Schrodinger equation solution when U>E

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SUMMARY

The discussion focuses on solving the time-independent Schrödinger equation (TISE) under the condition where potential energy (U) exceeds the total energy (E). The solution is represented as \(\Psi(x) = Be^{-\alpha x}\), with \(\alpha = \sqrt{\frac{2m[U(x)-E]}{\hbar^2}}\), indicating only transmission and no reflection as \(x\) approaches infinity. The participant also explores the scenario where \(U < E\), leading to a wave function of the form \(\Psi(x) = Ae^{i\beta x} + Be^{-i\beta x}\), with \(\beta = \sqrt{\frac{2m[E-U(x)]}{\hbar^2}}\). The discussion emphasizes the importance of understanding the implications of complex solutions in quantum mechanics.

PREREQUISITES
  • Understanding of the time-independent Schrödinger equation (TISE)
  • Familiarity with quantum mechanics concepts such as potential energy and total energy
  • Knowledge of second-order ordinary differential equations (ODEs)
  • Basic grasp of complex numbers and their physical interpretations in quantum mechanics
NEXT STEPS
  • Study the derivation and implications of the time-independent Schrödinger equation (TISE)
  • Learn about boundary conditions and their effects on wave functions in quantum mechanics
  • Explore the concept of complex wave functions and their physical significance
  • Review Griffiths' "Introduction to Quantum Mechanics" for a deeper understanding of quantum solutions
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Students of quantum mechanics, physicists working on wave functions, and anyone interested in the mathematical foundations of quantum theory.

td21
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Homework Statement



see attachment

Homework Equations


TISE

The Attempt at a Solution


a)When U>E
[tex]-\frac{\hbar^2}{2m}\frac{d^2\Psi(x,t)}{d x^2}+U(x)\Psi(x)=E\Psi(x)[/tex]
leads to blahblahblah...(there is only transmittion no reflection as x to infty)
[tex]\Psi(x)=Be^{-\alpha x}[/tex], where [tex]\alpha = \sqrt\frac{2m[U(x)-E]}{\hbar^2}[/tex]This is the solution i work out, but how to "demostrate" it is a solution?

b)
Actually i don't know 2nd order ODE, so i try here:
when U<E, there will be reflection and incident part:
so [tex]\Psi(x)=Ae^{+i\beta x}+Be^{-i\beta x}[/tex],where[tex]\beta =\sqrt\frac{2m[E-U(x)]}{\hbar^2}[/tex]or[tex]\beta =\sqrt\frac{2m[U(x)-E]}{\hbar^2}[/tex]?
but in any case, the exponential is "complex", yielding the Cosine function,(but where does sine go?)?So U<E there?

c)from part b, [tex]\beta=10^{9}[/tex]
but i don't know whether [tex]\alpha =\beta[/tex] ??
but i just assume so and i calculate: U-E=5.56^-21 J
I would appreciate it if you could help me out esp. the question mark part. Thx!
 

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Well put it in the equation and see it it will give you something useful (usually if you put the solution in an equation, the condition set on it will be satisfied...).

See the i in the exponentials. If you rewrite them you'll get something real and something imaginary, right? And what does an imaginary solution means in the real world? :D

See what the conditions must be satisfied when solving Schrödinger. And then try to see how they affect your solutions :)

I'm sure that exist in Griffiths or similar books...
 

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