Stationary States vs General Solution

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SUMMARY

The discussion centers on the interpretation of wave functions in quantum mechanics, specifically within the context of a simple infinite square well potential. The solutions to the Schrödinger equation are given by \(\Psi_n(x) = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})\), with time dependence introduced as \(\Psi_n(x,t) = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a}) e^{\frac{-iE_n t}{\hbar}}\). The actual wave function of a particle, represented as \(\Psi(x,t) = \sum c_n \Psi_n(x,t)\), describes a superposition of energy eigenstates, allowing for probabilistic outcomes when measuring energy. This indicates that the wave function does not yield a single quantized value but rather a distribution of possible states based on coefficients \(c_n\).

PREREQUISITES
  • Understanding of the Schrödinger equation and its solutions
  • Familiarity with quantum mechanics concepts such as wave functions and energy eigenstates
  • Knowledge of superposition principle in quantum systems
  • Basic grasp of probability theory as it applies to quantum measurements
NEXT STEPS
  • Study the implications of superposition in quantum mechanics
  • Learn about the time-independent Schrödinger equation and its applications
  • Explore the concept of probability amplitudes in quantum states
  • Investigate the role of coefficients \(c_n\) in determining measurement outcomes
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, wave function analysis, and energy state measurements will benefit from this discussion.

jaydnul
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Tell me if the following is correct. For a simple infinite square well potential, the solutions to the Schrödinger equation are [itex]\Psi_n(x)=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})[/itex], then you plug in the appropriate value for n and operate on the function accordingly to get your observables.

Then if you want to tack on the time dependence you can write it as [itex]\Psi_n(x,t)=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})e^{\frac{-iE_n t}{\hbar}}[/itex], but in this case we know it doesn't depend on time because the time variable will always cancel when calculating an observable. Fine.

What is confusing me is that [itex]\Psi(x,t)= \sum c_n\Psi_n(x,t)[/itex] is said to be the actual wave function of the particle. Wouldn't solving this summation result in one equation, and therefore one wave function that doesn't have distinct quantized values? In other words it doesn't have an n variable to input your state. Am I supposed to operate in this wave function to find my observables?

Thanks
 
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Jd0g33 said:
What is confusing me is that Ψ(x,t)=∑cnΨn(x,t)\Psi(x,t)= \sum c_n\Psi_n(x,t) is said to be the actual wave function of the particle. Wouldn't solving this summation result in one equation, and therefore one wave function that doesn't have distinct quantized values? In other words it doesn't have an n variable to input your state. Am I supposed to operate in this wave function to find my observables?

That general wave function describes a particle that is not in an energy eigenstate, but rather is in a superposition of energy eigenstates. If you measure its energy, you will get the result ##E_n## with probability ##|c_n^2|##.

Any arbitrary state, which may or may not be an eigenfunction of some other observable, can be written as a sum of the energy eigenstates found by solving the time-independent Schrödinger equation.
 

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