Considering a state that describes two subsystems $X$ and $Y$:

$$ \left|\psi \right\rangle =\sum_{i,k} \beta_{ik} \left| \theta_i^X \right\rangle \otimes \left| \lambda_i^Y \right\rangle$$

I am trying to show that it is possible to choose a basis for the space of states of $X$ and $Y$ such that the density matrix would reduce to the following form:

$$ \left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \alpha_{ij} \left| \phi_i^X \right\rangle \otimes \left| \psi_i^Y \right\rangle \left\langle\phi_j^X \right| \otimes \left\langle\psi_j^Y \right| \hspace{0,7cm}(1)$$

My attempt consisted in plugging in four identity operators that come from two new orthonormal basis: $\sum_i\left| \phi_i^X \right\rangle \left\langle\phi_i^X \right|= \hat{\mathbb{I}}$ and $\sum_j\left| \psi_j^Y \right\rangle \left\langle\psi_j^Y \right|= \hat{\mathbb{I}}$ such that:

$$ \left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \sum_{k,l} \beta_{ij} \beta_{kl}^* \hat{\mathbb{I}}\left| \theta_i^X \right\rangle \otimes \hat{\mathbb{I}}\left| \lambda_j^Y \right\rangle \left\langle\theta_k^X \right|\hat{\mathbb{I}} \otimes \left\langle\lambda_l^Y \right|\hat{\mathbb{I}}$$

$$ \left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \sum_{k,l} \beta_{ij} \beta_{kl}^* \left( \sum_n\left| \phi_n^X \right\rangle \left\langle\phi_n^X \right| \right)\left| \theta_i^X \right\rangle \otimes \left( \sum_{\alpha}\left| \psi_{\alpha}^Y \right\rangle \left\langle\psi_{\alpha}^Y \right| \right)\left| \lambda_j^Y \right\rangle \left\langle\theta_k^X \right| \left( \sum_n\left| \phi_n^X \right\rangle \left\langle\phi_n^X \right| \right) \otimes \left\langle\lambda_l^Y \right|\left( \sum_{\alpha}\left| \psi_{\alpha}^Y \right\rangle \left\langle\psi_{\alpha}^Y \right| \right)$$

$$ \left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \sum_{k,l} \beta_{ij} \beta_{kl}^* \sum_n\left| \phi_n^X \right\rangle \left\langle\phi_n^X | \theta_i^X \right\rangle \otimes \sum_{\alpha}\left| \psi_{\alpha}^Y \right\rangle \left\langle\psi_{\alpha}^Y | \lambda_j^Y \right\rangle \sum_n \left\langle\theta_k^X | \phi_n^X \right\rangle \left\langle\phi_n^X \right| \otimes \sum_{\alpha} \left\langle\lambda_l^Y | \psi_{\alpha}^Y \right\rangle \left\langle\psi_{\alpha}^Y \right|$$

I know that I can rearrange the terms a little and use the closure relations for the $\theta $ and $\lambda$ basis but these terms dont have the same indicies, therefore I am stuck. I can't obtain equation $(1)$.

How can I do this?

  • $\begingroup$ I think you've made a mistake: When you insert the identity operators, you can't use the same index twice. Instead, you have to use e.g. $n,n^\prime$ and $\alpha,\alpha^\prime$. To give you a hint how to solve the issue: You could try to use the Schmidt decomposition. $\endgroup$ Oct 26, 2021 at 6:13
  • $\begingroup$ If I use indicies $n$, $n'$ and $\alpha$ , $\alpha'$ and apply the method that was given in the answer I should get the correct result then, right? $\endgroup$ Oct 26, 2021 at 14:06
  • $\begingroup$ Well... try it... $\endgroup$ Oct 26, 2021 at 14:08
  • $\begingroup$ @Jakob Identity matrix is defined by $\sum_n |n\rangle \langle n |$, which is the sum of projection operators of all the eigenstates. So i don't think your definition is going to work. $\endgroup$
    – sslucifer
    Oct 27, 2021 at 17:16
  • 1
    $\begingroup$ @sslucifer Well, you're completely right. But $\mathbb I = \mathbb I\, \mathbb I \neq \sum\limits_n \, \sum\limits_n |n\rangle \langle n | |n\rangle \langle n|$...And yes, it won't lead to the desired result. In fact, it is just a change of basis and OP could've started from $|\psi\rangle\langle \psi| = \mathbb I|\psi\rangle\langle \psi| \mathbb I$ with the identity expressed in the new basis. The problem is that one has to show that there exists a basis in which the density operator takes the form of equation $(1)$ . In your case, it would hold for all bases, which is of course not true. $\endgroup$ Oct 27, 2021 at 17:32

1 Answer 1


I think you are almost there. We can write $|\theta_i^X\rangle$ as linear combination of $|\phi_n^X\rangle$ $$ |\theta_i^X\rangle = \sum_m c_m^i |\phi_m^X\rangle$$ Similarly, $$ |\lambda_j^Y\rangle = \sum_m d_m^j |\psi_m^Y\rangle$$ So the inner products become $$\langle \phi_n^X| \theta_i^X\rangle = c_n^i$$ $$\langle \psi_{\alpha}^Y| \lambda_j^Y\rangle = d_{\alpha}^j$$

Following your last equation,

$$ \left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \sum_{k,l} \beta_{ij} \beta_{kl}^* \sum_n\left| \phi_n^X \right\rangle c_n^i \otimes \sum_{\alpha}\left| \psi_{\alpha}^Y \right\rangle d_{\alpha}^j \sum_n c_{n}^{k*} \left\langle\phi_n^X \right| \otimes \sum_{\alpha} d_{\alpha}^{l*} \left\langle\psi_{\alpha}^Y \right|$$

Rearranging the terms, $$ = \sum_{n,\alpha} \underbrace{\left(\sum_i \beta_{ij}c_n^id_{\alpha}^j \sum_j \beta_{kl}^*c_n^{k*}d_{\alpha}^{l*}\right)}_{\chi_{n,\alpha}} \left| \phi_n^X \right\rangle \otimes \left| \psi_{\alpha}^Y \right\rangle \left\langle\phi_n^X \right| \otimes \left\langle\psi_{\alpha}^Y \right|$$

You can change the dummy variables and symbols to get equation (1)

  • $\begingroup$ I can't help but again point out (also for future readers) that this is wrong. The penultimate equation of your answer is basically correct, but the last equation is not: There are two sums missing (over $n$ and $\alpha$) and you cannot simply interchange these orders without relabeling the indices. Furthermore, in essence this is just a change of basis, for which in general we arrive at a form similar to the first equation in the question and thus this won't lead to the desired result. In fact, if this was true, it would hold for all bases... $\endgroup$ Oct 30, 2021 at 10:41
  • $\begingroup$ Could you, then, share the Schmidt decomposition method? I tried to do it, but to no avail, I can't obtain $(1)$ $\endgroup$ Nov 9, 2021 at 23:15
  • $\begingroup$ @MicrosoftBruh You should use @... such that the person is notified... Anyway, what are your attempts? You could edit the question accordingly. I mean once you've proven the existence of the Schmidt basis (which, basically is the proof given in Wikipedia or that should be given in any QI textbook) you just have to write the density matrix corresponding to a generic pure state... I could help you further if you'd show what exactly you're struggling with. $\endgroup$ Dec 9, 2021 at 8:55

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