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Suppose that $\left | m \right >=\int \frac {dk}{2\pi} G(k)\left | k \right >$ and $G(k)$ has the dimension of $[L]$. So both sides of the equation are dimensionless.

A density matrix is defined as $\rho=\sum_{mn} \rho_{mn}\left | m \right >\left < n\right |=\sum_{mn}\int \frac{dk_1dk_2}{(2\pi)^2}G_m(k_1)G^*_n(k_2) \rho_{mn}\left | k_1 \right >\left < k_2\right| $ and the dimension of $\rho_{mn}$ is zero which is consistent.

But if I use select a element by applying \begin{align}\rho_{mn}=\left < m \right | \rho \left | n \right >&=\int \frac{dk_3dk_4}{(2\pi)^2}G_m(k_3)G^*_n(k_4) \left < k_3\right|\rho \left | k_4 \right >\nonumber \\ &=\int \frac{dk_3dk_4}{(2\pi)^2}G_m(k_3)G^*_n(k_4) \sum_{m'n'}\int \frac{dk_1dk_2}{(2\pi)^2}G_{m'}(k_1)G^*_{n'}(k_2) \rho_{m'n'}\left < k_3\right|\left . k_1 \right >\left < k_2\right| \left . k_4 \right > \nonumber \\ &=\sum_{m'n'}\int \frac{dk_3dk_4}{(2\pi)^2}\int \frac{dk_1dk_2}{(2\pi)^2}G_m(k_3)G^*_n(k_4) G_{m'}(k_1)G^*_{n'}(k_2) \rho_{m'n'}(2\pi)^2\delta(k_3-k_1)\delta(k_4-k_2) \nonumber \\ &=\sum_{m'n'}\int \frac{dk_1dk_2}{(2\pi)^2}G_m(k_1)G^*_n(k_2) G_{m'}(k_1)G^*_{n'}(k_2) \rho_{m'n'} \nonumber \end{align}

Now the rhs of the equation has the dimension of $[L^2]$ which is inconsistent with the lhs.

Where did I make a mistake? I think the problem is somewhere arond the Dirac delta function, but I still have not find any useful information about that.

I know Dirac delta function has the inverse dimension of its parameters, but I could not see how this fit in here.

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I suppose you consider states $|k>$ in the continuous spectrum, otherwise you would not use the integration over $k$ but a summation. If so, the normalization of your $|k>$ probably looks like that $<k_i|k_j>=\delta(k_i-k_j)$ which means $|k>$ is not exactly dimensionless for the reasons you mentioned (Dirac delta is $[k]^{-1}$).

Actually, you don't need to calculate density matrix to notice that; just try $<m|n>$ and you"ll run into the same issue. One way out of this would be to say that the dimension of $|k>$ to be $L^{1/2}$ (assuming $k$ is sort of a wavevector) and thus consider instead $|m^{\prime}>=L^{-1/2}|m>$, in the spirit of eigenmodes in a box of a size $L$.

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  • $\begingroup$ Thanks, sleepy. I will try that. $\endgroup$
    – Haorong Wu
    May 6 at 9:37

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