# Change of basis for density operator

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?

• 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. Commented Oct 26, 2021 at 6:13
• 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? Commented Oct 26, 2021 at 14:06
• Well... try it... Commented Oct 26, 2021 at 14:08
• @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. Commented Oct 27, 2021 at 17:16
• @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. Commented Oct 27, 2021 at 17:32

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 $$\langle \phi_n^X| \theta_i^X\rangle = c_n^i$$ $$\langle \psi_{\alpha}^Y| \lambda_j^Y\rangle = d_{\alpha}^j$$

$$\left| \psi \right\rangle \left\langle\psi \right| = \sum_{i,j} \sum_{k,l} \beta_{ij} \beta_{kl}^* \sum_n\left| \phi_n^X \right\rangle c_n^i \otimes \sum_{\alpha}\left| \psi_{\alpha}^Y \right\rangle d_{\alpha}^j \sum_n c_{n}^{k*} \left\langle\phi_n^X \right| \otimes \sum_{\alpha} d_{\alpha}^{l*} \left\langle\psi_{\alpha}^Y \right|$$
Rearranging the terms, $$= \sum_{n,\alpha} \underbrace{\left(\sum_i \beta_{ij}c_n^id_{\alpha}^j \sum_j \beta_{kl}^*c_n^{k*}d_{\alpha}^{l*}\right)}_{\chi_{n,\alpha}} \left| \phi_n^X \right\rangle \otimes \left| \psi_{\alpha}^Y \right\rangle \left\langle\phi_n^X \right| \otimes \left\langle\psi_{\alpha}^Y \right|$$
• 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... Commented Oct 30, 2021 at 10:41
• Could you, then, share the Schmidt decomposition method? I tried to do it, but to no avail, I can't obtain $(1)$ Commented Nov 9, 2021 at 23:15