Understanding Quantum Wave Packets: How to Derive the Schrodinger Equation

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SUMMARY

The discussion centers on deriving the Schrödinger equation for a wave packet, specifically at time t = 0. The initial wave packet is defined as ψ(x,0) = ∫_{-∞}^{+∞} dk A(k) e^{i(kx)}, where A(k) = e^{-α(k - k₀)² / 2}. The transformation to q' = k - k₀ leads to the expression ψ(x,0) = e^{i k₀ x} e^{-x² / (2α)} ∫_{-∞}^{+∞} dq' e^{-αq'² / 2}. The key step involves completing the square in the integral and shifting variables, resulting in the e^{-x² / (2α)} term, which is crucial for understanding the wave packet's behavior.

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silimay
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Sorry... my message got posted by mistake before I started typing. Here is what I was going to say:

I'm having a problem just understanding something from my quantum book. They're deriving something to do with a wave packet with the Schrödinger equation, and they have the equation of a wave packet at time t = 0:

\psi(x,0) = \int_{- \infty}^{+ \infty} dk A(k) e^{i(kx)

where A(k) = e^{- \alpha (k - k_o)^2 / 2}

They change variables to q' = k - k_o and then they get

\psi(x,0) = e^{i k_o x} e^{- x^2 / {2 \alpha}} \int_{- \infty}^{+ \infty} dq' e^{- \alpha {q'}^2 / 2}

I don't understand how they got that (specifically, the e^{-x^2 / {2 \alpha}} term).
 
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They did the substitution and then completed the square in q' in the integral and brought the non-square stuff outside the integral. Finally they shifted q'->q'-ix/alpha.
 
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Thanks so much for your help =) =) =) I understand it now.
 

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