Consider that we are working with a joint system composed of system A with basis $|\alpha_j\rangle$ and system B with basis $|\beta_j\rangle$.

In my notes the density operator is denoted as follows:

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

whereby my notes state that $$ \rho_{jklm} = \langle\alpha_j| \langle\beta_k |\rho |\alpha_l\rangle |\beta_m\rangle $$

They also state the following equations for the Trace of A and the trace of B: $$\rho_\beta = Tr_\alpha(\rho) = \sum_{l,m}(\sum_{j} \rho_{j,l,j,m}) |\beta_l\rangle \langle\beta_m| $$

$$\rho_\alpha = Tr_\beta(\rho) = \sum_{j,k}(\sum_{l} \rho_{j,l,k,l}) |\alpha_j\rangle \langle\alpha_k| $$

My main question is how would one write out $\rho_{j,l,k,l}$ and $\rho_{j,l,j,m}$ explcitly as what I get do not seem to agree with a worked example in my book and so I am quite confused.


  • 1
    $\begingroup$ Can you elaborate on your confusion? You already wrote the expression for $\rho_{jk\ell m}$ where $j,k,\ell,m$ take arbitrary values. Setting the first and third (or second and fourth) indices equal to the same number is just a special case. $\endgroup$
    – J. Murray
    Nov 29, 2020 at 5:52
  • $\begingroup$ Well because if I were to do it myself I would write it as follows: $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\alpha_k\rangle |\beta_l\rangle$ However I am unsure because the worked examples I have seen suggest the following $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\beta_l\rangle |\alpha_k\rangle$ $\endgroup$
    – DJA
    Nov 29, 2020 at 14:55
  • 1
    $\begingroup$ @DJA I don't understand the confusion. The way you write $\rho_{j\ell k\ell}$ in the comment is consistent with how you write it in the post (which is how you say is written in your notes). The second way you write it in the comments can also be used, it's just a matter of conventions at the end of the day. You cannot mix the two conventions though (and I would usually use the former one) $\endgroup$
    – glS
    Nov 30, 2020 at 10:33
  • $\begingroup$ @DJA rather than a long answer, I think your confusion might be solved by the following line: $$|\alpha_k \rangle |\beta_l\rangle=(|\alpha_k \rangle\otimes \mathbb 1)(\mathbb 1\otimes |\beta_l\rangle)=(\mathbb 1\otimes |\beta_l\rangle)(|\alpha_k \rangle\otimes \mathbb 1)= |\beta_l\rangle|\alpha_k \rangle$$ $\endgroup$ Nov 30, 2020 at 12:56

2 Answers 2


Well because if I were to do it myself I would write it as follows: $\rho_{jlkl} =\langle \alpha_j|\langle \beta_l| \rho |\alpha_k\rangle |\beta_l\rangle $ However I am unsure because the worked examples I have seen suggest the following $\rho_{jlkl} =\langle \alpha_j|\langle \beta_l| \rho |\beta_l\rangle |\alpha_k\rangle $.

It seems you are misunderstanding the idea of a tensor product of states, so I'll review that briefly. Let $\mathcal H_A$ and $\mathcal H_B$ be Hilbert spaces, and let $\alpha \in \mathcal H_A$ and $\beta \in \mathcal H_B$. The tensor product of $\alpha$ and $\beta$ is the ordered pair $(\alpha,\beta)$ which has the following properties:

  • $(\alpha,\beta+\gamma)=(\alpha,\beta)+(\alpha,\gamma)$ for all $\alpha\in\mathcal H_A, \beta,\gamma \in \mathcal H_B$
  • $(\alpha+\delta,\beta)=(\alpha,\beta)+(\delta,\beta)$ for all $\alpha,\delta \in \mathcal H_A, \beta \in \mathcal H_B$
  • $\lambda (\alpha,\beta) = (\lambda \alpha,\beta) = (\alpha,\lambda \beta)$ for all $\lambda \in \mathbb C, \alpha\in\mathcal H_A, \beta \in \mathcal H_B$

Rather than write $(\alpha,\beta)$ for the tensor product, it is standard notation to write $\alpha \otimes \beta$.

The tensor product of Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$ is the space of all tensor products of the form $\alpha\otimes \beta$ with $\alpha\in\mathcal H_A$ and $\beta \in \mathcal H_B$, and all linear combinations thereof. The inner product on this space is taken to be

$$\bigg< (\alpha,\beta), (\gamma,\delta)\bigg>_{\mathcal H_A\otimes \mathcal H_B} := \left<\alpha,\gamma\right>_{\mathcal H_A} \cdot \left<\mathcal \beta ,\mathcal \delta\right>_{\mathcal H_B}$$

Therefore, an element $\psi \in \mathcal H_A \otimes \mathcal H_B$ might look like

$$\psi= \alpha\otimes \beta + 3\gamma \otimes \delta$$

It is clear from the definition that $\alpha$ and $\gamma$ belong to $\mathcal H_A$ while $\beta$ and $\delta$ belong to $\mathcal H_B$. Again per standard convention, we reuse the symbol $\otimes$ and denote the tensor product of Hilbert spaces by $\mathcal H_A \otimes \mathcal H_B$.

If you'd like to work with Dirac notation, then you can write something like $|\psi\rangle = |\alpha\rangle \otimes |\beta \rangle$. The corresponding bra would be $\langle \psi| = \langle \alpha| \otimes \langle \beta |$. If we let $|\phi\rangle = |\gamma\rangle \otimes |\delta \rangle$, then

$$\langle \psi|\phi\rangle = \bigg(\langle \alpha| \otimes \langle \beta|\bigg) \bigg( |\gamma \rangle \otimes |\delta \rangle\bigg) = \langle \alpha|\gamma\rangle \cdot \langle \beta|\delta\rangle$$

The convention is that whether you're talking about a bra or a ket, the first quantity in the tensor product belongs to $\mathcal H_A$ (or its dual space) and the second belongs to $\mathcal H_B$ (or its dual space).

With all that being said, your expression

$$\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\beta_l\rangle |\alpha_k\rangle$$

doesn't make sense to me, because the tensor product ket on the right is in the wrong order.

  • $\begingroup$ Oh okay so I see what your saying, and so this would suggest that $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\alpha_k\rangle |\beta_l\rangle$. $\endgroup$
    – DJA
    Nov 29, 2020 at 16:45
  • $\begingroup$ @DJA Yes. I assume you are following the notation of whatever source you're learning from, but my preference (as stated in my answer to your previous question) is to write $|\alpha_j\rangle \otimes |\beta_k\rangle \equiv |\alpha_j ,\beta_k\rangle$ rather than $|\alpha_j\rangle |\beta_k\rangle$ because the object in question is a single ket belonging to the tensor product space, and the latter naively seems to suggest otherwise. $\endgroup$
    – J. Murray
    Nov 29, 2020 at 16:57
  • $\begingroup$ However my confusion arises from the below worked example and perhaps you can reconcile this with the fact that $\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\alpha_k\rangle |\beta_l\rangle$. $\endgroup$
    – DJA
    Nov 29, 2020 at 17:11
  • $\begingroup$ Consider a system made up of a pair of two level atoms whereby $|g_{\alpha}\rangle $ and $|e_{\alpha}\rangle $ are the ground and excited states of atom $\alpha$ and $|g_{\beta}\rangle $ and $|e_{\beta}\rangle $ are the ground and excited states of atom $\beta$. How is it that we can denote the reduced density of $\beta$ as $\rho_{\beta} = \sum_{\nu, \mu = g_{\beta} ,e\beta} \bigg(\langle\nu| g_{\alpha} |\rho |g_{\alpha}\rangle |\mu\rangle + \langle\nu| e_{\alpha} |\rho |e_{\alpha}\rangle |\mu\rangle \bigg)|\nu \rangle \langle \mu| $ $\endgroup$
    – DJA
    Nov 29, 2020 at 17:11
  • $\begingroup$ @DJA It looks like your author uses a convention whereby they reverse the order of the states in the tensor product when they take the adjoint, so $(|\alpha\rangle|\beta\rangle)^\dagger = \langle \beta|\langle\alpha|$. I think that is a bad idea for several reasons, but my opinion is not universal. In any case, if you perform an actual calculation you will find that as long as you're consistent, you'll get the right answer. $\endgroup$
    – J. Murray
    Nov 29, 2020 at 17:22

First of all, it should be noted that the way you understand $\rho_{ijk\ell}$ is first and foremost a matter of convention. That said, some conventions are certainly more "natural" than others.

One way to think about it is that the matrix components of $\rho$ in a composite space $\mathcal H\equiv \mathcal X\otimes\mathcal Y$ are nothing but that: matrix components in some space. If you use the indices $I,J$ to label the elements of a basis of $\mathcal H$, you can write the matrix components as $$\rho_{I,J}\equiv \langle I|\rho|J\rangle, \qquad |I\rangle,|J\rangle\in\mathcal H.$$ However, this notation does not take into account the bipartite structure of $\mathcal H$. To do this, we observe that we can always find a basis of $\mathcal H$ that is built from bases of $\mathcal X$ and $\mathcal Y$. We can thus label the basis elements of $\mathcal H$ using two indices, denoting the corresponding basis elements of $\mathcal X$ and $\mathcal Y$. In other words, we can write $$\mathcal H = \mathrm{span}(\{|i,j\rangle\equiv|i\rangle\otimes|j\rangle : \quad |i\rangle\in\mathcal X, \,\,|j\rangle\in\mathcal Y\}).$$ Then, instead of an index $I$, we use a pair of indices, say $(i,j)$. The matrix elements of $\rho$ then become $$\rho_{(i,j),(k,\ell)} \equiv \langle i,j|\rho|k,\ell\rangle \equiv (\langle i|\otimes\langle j|)\rho(|k\rangle\otimes |\ell\rangle),$$ where I'm including different equivalent ways to write the expression. Note that I wrote the "input" and "output" indices of $\rho$ using pairs $(i,j)$ and $(k,\ell)$ here, to stress the different roles the indices have. For brevity, one does not usually do this, and simply writes $\rho_{ijk\ell}$ to mean $\rho_{(i,j),(k,\ell)}$.

Now, you can also decide to use $\rho_{ijk\ell}$ to mean something like $\langle \ell,j|\rho|k,i\rangle$. That would be quite an awkward notation though.


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