- #1
Reshma
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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?
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?
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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