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Eigenfunction of the Schrodinger wave equation

In the attachment, the question is shown in bold, and I've been able to produce the working out as shown.

However, the last line where the provided solution simply states, "From the definition of γ, all the x^2 terms cancel leaving..."

How do the x^2 terms cancel, and how is this last line arrived at?

Thanks
Reply 1
Original post by little pixie
In the attachment, the question is shown in bold, and I've been able to produce the working out as shown.

However, the last line where the provided solution simply states, "From the definition of γ, all the x^2 terms cancel leaving..."

How do the x^2 terms cancel, and how is this last line arrived at?

Thanks


The total coefficient of x^2 is 12mω2(4γ2×22m) \frac{1}{2} m \omega^2 - (4\gamma^2 \times \frac{\hbar^2}{2m})

If you substitue the value (definition) for γ=mω2\gamma = \frac{m\omega}{2\hbar}

then the total coefficient is 0
Reply 2
Original post by astro67
The total coefficient of x^2 is 12mω2(4γ2×22m) \frac{1}{2} m \omega^2 - (4\gamma^2 \times \frac{\hbar^2}{2m})

If you substitue the value (definition) for γ=mω2\gamma = \frac{m\omega}{2\hbar}

then the total coefficient is 0


Thanks you astro67.

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