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I Density of states with delta function

  1. Jun 23, 2016 #1
    Hello,
    I'm stuck with this exercise, so I hope anyone can help me.

    It is to prove, that the density of states of an unknown, quantum mechanical Hamiltonian ##\mathcal{H}##, which is defined by
    $$\Omega(E)=\mathrm{Tr}\left[\delta(E1\!\!1-\boldsymbol{H})\right]$$
    is also representable as
    $$\Omega(E)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\mathrm{d}k\ e^{ikE}\mathrm{Tr}\left[e^{-ik\boldsymbol{H}}\right]$$
    when using the definition of the Dirac delta function
    $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\mathrm{d}k\ e^{ikx}$$
    I really don't know what to do. Is it necessary to change to a dirac notation or is this just a simple representation of the Trace, which i don't know yet?

    It would be great if anyone can give me a hand with that.
    Thanks a lot,
    Arnd
     
  2. jcsd
  3. Jun 23, 2016 #2

    blue_leaf77

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    If you replace ##x## in the definition of Dirac delta function with ##EI-H## you get ##e^{ik(EI-H)}##. ##EI## commutes with ##H##, thus ##e^{ik(EI-H)}= e^{ikEI}e^{-ikH}##. Now simplify ##e^{ikEI}## such that the identity operator "goes down" out of the exponent.
     
  4. Jun 23, 2016 #3
    When multiplying $$e^{ikE1} \cdot e^{-ikH}$$ you get $$e^{ikE} e^{-ikH}$$ because ##e^{ikE1}## is diagonal, isn't it? And then you can pull it out of the trace?

    Thank you a lot, but now it got even worse:
    This Hamiltonian should be inserted
    $$H = \sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)$$
    and result in the density of states
    $$\Omega(E)=\frac{1}{2\pi}\int\mathrm{d}k\ e^{ikE}\left(\sum_{n=0}^{\infty}e^{-i\hbar k\omega(n+1/2)}\right)^N$$

    I don't know how to handle the ladder operators. Is there a good lecture?
    Thanks a lot.
     
  5. Jun 23, 2016 #4

    blue_leaf77

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    Yes.

    Note that
    $$
    e^{ikH} = \exp \left( ik\sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)\right) \\
    = \exp \left( ik\hbar\omega(a_1^{\dagger}a_1 + 1/2)\right) \otimes \exp \left( ik\hbar\omega(a_2^{\dagger}a_2 + 1/2)\right) \otimes \ldots \otimes \exp \left( ik\hbar\omega(a_N^{\dagger}a_N + 1/2)\right)
    $$
    Then use the property ##\textrm{Tr }(A\otimes B) = (\textrm{Tr }A)(\textrm{Tr }B)##.
     
  6. Jun 23, 2016 #5

    vanhees71

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    What are the tensor products doing there? Isn't this just the usual calculation of the microcanonical partition function in terms of a Fourier transform of the canonical one? As far as I can see, at the end of #3 you already have the correct result.
     
  7. Jun 23, 2016 #6
    The End of #3 is the correct result but I have to find the way to get there.
    Thank you! I will try the tensor products.
     
  8. Jun 23, 2016 #7

    blue_leaf77

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    I believe this is a multiparticle system where each particle obeys the ordinary harmonic oscillator Hamiltonian. A basis in the whole system is then a tensor product between the single particle state ##|n_1\rangle \otimes |n_2\rangle \otimes \ldots \otimes |n_N\rangle## and the corresponding symmetry operator must be a Kronecker product from each particle's symmetry operator. Moreover, the formal form of a symmetry generator of the system, such as the Hamiltonian, is
    $$
    H = H_1\otimes I_2 \otimes \ldots \otimes I_N + I_1\otimes H_2 \otimes \ldots \otimes I_N + \ldots + I_1\otimes I_2 \otimes \ldots \otimes H_N
    $$
    where ##H_i = \hbar\omega(a_i^{\dagger}a_i + 1/2)##. Using a hand-waving notation omitting those jumbled identity operators, the total Hamiltonian is rewritten as
    $$
    H = \sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)
    $$
    Yes, but the OP was trying to prove that.
     
  9. Jun 23, 2016 #8
    This is great, thank you very much! You are completely right with your assumption about the many particle system.
    Now the last part remains:

    When evaluating the geometric series, I get a density of states which looks like
    $$\Omega(\epsilon)=\frac{1}{2\pi}\int dk\exp\left(N\left[ik\epsilon-\ln \left(2i\sin(k\hbar\omega/2)\right)\right]\right)$$
    with ##\epsilon=E/N##

    I have to prove, that the average entropy per oscillator looks like
    $$s(\epsilon)=\lim_{N\rightarrow\infty}\frac{\ln\Omega(\epsilon)}{N}=\left(\frac{\epsilon}{\hbar\omega}+1/2\right)\ln\left(\frac{\epsilon}{\hbar\omega}+1/2\right)-\left(\frac{\epsilon}{\hbar\omega}-1/2\right)\ln\left(\frac{\epsilon}{\hbar\omega}-1/2\right)$$
    by using the saddle point approximation in the limit ##N\rightarrow\infty##.
    I'm familiar with the saddle point approximation but don't know how to evaluate the ln, how to find the ##x_0## and how to get the final result.

    I hope you have any advice. Thank you!
     
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