

 

The last term is zero , i.e.

 

- (2m/h'² ) E ∫ Ψ  dx =  - (2m/h'² ) E  Ψ( 0) (2ε ) → 0                           .     (15)

 

From (13) and (140 it follows that

 

                     d Ψ( +ε ) /dx )  = (-1/2)(2m/h'² ) η  Ψ(0)  =  -(m/h'² ) η  Ψ(0)  . (16)

The function near the origin ( x > 0) is thus of the form   

 

                 Ψ   ~     exp{-(m η /h'² ) x }  .                                                  (17)

To get the energy go back to (12) at x near the origin (x>0) and use (17)

 

      E Ψ  = - {h'²/(2m) } d² Ψ /dx² = - (1/2) (m/h'² ) η² Ψ                ,        (18) 

 

                   E = -m η² /(2 h'² )  ,

or                E = - η² /2     in units where m=1,h'=1.  

 

This is the only allowed energy eigenvalue by an attractive delta function of strength η.

Figure 4 shows a delta like potential. Its eigenvalue is E = - η² /2 = -50.