Delta-function potentials, scattering states

Click For Summary
SUMMARY

The discussion centers on delta-function potentials and their implications for scattering states in quantum mechanics, specifically referencing Griffiths' "Introduction to Quantum Mechanics" (2nd edition). It clarifies that for bound states where energy E is less than zero, the wave function exhibits exponential behavior with terms blowing up or decaying. In contrast, for scattering states where E is greater than zero, the wave function transitions to an oscillatory form, specifically \(\psi(x) = Ae^{-ikx} + Be^{ikx}\), which does not exhibit exponential growth or decay as x approaches negative infinity.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wave functions.
  • Familiarity with delta-function potentials and their role in quantum systems.
  • Knowledge of bound and scattering states in quantum mechanics.
  • Basic grasp of complex numbers and oscillatory functions.
NEXT STEPS
  • Study the mathematical derivation of delta-function potentials in quantum mechanics.
  • Explore the implications of bound states versus scattering states in quantum systems.
  • Learn about the role of complex wave numbers in quantum mechanics.
  • Investigate the physical interpretations of oscillatory wave functions in scattering scenarios.
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, as well as educators seeking to clarify concepts related to delta-function potentials and scattering states.

AntiStrange
Messages
18
Reaction score
1
I'm reading through the Introduction to Quantum Mechanics book by Griffiths (2nd edition)
and it is describing delta-function potential wells.
When it describes how to find bound states the energy is E < 0 (negative) in the region x < 0 (negative).

It says the general solution is:
\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}
It explains that as x goes to -∞ the first term blows up. Which makes sense because it will go to infinity, while the second term only goes to zero.

However, they then look at the situation when E > 0, x < 0
again the general solution is:
\psi (x) = Ae^{-\kappa x} + Be^{\kappa x}

but this time they say that neither term blows up. I don't understand why that is. Shouldn't the first term still go to infinity as x goes to -∞ ?
 
Physics news on Phys.org
No, for E > 0, x < 0 the general solution is
<br /> \psi(x) = Ae^{-ikx} + Be^{ikx}<br />
which oscillates but does not exponentially grow or decay (for real k).
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
983
  • · Replies 20 ·
Replies
20
Views
2K