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Second Quantization Density Matrix

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data Screen Shot 2015-09-17 at 11.29.35 AM.png

    2. Relevant equations and attempt at solution
    I think I got the ground state, which can be expressed as [itex]|\Psi \rangle = \prod_{k}^{N}\hat{a}_{k}^{\dagger} |0 \rangle[/itex] .

    Then for the density matrix I used:
    [itex]\langle 0|\prod_{k'}^{N}\hat{a}_{k}\hat{a}_{k}^{\dagger}\hat{a}_{l}\prod_{k'}^{N}\hat{a}_{k'}^{\dagger} |0 \rangle[/itex].
    Due to the commutation relations for fermions:
    [itex]\hat{a}_{l}\prod_{k}^{N}\hat{a}_{k}^{\dagger} |0 \rangle=\left ( -1 \right )^{\sum_{i=1}^{l-1}}\prod_{k\neq l}^{N}\hat{a}_{k}^{\dagger} |0 \rangle[/itex]
    So the density matrix becomes:
    [itex]\langle 0|\left ( -1 \right )^{\sum_{i=1}^{l-1}+{\sum_{i=1}^{k-1}}}\prod_{k'\neq k}^{N}\hat{a}_{k}\prod_{k'\neq l}^{N}\hat{a}_{k'}^{\dagger} |0 \rangle[/itex].
    I honestly don’t know what to do past this, or if I’m even on the right track. Any help is appreciated.
     
  2. jcsd
  3. Sep 17, 2015 #2
    The first questions seems to be answered correctly.

    In the second question you are forgetting an important part, what does a single-particle wavefunction look like?
    How many creation operators do you need and how to write the most general form of such a state?

    Furthermore you used the N-particle ground state. Nowhere is it mentioned you should work with the ground state.
    This last remark is related to the underlined part of my second (guiding) question.

    Finally, which texts are you using? It might be useful to point out important parts in such a text.
     
  4. Sep 17, 2015 #3
    Hi JoirsL, thank you for your response.

    I will first answer the ground state question, in the question above you can see that he defines the ground state as [itex]{\Psi}[/itex], so it appears to me as if we are supposed to be working within this state.

    The most general form of a state of a single particle wavefunction will be a superposition of the creation operators, correct? So really, in this basis of k states, there will be k operators in this state.

    I am being taught directly from the professor's notes, so unfortunately all of my knowledge (or lack of) comes from that. He has many recommended readings from Ashland and Leggett, which I have done, but it hasn't really been of much assistance to me so far.
     
  5. Sep 17, 2015 #4
    I agree the question is posed in a way that will lead to confusion.

    I am however convinced that he means what I asked (perhaps clarify this whenever you see him).
    It will indeed be a superposition (*)

    How would you proceed from there? In other words, how do we calculate the elements of a density matrix?

    However I'm curious about the validity of ##|\psi\rangle = \sum_k a^\dagger_k|0\rangle##.
    Because this is an infinite sum the expected energy ##\langle H\rangle## can become infinite (depending on what ##\epsilon_k## looks like).
    I'll try to find out if I can find the time
     
  6. Sep 18, 2015 #5
    Well, I suppose due to orthogonality of the basis states, the[itex]\langle\Psi|\hat{a}_{k}^{\dagger}\hat{a}_{k}|\Psi\rangle[/itex] would be [itex]\delta _{kl}\delta_{lk}[/itex]? Since it seems the only nonzero states would be the point in which the creation operator creates a particle in the one that the annihilation operator removed...
     
  7. Sep 18, 2015 #6
    likeLets try to make all of this more 'rigorous'.

    First we have the one-particle wave vector which I will call ##|\phi\rangle## to avoid any confusion.
    m hereThen ##|\phi\rangle = N\sum_k a^\dagger_k |0\rangle##
    To describe explicit density matrix elements we need a basis that spans our space.

    Lets call this basis for the one-particle states ##\left\{ |\epsilon_k\rangle \,|\, k\geq1 \text{ and } k\in \mathbb{N} \right\}##.
    Can you think of a basis?
    We want a basis that keeps the expression of ##|\phi\rangle## simple. Because we will be calculating elements relevant to the state vector

    Can you continue from here?

    If you can't, my first good/reasonable will show up on monday. (busy weekend!)
     
  8. Sep 19, 2015 #7
  9. Sep 21, 2015 #8
    Ok, to be honest, the rigorous part of the proof is what is baffling me so much. A basis that spans our state will simply be the set of the [itex]\hat{a}^{\dagger}[/itex]. But what next?
     
  10. Sep 21, 2015 #9
    Lets define ##|\epsilon_k\rangle = a^\dagger_k|0\rangle##.
    Do you agree this spans the space of one-particle wave vectors (and they are linearly independent by the hermitian property of the hamiltonian)?

    Now our most general one-particles state as defined in post #6 is clearly a mixture of such states.

    Do you see where I'm getting at?
     
  11. Sep 21, 2015 #10
    Ok, maybe. Let's see. So, let me know if this makes sense. Just working with the lowering operator:
    [itex]\hat{a}_{l} |\phi \rangle= \hat{a}_{l}N\sum_{k}|\epsilon _{k}\rangle[/itex] = [itex]\hat{a}_{l}N\sum_{k}\hat{a}^{\dagger}_{k}|0\rangle[/itex]
    From the anticommutation relations for fermions:
    [itex]\hat{a}_{l}\sum_{k}\hat{a}^{\dagger}_{k}=\delta _{lk}-\sum_{k}\hat{a}^{\dagger}_{k}\hat{a}_{l}[/itex]
    So that
    [itex]\hat{a}_{l} |\phi \rangle=N(\delta _{lk}-\sum_{k}\hat{a}^{\dagger}_{k}\hat{a}_{l})|0\rangle[/itex] is equal to [itex]\hat{a}_{l} |\phi \rangle=N\delta _{lk}|0\rangle[/itex] due to the lowering operator on the right returning a null. Since the bra version of this will return a [itex]\langle \phi|\hat{a}_{k}^{\dagger} =\langle0|N\delta _{kk}[/itex], we finally get the matrix elements of: [itex]N^{2}\delta _{kk}\delta _{lk}[/itex].

    Does this work?
     
  12. Sep 21, 2015 #11
    There are some minor issues with your notation.

    We know that for a system with wave vector ##|\phi\rangle## the elements of the density matrix are given by
    ##\rho_{mn} = \langle\epsilon_m|(|\phi\rangle\langle\phi|)|\epsilon_n\rangle## (in the basis defined before!).

    Furthermore ##N## can be an element of ##\mathbb{C}## so you get a factor ##|N|^2##, notice that we use the modulus squared.

    So recalling that ##|\epsilon_k\rangle = a^\dagger_k|0>## we can start calculating.

    ##\rho_{mn} = \langle0|a_m\left( N \sum_k a^\dagger_k |0\rangle\right) \left( N^* \sum_l \langle 0\rangle|a_l\right)a^\dagger_n|0\rangle##

    Then I get the result ##\rho_{mn} = |N|^2 \left( \sum_k \delta_{km} \right) \left(\sum_l \delta_{ln}\right)##.
    Now we know for a fact that the sums reduce to 1 identically.
    Meaning ##\rho_{mn} = |N|^2## for all ##m,\, n##.

    You got most of the steps right, you were however a little bit sloppy with your notation (you can see this from the fact that you have 3 k indices in your final result).

    I must correct myself, in post #2 I talked about the most general one particle wave vector. In post #6 however I gave a very special wave vector.
    You might want to try for a truly general wave vector ##|\alpha\rangle = \sum_k C_k a^\dagger_k |0\rangle##.
    Here the ##C_k## are (complex) coefficients.
     
  13. Sep 21, 2015 #12
    Ah, thank you, I guess I need to work on keeping my indices separated. I appreciate all of the help (and patience!), JorisL. I think I have a better grasp of this now.
     
  14. Sep 21, 2015 #13
    Good luck, and it gives me a break from my own studying so I'm happy to help :)
     
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