Probability distribution of stationary Schrodinger equation

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SUMMARY

The discussion centers on calculating the probability distribution of a particle described by the stationary Schrödinger equation in a potential well. The solutions are given as psi1(x)=e^-ax^2 with energy E1 and psi2(x)=xe^-ax^2 with energy E2. The initial state of the particle is a superposition of these two states, psi(x)=psi1(x)+psi2(x). The key issue raised is the omission of the time variable in the probability distribution calculation, which should incorporate the distinct energies E1 and E2 for accurate results.

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  • Understanding of the stationary Schrödinger equation
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  • Knowledge of complex exponentials and their application in quantum mechanics
  • Ability to perform modulus squared calculations for probability distributions
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ariana13
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Homework Statement


Stationary Schrödinger equation for a particle moving in a potential well has two solutions
psi1(x)=e^-ax^2 with energy E1 and
psi2(x)xe^-ax^2 with energy E2

At t=o, the particle is in the state psi(x)=psi1(x)+psi2(x)

Calculate the probability distribution as a function of time.


Homework Equations



psi(x,t)=e^iEt/h-bar * psi(x) and

probability distribution function=[mod(psi(x,t)]^2



The Attempt at a Solution



Well I tried subbing in psi(x,t)=e^iEt/h-bar * psi(x) for psi1(x) and psi2(x) to get psi(x) as a function of t.
But then when I try to do the modulus squared to get the probability distribution, the t disappears since I get (x+1)^2 * (e^-2ax^2) * (cos^2(Et/h-bar) + sin^2(Et/h-bar).
So I'm just left with (x+1)^2 * (e^-2ax^2).

I'm obviously making a stupid mistake here, unless I've got the theory part wrong? I definitely need a t term in the answer because the next question asks to figure out the time at which the probability distribution returns to the initial.
 
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Hi ariana13,

I think you did not take into account that the states have different energies; there is no energy E, just energies E1 and E2 when you construct psi1(x,t) and psi2(x,t).
 

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