Solutions to Time-dependent Schrodinger Equation

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Discussion Overview

The discussion revolves around the solutions to the time-dependent Schrödinger equation (TDSE) in quantum mechanics, particularly focusing on the distinction between discrete and continuous solutions based on the nature of the potential involved. Participants explore why bound states yield discrete energy levels while unbound states result in continuous energy spectra.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant questions why the general solution to the TDSE can be expressed as a continuous sum over wavefunctions rather than a discrete linear combination, as suggested by Griffiths.
  • Another participant explains that bound states, such as those in an infinite square well or harmonic oscillator, have a discrete energy spectrum, while unbound states, like free particles, have a continuous energy spectrum.
  • A participant seeks clarification on the reasons behind the discrete energy levels of bound states compared to the continuous levels of unbound states, indicating their status as a junior in high school.
  • A later reply provides an analogy of standing waves in open air versus waves in a pipe, suggesting that boundary conditions lead to discrete solutions in quantum systems, similar to how they affect wave propagation in classical systems.
  • One participant expresses understanding of the relationship between boundary conditions and the nature of wavefunction solutions after the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the distinction between bound and unbound states regarding energy spectra, but the initial question about the nature of the solutions (discrete vs. continuous) remains partially unresolved, with some participants seeking further clarification.

Contextual Notes

The discussion highlights the dependence on boundary conditions and the implications for the solutions of the Schrödinger equation, but does not resolve the deeper theoretical implications or assumptions underlying these concepts.

Who May Find This Useful

Students and individuals interested in quantum mechanics, particularly those exploring the foundational concepts of wavefunctions and the implications of boundary conditions on energy spectra.

Positron137
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I am reading David Griffiths' book on Quantum Mechanics, and he usually says that the general solution to the TDSE, given a potential V, can a DISCRETE linear combinations of the wavefunction solutions. However, in one section, he says that the linear discrete sum can be regarded as a continuous sum over continuous values of k, the wavenumber. Making a continuous transforms the discrete linear combination into an integral over all continuous values of k from negative to positive infinity, of all possible wavefunctions corresponding to the Hamiltonian provided? Why can't the linear combination be discrete instead of continuous? This is on page 46 of his book. What I don't understand, is why does he write it as an integral, and then proceed to use Fourier's trick, etc..? Why doesn't he keep it as a discrete sum, as is the usual idea with the "discrete linear combination of solutions"? Thanks. Sorry if the beginning is a bit confusing.
 
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Bound states (infinite square well, simple harmonic oscillator, etc.) have a discrete energy or momentum spectrum. The general solution is a linear combination (weighted sum) of eigenstates.

Unbound states (free particle, finite square well with E greater than the "top" of the well, potential barrier, etc.) have a continuous energy or momentum spectrum. The general solution is a Fourier-type integral.
 
Ah ok. By the way, even thought I'm reading the book, I don't completely understand why bound states have discrete energy, and why unbound states have continuous energy (I'm a junior in high school, so it's understandable that I won't comprehend everything :P). Could you explain to me why that is? Thanks.
 
Positron137 said:
Ah ok. By the way, even thought I'm reading the book, I don't completely understand why bound states have discrete energy, and why unbound states have continuous energy (I'm a junior in high school, so it's understandable that I won't comprehend everything :P). Could you explain to me why that is? Thanks.

The best analogy I have found is that of standing waves. Think about a big open piece of air--waves of any frequency at all can propagate through it (continuous solutions). However, if you put the air in a pipe with caps at the ends, resonance phenomena force the allowable spectrum of waves to a specific set of discrete pitches--the harmonic series. To get technical, the differential equation that describes wave propagation gets boundary conditions imposed on it, which force the solutions to take on a discrete set of forms, whereas a system with no boundary conditions can have a continuous infinity of solutions. The same thing happens in the quantum case, the only difference is that the differential equation in question is now the Schrödinger Equation.
 
Ah ok. So because the infinite 1-D potential well has those boundary conditions (also the SHM), the general wavefunction is a discrete linear combination. But for free particles, there is no boundary solution, so its quantum states form a continuous (integral) wavefunction solution. Ok thanks! I understand it now :)
 

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