Normalizing Wavefunction: Hard QM Question)

In summary: Why? The infinite square well has an infinite number of bound eigenstates.Regarding normalization: You will always be able to normalize the wave function as long as the norm is finite (which you have already used implicitly to discard the exponentially growing solutions). The issue is instead whether solutions exist for particular energies. To find out you need to use the matching conditions in the way described by @kuruman (also obtainable by dividing one matching condition by the other). To obtain the condition for no bound state solution (there will always be a continuum of scattering states in this problem), you need to find out when the equation thus obtained has no solutions for ##E < V##.
  • #1
drop_out_kid
34
2
Homework Statement
For a well left side is infinite potential and right side is finite potential V0 , and the energy is constant and less than V0(also constant). Now get the wave function , get the equation for eigen energy(the energy allowed) and the conditions for no solution
Relevant Equations
$$-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Phi(x)+V(x)\Phi(x)=0$$
1650421835423.png

So I have come up with my solution(attempt) which is:

1650421198982.png


where
1650421023842.png


(
$$\psi_ 1 \triangleq Asin(kx),0<x<L$$

$$\psi_ 2 \triangleq Be^{-sx},x>L$$

$$k \triangleq \sqrt{\frac{2mE}{\hbar^2}} $$

$$s \triangleq \sqrt{\frac{2m(V-E)}{\hbar^2}} $$)

But this has a serious problem about boundary: I think I should use that
1650421287866.png


But this is what I get:

1650421761245.png


So:
1. How can I get this normalisation in a easy way?
2. Can I use this for part (b) to solve eigen energy?Thought for hours , assignment is going to due (
 

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  • #2
And congratulation to myself, markdown finally worked.
 
  • #3
Some new progress:

I still didn't find way to normalise this.. Perhaps problem doesn't require it idk, and I am analyze this equation:

$$
\alpha \triangleq\sqrt{E} \ \
\gamma \triangleq\sqrt{\frac{2m}{\hbar^2}}L \ \
tan(\gamma\alpha)=-\sqrt{\frac{\alpha^2}{V-\alpha^2}}\ \ (\gamma,V\ \ are\ \ constant)
$$

1650423224325.png
 
  • #4
To normalize the piecewise continuous wavefunction, express ##\psi_1(x)## and ##\psi_2(x)## in terms of a single constant, ##A## or ##B##, and then normalize the usual way. The energy eigenvalues will have to be deternined graphically or numerically.
 
  • #5
kuruman said:
To normalize the piecewise continuous wavefunction, express ##\psi_1(x)## and ##\psi_2(x)## in terms of a single constant, ##A## or ##B##, and then normalize the usual way. The energy eigenvalues will have to be deternined graphically or numerically.
But I can determine the upper bound of V by Tangent function asymptotic line right?
Not sure but I get sth like V<=Pi^2*hbar^2/(8mL)
 
  • #6
drop_out_kid said:
But I can determine the upper bound of V by Tangent function asymptotic line right?
Not sure but I get sth like V<=Pi^2*hbar^2/(8mL)
What upper bound of V? How do you propose to find the eigenvalues? Have you studied how to find them in the case of the particle in a box and in the case of a square well? This is a hybrid that combines the two methods for finding the eigenvalues.
 
  • #7
kuruman said:
What upper bound of V? How do you propose to find the eigenvalues? Have you studied how to find them in the case of the particle in a box and in the case of a square well? This is a hybrid that combines the two methods for finding the eigenvalues.
If V is bigger than this , there's no possible energy.(from the two eqs of B/A)
 
  • #8
drop_out_kid said:
If V is bigger than this , there's no possible energy.(from the two eqs of B/A)
What is "this" that V cannot be bigger than? V is infinite at x = 0 and you already know how to deal with that. It is -U in the region 0 ≤ x ≤ L (the well) and zero for x > L. The two equations give you $$\frac{k\cos(kL)}{-se^{-sL}}=\frac{\sin(kL)}{e^{-sL}}.$$That's what you have to solve numerically for ##E## which is implicit in the definitions for ##k## and ##s##.
 
  • #9
drop_out_kid said:
If V is bigger than this , there's no possible energy.(from the two eqs of B/A)
Why? The infinite square well has an infinite number of bound eigenstates.
 
  • #10
Regarding normalization: You will always be able to normalize the wave function as long as the norm is finite (which you have already used implicitly to discard the exponentially growing solutions). The issue is instead whether solutions exist for particular energies. To find out you need to use the matching conditions in the way described by @kuruman (also obtainable by dividing one matching condition by the other). To obtain the condition for no bound state solution (there will always be a continuum of scattering states in this problem), you need to find out when the equation thus obtained has no solutions for ##E < V##.
 

1. What is the purpose of normalizing a wavefunction?

The purpose of normalizing a wavefunction is to ensure that the total probability of finding a particle in any position is equal to 1. This is a fundamental requirement in quantum mechanics and allows for accurate predictions of the behavior of particles.

2. How is the normalization condition expressed mathematically?

The normalization condition is expressed as the integral of the square of the wavefunction over all space, which must equal 1. Mathematically, this can be written as ∫|Ψ(x)|² dx = 1.

3. What happens if a wavefunction is not normalized?

If a wavefunction is not normalized, it means that the total probability of finding a particle in any position is not equal to 1. This can lead to inaccurate predictions and violates the fundamental principles of quantum mechanics.

4. How do you normalize a wavefunction?

To normalize a wavefunction, you must divide the original wavefunction by the square root of the integral of its absolute value squared. This ensures that the total probability of finding a particle in any position is equal to 1.

5. Can a wavefunction be normalized to a value other than 1?

No, the normalization condition requires that the total probability of finding a particle in any position must equal 1. Normalizing a wavefunction to any other value would violate this fundamental principle of quantum mechanics.

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