WKB method transmission coefficient

[B4cklfip]

Homework Statement

The Task is to calculate the Transmission coefficient with the WKB Approximation of following potential: V(x) = V_0(1-(x/a)²) |x|<a ; V(x) = 0 otherwise

Relevant Equations

ln|T|² = -2 ∫ p(x) dx

Homework Statement: The Task is to calculate the Transmission coefficient with the WKB Approximation of following potential: V(x) = V_0(1-(x/a)²) |x|<a ; V(x) = 0 otherwise
Homework Equations: ln|T|² = -2 ∫ p(x) dx

I have inserted the potential in the equation for p(x) and recieved

p(x) = 1/ħ ∫ √(2m(V_0(1-(x/a)²)-E)) dx from -a to a

But now there is the problem that i don´ t know how to solve this integral.
I would be glad if someone has some tips.

 

[haruspex]

Homework Statement: The Task is to calculate the Transmission coefficient with the WKB Approximation of following potential: V(x) = V_0(1-(x/a)²) |x|<a ; V(x) = 0 otherwise
Homework Equations: ln|T|² = -2 ∫ p(x) dx

I have inserted the potential in the equation for p(x) and recieved

p(x) = 1/ħ ∫ √(2m(V_0(1-(x/a)²)-E)) dx from -a to a

But now there is the problem that i don´ t know how to solve this integral.
I would be glad if someone has some tips.

In LaTeX, $p(x) =\frac 1ħ \int_{-a}^a \sqrt{ 2m(V_0(1-(\frac xa)^2-E)} dx$
Is that right?

 

[B4cklfip]

Not quite, there´ s missing one bracket. It should be:

$$p(x)=\frac{1}{\hbar} \int_{-a}^{a}\sqrt{2m(V_0(1-(\frac{x}{a})^2)-E)}dx$$

with the potential $$ V(x) = \begin{cases} V_0(1-(\frac{x}{a})^2) & \text{for -a $\leq x \leq $a} \\0 & \text{otherwise} \end{cases}$$

​

 

[haruspex]

Not quite, there´ s missing one bracket. It should be:

$$p(x)=\frac{1}{\hbar} \int_{-a}^{a}\sqrt{2m(V_0(1-(\frac{x}{a})^2)-E)}dx$$

with the potential $$ V(x) = \begin{cases} V_0(1-(\frac{x}{a})^2) & \text{for -a $\leq x \leq $a} \\0 & \text{otherwise} \end{cases}$$

​

Ok.
Did you try a trig substitution?

 

[B4cklfip]

Not yet. I´ m also not sure which one to use.
I´ ve now written the equation as

$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{V_0(a^2-x^2)-E} dx $$

but have now idea how to go further.

 

[hutchphd]

Is there a restriction on E ?

 

[B4cklfip]

E is smaller than the potential V(x).
I´ m at the moment trying to solve the integral by substituting $$x = sin(\Theta)$$

 

[haruspex]

E is smaller than the potential V(x).
I´ m at the moment trying to solve the integral by substituting $$x = sin(\Theta)$$

You need a substitution that gets rid of the factor a. But the resulting integral looks no better.

Isn't there a flaw here? Near the bounds of the integral the term inside the square root will go negative.

 

[B4cklfip]

No, everything should be correct.
After substituting x=sin(Theta) one gets

$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{y-V_0*sin^2(\Theta)}* cos(\theta)dx $$
that can be substituted by $$u = sin(\Theta)$$
which gets one
$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{y-V_0*u^2}du $$
finally one can substitute $$ u = \frac{sin(z)}{\sqrt{V_0}}*\sqrt{y} $$
which will lead into following integral:

$$ p(x) = \frac{1}{a} \frac{y}{\sqrt{V_0}} \int_{-a}^{a} cos(z)^2 dz $$

which is solveable, with y defined as $$ y = V_0a^2-E $$

Maybe someone can check if I made everything correct? :)

 

[haruspex]

After substituting x=sin(Theta) one gets
$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{y-V_0*sin^2(\Theta)}* cos(\theta)dx $$

With that substitution I get
$$p(x)=\frac{1}{\hbar} \int_{x=-a}^{a}\sqrt{2m(V_0(1-(\frac{\sin(\theta)}{a})^2)-E)}\cos(\theta)d\theta$$
Presumably you meant x=a sin(Theta), giving
$$p(x)=\frac{a}{\hbar} \int_{\theta=-\pi/2}^{+\pi/2}\sqrt{2m(V_0(1-\sin^2(\theta))-E)}\cos(\theta)d\theta$$
$$=\frac{a}{\hbar} \int_{\theta=-\pi/2}^{+\pi/2}\sqrt{2mV_0(\cos^2(\theta)-E)}\cos(\theta)d\theta$$
You still have the problem that at the limits of the integration the expression in the square root goes negative if E>0.

 

[hutchphd]

This is very confused...how can the definite integral still be a function of x?? Also I believe the limits need to somehow be the classical turning points.

 

[haruspex]

the limits need to somehow be the classical turning points.

Quite so.

 

[vela]

E is smaller than the potential V(x).

I think you mean $E < V_0$.

I agree with the others that you need to rethink the limits of the integral. There's no point in evaluating the wrong integral.