Normalizing Free Particle Homework: \psi(x,t)

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SUMMARY

The discussion focuses on normalizing the wave function \(\psi(x,0) = A e^{-ax^2}\) and deriving \(\psi(x,t)\). The normalization condition is established as \(1 = A^2 \int_{-\infty}^\infty e^{-2ax^2} dx\), leading to the solution \(A = \left(\frac{2a}{\pi}\right)^{1/4}\). The resulting time-dependent wave function is \(\psi(x,t) = \left(\frac{2a}{\pi}\right)^{1/4} e^{-2ax^2 - i\frac{\hbar k}{2m}t}\). The participant expresses concern that the solution appears overly simplistic and seeks clarification on the transition from \(\psi(x,0)\) to \(\psi(x,t)\).

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  • Understanding of quantum mechanics wave functions
  • Familiarity with normalization integrals
  • Knowledge of complex exponentials in quantum mechanics
  • Basic proficiency in calculus, particularly Gaussian integrals
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  • Study the derivation of time-dependent wave functions in quantum mechanics
  • Learn about the role of the Fourier transform in quantum mechanics
  • Explore the implications of the Schrödinger equation on wave function evolution
  • Investigate the significance of the parameters \(a\), \(\hbar\), and \(m\) in wave function behavior
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E92M3
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Homework Statement


[tex]\psi(x,0)=Ae^{-ax^2}[/tex]
Normalize and find:
[tex]\psi(x,t)[/tex]

Homework Equations


[tex]1=\int_{-\infty}^\infty\psi^*\psi dx[/tex]

The Attempt at a Solution


[tex]1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx[/tex]
let:
[tex]u=x\sqrt{2a}[/tex]

[tex]1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}[/tex]
Therefore:
[tex]A=(\frac{2a}{\pi})^{1/4}[/tex]

[tex]\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}[/tex]
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.
 
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E92M3 said:
[tex]A=(\frac{2a}{\pi})^{1/4}[/tex]

[tex]\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}[/tex]
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.

Your value of [itex]A[/itex] is correct; which leaves you with

[tex]\psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}[/tex]

How did you go from that, to your equation for [itex]\psi(x,t)[/itex]?
 

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