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Stationary States vs General Solution

  1. Oct 18, 2014 #1
    Tell me if the following is correct. For a simple infinite square well potential, the solutions to the Schrodinger 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
     
  2. jcsd
  3. Oct 18, 2014 #2

    Nugatory

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    Staff: Mentor

    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 Schrodinger equation.
     
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