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Zero from which the energy is measured

  1. Nov 17, 2014 #1

    ShayanJ

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    Consider the famous harmonic oscillator. Now imagine I use my freedom in choosing the origin for the potential and choose, instead of the usual thing, [itex] V(x)=\frac{1}{2}m \omega^2 x^2-\frac{1}{2} \hbar \omega [/itex], so the TISE becomes:
    [itex]
    -\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}+\frac 1 2 m\omega^2x^2 \psi=(E+\frac{1}{2} \hbar \omega) \psi
    [/itex]
    So its obvious, that the energy levels become [itex] E_n= n\hbar \omega [/itex]. So now the ground state has zero energy!
    But this can't be right. Where is zero-point energy? If it is a physical concept, then it shouldn't depend on our choices. So there should be a way that it shows up here too. But I fail to see that way. Any ideas?
    Thanks
     
  2. jcsd
  3. Nov 17, 2014 #2

    mfb

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    So what?
    Absolute energy has no meaning in nonrelativistic quantum mechanics. Only energy differences are meaningful, and they stay the same.
     
  4. Nov 17, 2014 #3

    dextercioby

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    It's actually simpler than that. A constant term to the classical Hamiltonian has no influence on the Hamilton's equations (remember, the Hamiltonian is under a differential operator wrt q,p) in the eom. The quantum case is simple. The constant produces a complex (unit modulus) time-dependent exponential, hence brings no involvement to the spectral equation which only considers stationary states.
     
  5. Nov 17, 2014 #4

    ShayanJ

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    So zero-point energy is meaningless?
    But that means its completely meaningless to talk about zero-point energy and say:" This is one difference between classical and quantum mechanics where the system can never have zero energy" which is something you can find in most introductory QM textbooks.
    So what's all the stuff in here ??? No meaning at all???!!!
    What about the minimum energy required by Uncertainty principle???
     
  6. Nov 17, 2014 #5

    dextercioby

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    It's actually not like that at all. The hbaromega/2 is mandatory in the quantum theory, can't get rid of it. It's due to the fact that q and p are non-commuting variables *operators/matrices*, The quantized harmonic oscillator has thus a 'residual' energy of purely quantum nature.
     
  7. Nov 17, 2014 #6

    ShayanJ

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    That's exactly my point. Every quantum system should contain that. But where is it in the problem I mentioned(with the potential I chose)?
     
  8. Nov 17, 2014 #7

    dextercioby

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    It's in the squared coordinate term of the potential. The constant you put by hand can be <gauged away/absorbed> into a time-dependent phase factor.
    About the comment <every quantum system should have that>. Well, not really. This <residue> is particular to the squared coordinate potential energy, hence to the harmonic oscillator (or to any Hamiltonian for which there's canonic transformation to bring it to the p^2 +q^2 form). This particularity is then carried forward to free field QFT.
     
    Last edited: Nov 17, 2014
  9. Nov 17, 2014 #8

    ShayanJ

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    I get it, thanks.
    And about that comment. I don't understand why you say this is particular to the squared potential. Uncertainty principles requires it to exist for all quantum systems.
     
  10. Nov 17, 2014 #9

    mfb

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    The zero-point energy is the difference to the lowest point in the potential (=the classical ground state), and that does not change with your energy shift.
     
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