Understanding how to reduce density matrices

[JD_PM]

TL;DR

I want to understand how to reduce the density matrix of a composite system to one describing a subsystem (method known as partial-tracing/tracing out/reduction) through (if possible) an explicit example.

Spoiler: Properties of density matrices

I have recently learned about one of the most powerful tools in quantum statistics: the density matrix $\rho \in \Bbb C^{n\times n}$ . Its most relevant properties are:

$$\text{Tr} (\rho) = 1 \tag{*}$$

$$\rho^{\dagger} = \rho \tag{**}$$

$$\langle u | \rho | u \rangle \geq 0 \tag{***}$$

Where $| u \rangle \in \Bbb C$ and $||u|| = \sqrt{\langle u,u \rangle}=1$

Then I learned that:

$1) \ \rho$ is used as a linear functional over Hermitian matrices (i.e. $\rho : \Bbb C^{n\times n} \rightarrow \Bbb R: A \mapsto \text{Tr} (\rho A)$, where $A=A^{\dagger}$)

$2) \ \rho$ has the following spectral representation

$$\rho = \sum_n \rho_n |\phi_n \rangle \langle \phi_n|$$

Then

$$\rho(A) = \sum_i p_i \langle e_i | A e_i \rangle$$

Where $p_i$ is the eigenvalue of $\rho$ with eigenvector $e_i$. If $A$ is diagonal (wrt the basis $\{e_i\}$) then $\rho(A) = \sum_i \lambda_i p_i$

Sidenote: I thought of writing $\rho, A \in \Bbb H^{n\times n}$ to make clear that $\rho, A$ are Hermitian. However, as it is not standard notation, I'll stick to $\rho, A \in \Bbb C^{n\times n}$ and $\rho = \rho^{\dagger}, A = A^{\dagger}$

I read that density matrices are useful in Physics mainly to describe a) mixtures; we do not know the wave function of the system so $\psi$ is random and b) entangled systems. I'd like to focus on the later.

Let us have a system $S_1$ entangled to the system $S_2$. Thus we start from a composite system with wave function $\psi(x_1,x_2) \in \mathscr{H}_1 \otimes \mathscr{H}_2$, where $\mathscr{H}_1, \mathscr{H}_2$ are Hilbert spaces. A priori there's no wave function belonging to only one of these Hilbert spaces.

I've studied that there's a method to 'reduce' the original density matrix $\rho$ (which describes the entangled system) to $\rho'$, which only describes one of the systems; let's say we want to only describe $S_1$. Then we 'trace out' $S_2$:

$$\hat D = \text{Tr}_2 |\psi \rangle \langle \psi| \tag{1}$$

The Neumann entropy $S$ measures the entanglement between $S_1, S_2$:

$$S(\hat D) := - \text{Tr} \hat D \log \hat D \tag{2}$$

I did not quite understand how the 'reducing/tracing out' (I think it is also known as partial-trace) method worked based on $(1)$ so I looked for an example:

But I still do not understand it.

Could you please explain how this 'reducing/tracing out' method works through an explicit example?

Thank you

PS: I am struggling to find a book explaining such a method; Ballentine explains density matrices but not how to reduce them (at least I did not find it). I started to check what looks like a promising source: Mathematical Fundations of Quantum Mechanics by Neumann (Princeton 2018); However I am still getting used to his notation; the method is probably in chapter III: The quantum Statistics (I still did not find it). If you know of any other book please let me know

 

[vanhees71]

A composite system consisting of two parts with Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ are described on the Hilbert space $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2$, the tensor or Kronecker product of the two Hilbert spaces.

If you have arbitrary orthonormal bases $|u_j \rangle$ of $\mathcal{H}_1$ and $|v_k \rangle$ of $\mathcal{H}_2$ then $\langle |U_{jk} \rangle = |u_j \rangle \otimes |v_k \rangle$ build an orthonormal basis of $\mathcal{H}$.

Now you can decompose the Statistical Operator of the composite system $\hat{\rho}$ wrt. this basis
$$\hat{\rho}=\sum_{jk,j'k'} \rho_{jk,j'k'} |U_{jk} \rangle \langle U_{jk}, \quad \rho_{jk,j'k'}=\langle U_{jk}|\hat \rho U_{j'k'} \rangle.$$
Then the reduced statistical operator describing the state of subsystem 1 is found by the partial trace wrt. system 2, i.e.,
$$\hat{\rho}^{(1)}=\sum_{j,j'} |u_j \rangle \langle u_{j'}| \rho_{jj'}^{(1)} \quad \text{with} \quad \rho_{jj'}^{(1)}=\sum_{k} \rho_{jk,j'k'}.$$
You can also write this as
$$\hat{\rho}^{(1)} = \sum_{jk,j'} |u_j \rangle \langle U_{jk}|\hat{\rho} U_{j'k} \rangle \langle u_{j'}|:=\mathrm{Tr}_2 \hat{\rho}.$$
In an analogous way you get $\hat{\rho}^{(2)}$ the reduced state describing system 2.

The interesting issue with entangled pure states is that the reduced statistical operators for the subsystems is a mixed and not a pure state.

The entire formalism above can of course be easily extended to composites with more than 2 subsystems.

 

[atyy]

The reduced density matrix is discussed in point #39 of https://pages.uoregon.edu/svanenk/solutions/Mixed_states.pdf.

 

[JD_PM]

Hi vanhees71

Now you can decompose the Statistical Operator of the composite system $\hat{\rho}$ wrt. this basis
$$\hat{\rho}=\sum_{jk,j'k'} \rho_{jk,j'k'} |U_{jk} \rangle \langle U_{jk}, \quad \rho_{jk,j'k'}=\langle U_{jk}|\hat \rho U_{j'k'} \rangle.$$
Then the reduced statistical operator describing the state of subsystem 1 is found by the partial trace wrt. system 2, i.e.,
$$\hat{\rho}^{(1)}=\sum_{j,j'} |u_j \rangle \langle u_{j'}| \rho_{jj'}^{(1)} \quad \text{with} \quad \rho_{jj'}^{(1)}=\sum_{k} \rho_{jk,j'k'}.$$
You can also write this as
$$\hat{\rho}^{(1)} = \sum_{jk,j'} |u_j \rangle \langle U_{jk}|\hat{\rho} U_{j'k} \rangle \langle u_{j'}|:=\mathrm{Tr}_2 \hat{\rho}.$$
In an analogous way you get $\hat{\rho}^{(2)}$ the reduced state describing system 2.

The interesting issue with entangled pure states is that the reduced statistical operators for the subsystems is a mixed and not a pure state.

The entire formalism above can of course be easily extended to composites with more than 2 subsystems.

I am struggling to follow the above due to my lack of knowledge. Would you recommend/know of any book deriving the whole process? Should I just re-read Ballentine's chapter 2? Thank you

I actually tried to work out an example following your guideline.

Given $|w \rangle = \frac{1}{\sqrt{2}} \Big( |0 1 \rangle - |1 0 \rangle \Big) \in \Bbb C^4$ show that its reduced density matrix is
$$\rho=
\begin{pmatrix}
1/2 & 0 \\
0 & 1/2 \\
\end{pmatrix}
$$

Step 1: Identify basis

We want a basis for $\Bbb C^4$. As $\Bbb C^4 = \Bbb C^2 \otimes \Bbb C^2$ we deal with the orthonormal basis $|0 \rangle \otimes |1 \rangle \in \Bbb C^2$ and $|1 \rangle \otimes |0 \rangle \in \Bbb C^2$. Thus the orthonormal basis for $\Bbb C^4$ is $|01 \rangle \otimes |10 \rangle=\langle |U_{jk} \rangle$

Step 2: get $\hat \rho$

I really get lost when trying to apply the density-matrix operator for this particular problem

$$\hat{\rho}=\sum_{jk,j'k'} \rho_{jk,j'k'} |U_{jk} \rangle \langle U_{jk}, \quad \rho_{jk,j'k'}=\langle U_{jk}|\hat \rho U_{j'k'} \rangle.$$

 

[Heidi]

What is interesting in your 2 by 2 density matrix is the null off diagonal terms.
they say that you will get no interference if you try to see them in interferometry.
that is why the density matrices are useful.

 

[vanhees71]

Hi vanhees71
I am struggling to follow the above due to my lack of knowledge. Would you recommend/know of any book deriving the whole process? Should I just re-read Ballentine's chapter 2? Thank you

I actually tried to work out an example following your guideline.

Given $|w \rangle = \frac{1}{\sqrt{2}} \Big( |0 1 \rangle - |1 0 \rangle \Big) \in \Bbb C^4$ show that its reduced density matrix is
$$\rho=
\begin{pmatrix}
1/2 & 0 \\
0 & 1/2 \\
\end{pmatrix}
$$

Step 1: Identify basis

We want a basis for $\Bbb C^4$. As $\Bbb C^4 = \Bbb C^2 \otimes \Bbb C^2$ we deal with the orthonormal basis $|0 \rangle \otimes |1 \rangle \in \Bbb C^2$ and $|1 \rangle \otimes |0 \rangle \in \Bbb C^2$. Thus the orthonormal basis for $\Bbb C^4$ is $|01 \rangle \otimes |10 \rangle=\langle |U_{jk} \rangle$

Step 2: get $\hat \rho$

I really get lost when trying to apply the density-matrix operator for this particular problem

$$\hat{\rho}=\sum_{jk,j'k'} \rho_{jk,j'k'} |U_{jk} \rangle \langle U_{jk}, \quad \rho_{jk,j'k'}=\langle U_{jk}|\hat \rho U_{j'k'} \rangle.$$

Matrices do not help too much for such calculations. Just stick to the Dirac formalism.
$$|\sigma_1 \rangle \otimes |\sigma_2 \rangle=|\sigma_1 \sigma_2 \rangle, \quad \sigma_1,\sigma_2 \in \{-1/2,1/2\}$$
is a basis of the 4D Hilbert space for two spins 1/2. Now first write
$$\hat{\rho} = |w \rangle \langle w|.$$
From this read off the matrix elements and then calculate the partial traces as given in the formulas in my previous posting.

 

[JD_PM]

I think I got it!

I am struggling to follow the above due to my lack of knowledge. Would you recommend/know of any book deriving the whole process? Should I just re-read Ballentine's chapter 2? Thank you

What I needed was to continue reading Ballentine's beautiful book! I feel more ready now, once I've checked #3, #4, and section 8.3.

Matrices do not help too much for such calculations. Just stick to the Dirac formalism.
$$|\sigma_1 \rangle \otimes |\sigma_2 \rangle=|\sigma_1 \sigma_2 \rangle, \quad \sigma_1,\sigma_2 \in \{-1/2,1/2\}$$
is a basis of the 4D Hilbert space for two spins 1/2. Now first write
$$\hat{\rho} = |w \rangle \langle w|.$$
From this read off the matrix elements and then calculate the partial traces as given in the formulas in my previous posting.

Absolutely, let's use Dirac notation while working out the example (which can be found in Ballentine's page 219).

Given two particles each having spin $s=1/2$. The vector $|\sigma_1, \sigma_2 \rangle = |\sigma_1 \rangle \otimes |\sigma_2 \rangle$ describes a state in which the $z$ component of the spin of particle $1$ is equal to $\hbar \sigma_1$ and that of particle $2$ is equal to $\hbar \sigma_2$.

The state vector of the two-particle system is

$$|\Psi \rangle = \sqrt{1/2} \Big( |1/2,-1/2 \rangle - |-1/2,1/2 \rangle \Big)$$

The spectral decomposition of the state operator $\hat \rho$ is

$$\hat \rho = \sum_m \rho_m |\phi_m \rangle \langle \phi_m | \tag{1}$$

The eigenvectors of $\hat \rho$ can be expanded in terms of its basis vectors

$$|\phi_k \rangle = \sum_{m,n} c_{m,n}^k | a_m b_n \rangle \tag{2}$$

Plugging $(2)$ into $(1)$ we get

$$\rho = \sum_k \rho_k \sum_{m,n} \sum_{m',n'} \Big( c_{m,n}^k \Big) \Big( c_{m',n'}^k \Big)^* | a_m b_n \rangle \langle a_{m'} b_{n'} | \tag{3}$$

By definition we know that

$$\rho^{(1)} := \text{Tr}^{(2)} \rho \tag{4}$$

The matrix elements of $\rho^{(1)}$ are (i.e. we sum over $n$)

$$\langle a_m | \rho^{(1)} | a_{m'} \rangle = \sum_n \langle a_m b_n| \rho | a_{m'} b_{n} \rangle \tag{5}$$

Based on $(3),(5)$ we get that the partial state $\rho^{(1)}$ has the following form

$$\rho^{(1)} = \sum_k \rho_k \sum_{m,m'} \sum_{n} \Big( c_{m,n}^k \Big) \Big( c_{m',n}^k \Big)^* | a_m \rangle \langle a_{m'}| \tag{6}$$

Analogously we get that $\rho^{(2)}$ has the following form

$$\rho^{(2)} = \sum_k \rho_k \sum_{n ,n'} \sum_{m} \Big( c_{m,n}^k \Big) \Big( c_{m,n'}^k \Big)^* | b_n \rangle \langle b_{n'}| \tag{7}$$Then, in this particular problem, the reduced states $\rho^{(1)}$ and $\rho^{(2)}$

$$ \rho^{(1)} =1/2 \Big( |1/2 \rangle - |-1/2 \rangle \Big)\Big( \langle 1/2| - \langle-1/2 | \Big)$$ $$=1/2 \Big( |1/2 \rangle \langle 1/2| + |-1/2 \rangle \langle -1/2| \Big)$$

$$ \rho^{(2)} =1/2 \Big( |-1/2 \rangle - |1/2 \rangle \Big)\Big( \langle -1/2| - \langle1/2 | \Big)$$ $$=1/2 \Big( |-1/2 \rangle \langle -1/2| + |1/2 \rangle \langle 1/2| \Big)$$

Equal to each other!

We see that $\rho^{(1)}$ and $\rho^{(2)}$ are no-pure states, as they can be written in function of other states. Besides, we realize that $\rho$ is entangled (i.e. it cannot be written as $\hat \rho = \rho^{(1)} \otimes \rho^{(2)}$)