Normalizing Wavefunction: Hard QM Question)

AI Thread Summary
The discussion revolves around normalizing a piecewise continuous wavefunction in quantum mechanics, specifically for a potential well scenario. The proposed wavefunctions, ψ1 and ψ2, need to be expressed in terms of a single constant for proper normalization. Participants highlight that energy eigenvalues can be determined graphically or numerically, and emphasize the importance of matching conditions for finding solutions. A critical point raised is that if the potential V exceeds a certain upper bound, no energy solutions exist, indicating the need for careful analysis of the system's parameters. The conversation underscores the complexity of finding eigenvalues in hybrid scenarios combining different quantum mechanics models.
drop_out_kid
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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:

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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
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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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And congratulation to myself, markdown finally worked.
 
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
 
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.
 
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)
 
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.
 
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)
 
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##.
 
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##.
 

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