Solving 1-D Potential TISE Homework: Get Solutions & Calculate Energies

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SUMMARY

The discussion focuses on solving the time-independent Schrödinger equation (TISE) for a particle trapped in a one-dimensional potential well. The allowed energies are derived from the transcendental equation k cot(kL) = -κ, where k² = (2mE/ħ²) and κ² = (2m(V₀ - E)/ħ²). Participants clarify the transition from the differential equation to the general solution ψ(x) = A sin(kx) + B cos(kx) and emphasize the importance of understanding this solution in the context of physics.

PREREQUISITES
  • Understanding of the time-independent Schrödinger equation (TISE)
  • Familiarity with quantum mechanics concepts, particularly potential wells
  • Knowledge of differential equations and their solutions
  • Basic grasp of quantum mechanics terminology, including terms like energy (E), potential (V₀), and Planck's constant (ħ)
NEXT STEPS
  • Study the derivation of the transcendental equation k cot(kL) = -κ in quantum mechanics
  • Learn about boundary conditions and their role in quantum mechanics
  • Explore differential equations, focusing on second-order linear differential equations
  • Review resources on quantum mechanics solutions, particularly for potential wells
USEFUL FOR

Students of quantum mechanics, physicists working with potential wells, and anyone seeking to deepen their understanding of the time-independent Schrödinger equation and its applications in quantum systems.

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Homework Statement


A particle of mass m and energy E<V0 is trapped in the 1-D potential well defined by,
V=[tex]\infty[/tex], x<0; V=0, 0[tex]\leq[/tex]x[tex]\leq[/tex]L; V=V0, x>L.

(b) Obtain solutions to the time independent Schrödinger equation for the three regions and by appropriate matching at the boundaries, show that the allowed energies are given by the transcendental equation, [tex]k cot(kL)=-\kappa[/tex], [tex]where[/tex][tex]{k^2}=\frac{2mE}{\hbar^2},{\kappa^2}=\frac{2m({V_0}-E)}{\hbar^2}[/tex].


Homework Equations


TISE


The Attempt at a Solution


For region 2,

[tex]\frac{{\partial^2}\psi}{\partial{x^2}}+\left(\frac{2mE}{\hbar^2}\right)\psi=0[/tex]

[tex]\frac{{\partial^2}\psi}{\partial{x^2}}+{k^2}\psi=0[/tex]

[tex]\psi(x)=Asin(kx)+Bcos(kx)[/tex]

I don't understand how to get from line two to line 3, and where does the potential term disappear to?
 
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8614smith said:
For region 2,

[tex]\frac{{\partial^2}\psi}{\partial{x^2}}+\left(\frac{2mE}{\hbar^2}\right)\psi=0[/tex]

[tex]\frac{{\partial^2}\psi}{\partial{x^2}}+{k^2}\psi=0[/tex]

[tex]\psi(x)=Asin(kx)+Bcos(kx)[/tex]

I don't understand how to get from line two to line 3, and where does the potential term disappear to?
Remember what the value of the potential is in region 2...

As for how to get from line 2 to line 3: first of all, I'll tell you directly that
[tex]\psi(x)=Asin(kx)+Bcos(kx)[/tex]
is the most general possible solution to the differential equation
[tex]\frac{{\partial^2}\psi}{\partial{x^2}}+{k^2}\psi=0[/tex]
This is a very useful thing for a physicist to know, because that equation comes up very frequently in physics. You're not going to want to go through the whole procedure of solving the equation every time you see it; it's awfully convenient to just say "the solution of [that equation] is..."

Since I gather that you're new to this equation, you should convince yourself that the solution is valid. The simple way is to just plug [itex]\psi(x)[/itex] into the equation and simplify it; you should find that everything cancels out, so the equation is satisfied. Showing that there are no other solutions is a little more complicated, and to learn about that I'd suggest taking a class on differential equations, if you have the opportunity. Otherwise, I'm sure there are good resources somewhere out there on the internet - a Google search or something should get you started.
 

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