The ground state of a one-dimensional Harmonic oscillator described by the Hamiltonian [itex]H = \frac{p^2}{2m} + \frac{kx^2}{2}[/itex] is of the form, [itex]\psi = Ae^{-ax^2}[/itex]. Determine 'A' and 'a' so that the wavefunction [itex]\psi[/itex] is a normalized eigenstate of the Hamiltonian. What is the energy eigenvalue of the wavefunction?(adsbygoogle = window.adsbygoogle || []).push({});

Well, I was able to normalize the wavefunction and obtained the value of 'A'.

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

[tex]A^2\int_{-\infty}^{\infty}e^{-2ax^2}dx =1[/tex]

[tex]A^2\sqrt{\frac{\pi}{2a}} =1[/tex]

[tex]A = (\frac{2a}{\pi})^{1/4}[/tex]

How do I determine 'a'? Any clues to obtain energy eigen value?

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# Homework Help: Energy eigen value

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