WKB method transmission coefficient

In summary, the task is to calculate the transmission coefficient using the WKB Approximation for a given potential. The potential is defined as V(x) = V_0(1-(x/a)²) for |x|<a and V(x) = 0 otherwise. The homework equation given is ln|T|² = -2 ∫ p(x) dx, where p(x) represents the integral √(2m(V_0(1-(x/a)²)-E)) from -a to a. However, there is a problem with solving this integral and a trig substitution is suggested. After multiple substitutions, the integral is transformed into a solvable form, but there are concerns about its correctness due to potential
  • #1
B4cklfip
18
0
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.
 
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  • #2
B4cklfip said:
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?
 
  • #3
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 $$ V(x) = \begin{cases} V_0(1-(\frac{x}{a})^2) & \text{for -a $\leq x \leq $a} \\0 & \text{otherwise} \end{cases}$$
 
  • #4
B4cklfip said:
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?
 
  • #5
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.
 
  • #6
Is there a restriction on E ?
 
  • #7
E is smaller than the potential V(x).
I´ m at the moment trying to solve the integral by substituting $$x = sin(\Theta)$$
 
  • #8
B4cklfip said:
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.
 
  • #9
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? :)
 
  • #10
B4cklfip said:
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.
 
  • #11
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.
 
  • #12
hutchphd said:
the limits need to somehow be the classical turning points.
Quite so.
 
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  • #13
B4cklfip said:
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.
 

Related to WKB method transmission coefficient

What is the WKB method transmission coefficient?

The WKB method transmission coefficient is a mathematical tool used in quantum mechanics to calculate the probability of a particle passing through a potential barrier. It takes into account the wave-like nature of particles and allows for a more accurate prediction of transmission probabilities.

How is the WKB method transmission coefficient calculated?

The WKB method transmission coefficient is calculated by solving the Schrödinger equation for a given potential barrier and then using the resulting wave function to calculate the transmission probability. This involves using the WKB approximation, which is based on the idea that the wave function can be approximated by a series of classical trajectories.

What is the significance of the WKB method transmission coefficient in quantum mechanics?

The WKB method transmission coefficient is significant because it allows us to understand the behavior of particles in the presence of potential barriers. It is used in various applications, such as tunneling phenomena, and helps us to better understand the wave-particle duality of quantum systems.

What are the limitations of using the WKB method transmission coefficient?

While the WKB method transmission coefficient is a useful tool in quantum mechanics, it has some limitations. One limitation is that it is only accurate for low-energy particles. It also does not take into account the effects of tunneling, which can occur for particles with high enough energies.

How is the WKB method transmission coefficient used in experimental studies?

The WKB method transmission coefficient is used in experimental studies by comparing the predicted transmission probabilities to actual measurements. This allows for the validation of the WKB method and helps to improve its accuracy. It is also used to analyze the behavior of particles in various experimental setups, such as scanning tunneling microscopy.

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