Wave function/Infinite square well confusion

In summary, the conversation discusses a particle of mass m in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a. The initial wave function is given by \Psi(x,0)=\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a}). The solution for a more general initial state is given by \sum_{n=1}^\infty |c_n|^2=1, with c_1=1 by normalization. The conversation also touches on the role of boundary conditions and initial values in determining the solution for first-order differential equations.
  • #1
Salviati
14
2

Homework Statement


"A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a.
[tex]\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})[/tex],
E1=[tex]\frac{h^2\pi ^2}{2ma^2}[/tex]*

What is the initial wave function [tex]\Psi(x,0)?[/tex]

*[tex]h[/tex] is supposed to be h bar, I just couldn't find it)

Homework Equations




The Attempt at a Solution


My attempt: If the general solution is a superposition of all stationary states, [tex]\Psi(x,t)=\sum c_n\psi_ne^\frac{-iE_nt}{h}[/tex], at t=0, [tex]\Psi(x,0)=\sum c_n\psi_n[/tex]. Also, at this time, the particle is in the ground state (n=1), so: [tex]\Psi(x,0)=c_1\psi_1[/tex]. Do I assume c1=1 at this point, because the wave function "collapses" once the energy becomes known? I'm just not sure if I understand exactly what happens when the known data is given.

The solution itself is supposed to be [tex]\Psi(x,0)=\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})[/tex].
 
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  • #2
c1=1 by normalization, assuming that [tex]\Psi(x,t)[/tex] is normalized. For a more general initial state,
[tex]\sum_{n=1}^\infty |c_n|^2=1.[/tex]
 
  • #3
Salviati said:
"A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a.
[tex]\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a})[/tex],
E1=[tex]\frac{h^2\pi ^2}{2ma^2}[/tex]*

What is the initial wave function [tex]\Psi(x,0)?[/tex]
You are right to be confused. You cannot know the answer to this question; that is the whole idea behind boundary conditions/initial values: they are INPUTS. For a first-order differential equation (like (d/dt)psi=iHpsi), you need one input BC for each degree of freedom.



Salviati said:
*[tex]h[/tex] is supposed to be h bar, I just couldn't find it)
Try \hbar.
 
  • #4
Thanks guys.

turin said:
Try \hbar.
[tex]\hbar[/tex] :redface:
 

FAQ: Wave function/Infinite square well confusion

What is a wave function?

A wave function is a mathematical representation of a quantum system that describes the probability of finding a particle in a certain position or state. It is used to study the behavior and properties of particles at the subatomic level.

How does the concept of an infinite square well relate to the wave function?

The infinite square well is a hypothetical potential energy barrier that is often used as a model for confined particles in quantum mechanics. The wave function for a particle in an infinite square well is a solution to the Schrödinger equation, which describes the behavior of quantum systems.

What causes confusion when studying the wave function and infinite square well?

One common cause of confusion is the concept of a particle's "position" in quantum mechanics. In classical mechanics, a particle's position is well-defined, but in quantum mechanics, it is described by a probability distribution represented by the wave function. This can be difficult to understand and can lead to confusion.

Can the wave function for a particle in an infinite square well have multiple energy states?

Yes, the wave function for a particle in an infinite square well can have multiple energy states. These states are quantized, meaning that they can only take on certain discrete values. The lowest energy state, known as the ground state, has the most probability of being occupied by the particle.

How does the wave function evolve over time in an infinite square well?

The wave function in an infinite square well does not evolve over time. This is because the potential energy inside the well is constant, so the particle does not gain or lose energy. The wave function remains the same, but the probability of finding the particle in certain positions may change over time due to the wave-like nature of quantum systems.

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