Consider that we are working with a joint system composed of system A with basis $|\alpha_j\rangle$ and system B with basis $|\beta_j\rangle$, we may write a general density matrix for the joint system with respect to tensor product basis $|\alpha_j\rangle |\beta_j\rangle$.

I want to understand then how we can deduce that the density operator can be written as follows.

$$\rho = \sum_{j,k,l,m} \langle\alpha_j| \langle\beta_k |\rho |\alpha_l\rangle |\beta_m\rangle |\alpha_j\rangle |\beta_k\rangle \langle\alpha_l| \langle \beta_m|$$

Any help to facilitate my understanding of this would be greatly appreciated.

  • $\begingroup$ I don't quite see the difference between this question and this other one physics.stackexchange.com/q/596995/58382 $\endgroup$
    – glS
    Commented Nov 30, 2020 at 10:45
  • $\begingroup$ The other question asks about how to find the reduced density operator by taking the trace. $\endgroup$
    – DJA
    Commented Nov 30, 2020 at 13:21
  • $\begingroup$ you might want to edit the title to reflect what you are actually asking then $\endgroup$
    – glS
    Commented Nov 30, 2020 at 16:27

2 Answers 2


If $\big\{|\alpha_j\rangle\big\}$ is a basis for the Hilbert space $\mathcal H_A$ and $\big\{|\beta_k\rangle\big\}$ is a basis for $\mathcal H_B$, then $\big\{|\alpha_j,\beta_k\rangle \big\}$ is a basis for $\mathcal H_A \otimes \mathcal H_B$, the natural Hilbert space for the composite system. To lighten the notation, I am defining $|\alpha_j,\beta_k\rangle \equiv |\alpha_j\rangle \otimes |\beta_k \rangle$.

From there, the identity operator on $\mathcal H_A \otimes \mathcal H_B$ can be written $$\mathbf 1 = \sum_{j,k} |\alpha_j,\beta_k\rangle\langle\alpha_j,\beta_k|$$

so an arbitrary operator $T$ can be written

$$T = \mathbf 1 \cdot T \cdot \mathbf 1 = \bigg(\sum_{j,k} |\alpha_j,\beta_k\rangle\langle\alpha_j,\beta_k|\bigg) T \bigg(\sum_{\ell,m} |\alpha_\ell,\beta_m\rangle\langle \alpha_\ell,\beta_m|\bigg)$$ $$ = \sum_{j,k,\ell,m}T_{jk\ell m} |\alpha_j,\beta_k\rangle\langle \alpha_\ell,\beta_m|$$

where $$T_{jk\ell m} \equiv \langle \alpha_j,\beta_k| T | \alpha_\ell,\beta_m\rangle$$


Short answer: apply both sides of the equation to an arbitrary ket basis vector, and things will simplify a lot.

The truth of that equation doesn't have anything to do with the fact that it's a joint system, or that it is a density operator. It would be true for any operator, and any orthonormal basis.

After you apply both sides of the equation to a basis vector, one way to proceed is to flip the two terms and use the resolution of identity.


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