How can I use determinants to find energy levels in a finite square potential?

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This discussion focuses on using determinants to find energy levels in a finite square potential well in quantum mechanics. The user has derived wave functions for three domains: psi1(x) = Ae^kx, psi2(x) = Ce^jkx + De^-jkx, and psi3(x) = He^-kx, with corresponding wave numbers k. The key method involves calculating the determinant of a system of four equations derived from boundary conditions, where setting the determinant to zero yields the energy levels. Additionally, the discussion highlights the necessity of plotting the determinant's imaginary component to identify zero crossings, which correspond to eigenvalues.

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pepsicola
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Hi, I am in the process of learning QM.
I am looking at this problem regarding to a finite square potential well.
I have derived psi(x) and the k's for the 3 domains,

psi1(x) = Ae^kx => k = sqrt(2m(V-E)/h^2)
psi2(x) = Ce^jkx + De^-jkx => k=sqrt(2mE)/h
psi3(x) = He^-kx => same k as in domain 1

and then what I did was to take the boundary conditions and substitue into the equations and make a system of 4 equations with so that I can solve the unknowns.

What I don't understand is the way I've been told to do this from the above.

I've been told to calculate the determinant of the system of the 4 equations, and by scanning E. When the determinant goes to 0, I will get the energy level for E.
Any one can explain to me why this will give the answer?

And also, it seems like I will need to get the answer by ploting the determinant from its imaginary component. Why?

And when I have the 3 different energy levels that satisfy the conditions, how do I find the unknowns?

Thanks for any help.
 
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The boundary conditions on the wavefunction and its first derivative give you the system of equations you are referring to.

This is a system of equations linear in the unknown coefficients. So the condition you stated is the condition for the existence of a nontrivial solution for this system.

And also, it seems like I will need to get the answer by ploting the determinant from its imaginary component. Why?

Not sure I understand this. The determinant is a complex number. Setting it equivalent to zero is equivalent to setting its real and imaginary parts to zero separately. Perhaps what you want to do is treat the determinant as a function of the parameter E and plot it as a function of E, to determine the zero crossings, which will give you the eigenvalues.
 


And when I get the energy states Es, how do I find out the unknowns?
 


pepsicola said:
And when I get the energy states Es, how do I find out the unknowns?

You must have obtained a transcendental equation..just solve for the eigenvalues numerically.
 

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