Penetration depth and step potential

In summary, the equation for the mean position is:$$\langle x \rangle=\frac{\int_{0}^{\infty} \psi^*~x~\psi~dx}{\int_0^\infty\psi^*\psi~dx}$$
  • #1
razidan
75
1

Homework Statement


I am trying to figure out the average penetration depth into a finite step potential, similar to this thread:
https://www.physicsforums.com/threads/qm-barrier-penetration.104641/
But instead of just estimating the depth as ##\frac{1}{\kappa}##, i would like to calculate it explicitly.

Homework Equations


##<x>=\int \psi^* x \psi##

The Attempt at a Solution


I took the solution to the step potential problem - ##\psi=t \cdot e^{\alpha x} \text{ where } \alpha=\sqrt{2m(V-E)}/\hbar^2## .
when I evaluated the integral i have: ##<x>=\frac{k^2}{\alpha^2} \frac{1}{k^2 + \alpha^2}## where ##k=\sqrt{2m(E)/\hbar^2}## .

this has unit of meter squared, and this is because the wavefunction is not normalized, which leads me to my question: Can i normalize my wavefunction only on [0,##+\infty##]? (on the negative half of x, it cannot be normalized.

If i do so, I get ##|c|^2=2\alpha##.
 
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  • #2
As explained in the thread you quoted, there is one scaling parameter and that is ##1/\alpha##. Any definition of penetration depth will be some constant times ##1\alpha##. For example, if by penetration depth you mean the distance over which the probability density drops to ##1/e## of its value at the boundary, then the penetration depth would be ##1/(2\alpha)##. Note that you don't need to normalize the wavefunction to get this answer.
 
  • #3
kuruman said:
As explained in the thread you quoted, there is one scaling parameter and that is ##1/\alpha##. Any definition of penetration depth will be some constant times ##1\alpha##. For example, if by penetration depth you mean the distance over which the probability density drops to ##1/e## of its value at the boundary, then the penetration depth would be ##1/(2\alpha)##. Note that you don't need to normalize the wavefunction to get this answer.
Thanks, but if I'm asked to find the mean position, isn't the rigorous way is to calculate <x>? And that's what I was trying to do.
 
  • #4
razidan said:
Thanks, but if I'm asked to find the mean position, isn't the rigorous way is to calculate <x>? And that's what I was trying to do.
I am not sure what "mean position" means if part of the wavefunction is reflected at the boundary and part is transmitted.
 
  • #5
kuruman said:
I am not sure what "mean position" means if part of the wavefunction is reflected at the boundary and part is transmitted.
The question was "find the mean distance an electron will travel into a step potential"
 
  • #6
razidan said:

Homework Statement


... when I evaluated the integral i have: ##<x>=\frac{k^2}{\alpha^2} \frac{1}{k^2 + \alpha^2}## where ##k=\sqrt{2m(E)/\hbar^2}## .
What is the integral you evaluated?
 
  • #7
kuruman said:
What is the integral you evaluated?
##<x>\int_0^\infty x|t|^2 e^{-2\alpha }##
 
  • #8
razidan said:
##<x>\int_0^\infty x|t|^2 e^{-2\alpha }##
And when I did normalize the wavefunction with ##|c|^2## and I plugged in the values of ##\alpha=k## I got ##\frac{1}{\alpha}##
 
  • #9
What is ##t## and why is it part of the wavefunction? For the step potential, the solution in the region ##V > E## (assumed at ##x>0##) is ##\psi(x)=Ce^{-\alpha x}## where ##C## is some normalization constant. Are ##t## and ##C## the same?
razidan said:
And when I did normalize the wavefunction with ##|c|^2## and I plugged in the values of ##\alpha=k## I got ##\frac{1}{\alpha}##
Why is ##\alpha## equal to ##k##? They have different definitions in post #1.
 
  • #10
kuruman said:
What is ##t## and why is it part of the wavefunction? For the step potential, the solution in the region ##V > E## (assumed at ##x>0##) is ##\psi(x)=Ce^{-\alpha x}## where ##C## is some normalization constant. Are ##t## and ##C## the same?

Why is ##\alpha## equal to ##k##? They have different definitions in post #1.
Sorry , I wasn't very specific.
t is the transmission coefficient, obtained from the full solution to the step potential problem.
α=k is just a special caae (the problem has numerical values).
 
  • #11
razidan said:
Sorry , I wasn't very specific.
t is the transmission coefficient, obtained from the full solution to the step potential problem.
α=k is just a special caae (the problem has numerical values).
For future reference, it would be easier for someone to help you if you posted the problem statement exactly as was given to you. Revealing it in successive postings is not very efficient. Yes, I believe you can find the mean distance using $$\langle x \rangle=\frac{\int_{0}^{\infty} \psi^*~x~\psi~dx}{\int_0^\infty\psi^*\psi~dx}$$
which is always the case whether the wavefunction is normalized or not. Obviously ##t## will drop out of the calculation.
 
  • #12
kuruman said:
For future reference, it would be easier for someone to help you if you posted the problem statement exactly as was given to you. Revealing it in successive postings is not very efficient. Yes, I believe you can find the mean distance using $$\langle x \rangle=\frac{\int_{0}^{\infty} \psi^*~x~\psi~dx}{\int_0^\infty\psi^*\psi~dx}$$
which is always the case whether the wavefunction is normalized or not. Obviously ##t## will drop out of the calculation.
I've never seen the formula written this way... Thanks, this was helpful.
 
  • #13
razidan said:
I've never seen the formula written this way... Thanks, this was helpful.
You must have seen it before but without realizing what you saw. The standard way to write an average of observable ##Q## using a distribution (or weighting) function ##\rho(x)## over variable ##x## is
$$\langle Q \rangle=\frac{\int\rho(x)Q~dx}{\int\rho(x)~dx}$$
In quantum mechanics the probability density ##\psi^*\psi## takes the place of the distribution function ##\rho(x)## and you have the expectation value (which is really an average) of ##Q##.
In classical mechanics, one application is to identify the distribution function ##\rho(x)## as the mass density in which case the x-coordinate averaged over mass is
$$\langle x \rangle=\frac{\int\rho(x)x~dV}{\int\rho(x)~dV}=\frac{\int x~dm}{\int dm}=X_{CM}$$
Another application is to do a similar mass-average of the square (##r^2##) of the distance from an axis in which case you get the moment of inertia.
In a calorimetry experiment in which you put in thermal contact ##N## masses where the heat capacity of the ith mass is ##C_i## an its initial temperatures is ##T_i##, ##i =1,2, ...N##, the weighting function is the distribution of heat capacities and the final temperature when equilibrium is reached is (what else?) the average
$$T_{final}=\frac{\sum C_i T_i } { \sum C_i}.$$
Gauss's law in integral form can be divided by the area ##S =\int dA## of the Gaussian surface to give
$$\frac{\int\vec E \cdot \hat n ~dA}{\int dA}=\frac{q_{encl}}{\epsilon_0 S}.$$ It becomes clear that the normal component of the electric field averaged over the Gaussian surface does not change when you deform the shape of the surface while keeping its surface area ##S## and the enclosed charge ##q_{encl}## the same.
 
  • #14
kuruman said:
You must have seen it before but without realizing what you saw. The standard way to write an average of observable ##Q## using a distribution (or weighting) function ##\rho(x)## over variable ##x## is
$$\langle Q \rangle=\frac{\int\rho(x)Q~dx}{\int\rho(x)~dx}$$
In quantum mechanics the probability density ##\psi^*\psi## takes the place of the distribution function ##\rho(x)## and you have the expectation value (which is really an average) of ##Q##.
In classical mechanics, one application is to identify the distribution function ##\rho(x)## as the mass density in which case the x-coordinate averaged over mass is
$$\langle x \rangle=\frac{\int\rho(x)x~dV}{\int\rho(x)~dV}=\frac{\int x~dm}{\int dm}=X_{CM}$$
Another application is to do a similar mass-average of the square (##r^2##) of the distance from an axis in which case you get the moment of inertia.
In a calorimetry experiment in which you put in thermal contact ##N## masses where the heat capacity of the ith mass is ##C_i## an its initial temperatures is ##T_i##, ##i =1,2, ...N##, the weighting function is the distribution of heat capacities and the final temperature when equilibrium is reached is (what else?) the average
$$T_{final}=\frac{\sum C_i T_i } { \sum C_i}.$$
Gauss's law in integral form can be divided by the area ##S =\int dA## of the Gaussian surface to give
$$\frac{\int\vec E \cdot \hat n ~dA}{\int dA}=\frac{q_{encl}}{\epsilon_0 S}.$$ It becomes clear that the normal component of the electric field averaged over the Gaussian surface does not change when you deform the shape of the surface while keeping its surface area ##S## and the enclosed charge ##q_{encl}## the same.
Thanks!
 

What is penetration depth?

Penetration depth refers to the distance that an electric current can travel through a material before it is significantly reduced in strength. It is typically measured in meters or centimeters.

How is penetration depth calculated?

The penetration depth of an electric current can be calculated using the material's resistivity and conductivity, as well as the frequency and strength of the current. It can also be affected by factors such as temperature and impurities in the material.

Why is penetration depth important in electrical safety?

Penetration depth is important in electrical safety because it determines the risk of electric shock for individuals working with or near electrical equipment. A deeper penetration depth means a higher risk of electric shock, while a shallower penetration depth means a lower risk.

What is step potential?

Step potential is the voltage difference that exists between a person's feet when standing near a source of electricity. It is caused by the difference in voltage between the ground at the person's feet and the ground at the source of electricity.

How does penetration depth affect step potential?

The deeper the penetration depth, the larger the step potential. This means that individuals working with or near high voltage equipment with a deeper penetration depth are at a higher risk of experiencing a large step potential and potential electric shock.

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