Quantum mechanics - several constant potentials

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SUMMARY

This discussion focuses on solving quantum mechanics problems involving constant potentials, specifically using the wave function's continuity and its derivatives. The equations presented include Asin(kb) = Be^{-\alpha b} and kAcos(kb) = -\alpha Be^{-\alpha b}, which are essential for determining the probability density. Participants emphasize the importance of including units in calculations and correcting standard form representation. The final answer discussed is α = 8.76 x 10⁸ m⁻¹, assuming SI units.

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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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