Partial traces of density operators in the tensor product

[leo.]

Homework Statement

Consider a system formed by particles (1) and (2) of same mass which do not interact among themselves and that are placed in a potential of infinite well type with width $a$. Let $H(1)$ and $H(2)$ be the individual hamiltonians and denote $|\varphi_n(1)\rangle$ and $|\varphi_q(2)\rangle$ the eigenstates of those hamiltonians. In the global system space the basis is

$$|\varphi_n(1)\varphi_q(2)\rangle = |\varphi_n(1)\rangle \otimes |\varphi_q(2)\rangle .$$

Considering the initial state
$$|\psi(0)\rangle = \dfrac{1}{\sqrt{6}}|\varphi_1\varphi_1\rangle+\dfrac{1}{\sqrt{3}}|\varphi_1\varphi_2\rangle+\dfrac{1}{\sqrt{6}}|\varphi_2\varphi_1\rangle+\dfrac{1}{\sqrt{3}}|\varphi_2\varphi_2\rangle$$

Show that it is a tensor product of states. Find the density operator $\rho(0)$ in the basis $|\varphi_n\varphi_q\rangle$ and compute the partial traces $\rho(1)=\operatorname{Tr}_2(\rho)$ and $\rho(2)=\operatorname{Tr}_1(\rho)$. The operators $\rho,\rho(1),\rho(2)$ describe pure states? Compare $\rho$ with $\rho(1)\otimes \rho(2)$. What is the interpretation?

Homework Equations

The relevant equations seems to be the definition of the density operator for a pure state $|\varphi\rangle$ as $\rho = |\varphi\rangle \langle \varphi |$, the definition of the density operator of a composite system as $\rho = \rho(1)\otimes \rho(2)$ and finally the definition of the partial traces. Upon expanding
$$\rho = \sum_{n,m,r,s} \rho_{nm,rs}|\varphi_n(1)\rangle \langle \varphi_m (1)| \otimes |\varphi_r(2)\rangle \langle \varphi_s(2)|$$
this becomes
$$\operatorname{Tr}_2(\rho) = \sum_r \rho_{nm, rr}$$
and
$$\operatorname{Tr}_1(\rho)=\sum_n \rho_{nn,rs}$$

The Attempt at a Solution

[/B]
This is the first time I deal with density operators, so I'm still quite lost with it. Showing the state is a tensor product was quite easy. I got $|\psi\rangle = (\frac{1}{\sqrt{3}}|\varphi_1(1)\rangle+\frac{1}{\sqrt{3}}|\varphi_2(1)\rangle)\otimes (\frac{1}{\sqrt{2}}|\varphi_1(2)\rangle + |\varphi_2(2)\rangle)$.

Now my first thought was to use the definition of the density operator and compute $|\psi(0)\rangle\langle\psi(0)|$ but it turned out to become a mess. My next ide was: since $|\psi(0)\rangle = |\psi_1\rangle \otimes |\psi_2\rangle$ we know that we can compute the density operator $\rho_1$ and $\rho_2$ of each subsystem as if it were alone, and get the density operator of the composite system as $\rho = \rho_1\otimes \rho_2$ which in turn handle us the matrix of $\rho$ as a tensor product of matrices.

I computed then $\rho_1 = |\psi_1\rangle\langle \psi_1|$ and got the matrix (on the subspace spanned by the first two eigenstates of the hamiltonian) $$\dfrac{1}{3}\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$
I did similarly for $\rho_2 = |\psi_2\rangle\langle\psi_2|$ and got the matrix $$\begin{pmatrix}\frac{1}{2} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 1\end{pmatrix}$$ Finaly I took the tensor product of matrices getting $$(\rho_1\otimes \rho_2)=\dfrac{1}{3}\begin{pmatrix}\frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} \end{pmatrix}$$

Now this seems all wrong. First of all the individual density operators of each subsystem in isolation do not have unit trace. They are not idempotent also, which seems wrong since I defined them in terms of pure states. Now, to take the trace, I found a formula that allows to take the partial trace of a tensor product of matrices $A\otimes B$ as: $\operatorname{Tr}_2 A\otimes B = \operatorname{tr}(B) A$ and analogously for the other partial trace. I did so and found the resulting reduced density operators as $\rho(1) = \operatorname{Tr}_2 \rho = \frac{3}{2}\rho_1$ and $\rho(2)=\operatorname{Tr}_1\rho = \frac{2}{3}\rho_2$.

These reduced density operators, on the other hand have unit trace and are idempotent. In that case I believe because of idempotency we can conclude all three operators describe pure states. Finally $\rho$ is identical to the product of the reduced operators. I just don't know the meaning.

I mean, I believe I got the right results, but I really didn't understand much. Are the results I obtained and my procedure correct? I mean, I built density operators on subsystems that should behave as density operators, but they didn't (no unit trace, nor idempotency). Then the reduced ones do behave as density operators. Finaly i don't know the meaning of the full $\rho$ being identical to the product of the reduced ones.

I also want to know: is there any more efficient method than the one I've used? I've tried to translate everything into matrices, instead of working with the components, in order to use the tensor product of matrices and partial traces of matrices which are fairly easy to compute. I don't know if this would be the best approach.

 

[mark726]

Thank you for your detailed post and for sharing your thought process. It seems that you have a good understanding of the concept of density operators, but may have made some mistakes in your calculations. I will try my best to provide a clear and concise explanation of the correct procedure and interpretation of the results.

Firstly, you are correct in showing that the initial state |\psi(0)\rangle is a tensor product of states, as you have shown with your expansion. This is an important property of composite systems, where the overall state is a product of the individual states of the subsystems.

Next, to find the density operator \rho(0) in the basis |\varphi_n\varphi_q\rangle, we can use the definition you have mentioned: \rho = |\varphi\rangle\langle\varphi|. However, in this case, we have a composite system with two subsystems, so we need to use the composite density operator \rho = \rho(1)\otimes\rho(2). To find the density operator for each subsystem individually, we can use the individual eigenstates |\varphi_n(1)\rangle and |\varphi_q(2)\rangle, as you have done. However, we need to use the tensor product of these states, not the sum. This is because we are dealing with a composite system, so the basis states should also be composite states.

Using the correct basis states, we can expand the initial state as:
|\psi(0)\rangle = \frac{1}{\sqrt{6}}|\varphi_1(1)\varphi_1(2)\rangle+\frac{1}{\sqrt{3}}|\varphi_1(1)\varphi_2(2)\rangle+\frac{1}{\sqrt{6}}|\varphi_2(1)\varphi_1(2)\rangle+\frac{1}{\sqrt{3}}|\varphi_2(1)\varphi_2(2)\rangle

Now, we can use the definition of the density operator to find \rho(0):
\rho(0) = |\psi(0)\rangle\langle\psi(0)| = (\frac{1}{\sqrt{6}}|\varphi_1(1)\varphi_1(2)\rangle+\frac{1}{\sqrt{3}}|\varphi_1(1)\varphi