Assume there is a N-particle state denoted as $|\Psi_N\rangle$, the density operator from its definition reads $\gamma_N = |\Psi_N\rangle\langle\Psi_N|$ and the density matrix elements take the form like \begin{equation} \langle\alpha_1,\alpha_2,\cdots,\alpha_N|\gamma_N|\beta_1,\beta_2,\cdots,\beta_N\rangle. \end{equation} The one-body reduced density matrix is defined as \begin{equation} \gamma_1(\alpha_1,\beta_1)=N\int\cdots\int \langle\alpha_1,\alpha_2,\cdots,\alpha_N|\gamma_N|\beta_1,\alpha_2,\cdots,\alpha_N\rangle d\alpha_2\cdots d\alpha_N. \end{equation} However, in some other books, someone writes down the one-body reduced density matrix using second quantization like \begin{equation} \gamma_1(\alpha_1,\beta_1) = \langle\Psi_N| \mathcal{c}^\dagger_{\alpha_1}\mathcal{c}_{\beta_1}|\Psi_N\rangle. \end{equation} So my question is do these two definitions of the one-body reduced density matrix equivalent? If they are, how to prove? Thank you!

  • $\begingroup$ All particles are identical, but the state $|\Psi_{N}\rangle$ in your question dose not lie in Fock space. It is neither symmetric or anti-symmetric for exchanging particles. $\endgroup$
    – Eric Yang
    Commented Jun 8, 2017 at 8:09
  • $\begingroup$ Sorry, can you provide more details? $|\Psi_N\rangle$ dose lie in Fock space. These two definitions both come from some books. I only changed the symbols. $\endgroup$
    – jwyan1126
    Commented Jun 8, 2017 at 8:45

1 Answer 1


First, $\alpha_1$ and $\beta_1$ in your second-quantized definition of the one-body density matrix should be interchanged (we will see below this is correct): $$ \gamma_1(\alpha_1,\beta_1)=\langle\Psi_N|c_{\beta_1}^+c_{\alpha_1}|\Psi_N\rangle $$ Then let's insert the unit operator $I=\sum_i|i\rangle\langle i|$, where $|i\rangle$ are all many-particle states, between $c_{\beta_1}^+$ and $c_{\alpha_1}$: $$ \gamma_1(\alpha_1,\beta_1)=\sum_i\langle\Psi_N|c_{\beta_1}^+|i\rangle\langle i|c_{\alpha_1}|\Psi_N\rangle. $$ Only the $(N-1)$-particle states $|i\rangle=|\alpha_2\ldots\alpha_N\rangle$ will survive in this sum: $$ \gamma_1(\alpha_1,\beta_1)=\int d\alpha_2\ldots d\alpha_N\langle\Psi_N|c_{\beta_1}^+|\alpha_2\ldots\alpha_N\rangle\langle \alpha_2\ldots\alpha_N|c_{\alpha_1}|\Psi_N\rangle=\\ =\int d\alpha_2\ldots d\alpha_N\langle \alpha_2\ldots\alpha_N|c_{\alpha_1}\gamma_Nc_{\beta_1}^+|\alpha_2\ldots\alpha_N\rangle. $$ Then, using the property of creation operators $$ c_{\beta_1}^+|\alpha_2\ldots\alpha_N\rangle=\sqrt{N}|\beta_1\alpha_2\ldots\alpha_N\rangle,\qquad\langle\alpha_2\ldots\alpha_N|c_{\alpha_1}=\sqrt{N}\langle\alpha_1\alpha_2\ldots\alpha_N|, $$ we get the first-quantized definition: $$ \gamma_1(\alpha_1,\beta_1)=N\int d\alpha_2\ldots d\alpha_N\langle\alpha_1\alpha_2\ldots\alpha_N|\gamma_N|\beta_1\alpha_2\ldots\alpha_N\rangle. $$

  • $\begingroup$ Thank you very much! you are right, the subscripts should be interchanged. $\endgroup$
    – jwyan1126
    Commented Jun 9, 2017 at 12:41
  • 2
    $\begingroup$ Hi Alexey, I have a question about the property of creation operators given by you. Where is the sqrt(N) from? $\endgroup$
    – Frank
    Commented Jun 28, 2022 at 5:29

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