Delta function potential I: Solving for the bound state | Summary and Q&A

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July 31, 2017
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Delta function potential I: Solving for the bound state

TL;DR

The video explains how to calculate the bound state energy for a Schrodinger equation with a delta function potential using integration and the discontinuity condition.

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Questions & Answers

Q: How is the wave function for the system determined?

The wave function is determined to be a continuous function that decays exponentially on both sides of the origin. It is symmetric and can be represented as e to the minus kappa |x|, where kappa is a constant.

Q: What is the significance of the delta function potential intensity appearing in the numerator of the discontinuity equation?

The appearance of the delta function potential intensity in the numerator is a positive sign as it indicates that the bound state becomes deeper and more strongly bound as the potential becomes stronger. If it appeared in the denominator, it would imply an unphysical behavior where the bound state becomes less bound as the potential deepens.

Q: How is the discontinuity condition for the delta function potential derived?

The discontinuity condition is derived by integrating the Schrodinger equation from a small positive value to zero and taking the limit as the interval approaches zero. This results in a discontinuity in the derivative at zero, which is proportional to the value of the wave function at zero.

Q: How is the bound state energy calculated for the system?

The bound state energy is calculated by solving for the value of kappa, which is equal to m alpha over h squared, where m is the mass of the particle and alpha is the intensity of the delta function potential. The energy is then determined to be minus one half times m alpha squared.

Summary & Key Takeaways

  • The wave function for the system is determined to be a continuous function that decays exponentially on both sides of the origin.

  • The delta function potential is analyzed by integrating the Schrodinger equation from a small positive value to zero and taking the limit as the interval approaches zero.

  • By evaluating the discontinuity in the derivative at zero, it is found that the bound state energy is equal to minus one half times the proportional constant.

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