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Density matrix elements, momentum basis, second quantization

  1. Mar 2, 2012 #1
    Hello everyone,

    I'm having some trouble, that I was hoping someone here could assist me with. I do hope that I have started the topic in an appropriate subforum - please redirect me otherwise.
    Specifically, I'm having a hard time understanding the matrix elements of the density matrix, [itex] \varrho[/itex]. For instance, I would like to determine the density matrix element [itex]\langle \boldsymbol{\mathrm{k}} | \varrho | \boldsymbol{\mathrm{k}}+\boldsymbol{\mathrm{q}} \rangle [/itex], i.e. matrix elements of [itex] \varrho[/itex] in a momentum basis.

    The reason for me wanting to do this, is that I am trying to understand an old paper by N.D. Mermin [1]. In this paper, the particle density, [itex]\rho(\boldsymbol{\mathrm{q}})[/itex], and current density, [itex]\boldsymbol{\mathrm{J}}(\boldsymbol{\mathrm{q}})[/itex], expectation values are introduced as (slightly rewritten - the essence remains the same):
    [tex] \langle \rho(\boldsymbol{\mathrm{q}})\rangle = \sum_{\boldsymbol{\mathrm{p}}} \langle \boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}} | \varrho | \boldsymbol{\mathrm{p}} \rangle \\
    \langle \boldsymbol{\mathrm{J}}(\boldsymbol{\mathrm{q}}) \rangle = \sum_{\boldsymbol{\mathrm{p}}} (\boldsymbol{\mathrm{p}}+\frac{1}{2} \boldsymbol{\mathrm{q}} )\langle \boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}} | \varrho | \boldsymbol{\mathrm{p}} \rangle [/tex]
    I'm used to the following second quantized forms of the particle density operator and particle currents:
    [tex] \rho(\boldsymbol{\mathrm{q}}) = \sum_{\boldsymbol{\mathrm{p}}} c_{\boldsymbol{\mathrm{p}}}^\dagger c_{\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}} \\
    \boldsymbol{\mathrm{J}}(\boldsymbol{\mathrm{q}}) = \sum_{\boldsymbol{\mathrm{p}}} (\boldsymbol{\mathrm{p}}+\frac{1}{2} \boldsymbol{\mathrm{q}} )c_{\boldsymbol{\mathrm{p}}}^\dagger c_{\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}} [/tex]
    My question is, how do I derive (preferably starting from the second-quantized form of the operators) the expectation values [itex]\langle \rho(\boldsymbol{\mathrm{q}})\rangle[/itex] and [itex]\langle \boldsymbol{\mathrm{J}}\boldsymbol{\mathrm{q}}) \rangle[/itex], expressed as sums over density matrix elements?
    I feel that, essentially, this should be a simple problem to do in a stringent manner - but I just can't seem to make the necessary connections.

    [1] Lindhard Dielectric Function in the Relaxation-Time Approximation

    EDIT: Edit since I had apparently not understood the use of [ tex ] [ /itex ].
    Last edited: Mar 2, 2012
  2. jcsd
  3. Mar 2, 2012 #2

    A. Neumaier

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    Use <A>=trace rho A = sum_{p,q} <p|rho|q><q|A|p> and the annihilation properties of the c's.

    Note that you use rho in two different fonts with two different meanings - not a very good choice of notation.
  4. Mar 2, 2012 #3
    Thanks for your reply Neumaier!
    Sadly, after having considered your suggestion for some time, I remain stuck with the same predicament as before. I must be confusing something basic. My intention with using [itex] \varrho [/itex] for the density matrix and [itex] \rho [/itex] for particle density operator, was that, in my mind they are different quantities and so deserve different notations (e.g. the density matrix evolve according to the Liouville-von Neumann equation while the particle density operator evolve according to the Heisenberg equation of motion)?

    Attempting to follow your suggestion to express the average of the particle density operator in momentum space, [itex] \langle \rho(\boldsymbol{\mathrm{q}}) \rangle [/itex], I make it as far as (again, I'm using two different notation for the density matrix and the particle density - I would be very interested in any explanation as to why this is inappropriate):
    \langle \rho(\boldsymbol{\mathrm{q}}) \rangle
    = \mathrm{tr}[\varrho \rho(\boldsymbol{\mathrm{q}})]
    = \sum_{\boldsymbol{\mathrm{p}}} \langle \boldsymbol{\mathrm{p}} | \varrho \rho(\boldsymbol{\mathrm{q}}) | \boldsymbol{\mathrm{p}} \rangle
    = \sum_{\boldsymbol{\mathrm{p\, k}}} \langle \boldsymbol{\mathrm{p}} | \varrho | \boldsymbol{\mathrm{k}} \rangle \langle \boldsymbol{\mathrm{k}} | \rho(\boldsymbol{\mathrm{q}}) | \boldsymbol{\mathrm{p}} \rangle
    = \sum_{\boldsymbol{\mathrm{p\, k}}} \langle 0 | c_{\boldsymbol{\mathrm{p}}} \varrho c_\boldsymbol{\mathrm{k}}^\dagger | 0 \rangle \langle 0 | c_\boldsymbol{\mathrm{k}} \rho(\boldsymbol{\mathrm{q}}) c_\boldsymbol{\mathrm{p}}^\dagger | 0 \rangle
    = \sum_{\boldsymbol{\mathrm{p\, k\, k}}'} \langle 0 | c_{\boldsymbol{\mathrm{p}}} \varrho c_\boldsymbol{\mathrm{k}}^\dagger | 0 \rangle \langle 0 | c_\boldsymbol{\mathrm{k}} c_{\boldsymbol{\mathrm{k}}'}^\dagger c_{\boldsymbol{\mathrm{k}}'} c_\boldsymbol{\mathrm{p}}^\dagger | 0 \rangle
    [/tex]... which does not appear a viable path. Could I possibly persuade you to give my yet a hint? What am I doing wrong here? Thanks again!
  5. Mar 2, 2012 #4

    A. Neumaier

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    You must aim at getting the result, hence not convert everything to c/a notation.
    Thus keep the bras and kets in the density matrix part, and replace the particle density by its definition. Then you get a triple sum in which most terms vanish.
  6. Mar 2, 2012 #5
    You was close to the result:

    \langle \rho(\boldsymbol{\mathrm{q}})\rangle=
    \sum_{\boldsymbol{\mathrm{p}}} \langle\boldsymbol{\mathrm{p}}| \rho (\boldsymbol{\mathrm{q}}) \varrho|\boldsymbol{\mathrm{p}}\rangle=
    \sum_{\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{k}}} \langle\boldsymbol{\mathrm{p}}| c^\dagger_\boldsymbol{\mathrm{k}}c_{\boldsymbol{ \mathrm{k} } + \boldsymbol{\mathrm{q}}} \varrho|\boldsymbol{\mathrm{p}}\rangle=
    \sum_{\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{k}}, \boldsymbol{\mathrm{k'}}'} \langle\boldsymbol{\mathrm{p}}| c^\dagger_\boldsymbol{\mathrm{k}}c_{\boldsymbol{ \mathrm{k} } + \boldsymbol{\mathrm{q}}} |\boldsymbol{\mathrm{k}}'\rangle\langle\boldsymbol{\mathrm{k}}'|\varrho|\boldsymbol{\mathrm{p}} \rangle=
    \sum_{\boldsymbol{\mathrm{k}}} \langle\boldsymbol{\mathrm{k}} + \boldsymbol{\mathrm{q}}|\varrho| \boldsymbol{\mathrm{k}} \rangle ,

    where in the last step I used

    \langle\boldsymbol{\mathrm{p}}| c^\dagger_\boldsymbol{\mathrm{k}}c_{\boldsymbol{ \mathrm{k} } + \boldsymbol{\mathrm{q}}} |\boldsymbol{\mathrm{k}}'\rangle = \delta_{\boldsymbol{ \mathrm{k} }, \boldsymbol{ \mathrm{p} }}\delta_{\boldsymbol{ \mathrm{k} } + \boldsymbol{ \mathrm{q} }, \boldsymbol{ \mathrm{k} }'}.

    With analogous calculation you can easly obtain the second equation you needed.

    I hope this helps.

  7. Mar 3, 2012 #6
    Thank you both a ton - your help is very much appreciated. Just couldn't get that bit right! Thanks alot, this has saved my week :).
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