Solving QM Griffiths 2.28: Transmission Coefficient

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In summary, the conversation was about finding the transmission coefficient for a wave packet passing through a potential function with dirac delta wells. The process involved splitting the region into three intervals and applying boundary conditions to reduce the unknowns. The final solution was found to be the square of (E/A), where beta is equal to (m alpha)/(hbar k). The speaker was hoping for validation of their answer, but ultimately concluded that their method was correct.
  • #1
genxhis
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The problem asks for the transmission coefficient for a wave packet with energy approximately E passing through a potential function with dirac delta wells of strength alpha at x = -a and x = +a. To solve the problem I split the region into the three obvious intervals [-inf, -a], [-a, a], and [a, +inf]. For the first two regions I expressed the solutions as (A or C) exp(i k x) + (B or D) exp(-i k x) and for the last as E exp(i k x) where k = sqrt(2 m E)/hbar. I then applied the two contraints at the two boundary conditions to reduce the five unknowns to just one. Finally I did some more algebra to find the trasmission coefficient as the square of (E/A). But the entire process was lengthy and tedious. I was wondering if someone could validate this answer:

[tex] T = \frac{1}{1 + 2 \beta^2[ (1+\beta^2) + (1-\beta^2)\cos 4ka -2\beta \sin4ka ]} [/tex]​


where [tex] \beta = (m \alpha)/(\hbar k) [/tex].
 
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  • #2
There's no better checking than the one you can do it yourself by making sure your method & calculations were correct.

I think no one around here will do the calculations at this problem,just to agree or disagree with your answer.

Daniel.
 
  • #3
your right, sorry. i was hoping someone could recognize this as a special case of a more general problem or simply point out that the answer is unviable. but, I've looked it over more carefully, and i think it is. in any case, recapitulating what i did helps me understand it a little better.
 

FAQ: Solving QM Griffiths 2.28: Transmission Coefficient

1. What is the purpose of solving QM Griffiths 2.28?

The purpose of solving QM Griffiths 2.28 is to calculate the transmission coefficient for a one-dimensional potential barrier in quantum mechanics. This coefficient represents the probability of a particle passing through the barrier, and it is an important factor in understanding the behavior of quantum systems.

2. How do I solve QM Griffiths 2.28?

To solve QM Griffiths 2.28, you will need to use the principles of quantum mechanics, specifically the Schrödinger equation and the wave function. You will also need to apply boundary conditions and use mathematical techniques such as integration and solving differential equations. It is recommended to follow a step-by-step approach and consult resources such as textbooks or online tutorials for guidance.

3. What is the significance of the transmission coefficient in quantum mechanics?

The transmission coefficient is significant in quantum mechanics because it helps us understand the behavior of particles at the quantum level. It tells us the probability of a particle passing through a potential barrier, and this information is crucial in various applications such as studying the transmission of electrons in semiconductors or designing quantum devices.

4. Can I use QM Griffiths 2.28 to solve other similar problems?

Yes, you can use the same principles and techniques used in solving QM Griffiths 2.28 to solve other similar problems involving potential barriers in quantum mechanics. However, it is important to note that the specific values and parameters may vary, so you will need to adapt your approach accordingly.

5. Are there any limitations to using QM Griffiths 2.28 to solve transmission coefficient problems?

While QM Griffiths 2.28 provides a useful framework for solving transmission coefficient problems, it does have some limitations. It assumes a one-dimensional potential barrier and does not take into account other factors such as the spin of particles. Additionally, it may not be applicable in more complex systems, so it is important to consider the specific context and limitations of the problem at hand.

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