Transmission over a finite square well

In summary, the conversation discusses the solution for 'T', the transmission coefficient of a wave function, in terms of the reflection coefficient 'R'. The textbook offers two equations for T in terms of R and the conversation explores the gap between the two equations and how to make the derivation match the textbook formula. There is a discrepancy in one of the equations that may be causing the issue.
  • #1
T-7
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Hi,

My Quantum textbook loves to skip the algebra in its derivations. It claims that the solution for 'T', the transmission coefficient of the wave function (E>0, ie. unbound) is

[tex]T = \frac{2qke^{-2ika}}{2qkcos(2qa)-i(k^{2}+q^{2})sin(2qa)}[/tex]

Its prior step is to offer two equations (which match my own derivation) for T in terms of R (the reflection coefficient):

[tex]
e^{-ika}+R^{ika} = \frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}+(1-\frac{k}{q})e^{ika+2iqa})
[/tex]

[tex]
e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})
[/tex]

My own attempt to fill in the gap between the two proceeds as follows, but I didn't quite arrive at that final expression...

[tex]
2e^{-ika} = T/2.e^{ika}[(1+k/q).e^{-2iqa} + (1-k/q).e^{2iqa} + (q/k-1).e^{-2iqa} - (q/k-1).e^{2iqa}]
[/tex]

[tex]
4e^{-2ika} = T.[(1+k/q+q/k-1).e^{-2iqa} + (1-k/q-q/k+1).e^{2iqa}]
[/tex]

[tex]
4qke^{-2ika} = T.[(k^{2}+q^{2}).e^{-2iqa} + (2kq-k^{2}-q^{2}).e^{2iqa}]
[/tex]

[tex]
4qke^{-2ika} = T.[(k^{2}+q^{2}).(e^{-2iqa}- + e^{2iqa}) + 2kq.e^{2iqa}]
[/tex]

[tex]
4qke^{-2ika} = T.[(k^{2}+q^{2}).(-2isin(2qa)) + 2kq.(cos 2qa + isin sqa)]
[/tex]

[tex]
T = \frac{4qke^{-2ika}}{2kq.(cos 2qa + isin 2qa)-2i(k^{2}+q^{2})sin(2qa)}
[/tex][tex]
T = \frac{2qke^{-2ika}}{kq.(cos 2qa + isin 2qa)-i(k^{2}+q^{2})sin(2qa)}
[/tex]

(Hopefully I've copied all that down correctly!).

Can anyone spot a false move, or what it is I need to do to make my derivation match the textbook formula.

Cheers!
 
Last edited:
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  • #2
It seems like your answer would match theirs if:

[tex]e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})[/tex]

was changed to:

[tex]e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})[/tex]

which also matches the other line better. Are you sure you derived/copied this down right?
 

1. What is "Transmission over a finite square well"?

"Transmission over a finite square well" refers to the transfer of a particle (such as an electron) through a potential barrier that is in the shape of a square well with finite boundaries.

2. How is the transmission probability calculated for a finite square well?

The transmission probability for a finite square well is calculated using the Schrödinger equation, which describes the behavior of quantum particles. The solution to this equation yields the transmission coefficient, which is used to calculate the transmission probability.

3. What factors affect the transmission over a finite square well?

The transmission over a finite square well is affected by several factors, including the height and width of the potential barrier, the energy of the particle, and the shape of the barrier. Additionally, the mass and charge of the particle can also impact the transmission probability.

4. What are some practical applications of "Transmission over a finite square well"?

"Transmission over a finite square well" has various applications in the field of nanotechnology, such as in the design of quantum dots and nanowires. It is also used in the development of electronic devices, such as transistors and diodes.

5. How does the transmission over a finite square well differ from transmission through an infinite potential barrier?

The main difference between transmission over a finite square well and transmission through an infinite potential barrier is that in the former, there is a finite probability for the particle to pass through the barrier, while in the latter, the particle must have enough energy to overcome the infinite barrier. Additionally, the transmission probability for a finite square well is dependent on the shape and size of the well, while for an infinite barrier, it is solely dependent on the energy of the particle.

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