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Phonons: For oscillator wave function

  1. Mar 22, 2014 #1
    For oscillator wave function

    ##\frac{1}{\sqrt 2}(y-\frac{d}{dy})\psi_n(y)=\sqrt{n+1}\psi_{n+1}(y)##
    ##\frac{1}{\sqrt 2}(y+\frac{d}{dy})\psi_n(y)=\sqrt{n}\psi_{n-1}(y)##
    and I interpretate ##n## as number of phonons.
    Of course ##\psi_n(y)=C_ne^{-\frac{y^2}{2}}H_n(y)##.
    And ##C_n=f(n)##.
    Define ##\frac{1}{\sqrt 2}(y-\frac{d}{dy})=\hat{a}^+##, ##\frac{1}{\sqrt 2}(y+\frac{d}{dy})=\hat{a}##.
    Why in case of infinite square well
    ##\psi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}##
    you can not define
    ##\hat{a}^+\psi_n(x) \propto \psi_{n+1}(x)##
    ##\hat{a}\psi_n(x) \propto \psi_{n-1}(x)##
    and why quants of energy in problem of infinite square well do not have a name.
     
  2. jcsd
  3. Mar 22, 2014 #2

    ShayanJ

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    Because the situations are essentially different. In the case of electromagnetic or crystal waves, we have some physically observable thing vibrating but in the case of infinite square well, it is only the wave function which is vibrating.
     
  4. Mar 22, 2014 #3
    Also I see big difference between phonon oscillation in the lattice because in the lattice phonons are consequence of heating and in case of Shroedinger eq. ##T=0##.
     
  5. Mar 22, 2014 #4

    ShayanJ

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    No, that's wrong. At first, in the problem of infinite well, we're talking about only a few number of particles, so temperature has no meaning. But if you consider millions of particles for that problem, you can have temperature.
    And another point is, Schrodinger equation is valid for any temperature!
     
  6. Mar 22, 2014 #5
    As far as I know we take only one particle for a problem of infinite potential well. Also like in the problem of oscillator. In case of S. eq. you do not take in account temperature. So you will get the same result for any temperature. It is the same like you say I take ##T=0##. Could you really measure the same result of energy in any temperature for any single problem?
     
  7. Mar 22, 2014 #6

    ShayanJ

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    For now you take one particle. You can later solve if for more than one particle. And in statistical mechanics, you will do it for a huge number of particles. Temperature has meaning only when the number of particles is huge.So when you're doing it for only one particle, there is nothing called temperature and so you can't say in which temperature you're solving the problem.
    But in the case of having a huge number of particles, temperature is only a measure of average energy so when you say high or low temperatures, you're talking about the average energy of the system which depends on the number of particles in each level and so can be considered if needed.
     
  8. Mar 22, 2014 #7

    ChrisVer

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    Temperature does indeed have a meaning for many particles, in the sense that it's connected to the system's entropy - i.e. the number of microstates the system can exist.
    But couldn't we model an infinite well with (let's say) N fermions inside?
    I guess you don't need N→∞ to speak about temperature... even 3 can make the deal?
     
  9. Mar 22, 2014 #8
    Ok. It makes sence. But even in the case of large number of particles if the particles are in the same state and don't interact with each other you could say that they have one wave function and so one Sroedinger equation. Right?
     
  10. Mar 22, 2014 #9
    Interesting. Why 3 fermions? Why not 3 bosons? Or perhaps 2 fermions or to bosons?
     
  11. Mar 22, 2014 #10

    ChrisVer

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    I avoided bosons to avoid a condensate ....
    I said 3 instead of 2, because.... I don't really get the idea of how 2 fermions can cause a disordering...
     
  12. Mar 22, 2014 #11

    ShayanJ

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    Well, maybe you can do that, but there isn't much use of it and you will abandon it soon. Because the machinery of statistical mechanics is designed for large number of particles.

    Of course you can have any number of fermions in the well, but for them to have temperature, they should be a lot.
     
  13. Mar 22, 2014 #12

    ShayanJ

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    You always have one Schrodinger equation and one wave function and the number of particles or whether they have interaction or not doesn't matter. The difference appear in the dimensionality of the wave function(number of particles) and the Hamiltonian(Interaction between them).
     
  14. Mar 22, 2014 #13

    ChrisVer

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    I proposed that in the same way you can do it for nuclei ... assign fermions in a step potential (just generalize it for a closed box one, like an infinite well).
    I guess we can define temperature for nuclei, altough the number of nucleons N is finite? (I may be wrong)
     
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