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I Harmonic Oscillator in 3D, different values on x, y and z

  1. Sep 12, 2016 #1
    Hi,

    For a harmonic oscillator in 3D the energy level becomes En = hw(n+3/2) (Note: h = h_bar and n = nx+ny+nz) If I then want the 1st excited state it could be (1,0,0), (0,1,0) and (0,0,1) for x, y and z.

    But what happens if for example y has a different value from the beginning? Like this: V(x,y,z) = 1/2mw2(x2+4y2+z2) and for this decide the energy level AND degeneracy for the 1st excited state. I can only find simple examples when x, y and z are equal and 1.

    Best regards
     
  2. jcsd
  3. Sep 12, 2016 #2

    Demystifier

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    In such a more general case you have
    $$E_{n_1n_2n_3}=\hbar \omega_1 \left( n_1+\frac{1}{2} \right) + \hbar \omega_2 \left( n_2+\frac{1}{2} \right) + \hbar \omega_3 \left( n_3+\frac{1}{2} \right)$$
     
  4. Sep 12, 2016 #3
    Can you elaborate on that? Is 4y2 just n2 here? And in that case you will get three different energy values:

    E100 = 3hw1/2
    E010 =6hw2
    E001 =3hw3/2

    Which one is the 1st excited state? Is it E010?
     
  5. Sep 12, 2016 #4

    Vanadium 50

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    No, it enters in as the frequency.
     
  6. Sep 12, 2016 #5

    Demystifier

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    In your case
    $$\omega_1=w$$
    $$\omega_2=2w$$
    $$\omega_3=w$$
    Therefore
    $$E_{000}=2\hbar w$$
    $$E_{100}=E_{001}=3\hbar w$$
    $$E_{010}=4\hbar w$$
    Hence the first excited states are ##E_{100}=E_{001}##.
     
  7. Sep 12, 2016 #6
    Thank you, now I understand that part.

    What will the degeneracy become for the 1st excited state then? Can I use the same formula gn = 1/2(n+1)(n+2) for this case?
     
  8. Sep 12, 2016 #7

    Demystifier

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    It's 2.

    No.
     
  9. Sep 12, 2016 #8
    Why is it 2? What formula do you use to calculate that? (Sorry for all the questions..)
     
  10. Sep 12, 2016 #9

    Demystifier

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    It follows from the last line of post #5. There you see that there are 2 "first excited states" with equal energies. Hence the degeneracy of first excited state is 2.
     
  11. Sep 12, 2016 #10
    Oh! Thank you so much for the answers, this has been bugging me for a while now.

    Best regards
     
  12. Sep 22, 2016 #11
    Late questions.. but why is w2=2w and not 4w?
     
  13. Sep 22, 2016 #12

    Demystifier

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    Because, by definition,
    $$V(x)=\frac{1}{2}m\omega^2 x^2$$
     
  14. Sep 22, 2016 #13
    Right, of course. Thank you.
     
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