Ground state of wavefunction

  • Thread starter Reshma
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  • #1
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A particle has mass 'm' and potential [itex]V(x) = x^2[/itex]. Find the ground state of this wavefunction.

Should I use the Hamiltonian operator here?
 

Answers and Replies

  • #2
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Yes. You should get something very similiar to the harmonic oscillator.
 
  • #3
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Thanks, inha! I compared it to the PE function of a harmonic oscillator.

[tex]V(x) = {1\over 2} m \omega^2 x^2[/tex]

Comparing it with [itex]V(x) = x^2[/tex],

[tex]{1\over 2} m \omega^2 = 1[/tex]

[tex]\omega = \sqrt{2\over m}[/tex]

And I substitute this value in the formula:
[tex]E_n = \hbar \omega \left(n + {1\over 2}\right)[/tex]

Am I going right? :confused:
 
  • #4
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Reshma said:
Thanks, inha! I compared it to the PE function of a harmonic oscillator.

[tex]V(x) = {1\over 2} m \omega^2 x^2[/tex]

Comparing it with [itex]V(x) = x^2[/tex],

[tex]{1\over 2} m \omega^2 = 1[/tex]

[tex]\omega = \sqrt{2\over m}[/tex]

And I substitute this value in the formula:
[tex]E_n = \hbar \omega \left(n + {1\over 2}\right)[/tex]

Am I going right? :confused:

It looks good (Of course, n=0 for the ground state, don't forget that!), but make sure you know how to do the problem from scratch. I'm guessing that's what your professor wanted you to do.

-Dan
 

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