Consider the double delta-function potential$\mathit{V}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathit{\alpha }\left[\delta \left(x+a\right)+\delta \left(x-a\right)\right]$Whereand are positive constants

(a) Sketch this potential.

(b) How many bound states does it possess? Find the allowed energies, for$\mathit{\alpha }\mathbf{=}\mathit{ħ}\mathbf{/}\mathit{m}\mathit{a}$and for$\mathit{\alpha }\mathbf{=}{\mathit{ħ}}^{\mathbf{2}}\mathbf{/}\mathbf{4}\mathit{m}\mathit{a}$, and sketch the wave functions.

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer relates to a particle's probability density in space and time.

(a)

The image represents a potential given in the problem for values $\alpha =1$and $a=2$

(b)

Schrodinger equation can be written as:

$$\begin{gathered} -\frac{{ħ}^2}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)+V\left(x\right)\psi \left(x\right)=E\psi \left(x\right) \\ {\int }_{a-\epsilon }^{a+\epsilon }\left(-\frac{{ħ}^2}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)+V\left(x\right)\psi \left(x\right)\right)dx={\int }_{a-\epsilon }^{a+\epsilon }E\psi \left(x\right)dx \end{gathered}$$

When we integrate the Schrodinger equation around the small neighborhoodlocalid="1658224661226" $\epsilon$around$x=a$:

Using the linearity of the integral,

localid="1658224671672" $$\begin{gathered} -{\int }_{a-\epsilon }^{a+\epsilon }\frac{ħ}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)dx-{\int }_{a-\epsilon }^{a+\epsilon }a\delta \left(x\right)\psi \left(x\right)dx={\int }_{a-\epsilon }^{a+\epsilon }E\psi \left(x\right)dx \\ {\int }_{-\epsilon }^{+\epsilon }\frac{{ħ}^2}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)dx=-a\psi \left(a\right) \end{gathered}$$

When the arbitrary region is equal to zero.

So, the fundamental theorem of calculus:

localid="1658224677894" $\frac{d}{dx}\psi \left(-a\right)\underline{+}\frac{d}{dx}\psi \left(-a\right)=-\frac{2m}{{ħ}^2}a\psi \left(-a\right))$

Now we need to solve. Schrodinger equation outside the delta function we shall separate in 3 regionlocalid="1658224684056" $\left(-\infty ,-a\right),\left(-a,a\right),\left(a,\infty \right)$

localid="1658224689367" $\frac{{ħ}^2}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)=-E\psi \left(x\right)$

Letlocalid="1658224699837" $k^2=\frac{2mE}{{ħ}^2}$, and then the expression can be written as:

localid="1658224705643" ${\psi }^{"}\left(x\right)-k^2\psi \left(x\right)=0$

The solution to the above equation can be written as follows:

localid="1658224713016" $\psi \left(x\right)=A{e}^{Kx}+B{e}^{-Kx}$

So, the solutions in the 3 regions:

localid="1658224718193" $$\begin{gathered} {\psi }_l\left(x\right)=A_1{e}^{kx}+B_1{e}^{-kx} \\ {\psi }_{ll}\left(x\right)=C\left(e^{kx}+e^{-kx}\right) \\ {\psi }_{lll}\left(x\right)=A_2{e}^{kx}+B_2{e}^{-kx} \end{gathered}$$

We are assuming bound states ( for even wave function ):

localid="1658224723913" $B_1=A_2=0$,localid="1658224729834" $A_1=B_2=A$,localid="1658224738413" $C=B$

Then the solutions in the 3 regions:

localid="1658224746652" $$\begin{gathered} {\psi }_l\left(x\right)=A{e}^{kx} \\ {\psi }_{ll}\left(x\right)=B\left(e^{kx}+e^{-kx}\right) \\ {\psi }_{lll}\left(x\right)=A{e}^{-kx} \end{gathered}$$

From the definition of the wave function and its first derivative will be continuous across the boundary (at a), thus:

The equations can be solved as follows:

localid="1658224831570" $$\begin{gathered} A{e}^{-ka}=B\left(e^{ka}+e^{-ka}\right) \\ A=B\left(1+e^{2ka}\right) \end{gathered}$$

And

localid="1658224843405" $$\begin{gathered} -B\left(1+e^{2ka}\right)k{e}^{-ka}B(k{e}^{ka}-k{e}^{-ka})=-B\left(1+e^{2ka}\right)\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ -\left(1+e^{2ka}\right)k{e}^{-ka}+\left(k{e}^{ka}-k{e}^{-ka}\right)=-\left(1+e^{2ka}\right)\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ \left(1+e^{2ka}\right)k{e}^{-ka}-\left(e^{ka}-e^{-ka}\right)k=\left(1+e^{2ka}\right)\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ \left(1+e^{2ka}\right)k{e}^{-ka}+\left(e^{2ka}-1\right)e^{-ka}k=\left(1+e^{2ka}\right)\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ \left(e^{2ka}-1\right)=\left(1+e^{2ka}\right)\frac{2m\alpha }{k{ħ}^2}-\left(1+e^{2ka}\right) \\ \left(e^{2ka}-1\right)=\left(1+e^{2ka}\right)\left(\frac{2m\alpha }{k{ħ}^2}-1\right) \end{gathered}$$

Further solving the above expression as:

localid="1658224853860" $$\begin{gathered} \left(e^{2ka}-1\right)=\left(\frac{2m\alpha }{k{ħ}^2}-1\right)+e^{2ka}\left(\frac{2m\alpha }{k{ħ}^2}-1\right) \\ \frac{k{ħ}^2}{m\alpha }=1+e^{-2ka} \\ {e}^{-2ka}=\frac{k{ħ}^2}{m\alpha }-1 \end{gathered}$$

This is a transcendental equation for k. Let's convert it into a simpler form:

Let localid="1658224882054" $z=2ka$, and localid="1658224875000" $c=\frac{{ħ}^2}{2am\alpha }$Thus, the equation will become:

localid="1658224888748" $e^{-z}=cz-1$

Plotting both curves as:

As we can see that only one solution is available here, which is:

If localid="1658224895973" $\alpha =\frac{{ħ}^2}{2ma}$then c = 1, then z = 1.278.

Energy can be calculated as:

localid="1658224906389" $$\begin{gathered} k^2=-\frac{2mE}{{ħ}^2}=\frac{z^2}{{\left(2a\right)}^2} \\ E=-0.204\left(\frac{{ħ}^2}{m{a}^2}\right) \end{gathered}$$

Now, We are assuming bound states ( for odd wave function ) and solve the equations as:

localid="1658224917213" $$\begin{gathered} {\psi }_l\left(x\right)=A{e}^{-kx} \\ {\psi }_{ll}\left(x\right)=B(e^{kx}-e^{-kx}) \\ {\psi }_{lll}\left(x\right)=-A{e}^{kx} \end{gathered}$$

From the definition of the wave function and its first derivative will be continuous across the boundary (at a), thus:

localid="1658224930140" $$\begin{gathered} A{e}^{-ka}=B(e^{ka}-e^{-ka}) \\ -Ak{e}^{-ka}+B\left(k{e}^{ka}-k{e}^{-ka}\right)=-A\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \end{gathered}$$

The equations can be solved as follows:

localid="1658224942600" $$\begin{gathered} A{e}^{-ka}=B(e^{ka}+e^{-ka}) \\ A=B(1+e^{2ka}) \end{gathered}$$

And

localid="1658224951140" $$\begin{gathered} -B\left(1+e^{2ka}\right)k{e}^{-ka}+B(k{e}^{ka}-k{e}^{-ka})=-B(1+e^{2ka})\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ -\left(1+e^{2ka}\right)k{e}^{-ka}+(k{e}^{ka}-k{e}^{-ka})=-(1+e^{2ka})\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ \left(1+e^{2ka}\right)k{e}^{-ka}-(e^{ka}-e^{-ka})k=(1+e^{2ka})\frac{2m\alpha }{{ħ}^2}{e}^{-ka} \\ \left(1+e^{2ka}\right)k{e}^{-ka}+(e^{ka}-1)e^{-ka}k=(1+e^{2ka})\frac{2m\alpha }{{ħ}^2} \\ \left(e^{2ka}-1\right)=\left(1+e^{2ka}\right)\frac{2m\alpha }{k{ħ}^2}-\left(1+e^{2ka}\right) \\ \left(e^{2ka}-1\right)=\left(1+e^{2ka}\right)\left(\frac{2m\alpha }{k{ħ}^2}-1\right) \end{gathered}$$

Further solving the above expression as:

localid="1658224964026" $$\begin{gathered} \left(e^{2ka}-1\right)=\left(\frac{2m\alpha }{k{ħ}^2}-1\right)+e^{2ka}\left(\frac{2m\alpha }{k{ħ}^2}-1\right) \\ \frac{k{ħ}^2}{m\alpha }=1+e^{-2ka} \\ {e}^{-2ka}=\frac{k{ħ}^2}{m\alpha }-1 \end{gathered}$$

This is a transcendental equation for k. Let's convert it into a simpler form:

Let localid="1658224983944" $z=2ka$, and localid="1658224975365" $c=\frac{{ħ}^2}{2am\alpha }$Thus, the equation will become:

localid="1658224990880" $e^{-z}=cz-1$

Plotting both curves as:

Both curves intercept at the y-axis at 1. The solution now depends on the values of c:

1. If the value of c is large, there might be no intersection ( $\alpha$is too small)
2. If the value of c is small, there might be an intersection ( pink line)

The slope of curve 1: $e^{-z}$ is –1,and curve 2:$1-cz$is – c (at z=0). So there is an odd solution which is:

$c<1$

$$\begin{gathered} \frac{{ħ}^2}{2am\alpha }<1 \\ \frac{{ħ}^2}{2am}<\alpha \end{gathered}$$

Thus there are :

One bound state if$\alpha \le \frac{{ħ}^2}{2ma}$

Two bound state $\alpha \le \frac{{ħ}^2}{2ma}$

Allowed energies:

For $$\begin{gathered} \alpha =\frac{{ħ}^2}{ma} \\ c=\frac{{ħ}^2}{2am\alpha } \\ =1/2 \end{gathered}$$

Then the value of z can be calculated :

For even:

$$\begin{gathered} e^{-z}=\frac{1}{2}z-1 \\ z=2.21 \end{gathered}$$

For odd:

$$\begin{gathered} e^{-z}=1-\frac{1}{2}z \\ z=1.59 \end{gathered}$$

Now the energy can be calculated as:

$E=-\frac{{ħ}^2{z}^2}{{\left(2a\right)}^{2}2m}$

For even, substituting the value of z and we get,

$E=-\frac{0.6105{ħ}^2}{a^{2}m}$

For odd, substituting the value of z and we get,

$E=-\frac{0.316{ħ}^2}{a^{2}m}$

For $\alpha =\frac{{ħ}^2}{4ma}$

$$\begin{gathered} c=\frac{{ħ}^2}{2am\alpha } \\ =2 \end{gathered}$$

Then the value of z can be calculated :

For even:

$$\begin{gathered} e^{-z}=2z-1 \\ z=0.7388 \end{gathered}$$

Now the energy can be calculated as:

$E=-\frac{{ħ}^2{z}^2}{{\left(2a\right)}^{2}2m}$$E=-\frac{{ħ}^2{z}^2}{{\left(2a\right)}^{2}2m}$

For even, substituting the values of z and we get,

$E=-\frac{0.0682{ħ}^2}{a^{2}m}$

Allowed energies are :

For $\alpha =\frac{{ħ}^2}{ma}$For even:$E=-\frac{0.6105{ħ}^2}{a^{2}m}$, For odd:$E=-\frac{0.316{ħ}^2}{a^{2}m}$

For $\alpha =\frac{{ħ}^2}{4ma}$ For even: $E=-\frac{0.0682{ħ}^2}{a^{2}m}$

Sketch of the wave function: