Quantum mechanics - several constant potentials

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Homework Help Overview

The discussion revolves around a quantum mechanics problem involving constant potentials and the continuity of wave functions. Participants are analyzing equations related to wave functions and probability density, specifically focusing on boundary conditions and normalization integrals.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the continuity of the wave function and its derivative, raising questions about the complexity of the equations involved. Some express uncertainty about the approach and suggest that fewer equations might suffice. Others inquire about the calculation of a specific value and the absence of units in the original problem.

Discussion Status

There is an ongoing exploration of the problem, with participants providing hints and questioning the assumptions made regarding the equations. Some guidance has been offered regarding the sufficiency of the equations needed for the solution, but no consensus has been reached on the final answer or methodology.

Contextual Notes

Participants note the lack of units provided in the original question, which raises concerns about the clarity and completeness of the problem setup. There is also discussion about the appropriateness of using certain units in the context of the calculations.

Eitan Levy
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Homework Statement
A particle with mass m is in a one dimensional potential, as seen below.

The wave function in [itex]x<0[/itex] is 0.

The wave function in [itex]0<x<b[/itex] is: [itex]Asin(kx)[/itex]

The wave function in [itex]x>b[/itex] is: [itex]Be^{-\alpha x}[/itex]

It is known that [itex]k=3*10^{10}[/itex] and [itex]b=0.5333333*10^{-10}[/itex]

Find [itex]\alpha[/itex]
Relevant Equations
Schrodinger stationary equation
Capture.PNG


What I tried to do was using the fact that the wave function should be continuous.

Asin(kb)=Be^{-\alpha b}

The derivative also should be continuous:

kAcos(kb)=-\alpha Be^{-\alpha b}

And the probability to find the particle in total should be 1:

\int_0^b A^2sin^2(kx) dx + \int_b^{\infty} B^2e^{-2\alpha x} dx =1

This set of equations is to hard to deal with, the equations should be solved with calculator only so I think I did something wrong.

Also, there may be a better way to approach this problem, but I'm not seeing it.
 
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Eitan Levy said:
Asin(kb)=Be^{-\alpha b}

kAcos(kb)=-\alpha Be^{-\alpha b}

\int_0^b A^2sin^2(kx) dx + \int_b^{\infty} B^2e^{-2\alpha x} dx =1
Hints: It's much easier than you think. You only need the first 2 equations.
 
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Steve4Physics said:
Hints: It's much easier than you think. You only need the first 2 equations.
Is the answer 0.08763*10^(10)?
 
Eitan Levy said:
Is the answer 0.08763*10^(10)?
I haven't done the calculation. How did you get that value?
Also, have you forgotten units or were there really none supplied in the original question?
 
Steve4Physics said:
I haven't done the calculation. How did you get that value?
Also, have you forgotten units or were there really none supplied in the original question?
I divided the first equation by the second:

\frac{1}{k}tan(kb)=-\frac{1}{\alpha}

And I plugged in the numbers.

No units were provided.
 
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Eitan Levy said:
I divided the first equation by the second:

\frac{1}{k}tan(kb)=-\frac{1}{\alpha}

And I plugged in the numbers.

No units were provided.
Agreed. Well done! A couple of points:
1) You haven't correctly used standard form (mantissa in wrong range).
2) You have rounded the last digit incorrectly when giving the answer to 4 significant figures.

If working in SI units, what do you think the unit would be? (If I were handing this in, I would make a comment about the appropriate SI units below my answer.)
 
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Steve4Physics said:
Agreed. Well done! A couple of points:
1) You haven't correctly used standard form (mantissa in wrong range).
2) You have rounded the last digit incorrectly when giving the answer to 4 significant figures.

If working in SI units, what do you think the unit would be? (If I were handing this in, I would make a comment about the appropriate SI units below my answer.)
(Angstram)^{-1}
 
Eitan Levy said:
(Angstram)^{-1}
The ångström (Å) (note spelling and capitalisation) is a bit old-fashioned, but is still used in some areas. It is not an SI unit.

1 Å = 10⁻¹⁰ m
That means
1 Å⁻¹ =10¹⁰ m⁻¹

I would have given the answer as:
α = 8.76x10⁸ (m⁻¹ assuming we are working in SI units)

EDIT: It's very poor practice to give values without units. The question should have given the units for k and b.
 

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